Ch 6 Energy and Chemical Change

Brady amp Senese 5th Ed

2

Energy Is The Ability To Do Work

bull Energy is the ability to do work (move mass over a distance) or transfer heat

bull Types kinetic and potentialbull kinetic the energy of motionbull potential the stored energy in matter

bull Internal energy (E) the sum of the kinetic and potential energy for each particle in the system

3

Kinetic Energy The Energy Of Motionbull KE=frac12mv2

bull Energy can be transferred by moving particles

bull Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slowsbull this is why hot water cools in contact with cool

air

4

Potential Energy Depends on Position

bull Potential energy increases when objects that attract move apart or objects that repel move toward each other

bull Stored energy that can be converted to kinetic energy

5

Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a deskB A pencil is sharpenedC A pencil is heatedD All are forms of kinetic energyE None are forms of kinetic energy

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

2

Energy Is The Ability To Do Work

bull Energy is the ability to do work (move mass over a distance) or transfer heat

bull Types kinetic and potentialbull kinetic the energy of motionbull potential the stored energy in matter

bull Internal energy (E) the sum of the kinetic and potential energy for each particle in the system

3

Kinetic Energy The Energy Of Motionbull KE=frac12mv2

bull Energy can be transferred by moving particles

bull Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slowsbull this is why hot water cools in contact with cool

air

4

Potential Energy Depends on Position

bull Potential energy increases when objects that attract move apart or objects that repel move toward each other

bull Stored energy that can be converted to kinetic energy

5

Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a deskB A pencil is sharpenedC A pencil is heatedD All are forms of kinetic energyE None are forms of kinetic energy

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

3

Kinetic Energy The Energy Of Motionbull KE=frac12mv2

bull Energy can be transferred by moving particles

bull Collision of fast particles with slower particles causes the slow particle to speed up while the fast molecule slowsbull this is why hot water cools in contact with cool

air

4

Potential Energy Depends on Position

bull Potential energy increases when objects that attract move apart or objects that repel move toward each other

bull Stored energy that can be converted to kinetic energy

5

Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a deskB A pencil is sharpenedC A pencil is heatedD All are forms of kinetic energyE None are forms of kinetic energy

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

4

Potential Energy Depends on Position

bull Potential energy increases when objects that attract move apart or objects that repel move toward each other

bull Stored energy that can be converted to kinetic energy

5

Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a deskB A pencil is sharpenedC A pencil is heatedD All are forms of kinetic energyE None are forms of kinetic energy

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

5

Your Turn

Which of the following is not a form of kinetic energy

A A pencil rolls across a deskB A pencil is sharpenedC A pencil is heatedD All are forms of kinetic energyE None are forms of kinetic energy

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

6

Law Of Conservation Of Energy

bull Energy cannot be created or destroyed but can be transformed from one form of energy to another

bull Also known as the first law of thermodynamics

bull How does water falling over a waterfall demonstrate this law

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

7

Problems 3 5 9

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

8

Heat And Temperature Are Not The Samebull The temperature of an object is proportional

to the average kinetic energy of its particlesmdashthe higher the average kinetic energy the higher the temperature

bull Heat is energy (also called thermal energy) transferred between objects caused by differences in their temperatures until they reach thermal equilibrium

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

9

Units Of Energy

bull SI unit is the Joule Jbull J = kgms2

bull If the calculated value is greater than 1000 J use the kJ

bull British unit is the calorie calbull cal = 4184 J (exact)

bull Nutritional unit is the Calorie (note capital c) which is one kilocaloriebull 1 Cal = 1 kcal = 4184 kJ

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

10

bull 1st Law of Thermodynamics For an isolated system the internal energy (E) is constant

Δ E = Ef - Ei = 0 Δ E = Eproduct - Ereactant = 0

bull We canʼt measure the internal energy of anything so we measure the changes in energy

bull E is a state functionbull E = work + heat

Internal Energy is Conserved

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

11

bull Temperature (T) is proportional to the average kinetic energy of all particle units degC degF Kbull Avg KE= frac12 mvavg

