Chapter 5: Chapter 5: ThermochemistryThermochemistry

AP ChemistryAP Chemistry

EnergyEnergyThe capacity to do WORK or The capacity to do WORK or

produce HEATproduce HEAT

No work

work

Forms of EnergyForms of Energy

Potential EnergyPotential Energy

Stored energy due to Stored energy due to

positionposition or or compositioncomposition

Gravitational Potential Energy Gravitational Potential Energy (E(Egg))

Stored energy due to an object’s height above the surface of the earth

Chemical Potential Energy Chemical Potential Energy (E(Echemchem))

MethaneCH4

Landfill Gas FlareEnergy Stored in the bonds of molecules

Forms of EnergyForms of Energy

Kinetic EnergyKinetic Energy

Energy due to Energy due to motionmotion

Mechanical Kinetic Energy (Ek)

Ek = ½ mv2

Energy due to an object’s motion

(Add to your notes)

Thermal Kinetic Thermal Kinetic EnergyEnergy Energy due to molecular motion Energy due to molecular motion

First Law of First Law of ThermodynamicsThermodynamics

AKA The Law of Conservation of AKA The Law of Conservation of EnergyEnergy

Energy cannot be created nor Energy cannot be created nor destroyed; destroyed;

it can only change forms it can only change forms

James Prescott JouleJames Prescott JouleWealthy English Beer Baron (1818 - 1889)

JouleJoule’’s Chamonix Falls s Chamonix Falls ExperimentExperiment

Potential Energy (mechanical)P.E. = m g h

Kinetic Energy (mechanical and

thermal)K.E. = ½ m v2

Q = m c TLaw of Conservation of Energy

Energy cannot be created or destroyed in only changes forms

How we look at How we look at reactionsreactions

systemsystem: The Reactants and : The Reactants and Products Products

surroundings:surroundings: The room, The room, container, etc.container, etc.

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Internal Energy, EInternal Energy, E

The TOTAL energy of the system The TOTAL energy of the system (potential + kinetic energy)(potential + kinetic energy)

E

R P

E = Efinal - Einitial

R P

EE E

Energy GAINED by system Energy LOST by system

Temperature vs. Temperature vs. HeatHeat

Temperature— Measure of the average kinetic energy of molecules

100, 300, 700K Cl2, N2, He

Heat (q) Heat (q)

Transfer of thermal energy Transfer of thermal energy between the system and between the system and surroundings from surroundings from highhigh to to lowlow temps temps

Work (w) Work (w) A force exerted over a distanceA force exerted over a distance

w = F•w = F•dd

Examples: syringe, hatchback

Internal EnergyInternal Energy The energy of a system can be The energy of a system can be

changed by a flow of changed by a flow of heat (q)heat (q), , work (w) work (w) or or both both

E = q + wE = q + wNote: q and w must have Note: q and w must have a magnitude (#) and a sign (+ or -)a magnitude (#) and a sign (+ or -)System

surroundings

Internal EnergyInternal Energy

So when heat is added to a So when heat is added to a system or work is done on a system or work is done on a system, its internal energy system, its internal energy increases.increases.

systemsystem

Energy transfer of heat Energy transfer of heat only (no work)only (no work)

heat flows INTO the systemheat flows INTO the system heat EXITS the systemheat EXITS the system

the temperature _________the temperature _________the temperature ________the temperature ________

surroundings

ENDOTHERMIC EXOTHERMIC

Energy transfer of heat Energy transfer of heat onlyonly

E

R P

ENDOTHERMIC

R P

EE E

Energy GAINED by system (+q) Energy LOST by system (-q)

EXOTHERMIC

When work is done by a When work is done by a gasgas

w = Fd P = FA V = Ad

F = PA

wout = PAd

P =1 atm cnst

d

wout = PVSign of w? Why?

-w = PV

w = -PVSystem

surroundings

Sign Conventions for q & Sign Conventions for q & w—w—

Take the point of view of the Take the point of view of the system!system!

