Ch. 6: Thermochemistry 6.1 The Nature of Energy. Energy Energy- Law of conservation of energy-...

Ch. 6: Ch. 6: Thermochemistry Thermochemistry

6.1 The Nature of Energy6.1 The Nature of Energy

EnergyEnergy

Energy-Energy-

Law of conservation of energy-Law of conservation of energy- energy energy cancan be converted but be converted but notnot created or destroyedcreated or destroyed

Types of EnergyTypes of Energy

potential energy-potential energy- (PE) (PE)

ex. attractive or repulsive forcesex. attractive or repulsive forces kinetic energy-kinetic energy- (KE) (KE)

KE = ½mvKE = ½mv2 2 :depends on mass and :depends on mass and volumevolume

Types of EnergyTypes of Energy

(a): PE(a): PEAA > PE > PEBB

(b):(b): ball A has ball A has

rolled down rolled down the hillthe hill

has lost PE to has lost PE to friction and PE friction and PE in ball Bin ball B

Transfer of EnergyTransfer of Energy

Temperature-Temperature-

Two Ways to Transfer Energy:Two Ways to Transfer Energy: Heat-Heat- (q) transfer of energy between (q) transfer of energy between

two objects because of a temperature two objects because of a temperature differencedifference

Work-Work- (w) force acting over a distance (w) force acting over a distance

PathwayPathway

the specific conditions of energy the specific conditions of energy transfertransfer

state function-state function- depends only on depends only on current conditions, not past or futurecurrent conditions, not past or future

energy change is independent of energy change is independent of pathway because it is a pathway because it is a state functionstate function

work and heat depend on pathway so work and heat depend on pathway so are not state functionsare not state functions

Transfer of EnergyTransfer of Energy

Two parts of the UniverseTwo parts of the Universe system-system-

surroundings-surroundings-

usuallyusually system: what is inside the containersystem: what is inside the container surroundings: room, container, etc.surroundings: room, container, etc.

Transfer of EnergyTransfer of Energy

exothermic- energy is produced in reactionenergy is produced in reaction flows out of systemflows out of system container feels hot to the touchcontainer feels hot to the touch

endothermic-endothermic- energy is consumed by the reactionenergy is consumed by the reaction flows into the systemflows into the system container feels cold to the touchcontainer feels cold to the touch

Transfer of EnergyTransfer of Energy

Combustion of Methane Gas is exothermic

Transfer of EnergyTransfer of Energy

Reaction between nitrogen and oxygen is endothermic

Transfer of EnergyTransfer of Energy

the energy comes from the the energy comes from the ________________________ between the ________________________ between the reactants and productsreactants and products

energy produced (or absorbed) by energy produced (or absorbed) by reaction must equal the energy absorbed reaction must equal the energy absorbed (or produced) by surroundings (or produced) by surroundings

usually the molecules with usually the molecules with higher higher potential energypotential energy have have weaker bondsweaker bonds than molecules with lower potential than molecules with lower potential energyenergy

ThermodynamicsThermodynamics

Thermodynamics-Thermodynamics- study of energy study of energy and its transfersand its transfers

First Law of Thermodynamics-First Law of Thermodynamics-

Internal EnergyInternal Energy

(E) sum of potential and kinetic (E) sum of potential and kinetic energy in system energy in system

can be changed by work (w), heat can be changed by work (w), heat (q), or both(q), or both

SignsSigns signs are very importantsigns are very important signs will always reflect the system’s point signs will always reflect the system’s point

of view unless otherwise statedof view unless otherwise stated

∆∆EE qq ww

change change in in

internal internal energyenergy

heatheat workwork

exothermicexothermic

endothermiendothermicc

SignsSigns

WorkWork

common types of common types of workwork

expansion-expansion-

compression-compression-

expansionexpansion

compressioncompression

P is external P is external pressure – not pressure – not internal like we internal like we normally refer tonormally refer to