2

bull At a high temperature most molecules are moving at higher average

What Is Temperature

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

12

State Functionbull A property whose value depends only on the

present state of the system not on the method or mechanism used to arrive at that state

bull Position is a state function both train and car travel to the same locations although their paths vary

bull The actual distance traveled does vary with pathNew York

Los Angeles

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

13

Problems 41 43

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

14

Heat Transfer q

bull Heat (q) = the transfer of energy from regions of high temperature to regions of lower temperature

bull units J cal kgmiddotm2s2

bull A calorie is the amount of energy needed to raise the temperature of 100 g water from 145 to 155 degC

bull A metal spoon at 25degC is placed in boiling water What happens

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

15

Surroundings System Universe

bull System the reaction or area under studybull Surroundings the rest of the universebull Open systems can gain or lose mass and

energy across their boundariesbull ie the human body

bull Closed systems can absorb or release energy but not mass across the boundary bull ie a light bulb

bull Isolated systems cannot exchange matter or energy with their surroundings Adiabaticbull ie a stoppered Thermos bottle

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

16

The Sign Convention

bull Endothermic systems require energy to be added to the system thus the q is (+)

bull Exothermic reactions release energy to the surroundings Their q is (-)

bull Energy changes are measured from the point of view of the system

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

17

Your Turn

A cast iron skillet is moved from a hot oven to a sink full of water Which of the following is not true

A The water heatsB The skillet coolsC The heat transfer for the skillet has a (-) signD The heat transfer for the skillet is the same

as the heat transfer for the waterE None of these are untrue

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

18

Heat Capacity and Transfer

bull Heat capacity (C)- the (intensive) ability of an object with constant mass to absorb heat bull aka calorimeter constantbull Varies with the sample mass and the identity of

the substancebull Units JdegC

bull q=CtimesΔtbull q= heat transferredbull C= heat capacity of objectbull Δt= Change in Temperature (tfinal-tinitial)

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

19

Learning Check

A cup of water is used in an experiment Its heat capacity is known to be 720 J degC How much heat will it absorb if the experimental temperature changed from 192 degC to 235 degC

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

20

Heat Transfer and Specific Heat

bull Specific heat (s)- The (extensive) ability of a substance to store heatbull C = mtimessbull Units JgmiddotdegC Jg middot K Jmol middot K

bull q=mtimesΔttimessbull q= heat transferredbull m= mass of objectbull Δt= Change in Temperature (tfinal-tinitial)

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

21

Specific Heats Substance

Specific Heat J g degC (25 degC)

Carbon (graphite) 0711

Copper 0387

Ethyl alcohol 245

Gold 0129 Granite 0803 Iron 04498 Lead 0128 Olive oil 20 Silver 0235 Water (liquid) 418

bull Substances with high specific heats resist temperature changes

bull Note that water has a very high specific heatbull (this is why coastal

temperatures are different from inland temperatures)

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

22

Learning Check

Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a 3291 g sample by 253degC

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

23

Problems 45 47 49

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

24

The First Law Of Thermodynamics Explains Heat Transferbull If we monitor the heat transfers (q) of all

materials involved and all work processes we can predict that their sum will be zero

bull By monitoring the surroundings we can predict what is happening to our system

bull Heat transfers until thermal equilibrium thus the final temperature is the same for all materials

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

25

Learning Check

A 4329 g sample of solid is transferred from boiling water (T=998degC) to 152 g water at 225degC in a coffee cup The Twater rose to 243degC Calculate the specific heat of the solid

qsample+ qwater + qcup=0qcup is neglected in problemqsample = - qwater

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

26

Your Turn

What is the heat capacity of the container if 100g of water (s=4184 JgmiddotdegC) at 100degC are added to 100 g of water at 25degC in the container and the final temperature is 61degC

A 870 JdegC

B 35 JdegC

C -35 JdegC

D -870 JdegC

E None of these

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

27

bull Chemical bond net attractive forces that bind atomic nuclei and electrons together

bull Exothermic reactions form stronger bonds in the product than in the reactant and release energy (-E)

bull Endothermic reactions break stronger bonds than they make and require energy (+E)

Chemical Potential Energy

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

28

Work and Pistonsbull Pressure = force

areabull If the container

volume changes the pressure changes

bull Work = -PtimesΔVbull units L bull atmbull 1 L bull atm = 101 J

bull In expansion ΔVgt0 and is exothermic

bull Work is done by the system in expansion

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

29

How does work relate to reactions

bull Work = Force middot Distance bull is most often due to the expansion or contraction of

a system due to changing moles of gasbull Gases push against the atmospheric pressure so

Psystem = -Patm

bull w = -PatmtimesΔVbull The deployment of an airbag is one example of this

process1 C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g)

