Unit 3: Thermochemistry - NLESDschool.nlesd.ca/~glennlegge/listing folder/chem 3202/unit 3... ·...

Unit 3:

Thermochemistry

Chemistry 3202

Unit Outline

Temperature and Kinetic Energy

Heat/Enthalpy Calculation

Temperature changes (q = mc∆T)

Phase changes (q = n∆H)

Heating and Cooling Curves

Calorimetry (q = C∆T & above

formulas)

Unit Outline

Chemical Reactions

PE Diagrams

Thermochemical Equations

Hess’s Law

Bond Energy

STSE: What Fuels You?

Temperature and Kinetic Energy

Thermochemistry is the study of energy

changes in chemical and physical changes

eg. dissolving a solid

burning

phase changes

Temperature - a measure of the average

kinetic energy of particles in a substance

- a change in temperature means

particles are moving at different speeds

- measured in either Celsius degrees or

Kelvins

Kelvin = Celsius + 273.15

p. 628

K 50.15 450.15

°C 48 -200

# o

f part

icle

s

500 K

300 K

Kinetic Energy

The Celsius scale is based on the

freezing and boiling point of water

The Kelvin scale is based on absolute

zero - the temperature at which

particles in a substance have zero

kinetic energy.

Heat/Enthalpy Calculations

system - the part of the universe being studied

and observed

surroundings - everything else in the universe

open system - a system that can exchange

matter and energy with the surroundings

eg. an open beaker of water

a candle burning

closed system - allows energy transfer but is

closed to the flow of matter.

isolated system – a system completely

closed to the flow of matter and energy

heat - refers to the transfer of kinetic

energy from a system of higher

temperature to a system of lower

temperature.

- the symbol for heat is q

WorkSheet: Thermochemistry #1

Part A: Thought Lab (p. 631)

Part B: Thought Lab

specific heat capacity - the amount of

energy , in Joules (J), needed to change

the temperature of one gram (g) of a

substance by one degree Celsius (°C).

The symbol for specific heat capacity is

a lowercase c

Heat/Enthalpy Calculations

A substance with a large value of c can

absorb or release more energy than a

substance with a small value of c.

ie. For two substances, the substance

with the larger c will undergo a smaller

temperature change with the same

amount of heat applied

FORMULA

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti

eg. How much heat is needed to raise the

temperature of 500.0 g of water from

20.0 °C to 45.0 °C?

Solve q = m c ∆T

for c, m, ∆T, T2 & T1

p. 634 #’s 1 – 4

p. 636 #’s 5 – 8

WorkSheet: Thermochemistry #2

heat capacity - the quantity of energy , in

Joules (J), needed to change the

temperature of a substance by one

degree Celsius (°C)

The symbol for heat capacity is

uppercase C

The unit is J/ °C or kJ/ °C

FORMULA

C = mc

q = C ∆T

C = heat capacity

c = specific heat

capacity

m = mass

∆T = T2 – T1

Your Turn p.637 #’s 11-14

WorkSheet: Thermochemistry #3

Enthalpy Changes

The difference between the potential energy of the reactants and the products during a physical or chemical change is the Enthalpy change or ∆H.

AKA: Heat of Reaction

Reaction Progress

PE

Reactants

Products

∆H

Endothermic Reaction

Reaction Progress

PE

Reactants

Products

∆H ∆H Enthalpy

Endothermic Reaction

∆H is + Enthalpy

Reactants

Products

Endothermic

Enthalpy

products ∆H is -

Exothermic

reactants

Enthalpy Changes in Reactions

All chemical reactions require bond

breaking in reactants followed by

bond making to form products

Bond breaking requires energy

(endothermic) while bond formation

releases energy (exothermic)

see p. 639

Enthalpy Changes in Reactions

endothermic reaction - the energy

required to break bonds is greater than

the energy released when bonds form.

exothermic reaction - the energy

required to break bonds is less than the

energy released when bonds form.

Enthalpy Changes in Reactions

∆H can represent the enthalpy change for

a number of processes

1. Chemical reactions

∆Hrxn – enthalpy of reaction

∆Hcomb – enthalpy of combustion

(see p. 643)

2. Formation of compounds from elements

∆Hof – standard enthalpy of formation

The standard molar enthalpy of formation

is the energy released or absorbed when

one mole of a compound is formed

directly from the elements in their

standard states.

( see p. 642)

3. Phase Changes (p.647)

∆Hvap – enthalpy of vaporization

∆Hfus – enthalpy of melting

∆Hcond – enthalpy of condensation

∆Hfre – enthalpy of freezing

4. Solution Formation

∆Hsoln – enthalpy of solution

There are three ways to represent any enthalpy change:

1. thermochemical equation - the energy term written into the equation.

2. enthalpy term is written as a separate expression beside the equation.