WorkWork

Example 1Example 1

Find the ∆E for endothermic Find the ∆E for endothermic process where 15.6 kJ of heat process where 15.6 kJ of heat flows and 1.4 kJ of work is done flows and 1.4 kJ of work is done on systemon system Since it is endothermic, q is ___ Since it is endothermic, q is ___

and w is ___and w is ___

Example 2Example 2

Calculate the work of expansion of a Calculate the work of expansion of a gas from 46 L to 64 L at a constant gas from 46 L to 64 L at a constant pressure of 15 atm.pressure of 15 atm. Since it is an expansion, ∆V is ___ and Since it is an expansion, ∆V is ___ and

w is ___w is ___

Example 3Example 3 A balloon was inflated from 4.00 x A balloon was inflated from 4.00 x

101066 L to 4.50 x 10 L to 4.50 x 1066 L by the addition L by the addition of 1.3 x 10of 1.3 x 1088 J of heat. Assuming the J of heat. Assuming the pressure is 1.0 atm, find the ∆E in pressure is 1.0 atm, find the ∆E in Joules. Joules.

(1 L∙atm=101.3 J)(1 L∙atm=101.3 J) Since it is an expansion, ∆V is ___ and Since it is an expansion, ∆V is ___ and

w is ___w is ___

Ch. 6: Ch. 6: Thermochemistry Thermochemistry

6.2 Enthalpy and Calorimetry6.2 Enthalpy and Calorimetry

EnthalpyEnthalpy

definition:definition: since E, P and V are all state functions, then H since E, P and V are all state functions, then H

is toois too for the following, the process is at constant P for the following, the process is at constant P

and the only type of work allowed is PV workand the only type of work allowed is PV work

∆∆E = qE = qPP + w = q + w = qPP - P∆V - P∆V qqPP = ∆ E +P∆V = ∆ E +P∆V

H = E + PV H = E + PV ∆H = ∆E+P∆V ∆H = ∆E+P∆V

so at constant Pso at constant P

EnthalpyEnthalpy

heat of reaction and change in heat of reaction and change in enthalpy are used interchangeably enthalpy are used interchangeably for a reaction at constant Pfor a reaction at constant P

∆∆H = HH = Hproductsproducts - H - Hreactantsreactants

endo: endo: exo: exo:

CalorimetryCalorimetry

science of measuring science of measuring heatheat

calorimeter-calorimeter- device used to device used to experimentally find the heat experimentally find the heat associated with a chemical reactionassociated with a chemical reaction

substances respond differently when substances respond differently when heatedheated

Heat CapacityHeat Capacity

(C) how much heat it takes to raise a (C) how much heat it takes to raise a substance’s T by ________________substance’s T by ________________

the amount of energy depends on the amount of energy depends on the amount of substancethe amount of substance

K)or C(in Tin increase

J)(in absorbedheat Ccapacityheat

Heat CapacityHeat Capacity

specificspecific heat capacity heat capacity (s) heat (s) heat

capacity_________capacity_________ in J/°C*g or J/K*g in J/°C*g or J/K*g

molarmolar heat capacity heat capacity heat capacity __________heat capacity __________ in J/°C*mol or J/K*mol in J/°C*mol or J/K*mol

Constant-Pressure Calorimetry Constant-Pressure Calorimetry

uses simplest calorimeter (like coffee-uses simplest calorimeter (like coffee-cup calorimeter) since it is open to aircup calorimeter) since it is open to air

used to find changes in enthalpy used to find changes in enthalpy (heats of reaction) for reactions (heats of reaction) for reactions occurring in a solution since qoccurring in a solution since qPP = ∆H = ∆H

heat of reaction is an extensive heat of reaction is an extensive property, so we usually write them per property, so we usually write them per mole so they are easier to usemole so they are easier to use

Constant-Pressure Calorimetry Constant-Pressure Calorimetry when 2 reactants are when 2 reactants are

mixed and T mixed and T increasesincreases, , the chemical reaction the chemical reaction must be releasing heat must be releasing heat so is so is ______________________________

the released energy the released energy from the reaction from the reaction increases the _________ increases the _________ of molecules, which in of molecules, which in turn increases the Tturn increases the T

Constant-Pressure Calorimetry Constant-Pressure Calorimetry

If we assume that the calorimeter did If we assume that the calorimeter did not leak energy or absorb any itself not leak energy or absorb any itself (that all the energy was used to (that all the energy was used to increase the T), we can find the increase the T), we can find the energy released by the reaction:energy released by the reaction:

E released by rxn = E absorbed by E released by rxn = E absorbed by solnsoln

Constant-Volume Calorimetry Constant-Volume Calorimetry

uses a bomb calorimeteruses a bomb calorimeter weighed reactants are placed inside weighed reactants are placed inside

the rigid, steel container and ignitedthe rigid, steel container and ignited water surrounds the reactant water surrounds the reactant

container so the T of it and other container so the T of it and other parts are measured before and after parts are measured before and after reactionreaction

Constant-Volume Calorimetry Constant-Volume Calorimetry

Here, the ∆V = 0 Here, the ∆V = 0 so -P∆V = w = 0 so -P∆V = w = 0

∆∆E = q + w = qE = q + w = qVV for constant volume for constant volume

EErxnrxn = ∆T x C = ∆T x Ccalorimetercalorimeter

Example 1Example 1

When 1 mol of CHWhen 1 mol of CH44 is burned at is burned at constant P, 890 kJ of heat is released. constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CHFind ∆H for burning of 5.8 g of CH44 at at constant P.constant P.

890 kJ is released per mole of CH890 kJ is released per mole of CH44

Example 2Example 2

When 1.00 L of 1.00 M Ba(NOWhen 1.00 L of 1.00 M Ba(NO33))22 solution at 25.0°C is mixed with 1.00 L solution at 25.0°C is mixed with 1.00 L of 1.00 M Naof 1.00 M Na22SOSO44 solution at 25.0°C in solution at 25.0°C in a coffee-cup calorimeter, solid BaSOa coffee-cup calorimeter, solid BaSO44 forms and the T increases to 28.1°C. forms and the T increases to 28.1°C. The specific heat capacity of the The specific heat capacity of the solution is 4.18 J/g*°C and the density solution is 4.18 J/g*°C and the density is 1.0 g/mL. Find the enthalpy change is 1.0 g/mL. Find the enthalpy change per mole of BaSOper mole of BaSO44 formed. formed.

Example 2Example 2

Write the net ionic equation for the Write the net ionic equation for the reaction:reaction:

Is the energy released or absorbed? What Is the energy released or absorbed? What does that mean about ∆H and q?does that mean about ∆H and q?

How can we calculate ∆H or heat?How can we calculate ∆H or heat?

How can we find the m?How can we find the m?

Example 2Example 2 Find the mass:Find the mass:

Find the change in T:Find the change in T:

Calculate the heat created:Calculate the heat created:

Example 2Example 2

since it is a one-to-one ratio and the since it is a one-to-one ratio and the moles of reactants are the same, moles of reactants are the same, there is no limiting reactantthere is no limiting reactant

Example 3Example 3

Compare the energy released in the Compare the energy released in the combustion of Hcombustion of H22 and CH and CH44 carried out carried out in a bomb calorimeter with a heat in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T of 1.50 g of methane produced a T change of 7.3°C while the combustion change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of change of 14.3°C. Find the energy of combustion per gram for each. combustion per gram for each.

Example 3Example 3 methane: CHmethane: CH44

hydrogen: Hhydrogen: H22

Ch. 6: Ch. 6: Thermochemistry Thermochemistry

6.3 Hess’ Law6.3 Hess’ Law

Hess’ LawHess’ Law

since H is a state function, the since H is a state function, the change in H is independent of change in H is independent of pathwaypathway

Hess’ Law-Hess’ Law- when going from a set of when going from a set of reactants to a set of products, the ∆H reactants to a set of products, the ∆H is the same whether it happens in is the same whether it happens in __________________________________________________________________________

Example 1Example 1

Example 1Example 1

NN22(g) + 2O(g) + 2O22(g) (g) 2NO 2NO22(g)(g) ∆H = 68 kJ∆H = 68 kJ

OROR

NN22(g) + O(g) + O22(g) (g) 2NO(g) 2NO(g) ∆H = 180 kJ∆H = 180 kJ

2NO(g) + O2NO(g) + O22(g) (g) 2NO 2NO22(g)(g) ∆H = -112 kJ∆H = -112 kJ

NN22(g) + 2O(g) + 2O22(g) (g) 2NO 2NO22(g)(g) ∆H = 68 kJ∆H = 68 kJ

RulesRules

1.1. If a reaction is reversed, the sign of If a reaction is reversed, the sign of ∆H must be reversed as well.∆H must be reversed as well.

because the sign tells us the direction because the sign tells us the direction of heat flow as constant Pof heat flow as constant P

2.2. The magnitude of ∆H is directly The magnitude of ∆H is directly proportional to quantities of proportional to quantities of reactants and products in reaction. reactants and products in reaction.