6 moles of gas rarr 7 moles of gas

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

30

Learning Check P-V work

bull Ethyl chloride is prepared by reaction of ethylene with HCl How much PV work (in J) is done if 895 g ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

bull Calculate the work (in kilojoules) done during a synthesis of ammonia in which the volume contracts from 86 L to 43L at a constant external pressure of 44 atm

w = 00019 kJ

w = 0708 J

w = -44atm times (43-86)L =1892Lmiddotatm

w = -1atm times -715L =715 Lmiddotatm

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

31

Energy Can Be Transferred as Heat and Work

bull ΔE= q + wbull internal

energy changes are state functions

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

32

Your TurnWhen TNT is combusted in air it is according to the

following reaction4C6H2(NO2)3CH3(s) + 17 O2(g) rarr24 CO2(g) + 10H2O(l) + 6N2(g)

The reaction will do work for all of these reasons exceptA The moles of gas increaseB The volume of gas increasesC The pressure of the gas increasesD None of these

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

33

Problems 23 27

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

34

Calorimetry Is Used To Measure Heats Of Reactionbull Heat of reaction the amount of heat absorbed

or released in a chemical reactionbull Calorimeter an apparatus used to measure

temperature changes in materials surrounding a reaction that result from a chemical reaction

bull From the temperature changes we can calculate the heat of the reaction qbull qv heat measured under constant volume conditionsbull qp heat measured under constant pressure

conditions

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

35

Internal Energy Is Measured With A Bomb Calorimeter

bull Used for reactions in which there is change in the number of moles of gas present

bull Measures qv

bull Immovable walls mean that work is zero

bull ΔE = qv

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

36

Learning Check Bomb Calorimeter500 mg naphthalene (C10H8) is combusted in a

bomb calorimeter containing 1000 g of water The temperature of the water increases from 2000degC to 2437degC The calorimeter constant is 420 JdegC What is the change in internal energy for the reaction swater = 4184 JgdegC

qcal= 420 Jg times(2437-2000)degCqwater= 1000 g times(2437-2000)degCtimes 4184 JgdegC

qv reaction + qwater + qcal = 0 by the first law

qv = ΔE = -20times104 J

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

37

Enthalpy Of Combustion

bull When one mole of a fuel substance is reacted with elemental oxygen a combustion reaction can be written whose enthalpy is ΔΗc˚

bull Is always negative bull Learning Check What is the equation

associated with the enthalpy of combustion of C6H12O6(s)

C6H12O6(s) + 9O2(g) rarr6CO2(g) + 6H2O(l)

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

38

Your Turn

252 mg of benzoic acid C6H5CO2H is combusted in a bomb calorimeter containing 814 g water at 2000ordmC The reaction increases the temperature of the water to 2170 What is the internal energy released by the process

A -711 JB -285 JC +711 JD +285 JE None of these

swater = 4184 JgdegC

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

39

Enthalpy Change (ΔH)

bull enthalpy is the heat transferred at constant pressure

bull ΔH = qp

bull ΔE = qp - P ΔV = ΔH ndash PΔVbull ΔH = ΔHfinal - ΔHinitialbull ΔH= ΔHproduct- ΔHreactant

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

40

Enthalpy Measured in a Coffee Cup Calorimeter

bull when no change in moles of gas is expected we may use a coffee cup calorimeter

bull the open system allows the pressure to remain constant

bull thus we measure qp

bull ΔE=q + wbull since there is no change in the

moles of gas present there is no work

bull thus we also are measuring ΔE

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

41

Learning Check Coffee Cup CalorimeteryWhen 500 mL of 987 M H2SO4 is added to 500 mL of

100 M NaOH at 250 degC in a calorimeter the temperature of the aqueous solution increases to 317 degC Calculate heat for the reaction per mole of limiting reactant

Assume that the specific heat of the solution is 418 JgdegC the density is 100 gmL and that the calorimeter itself absorbs a negligible amount of heat

H2SO4(aq) + 2 NaOH(aq) rarr 2H2O(l) + Na2SO4(aq)

moles NaOH = 00500 is limitingmoles H2SO4 = 004935

qsoln = 100gsolntimes(317-250)degC times418JgdegCqv rxn = -56times104 J

qv rxn + qcal + qsoln = 0

qv rxn = -28times103 J

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

42

Your TurnA sample of 5000mL of 0125M HCl at 2236 ordmC

is added to a 5000mL of 0125M Ca(OH)2 at 2236 ordmC The calorimeter constant was 72 Jg ordmC The temperature of the solution (s=4184 Jg ordmC d=100 gmL) climbed to 2330 ordmC Which of the following is not true