3. enthalpy diagram.

eg. the formation of water from the elements

produces 285.8 kJ of energy.

1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ

2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol

thermochemical

equation

3.

H2O(l)

H2(g) + ½ O2(g)

∆Hf = -285.8 kJ/mol Enthalpy

(H)

enthalpy

diagram

examples: pp. 641-643

questions p. 643 #’s 15-18

WorkSheet: Thermochemistry #4

Calculating Enthalpy Changes

FORMULA:

q = n∆H

q = heat (kJ)

n = # of moles

∆H = molar enthalpy

(kJ/mol)

n m

M

eg. How much heat is released when

50.0 g of CH4 forms from C and H ?

p. 642

n 50.0 g

16.05 g / mol

3.115 mol

q = nΔH

= (3.115 mol)(-74.6 kJ/mol)

= -232 kJ

eg. How much heat is released when

50.00 g of CH4 undergoes complete

combustion?

n

50.0 g

16.05 g / mol

3.115 mol

q = nΔH

= (3.115 mol)(-965.1 kJ/mol)

= -3006 kJ

eg. How much energy is needed to change 20.0

g of H2O(l) at 100 °C to steam at 100 °C ?

Mwater = 18.02 g/mol ΔHvap = +40.7 kJ/mol

n 20.0 g

18.02 g / mol

1.110 mol

q = nΔH

= (1.110 mol)(+40.7 kJ/mol)

= +45.2 kJ

∆Hfre and ∆Hcond have the opposite sign

of the above values.

eg. The molar enthalpy of solution for

ammonium nitrate is +25.7 kJ/mol. How

much energy is absorbed when 40.0 g of

ammonium nitrate dissolves?

n 40.0 g

80.06 g / mol

0.4996 mol

q = nΔH

= (0.4996 mol)(+25.7 kJ/mol)

= +12.8 kJ

What mass of ethane, C2H6, must be

burned to produce 405 kJ of heat?

ΔHcomb = -1250.9 kJ/mol

q = - 405 kJ

H

q n

q = nΔH

kJ/mol 1250.9

kJ 405- n

n = 0.3238 mol

m = n x M

= (0.3238 mol)(30.08 g/mol)

= 9.74 g

Complete: p. 645; #’s 19 – 23

pp. 648 – 649; #’s 24 – 29

19. (a) -8.468 kJ (b) -7.165 kJ

20. -1.37 x103 kJ

21. (a) -2.896 x 103 kJ

21. (b) -6.81 x104 kJ

21. (c) -1.186 x 106 kJ

22. -0.230 kJ

23. 3.14 x103 g

24. 2.74 kJ

25.(a) 33.4 kJ (b) 33.4 kJ

26.(a) absorbed (b) 0.096 kJ

27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq)

(b) 1.69 kJ

(c) cool; heat absorbed from water

28. 819.2 g

29. 3.10 x 104 kJ

p. 638 #’ 4 – 8

pp. 649, 650 #’s 3 – 8

p. 657, 658 #’s 9 - 18

WorkSheet: Thermochemistry #5

Heating and Cooling Curves

Demo: Cooling of p-dichlorobenzene Time (s) Temperature (°C) Time (s) Temperature (°C)

0 100 80 20

10 80

20 60

30 50

40 50

50 50

60 40

70 30

Cooling curve for p-dichlorobenzene

Temp.

(°C )

50

80

KE

PE

KE

Time

solid

freezing

liquid

20

Heating curve for p-dichlorobenzene

Temp.

(°C ) 50

20

80

KE

KE

PE

Time

What did we learn from this demo??

During a phase change temperature

remains constant and PE changes

Changes in temperature during

heating or cooling means the KE of

particles is changing

p. 651

p. 652

p. 656

Heating Curve for H20(s) to H2O(g)

A 40.0 g sample of ice at -40.0 °C is heated until it changes to steam and is heated to 140. °C.

1. Sketch the heating curve for this change.

2. Calculate the total energy required for this transition.

Time

Temp.

(°C )

-40

0

100

140

q = mc∆T

q = n∆H

q = mc∆T

q = mc∆T

q = n∆H

Data:

cice = 2.03 J/g.°C

cwater = 4.184 J/g.°C

csteam = 2.01 J/g.°C

ΔHfus = +6.02 kJ/mol

ΔHvap = +40.7 kJ/mol

warming ice:

q = mc∆T

= (40.0)(2.03)(0 - -40.0)

= 3248 J

warming water:

q = mc∆T

= (40.0)(4.184)(100. – 0)

= 16736 J

warming steam:

q = mc∆T

= (40.0)(2.01)(140.-100.)