+q:+q: - q: - q: +w: +w: -w:-w:

heat flows into the systemheat flows into the system

heat flows out of the systemheat flows out of the systemwork is done ON the systemwork is done BY the system

Sample Problem:Sample Problem:

Calculate Calculate E for an exothermic E for an exothermic reaction in which 15.6 kJ of heat reaction in which 15.6 kJ of heat flows and where 1.4 kJ of work is flows and where 1.4 kJ of work is done on the system.done on the system.

E = q + w

E = -15.6 kJ + 1.4 kJ = -14.2 kJ

Units of EnergyUnits of Energy

calorie calorie Joule Joule

1 cal = 4.18 J 1 cal = 4.18 J

State Function—UPPER State Function—UPPER CASE CASE depends on the initial and final positions; depends on the initial and final positions; not the pathnot the path depends on the current state; not how it got theredepends on the current state; not how it got there

Pan

thers

Vis

itors

Explain why Energy is a state function

6 pts either way

start

finish

Path Function—Path Function—lower caselower case

depends on the depends on the pathpath because they have to do with because they have to do with a a transitiontransition..

Explain why work is a path function

Explain why heat is a path function

STATESTATE pathpath

TemperatureTemperature

PressurePressure VolumeVolume Heat CapacityHeat Capacity DensityDensity EnergyEnergy DisplacementDisplacement AltitudeAltitude

HeatHeat WorkWork DistanceDistance

Demo: State or Path Function?Demo: State or Path Function?control add KI add MnO4

H2O2

H2O + O2

Ep

How can we calculate heat How can we calculate heat (q)?(q)?

1)1) Heat Capacity (C): Heat Capacity (C):

quantity of heat needed to change itquantity of heat needed to change it’’s s temperature by 1 temperature by 1 ooCC

an extensive property—an extensive property—

depends on amount of the substancedepends on amount of the substance

C =Heat absorbed

T=

q

T

J oC

2) specific heat capacity (c)2) specific heat capacity (c)

intensive propertyintensive property

does not depend on amountdoes not depend on amount

can be used to identify a can be used to identify a substance substance

How can we calculate heat How can we calculate heat (q)?(q)?

The amount of heat needed to raise the temperature of 1 g of substance 1oC

cH2O(l) = 4.18 J/g • oC

c =Heat

m • T=

q

m •T

J

g • oC

Heat transferHeat transfer

In an isolated system the sum of In an isolated system the sum of the energy (heat) changes within the energy (heat) changes within the system must be zerothe system must be zero

qq11 + q + q22 + q + q33 + ……. = 0 + ……. = 0

ExampleExample

Hot metal block in water:Hot metal block in water: q q metalmetal + q + qwaterwater = 0 = 0

(C(CmmmmmmTTfinalfinal-T-Tintialintial)+(C)+(CwwmmwwTTfinalfinal-T-Tinitialinitial) = ) = 00

EnthalpyEnthalpy a property defined as internal energy + a property defined as internal energy +

product of pressure and volumeproduct of pressure and volumeH = E + PH = E + PVV

H = H = E + PE + PVVEnthalpy (H)

Energy = q + w

Pressure • volume = work

Comparing E and H

1) Reactions that do not involve gases2KOH (aq) + H2SO4(aq) K2SO4 (aq) + 2H2O(l)

H = H = E + PE + PVVWhat is What is V?V?

With no gases, With no gases, V = 0V = 0

sooooooosooooooo

H = H = E E

Comparing E and H

1) Reactions in which the moles of gas does NOT change

N2 (g) + O2 (g) 2NO(g)

H = H = E + PE + PVVWhat is What is V?V?

same moles of gas, same moles of gas, V = 0V = 0

sooooooosooooooo

H = H = E E

Comparing E and H—redo this; add derivation1) Reactions in which the moles of gas

DOES change 2H2 (g) + O2(g) 2H2O (g)

H = H = E + PE + PVVWhat is What is V?V?

waaaayyyy smaller than waaaayyyy smaller than H so H so that itthat it’’s insignificant and s insignificant and