A qczl=679 JB qsolution= 3933JC qrxn = 4610 JD qrxn = -4610 JE None of these

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

43

Calorimetry Overview

bull The equipment used depends on the reaction type

bull If there will be no change in the moles of gas we may use a coffee-cup calorimeter or a closed system Under these circumstances we measure qp

bull If there is a large change in the moles of gas we use a bomb calorimeter to measure qv

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

44

Problems 51 53 55

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

45

Thermochemical Equations

bull Relate the energy of a reaction to the quantities involved

bull Must be balanced but may use fractional coefficients

bull quantities are presumed to be in moles

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

46

bull 2C2H2(g) + 5 O2(g) rarr 4 CO2(g) + 2 H2O(g)

ΔE = -2511 kJbull The reactants (acetylene and oxygen) have 2511

kJ more energy than the products bull How many KJ are released for 1 mol C2H2

Learning Check

1256 kJ

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

47

Learning Check

bull 6CO2(g) + 6H2O(l) rarr C6H12O6(s) + 6O2(g) Δ H = 2816 kJ

bull How many kJ are required for 44 g CO2 (MM=4400986 gmol)

bull If 100 kJ are provided what mass of CO2 can be converted to glucose

938 g

470 kJ

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

48

Learning Check Calorimetry of Chemical ReactionsThe meals-ready-to-eat (MRE) in the military can be

heated on a flameless heater Assume the reaction in the heater is Mg(s) + 2H2O(l) rarr Mg(OH)2(s) + H2(g)

ΔH = -353kJWhat quantity of magnesium is needed to supply the

heat required to warm 25 mL of water from 25 to 85 degC Specific heat of water = 4179 Jg degC Assume the density of the solution is the same as for water at 25degC 100 gmL

qsoln = 25 g times(85-25) degC times4179 Jg degC= 6269 times103 J

masssoln = 25 mLtimes100gmL= 25 g

massMg =043 g

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

49

Your Turn

Consider the thermite reaction The reaction is initiated by the heat released from a fuse or reaction The enthalpy change is ndash852 kJmol Fe2O3 at 298K

2 Al(s) + Fe2O3(s) rarr2Fe(s) + Al2O3(s)

What mass of Fe (MM=55847 gmol) is made when 500 kJ are released

A 655 gB 0587 gC 328 gD None of these

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

50

Learning Check Ethyl Chloride Reaction Revisitedbull Ethyl chloride is prepared by reaction of

ethylene with HClbull C2H4(g) + HCl(g) rarr C2H5Cl(g) ΔHdeg = -723kJbull What is the value of ΔE if 895 g ethylene and

125 g of HCl are allowed to react at atmospheric pressure and the volume change is -715 L

MM ethylene=28054 gmol MM HCl=364611 gmol

mol C2H4 31903 is Limitingmol HCl 34283

ΔHrxn =-23066 kJ

w = -1atm times -715L =715 LatmΔE= -230 kJ

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

51

Enthalpy Diagram

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

52

Problems 57 59

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

53

Hessrsquos Law

bull The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction

bull For examplebull 2 Fe + 32 O2 rarrFe2O3(s)amp ΔH= -8222 kJbull Fe2O3(s) + 2Al(s) rarrAl2O3(s) + 2Fe ΔH= -852 kJbull 32 O2 + 2Al(s) rarrAl2O3(s ΔH= -8222 kJ + -852 kJ

-1674 kJ

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

54

Rules for Adding Thermochemical Reactions

1 When an equation is reversedmdashwritten in the opposite directionmdashthe sign of H must also be reversed

2 Formulas canceled from both sides of an equation must be for the substance in identical physical states

3 If all the coefficients of an equation are multiplied or divided by the same factor the value of H must likewise be multiplied or divided by that factor

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

55

Strategy for Adding Reactions Together

1 Choose the most complex compound in the equation (1)

2 Choose the equation (2 or 3 orhellip) that contains the compound

3 Write this equation down so that the compound is on the appropriate side of the equation and has an appropriate coefficient for our reaction

4 Look for the next most complex compound

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

56

Hessrsquos Law (Cont)