= 3216 J

melting ice:

q = n∆H

= (2.220 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water:

q = n∆H

= (2.220 mol)(40.7 kJ/mol)

= 90.354 kJ

n = 40.0 g

18.02 g/mol

= 2.220 mol

moles of water:

Total Energy

90.354 kJ

13.364 kJ

3248 J

3216 J

16736 J

126.9 kJ

Practice p. 655: #’s 30 – 34

pp. 656: #’s 1 - 9

p. 657 #’s 2, 9

p. 658 #’s 10, 16 – 20

30.(b) 3.73 x103 kJ

31.(b) 279 kJ

32.(b) -1.84 x10-3 kJ

33.(b) -19.7 kJ -48.77 kJ

34. -606 kJ

WorkSheet:

Thermochemistry #6

Law of Conservation of Energy (p. 627)

The total energy of the universe is constant

∆Euniverse = 0

Universe = system + surroundings

∆Euniverse = ∆Esystem + ∆Esurroundings

∆Euniverse = ∆Esystem + ∆Esurroundings = 0

OR ∆Esystem = -∆Esurroundings

OR qsystem = -qsurroundings

First Law of Thermodynamics

Calorimetry (p. 661)

calorimetry - the measurement of heat

changes during chemical or physical

processes

calorimeter - a device used to measure

changes in energy

2 types of calorimeters

1. constant pressure or simple

calorimeter (coffee-cup calorimeter)

2. constant volume or bomb calorimeter.

Simple

Calorimeter

p.661

a simple calorimeter consists of an

insulated container, a thermometer, and

a known amount of water

simple calorimeters are used to measure

heat changes associated with heating,

cooling, phase changes, solution

formation, and chemical reactions that

occur in aqueous solution

to calculate heat lost or gained by a chemical or physical change we apply the first law of thermodynamics:

qsystem = -qcalorimeter

Assumptions:

- the system is isolated

- c (specific heat capacity) for water is not affected by solutes

- heat exchange with calorimeter can be ignored

eg.

A simple calorimeter contains 150.0 g of

water. A 5.20 g piece of aluminum alloy at

525 °C is dropped into the calorimeter

causing the temperature of the calorimeter

water to increase from 19.30°C to

22.68°C.

Calculate the specific heat capacity of the

alloy.

eg.

The temperature in a simple calorimeter with a heat capacity of 1.05 kJ/°C changes from 25.0 °C to 23.94 °C when a very cold 12.8 g piece of copper was added to it.

Calculate the initial temperature of the copper. (c for Cu = 0.385 J/g.°C)

Homework

p. 664, 665 #’s 1b), 2b), 3 & 4

p. 667, #’s 5 - 7

(60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3)

26.818(T2 – 98.0) = -523.84(T2 – 22.3)

26.818T2 - 2628.2 = -523.84T2 + 11681

550.66T2 = 14309.2

T2 = 26.0 °C

6. System (Mg)

m = 0.50 g or 0.02057 mol

Find ΔH

Calorimeter

100 ml or m = 100 g

c = 4.184

T2 = 40.7

T1 = 20.4

7. System

ΔH = -53.4 kJ/mol

n = CV = (0.0550L)(1.30 mol/L) = 0.0715 mol

Calorimeter

110 ml or m = 110 g

c = 4.184

T1 = 21.4

Find T2

6.

qMg = -qcal

nΔH = -mcΔT

n = 0.50 g

24.31 g / mol 0.02057 mol

Bomb

Calorimeter

Bomb Calorimeter

used to accurately measure enthalpy

changes in combustion reactions

the inner metal chamber or bomb contains

the sample and pure oxygen

an electric coil ignites the sample

temperature changes in the water

surrounding the inner “bomb” are used to

calculate ΔH

to accurately measure ΔH you need to

know the heat capacity (kJ/°C) of the

calorimeter.

must account for all parts of the

calorimeter that absorb heat

Ctotal = Cwater + Cthermom.+ Cstirrer + Ccontainer

NOTE: C is provided for all bomb

calorimetry calculations

eg. A technician burned 11.0 g of octane in a

steel bomb calorimeter. The heat capacity of

the calorimeter was calibrated at 28.0 kJ/°C.

During the experiment, the temperature of

the calorimeter rose from 20.0 °C to 39.6 °C.

What is the enthalpy of combustion for

octane?

Eg.

1.26 g of benzoic acid, C6H5COOH(s)

(∆Hcomb = -3225 kJ/mol), is burned in a

bomb calorimeter. The temperature of the

calorimeter and its contents increases

from 23.62 °C to 27.14 °C. Calculate the

heat capacity of the calorimeter.

Homework

p. 675 #’s 8 – 10

WorkSheet:

Thermochemistry #7

Hess’s Law of Heat Summation

the enthalpy change (∆H) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products)

∆H is independent of the pathway and/or the number of steps in the process

∆H is the sum of the enthalpy changes of all the steps in the process

eg. Enthalpy of the production of

carbon monoxide

Pathway #1: 2-step mechanism

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ

CO(g) + ½ O2(g) → CO2(g) ∆H = -283.0 kJ

C(s) + ½ O2(g) → CO(g) ∆H = -110.5 kJ

eg. Enthalpy of the production of

carbon monoxide

Pathway #2: formation from the elements

C(s) + ½ O2(g) → CO(g) ∆H = -110.5 kJ

Using Hess’s Law

We can manipulate equations with

known ΔH to determine the enthalpy

change for other reactions.