H ≈ H ≈ E E

Enthalpy changes for Enthalpy changes for chemical reactionschemical reactions Given as: ΔGiven as: ΔrrHHoo = standard = standard

reaction enthalpy; unit kJ/mol-rxnreaction enthalpy; unit kJ/mol-rxn

HH22(g) + 1/2 O(g) + 1/2 O22(g) (g) HH22O(g)O(g)

– ΔΔrrHHoo = -241.8 kJ/mol-rxn = -241.8 kJ/mol-rxn

Enthalpy change Enthalpy change

Reverse reaction: opposite signReverse reaction: opposite sign CHCH44(g) + 2O(g) + 2O22(g) (g) CO CO22(g) + (g) +

2H2H22O(l)O(l) ΔH = -890 kJΔH = -890 kJ COCO22(g) + 2H(g) + 2H22O(l) O(l) CHCH44(g) + (g) +

2O2O22(g)(g) ΔH = +890 kJΔH = +890 kJ

Enthalpy changeEnthalpy change

ΔH depends on state of reactant ΔH depends on state of reactant and productsand products

2H2H22O(l) O(l) 2H 2H22O(g): ΔH = +88 kJO(g): ΔH = +88 kJ

Types of Reaction Types of Reaction systemssystems

OPEN CLOSED INSULATED

HARD! Engineering/Thermo Classes

Enthalpy; Endo/Exo Problems

Calorimetry

Calorimetry:Calorimetry:

A A calorimetercalorimeter is a is a device used to device used to experimentally experimentally determine the determine the amount of heat amount of heat associated with a associated with a chemical reaction chemical reaction (a device that (a device that measures measures caloriescalories))

Consider a system at a constant Consider a system at a constant pressure (an insulated system)pressure (an insulated system)

Recall: E = q + wRecall: E = q + w andand E = H -PE = H -PVV

E = q - PV

q - PV = H - PVH = q

Conclusion: Enthalpy = heat AT A CONSTANT PRESSURE

H = qp

q - PV = H - PV

Bomb Calorimetry: Bomb Calorimetry:

used to determine heat content in used to determine heat content in food/fuels food/fuels

constant __________

E = q + w

w = 0 since constant volume

E = qv

volume

E = q - PV

Calculation of energy transfer:Calculation of energy transfer: qqrr + q + qbombbomb + q + qwaterwater = 0, where q = 0, where qrr is is

energy released as heat by energy released as heat by reaction.reaction.

HessHess’’s Law:s Law:

In going from a set of reactants to a set of In going from a set of reactants to a set of products, the change in enthalpy (products, the change in enthalpy (H) is H) is the same whether the reaction takes the same whether the reaction takes place in one step or in a series of steps.place in one step or in a series of steps.

Enthalpy diagrams helps to visualize this.Enthalpy diagrams helps to visualize this.

ExampleExample

Given:Given: C(s) + OC(s) + O22(g) (g) CO CO22(g); ΔH = -393.5 kJ(g); ΔH = -393.5 kJ

CO(g) + ½OCO(g) + ½O22(g) (g) CO CO22(g); ΔH = -283.0 kJ(g); ΔH = -283.0 kJ Calculate ΔH for:Calculate ΔH for: C(s) + ½OC(s) + ½O22(g) (g) CO(g) CO(g)

ΔH = -110.5 kJΔH = -110.5 kJ

Things to RememberThings to Remember

1) If you reverse a reaction, you 1) If you reverse a reaction, you must change the sign of must change the sign of HH

2) If you multiply the coefficients by 2) If you multiply the coefficients by an integer, do the same to an integer, do the same to H H

HessHess’’s Law Problemss Law Problems Use the thermochemical equations to Use the thermochemical equations to

determine the enthalpy for the reaction:determine the enthalpy for the reaction:

CC22HH44O(l) O(l) + 5/2O+ 5/2O22(g)(g) 2CO2CO22(g) (g) + + 2H2H22O(l)O(l)

2C2C22HH44O(l) + 2HO(l) + 2H22O(l)O(l) 2C2C22HH66O(l) + OO(l) + O22(g) (g) H=610.5KJ H=610.5KJ