5 Choose an equation that allows you to cancel intermediates and multiply by an appropriate coefficient

6 Add the reactions together and cancel like terms

7 Add the energies together modifying the enthalpy values in the same way that you modified the equation

bull if you reversed an equation change the sign on the enthalpy

bull if you doubled an equation double the energy

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

57

Learning Check

bull How can we calculate the enthalpy change for the reaction 2 H2(g) + N2(g) rarr N2H4(g) using these equations bull N2H4(g) + H2(g) rarr 2 NH3(g) ΔHdeg = -1876 kJbull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull Reverse the first reaction (and change sign)bull 2 NH3(g) rarr N2H4(g) + H2(g) ΔHdeg = +1876 kJ

bull Add the second reaction (and add the enthalpy)bull 3H2(g) + N2(g) rarr 2NH3(g) ΔHdeg = -922 kJ

bull 2 NH3(g)+ 3H2(g) + N2(g) rarr N2H4(g) + H2(g)+ 2NH3(g) bull 2 H2(g) + N2(g) rarr N2H4(g) (1876-922)= +954 kJ

2

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

58

Learning Check

Calculate ΔH for 2C(s) + H2(g) rarr C2H2(g) usingbull 2 C2H2(g) + 5O2(g) rarr 4 CO2(g) + 2H2O(l) ΔHdeg = -25992bull C(s) + O2(g) rarr CO2(g) ΔHdeg = -3935 kJbull H2O(l) rarr H2(g) + frac12 O2(g) ΔHdeg = +2859 kJ

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

59

Your Turn

What is the energy of the following process6A + 9B + 3D + 1 Frarr2 GGiven that bull C rarr A + 2B ∆H=202 kJmolbull 2C + D rarrE + B ∆H=301 kJmolbull 3E + F rarr2G ∆H=-801 kJmolA 706 kJB -298 kJC -1110 kJD None of these

(202times-6)+(301times3)+(-801) kJ

=-1134

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

60

State matters

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(g) bull ΔH= -2043 kJ

bull C3H8(g) + 5 O2(g) rarr 3 CO2(g) + 4 H2O(l) bull ΔH= -2219 kJ

bull note that there is a difference in energy because the states do not match

bull If H2O(l) rarr H2O(g) ΔH = 44 kJmolbull 4H2O(l) rarr 4H2O(g) ΔH = 176 kJmolbull -2219 +176 kJ = -2043 kJ

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

61

Most stable form of the pure substance atbull 1 atm pressurebull Stated temperature If temperature is not

specified assume 25 degCbull Solutions are 1M in concentrationbull Measurements made under standard state

conditions have the deg mark ΔHdegbull Most ΔH values are given for the most stable

form of the compound or element

Standard State

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

62

Determining the Most Stable Statebull The most stable form of a substance

bull below the melting point is solidbull above the boiling point is gasbull between these temperatures is liquid

bull What is the standard state of GeH4 bull mp -165 degC bp -885 degCbull What is the standard state of GeCl4 bull mp -495 degC bp 84 degC

gas

liquid

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

63

Problems 61 63 65 67 69

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

64

Allotropes

bull Are substances that have more than one form in the same physical state

bull You should know which form is the most stable

bull C P O and S all have multiple allotropes Which is the standard state for each

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

65

Enthalpy Of Formation

bull Is the enthalpy change ΔHdegf for the formation of 1 mole of a substance in its standard state from elements in their standard states

bull Note ΔHdegf = 0 for an element in its standard state

bull Learning Checkbull What is the equation that describes the formation

of CaCO3(s)

Ca(s) + C(gr) +32 O2(g) rarrCaCO3(s)

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

66

Calculating ΔH For Reactions Using ΔHdegf

ΔHdegrxn = [sum of ΔHdegf of all products]

ndash [sum of ΔHdegf of all reactants]bull 2Fe(s) + 6H2O(l) rarr 2Fe(OH)3(s) + 3H2(g)

0 -2858 -6965 0 ΔHdegf

bull CO2(g) + 2H2O(l) rarr 2O2(g) + CH4(g) -3935 -2858 0 -748 ΔHdegf

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

Your Turn

What is the enthalpy for the following reaction

A -75628 kJB 75628 kJC -154574 kJD 154574 kJE None of these

2 HCH3CO2(aq) + 2 OH-(aq) rarr 2H2O(l) + 2 C2H3O2(aq)

∆Hfordm -487 kJmol

-22994 kJmol

-28584kJmol

-54296kJmol

68

Problems 71 73 75 83 85

68

Problems 71 73 75 83 85