NOTE:

Reversing an equation changes the sign

of ΔH.

If we multiply the coefficients we must

also multiply the ΔH value.

eg.

Determine the ΔH value for:

H2O(g) + C(s) → CO(g) + H2(g)

using the equations below.

C(s) + ½ O2(g) → CO(g) ΔH = -110.5 kJ

H2(g) + ½ O2(g) → H2O(g) ΔH = -241.8 kJ

eg.

Determine the ΔH value for:

4 C(s) + 5 H2(g) → C4H10(g)

using the equations below.

ΔH (kJ)

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g) -110.5

H2(g) + ½ O2(g) → H2O(g) -241.8

C(s) + O2(g) → CO2(g) -393.5

Switch

Multiply

by 5 Multiply

by 4

4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) +110.5

5(H2(g) + ½ O2(g) → H2O(g) -241.8)

4(C(s) + O2(g) → CO2(g) -393.5)

Ans: -2672.5 kJ

4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) +110.5

5 H2(g) + 2½ O2(g) → 5 H2O(g) -1209.0

4C(s) + 4 O2(g) → 4 CO2(g) -1574.0

Practice

pg. 681 #’s 11-14

WorkSheet:

Thermochemistry #8

Review

∆Hof (p. 642, 684, & 848)

The standard molar enthalpy of formation is

the energy released or absorbed when one

mole of a substance is formed directly from

the elements in their standard states.

∆Hof = 0 kJ/mol

for elements in the standard state

The more negative the ∆Hof, the more

stable the compound

Determine the ΔH value for:

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)

using the formation equations below.

ΔHf (kJ/mol)

4 C(s) + 5 H2(g) → C4H10(g) -2672.5

H2(g) + ½ O2(g) → H2O(g) -241.8

C(s) + O2(g) → CO2(g) -393.5

Using Hess’s Law and ΔHf

Using Hess’s Law and ΔHf

ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants)

eg. Calculate the enthalpy of reaction for

the complete combustion of glucose.

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)

Use the molar enthalpy’s of formation to

calculate ΔH for the reaction below

Fe2O3(s) + 3 CO(g) → 3 CO2(g) + 2 Fe(s)

p. 688 #’s 21 & 22

Eg.

The combustion of phenol is represented by

the equation below:

C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(g)

If ΔHcomb = -3059 kJ/mol, calculate the heat of

formation for phenol.

Bond Energy Calculations (p. 688)

The energy required to break a bond

is known as the bond energy.

Each type of bond has a specific

bond energy (BE).

(table p. 847)

Bond Energies may be used to

estimate the enthalpy of a reaction.

Bond Energy Calculations (p. 688)

ΔHrxn = ∑BE(reactants) - ∑BE (products)

eg. Estimate the enthalpy of reaction for the

combustion of ethane using BE.

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

Hint: Drawing the structural formulas for all

reactants and products will be useful here.

+ 7 O = O C C

[2(347) + 2(6)(338) + 7(498)]

→ 4 O=C=O + 6 H-O-H

- [4(2)(745) + 6(2)(460)]

p. 690 #’s 23,24,& 26

p. 691 #’s 3, 4, 5, & 7

= -3244 kJ

2

8236 - 11480

Energy Comparisons

Phase changes involve the least amount

of energy with vaporization usually

requiring more energy than melting.

Chemical changes involve more energy

than phase changes but much less than

nuclear changes.

Nuclear reactions produce the largest ΔH

eg. nuclear power, reactions in the sun

STSE

What fuels you? (Handout)

#’s 1 – 4, 7, 9, 10, 13, 14, & 16

(List only TWO advantages for #2)

aluminum alloy water

m = 5.20 g m = 150.0 g

T1 = 525 ºC T1 = 19.30 ºC

T2 = ºC T2 = 22.68 ºC

FIND c for Al c = 4.184 J/g.ºC

qsys = - qcal

mc T = - mc T

(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)

-2612 c = -2121

c = 0.812 J/g.°C

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mc T = - C T

(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)

4.928 (23.94 – T1) = 1113

23.94 – T1= 1113/4.928

23.94 – T1= 225.9

T1= -202 ºC

calorimeter

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

q heat J or kJ

c Specific heat

capacity

J/g.ºC

C Heat capacity kJ/ ºC or

J/ ºC

ΔH Molar heat or

molar enthalpy

kJ/mol

Group 2 Total Mass = 2.05 g

Group 5 Total Mass = 1.86 g