2CO2CO22(g) + 3H(g) + 3H22O(l)O(l) CC22HH66O(l) + 3OO(l) + 3O22(g) (g) H= H= 2056.5KJ2056.5KJ

NONO22(g) + 2H(g) + 2H22(g) (g) 1/2N1/2N22(g) + 2H(g) + 2H220(l)0(l)

2NH2NH33(g)(g) NN22(g) + 3H(g) + 3H22(g) (g) H= H= 23KJ 23KJ

2H2H22O(l)+NHO(l)+NH33(g)(g)NONO22(g) + 7/2H(g) + 7/2H22(g) (g) H=54KJH=54KJ

Standard Heat of Reaction Standard Heat of Reaction ((HHoo

rxnrxn))

The enthalpy taken in / The enthalpy taken in / given off in a reactiongiven off in a reaction

Standard StateStandard State

Symbol = Symbol = oo

gases are at 1 atmgases are at 1 atm ex: NOex: NO(g)(g)

solutions are at 1.0 Msolutions are at 1.0 M ex: NaClex: NaCl(aq)(aq), , BaBa2+2+

Pure liquids and solids are in their Pure liquids and solids are in their condensed statescondensed states ex: Hex: H22O O (l)(l)

Standard Enthalpies of Standard Enthalpies of Formation (Formation (HHoo

ff))

WhenWhen H values cannot be obtained H values cannot be obtained from constant pressure calorimetry from constant pressure calorimetry experimentsexperiments

The change in enthalpy to form The change in enthalpy to form one moleone mole of a compound of a compound from its elements from its elements with all substances at their with all substances at their standard states.standard states.

Standard Enthalpies of Standard Enthalpies of Formation (Formation (HHoo

ff))

HHooff for all for all elementselements at standard state at standard state

= 0 = 0

(1 atm, the state they are at 25(1 atm, the state they are at 25ooC)C)

HHooff of most of most compoundscompounds is negative— is negative—

find on Thermochemical Ref Chart: find on Thermochemical Ref Chart: Appendix LAppendix L

OO22 (g) , Hg ( (g) , Hg (ll) , Fe (s)) , Fe (s)

Heat of Reaction, Heat of Reaction, HHrxnrxn

HHoorxnrxn = = ((HHoo

ff))productsproducts - - ((HHooff))reactantsreactants

Sample Problem 1:

The thermite reaction occurs when a mixture of The thermite reaction occurs when a mixture of powdered aluminum and iron (III) oxide is ignited with powdered aluminum and iron (III) oxide is ignited with a magnesium fuse. It is used to weld massive steel a magnesium fuse. It is used to weld massive steel objects such as shipsobjects such as ships’’ propellers underwater. Calculate propellers underwater. Calculate the standard change in enthalpy for the thermite the standard change in enthalpy for the thermite reaction:reaction: 2Al(s) + Fe2Al(s) + Fe22OO33(s) (s) Al Al22OO33(s) + 2Fe(s) (s) + 2Fe(s)

orxn = (-1675.7 + 0) – (0 + (-825.5)) = -850.2 kJ/mol

H of reverse reaction?

Sample Problem 2:Sample Problem 2:

Calculate Calculate HHrxnrxnoo for the following for the following

reaction:reaction:

CaCOCaCO33(s) (s) CaO(s) + CO CaO(s) + CO22(g)(g)

ΔΔffHHoo’s (in kJ/mol): ’s (in kJ/mol):

CaCOCaCO33 = -1207.6; CaO = -635.1; CO = -1207.6; CaO = -635.1; CO22 = - = -393.5393.5

The First Law of The First Law of ThermodynamicsThermodynamicsEnergy is neither created nor destroyedEnergy is neither created nor destroyed

E= q + w w = F•d = PV E = energyEsystem = heat in – work out P = pressureEsystem = q - PV V = volume

H = enthalpyRearranging: q = E+ PV w = work

q = heat

Enthalpy definition: H = E + PVFor a given reaction: H = E+PV)

At constant pressure: H = E+PV H = q

PATMPATM

V

System

surroundings