If coefficients are multiplied by an If coefficients are multiplied by an integer, the ∆H must be multiplied integer, the ∆H must be multiplied in the same way.in the same way.

because ∆H is an extensive propertybecause ∆H is an extensive property

Example 2Example 2

Using the enthalpies of combustion Using the enthalpies of combustion for graphite (-394 kJ/mol) and for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to for the conversion of graphite to diamond.diamond.

CCgraphitegraphite (s) (s) C Cdiamonddiamond (s) (s)∆H=?∆H=?

Example 2Example 2CCgraphitegraphite (s) (s) C Cdiamonddiamond (s) (s)∆H=?∆H=?

(1)(1)

(2)(2)

to get the desired equation, we must reverse 2to get the desired equation, we must reverse 2ndnd equation:equation:

(1) C(1) Cgraphitegraphite(s) + O(s) + O22(g) (g) CO CO22(g) (g) ∆H=-394kJ/mol∆H=-394kJ/mol

(2) (2) ∆H=___________∆H=___________

CCgraphitegraphite (s) (s) C Cdiamonddiamond (s) (s) ∆H=∆H=

∆∆H=H=

Example 3Example 3

Find ∆H for the synthesis of BFind ∆H for the synthesis of B22HH6, 6, diborane:diborane:

2B(s) + 3H2B(s) + 3H22(g) (g) B B22HH66(g)(g)

Given:Given:(1)(1) 2B(s) + 3/2O2B(s) + 3/2O22(g) (g) B B22OO33(s) (s) ∆H∆H11=-1273kJ=-1273kJ

(2)(2) BB22HH66(g) + 3O(g) + 3O22(g) (g) B B22OO33(s) + 3H(s) + 3H22O(g) O(g) ∆H∆H22=-2035kJ=-2035kJ

(3)(3) HH22(g) + (g) + ½O½O22(g) (g) H H22O(l) O(l) ∆H∆H33=-286kJ=-286kJ

(4)(4) 3H3H22O(l) O(l) 3H 3H22O(g) O(g) ∆H∆H44=44 kJ=44 kJ

Example 3Example 3

Start by paying attention to what needs Start by paying attention to what needs to be on reactants and products sideto be on reactants and products side

(1)(1)

(2)(2)

(3)(3)

(4)(4)

Example 3Example 3

Underline what you want to keep- that will help Underline what you want to keep- that will help you figure out how to cancel everything else:you figure out how to cancel everything else:

(1)(1)

(2)(2)

(3)(3)

(4)(4)

Example 3Example 3 Need 3 HNeed 3 H22 (g) so 3 x (3) (g) so 3 x (3) Need 3 HNeed 3 H22O to cancel so 3 x (4)O to cancel so 3 x (4)

(1)(1)(2)(2) (3)(3)(4)(4)

2B(s) + 3H2B(s) + 3H22(g) (g) B B22HH66(g)(g)∆∆H = -1273 + -(-2035) + 3(-286) + 3(44) = H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ36kJ

Ch. 6: Ch. 6: Thermochemistry Thermochemistry

6.4 Standard Enthalpies of 6.4 Standard Enthalpies of FormationFormation

Standard Enthalpy of Standard Enthalpy of FormationFormation

∆∆HHff°° change in enthalpy that accompanies change in enthalpy that accompanies

the formation of one mole of a the formation of one mole of a compound from its elements in compound from its elements in standard statesstandard states

° means that the process happened ° means that the process happened under standard conditions so we can under standard conditions so we can compare more easily compare more easily

Standard StatesStandard States For a COMPOUND:For a COMPOUND:

for gas: P = 1 atmfor gas: P = 1 atm pure liquid or solid statepure liquid or solid state in solution: concentration is 1 Min solution: concentration is 1 M

For an ELEMENT:For an ELEMENT: form that it exists in at 1 atm and 25°Cform that it exists in at 1 atm and 25°C

O: OO: O22(g)(g) K: K(s)K: K(s) Br: Br: BrBr22(l)(l)

Writing Formation EquationsWriting Formation Equations always write equation where 1 mole of always write equation where 1 mole of

compound is formed (even if you must compound is formed (even if you must use non-integer coefficients)use non-integer coefficients)

NONO22(g):(g):

½N½N22(g) + O(g) + O22(g) (g) NO NO22(g)(g)

∆∆HHff°= 34 kJ/mol°= 34 kJ/mol

CHCH33OH(l):OH(l):

C(s) + HC(s) + H22(g) + O(g) + O22(g) (g) CHCH33OH(l) OH(l)

∆∆HHff°= -239 kJ/mol°= -239 kJ/mol

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

where where n =n = ∑ ∑ means “sum of”means “sum of” ∆∆HHff° is the standard enthalpy of formation for ° is the standard enthalpy of formation for

reactants or productsreactants or products

∆∆HHff° for any ______________ in standard ° for any ______________ in standard state is ________ so elements are not state is ________ so elements are not included in the summationincluded in the summation

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

since ∆H is a state function, we can since ∆H is a state function, we can use any pathway to calculate ituse any pathway to calculate it

one convenient pathway is to break one convenient pathway is to break reactants into elements and then reactants into elements and then recombine them into productsrecombine them into products

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

0)()()( 422

CHfOHfCOf HHHH

Using Standard Enthalpies of Using Standard Enthalpies of FormationFormation

Example 1Example 1

Calculate the standard enthalpy Calculate the standard enthalpy change for the reaction that occurs change for the reaction that occurs when ammonia is burned in air to when ammonia is burned in air to make nitrogen dioxide and watermake nitrogen dioxide and water

4NH4NH33(g) + 7O(g) + 7O22(g) (g) 4NO 4NO22(g) + 6H(g) + 6H22O(l)O(l)

break them apart into elements and break them apart into elements and then recombine them into productsthen recombine them into products

Example 1Example 1

Example 1Example 1

can be solved using Hess’ Law:can be solved using Hess’ Law:

(1)(1) 4NH4NH33(g) (g) 2N 2N22(g) + 6H(g) + 6H22(g)(g) -4∆H-4∆Hff°°NH3NH3

(2)(2) 7O7O22(g) (g) 7O 7O22(g)(g) 00

(3)(3) 2N2N22(g) + 4O(g) + 4O22(g) (g) 4NO 4NO22(g)(g) 4 ∆H4 ∆Hff°°NO2NO2

(4)(4) 6H6H22(g) + 3O(g) + 3O22(g) (g) 6H 6H22O(l)O(l) 6 ∆H6 ∆Hff°°H2OH2O

Example 1Example 1

can also be solved using enthalpy of can also be solved using enthalpy of formation equation:formation equation:

values are in Appendix 4: p. A21-A23values are in Appendix 4: p. A21-A23

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

Example 2Example 2

Calculate the standard enthalpy change Calculate the standard enthalpy change for the following reaction:for the following reaction:

2Al(s) + Fe2Al(s) + Fe22OO33(s) (s) Al Al22OO33(s) + 2Fe(s)(s) + 2Fe(s)

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

Example 3Example 3

Compare the standard enthalpy of Compare the standard enthalpy of combustion per gram of methanol combustion per gram of methanol (CH(CH33OH) with per gram of gasoline OH) with per gram of gasoline (C(C88HH1818).).

Write equations:Write equations:

Example 3Example 3

Calculate the enthalpy of combustion per Calculate the enthalpy of combustion per mole:mole:

]2[]42[ )()()(3 322

OHCHfOHfCOfOHCH HHHH

]2[]1816[ )()()( 18822188

HCfOHfCOfHC HHHH

Example 3Example 3

Convert to per gram using molar mass:Convert to per gram using molar mass: