Thermochem

1THERMOCHEMISTRYTHERMOCHEMISTRYThermodynamicsThermodynamics

The study of Heat and The study of Heat and Work and State FunctionsWork and State Functions

2Energy & ChemistryEnergy & ChemistryENERGYENERGY is the capacity to do is the capacity to do

work or transfer heat.work or transfer heat.HEATHEAT is the form of energy that is the form of energy that

flows between 2 objects flows between 2 objects because of their difference in because of their difference in temperature.temperature.

Other forms of energy —Other forms of energy —• lightlight• electricalelectrical• kinetic and potentialkinetic and potential

3Energy & ChemistryEnergy & Chemistry

• Burning peanuts supply sufficient peanuts supply sufficient energy to boil a cup of water.energy to boil a cup of water. • Burning sugar Burning sugar

(sugar reacts with (sugar reacts with KClOKClO33, a strong , a strong oxidizing agent)oxidizing agent)

4Energy & ChemistryEnergy & Chemistry

• These reactions are These reactions are PRODUCT FAVOREDPRODUCT FAVORED• They proceed almost completely from They proceed almost completely from

reactants to products, perhaps with some reactants to products, perhaps with some outside assistance.outside assistance.

http://xbeams.chem.yale.edu/~batista/113/movies/06z01vd1.mov

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Energy & ChemistryEnergy & Chemistry2 H2 H22(g) + O(g) + O22(g) --> (g) -->

2 H2 H22O(g) + heat and lightO(g) + heat and light

This can be set up to provide This can be set up to provide ELECTRIC ENERGYELECTRIC ENERGY in a in a fuel fuel cellcell..

Oxidation:Oxidation:2 H2 H22 ---> 4 H ---> 4 H++ + 4 e + 4 e--

Reduction: Reduction: 4 e4 e-- + O + O22 + 2 H + 2 H22O ---> 4 OHO ---> 4 OH--

CCR, page 845

6Potential & Kinetic Potential & Kinetic EnergyEnergy

Potential energy Potential energy — energy a — energy a motionless motionless body has by body has by virtue of its virtue of its position.position.

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy

Potential EnergyPotential Energyon the Atomic on the Atomic

ScaleScale

NaCl — composed of NaCl — composed of NaNa++ and Cl and Cl-- ions. ions.

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy

Potential EnergyPotential Energyon the Atomic on the Atomic

ScaleScale


Kinetic energy Kinetic energy — energy of — energy of motionmotion

• • TranslationTranslation


Kinetic energy Kinetic energy — — energy of energy of motion.motion.

translate

rotate

vibrate

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Internal Energy (E)Internal Energy (E)

• PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)• Int. E of a chemical system Int. E of a chemical system

depends ondepends on• number of particlesnumber of particles• type of particlestype of particles• temperaturetemperature

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Internal Energy (E)Internal Energy (E)• PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)

QuickTime™ and aGraphics decompressor

are needed to see this picture.

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Internal Energy (E)Internal Energy (E)• The higher the T The higher the T

the higher the the higher the internal energyinternal energy• So, use changes So, use changes

in T (∆T) to in T (∆T) to monitor changes monitor changes in E (∆E).in E (∆E).

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ThermodynamicsThermodynamics• Thermodynamics is the science of heat

(energy) transfer.

Heat energy is associated Heat energy is associated with molecular motions.with molecular motions.

Heat transfers until thermal equilibrium is established.

15Directionality of Heat TransferDirectionality of Heat Transfer• Heat always transfer from hotter object to

cooler one.•EXOthermic: heat transfers from SYSTEM

to SURROUNDINGS.

T(system) goes downT(system) goes downT(surr) goes upT(surr) goes up

16Directionality of Heat TransferDirectionality of Heat Transfer• Heat always transfer from hotter object to

cooler one.•ENDOthermic: heat transfers from

SURROUNDINGS to the SYSTEM.

T(system) goes upT(system) goes upT (surr) goes downT (surr) goes down

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Energy & ChemistryAll of thermodynamics depends on All of thermodynamics depends on

the law of the law of CONSERVATION OF ENERGYCONSERVATION OF ENERGY..• The total energy is unchanged in The total energy is unchanged in

a chemical reaction.a chemical reaction.• If PE of products is less than If PE of products is less than

reactants, the difference must be reactants, the difference must be released as KE.released as KE.

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Energy Change in Energy Change in Chemical ProcessesChemical Processes

Reactants

Products

Kinetic Energy

PE

PE of system dropped. KE increased. Therefore, PE of system dropped. KE increased. Therefore, you often feel a T increase.you often feel a T increase.

19UNITS OF ENERGYUNITS OF ENERGY1 calorie = heat required to 1 calorie = heat required to

raise temp. of 1.00 g of Hraise temp. of 1.00 g of H22O O by 1.0 by 1.0 ooC.C.

1000 cal = 1 kilocalorie = 1 1000 cal = 1 kilocalorie = 1 kcalkcal

1 kcal = 1 Calorie (a food 1 kcal = 1 Calorie (a food “calorie”)“calorie”)

But we use the unit called the But we use the unit called the JOULEJOULE

1 cal = 4.184 joules1 cal = 4.184 joules James JouleJames Joule1818-18891818-1889

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HEAT CAPACITYHEAT CAPACITYThe heat required to raise an

object’s T by 1 ˚C.

Which has the larger heat capacity?Which has the larger heat capacity?

21SpecificSpecific Heat Heat CapacityCapacity

How much energy is transferred due to T difference?How much energy is transferred due to T difference?The heat The heat (q)(q) “lost” or “gained” is related to “lost” or “gained” is related to a)a)sample masssample massb) b) change in T andchange in T andc) c) specific heat capacityspecific heat capacity

Specific heat capacity =

heat lost or gained by substance (J)(mass, g)(T change, K)

22Specific Heat Specific Heat CapacityCapacity

SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K)HH22OO 4.1844.184Ethylene glycolEthylene glycol 2.392.39AlAl 0.8970.897glassglass 0.840.84

AluminumAluminum


If 25.0 g of Al cool from If 25.0 g of Al cool from 310 310 ooC to 37 C to 37 ooC, how C, how many joules of heat many joules of heat energy are lost by energy are lost by the Al?the Al?

Specific heat capacity =

heat lost or gained by substance (J)(mass, g)(T change, K)


If 25.0 g of Al cool from 310 If 25.0 g of Al cool from 310 ooC to 37 C to 37 ooC, how many joules of heat energy are lost by the Al?C, how many joules of heat energy are lost by the Al?

heat gain/lose = q = (sp. ht.)(mass)(∆T)

where ∆T = Twhere ∆T = Tfinalfinal - T - Tinitialinitial

q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = - 6120 Jq = - 6120 J

Notice that the negative sign on q signals Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.

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Heat TransferHeat TransferNo Change in StateNo Change in State

q transferred = (sp. ht.)(mass)(∆T)

26Heat Transfer with Heat Transfer with Change of StateChange of State

Changes of state involve energy Changes of state involve energy (at constant T)(at constant T)Ice + 333 J/g (heat of fusion) -----> Liquid waterIce + 333 J/g (heat of fusion) -----> Liquid water

q = (heat of fusion)(mass)q = (heat of fusion)(mass)

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Heat Transfer and Heat Transfer and Changes of StateChanges of State

Requires energy (heat).Requires energy (heat).This is the reasonThis is the reasona)a) you cool down after you cool down after

swimming swimming b)b) you use water to put you use water to put

out a fire.out a fire.

+ energy

Liquid ---> VaporLiquid ---> Vapor

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Heat waterHeat water

Evaporate waterEvaporate water

Melt iceMelt ice

Heating/Cooling Curve for Water

Note that T is Note that T is constant as ice meltsconstant as ice melts

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Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

What quantity of heat is required to melt What quantity of heat is required to melt 500. g of 500. g of iceice and heat the water to and heat the water to steamsteam at 100 at 100 ooC?C?

Heat & Changes of StateHeat & Changes of State

+333 J/g+333 J/g +2260 J/g+2260 J/g

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How much heat is required to melt 500. g of ice How much heat is required to melt 500. g of ice and heat the water to steam at 100 and heat the water to steam at 100 ooC?C?

1. 1. To melt iceTo melt ice q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 1055 J J2.2. To raise water from 0 To raise water from 0 ooC to 100 C to 100 ooCC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 10q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 1055 J J3.3. To evaporate water at 100 To evaporate water at 100 ooCC q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 J J

4. 4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 1066 J = 1510 kJ J = 1510 kJ

Heat & Changes of StateHeat & Changes of State

31ChemicalChemical Reactivity ReactivityWhat drives chemical reactions? How do they What drives chemical reactions? How do they

occur?occur?The first is answered by The first is answered by THERMODYNAMICSTHERMODYNAMICS

and the second by and the second by KINETICSKINETICS..Have already seen a number of “driving forces” Have already seen a number of “driving forces”

for reactions that are for reactions that are PRODUCT-FAVOREDPRODUCT-FAVORED..•• formation of a precipitateformation of a precipitate•• gas formationgas formation•• HH22O formation (acid-base reaction)O formation (acid-base reaction)•• electron transfer in a batteryelectron transfer in a battery

32ChemicalChemical Reactivity ReactivityBut energy transfer also allows us to predict But energy transfer also allows us to predict

reactivity.reactivity.In general, reactions that transfer In general, reactions that transfer

energy to their surroundings are energy to their surroundings are product-favored.product-favored.

So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.

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Heat Energy Transfer Heat Energy Transfer in a Physical Processin a Physical Process

COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2 (g, -78 (g, -78 ooC)C)

Heat transfers from surroundings to system in endothermic process.

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Heat Energy Transfer Heat Energy Transfer in a Physical Processin a Physical Process• COCO2 2 (s, -78 (s, -78 ooC) ---> C) --->

COCO2 2 (g, -78 (g, -78 ooC)C)

• A regular array of A regular array of molecules in a solid molecules in a solid -----> gas phase -----> gas phase molecules. molecules.

• Gas molecules have Gas molecules have higher kinetic higher kinetic energy.energy.

35Energy Level Energy Level Diagram for Heat Diagram for Heat Energy Energy TransferTransfer

∆E = E(final) - E(initial) = E(gas) - E(solid)

COCO22 solid solid

COCO22 gas gas

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Heat Energy Transfer in Heat Energy Transfer in Physical ChangePhysical Change

• Gas molecules have higher kinetic Gas molecules have higher kinetic energy.energy.• Also, Also, WORKWORK is done by the system is done by the system

in pushing aside the atmosphere.in pushing aside the atmosphere.

COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2 (g, -78 (g, -78 ooC)C)Two things have happened!Two things have happened!

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FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS

∆∆E = q + wE = q + w

heat energy transferredheat energy transferred

energyenergychangechange

work donework doneby the by the systemsystem

Energy is conserved!Energy is conserved!

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heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEM

∆E = q + w

w transfer inw transfer in(+w)(+w)

w transfer outw transfer out(-w)(-w)

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ENTHALPYENTHALPYMost chemical reactions occur at constant P, soMost chemical reactions occur at constant P, so

and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small)∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E∆∆H = change in H = change in heat content heat content of the systemof the system∆∆H = HH = Hfinalfinal - H - Hinitialinitial

Heat transferred at constant P = qHeat transferred at constant P = qpp

qqpp = = ∆H ∆H where where H = enthalpyH = enthalpy

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If If HHfinalfinal < H < Hinitialinitial then ∆H is negative then ∆H is negativeProcess is Process is EXOTHERMICEXOTHERMIC

If If HHfinalfinal > H > Hinitialinitial then ∆H is positive then ∆H is positiveProcess is Process is ENDOTHERMICENDOTHERMIC

ENTHALPYENTHALPY∆∆H = HH = Hfinalfinal - H - Hinitialinitial

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Consider the formation of waterConsider the formation of waterHH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) + O(g) + 241.8 kJ241.8 kJ

USING ENTHALPYUSING ENTHALPY

Exothermic reaction — heat is a “product” Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJand ∆H = – 241.8 kJ

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Making Making liquidliquid H H22O from HO from H22 + + OO22 involves involves twotwo exoexothermic thermic steps. steps.


H2 + O2 gas

Liquid H2OH2O vapor

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Making HMaking H22O from HO from H22 involves two steps. involves two steps.

HH22(g) + 1/2 O(g) + 1/2 O22(g) ---> H(g) ---> H22O(g) + 242 kJO(g) + 242 kJ

HH22O(g) ---> HO(g) ---> H22O(liq) + 44 kJ O(liq) + 44 kJ -----------------------------------------------------------------------

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liq) + 286 kJO(liq) + 286 kJ

Example of Example of HESS’S LAWHESS’S LAW——If a rxn. is the sum of 2 or more If a rxn. is the sum of 2 or more

others, the net ∆H is the sum of others, the net ∆H is the sum of the ∆H’s of the other rxns.the ∆H’s of the other rxns.


44Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Forming H2O can occur in a single step or in a two steps.

∆Htotal is the same no matter which path is followed.

45Hess’s Law Hess’s Law & Energy Level Diagrams& Energy Level Diagrams

Forming CO2 can occur in a single step or in a two steps.

∆Htotal is the same no matter

which path is followed.

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• This equation is valid because This equation is valid because ∆H is a ∆H is a STATE FUNCTIONSTATE FUNCTION

• These depend only on the state These depend only on the state of the system and of the system and notnot on how on how the system got there.the system got there.

• V, T, P, energy — and your bank V, T, P, energy — and your bank account!account!

• Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot measure absolute H. Can only measure absolute H. Can only measure ∆H.measure ∆H.

∆∆H along one path =H along one path = ∆∆H along another pathH along another path

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Standard Enthalpy ValuesStandard Enthalpy ValuesMost ∆H values are labeled Most ∆H values are labeled ∆H∆Hoo Measured under Measured under standard conditionsstandard conditions

P = 1 bar = 10P = 1 bar = 1055 Pa = 1 atm /1.01325 Pa = 1 atm /1.01325 Concentration = 1 mol/LConcentration = 1 mol/L

T = usually 25 T = usually 25 ooCCwith all species in standard stateswith all species in standard statese.g., C = graphite and Oe.g., C = graphite and O22 = gas = gas

48Enthalpy ValuesEnthalpy Values

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g)

∆∆H˚ = -242 kJH˚ = -242 kJ2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g)O(g)

∆∆H˚ = -484 kJH˚ = -484 kJHH22O(g) ---> HO(g) ---> H22(g) + 1/2 O(g) + 1/2 O22(g) (g)

∆∆H˚ = +242 kJH˚ = +242 kJHH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liquid)O(liquid)

∆∆H˚ = -286 kJH˚ = -286 kJ

Depend on Depend on how the reaction is writtenhow the reaction is written and on phases and on phases of reactants and productsof reactants and products

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Standard Enthalpy ValuesStandard Enthalpy ValuesNIST (Nat’l Institute for Standards and NIST (Nat’l Institute for Standards and

Technology) gives values ofTechnology) gives values of ∆∆HHff

oo = standard molar enthalpy of = standard molar enthalpy of formationformation

— — the enthalpy change when 1 mol of the enthalpy change when 1 mol of compound is formed from elements under compound is formed from elements under standard conditions.standard conditions.

See Table 6.2See Table 6.2

50∆∆HHffoo, standard molar , standard molar

enthalpy of formationenthalpy of formationEnthalpy change when 1 mol of compound is Enthalpy change when 1 mol of compound is

formed from the corresponding elements under formed from the corresponding elements under standard conditionsstandard conditions

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g)

∆∆HHffoo (H (H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol

By definition, By definition, ∆∆HHff

oo = 0 for elements in their standard states.= 0 for elements in their standard states.

51Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Use ∆H˚’s to calculate Use ∆H˚’s to calculate enthalpy changeenthalpy change for for HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

(product is called “(product is called “water gaswater gas”)”)


HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

From reference books we findFrom reference books we find

• HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) ∆HO(g) ∆Hff˚ = - 242 kJ/mol˚ = - 242 kJ/mol

• C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H ∆Hff˚ = - 111 kJ/mol˚ = - 111 kJ/mol


HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) ∆H(g) ∆Hoo = +242 kJ = +242 kJ

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H∆Hoo = -111 kJ = -111 kJ

--------------------------------------------------------------------------------

To convert 1 mol of water to 1 mol each of HTo convert 1 mol of water to 1 mol each of H22 and CO and CO requiresrequires 131 kJ of energy. 131 kJ of energy. The “water gas” reaction is The “water gas” reaction is ENDOENDOthermic.thermic.

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) ∆∆HHoo

netnet = +131 kJ = +131 kJ

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Using Standard Enthalpy Using Standard Enthalpy ValuesValues

In general, when In general, when ALLALL enthalpies of formation are enthalpies of formation are known: known:

Calculate ∆H of Calculate ∆H of reaction?reaction?

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (products) - (products) - ∆H ∆Hff

oo (reactants)(reactants)

Remember that ∆ always = final – initial

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Using Standard Enthalpy Using Standard Enthalpy ValuesValues

Calculate the heat of combustion of Calculate the heat of combustion of methanol, i.e., ∆Hmethanol, i.e., ∆Hoo

rxnrxn for for

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)


∆∆HHoorxnrxn = ∆H = ∆Hff

oo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O) O)

- {3/2 ∆H- {3/2 ∆Hffoo

(O(O22) + ∆H) + ∆Hffoo

(CH(CH33OH)} OH)} = (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}∆∆HHoo

rxnrxn = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g) ∆∆HHoo

rxnrxn = = ∆H ∆Hffoo

(prod) - (prod) - ∆H ∆Hffoo

(react)(react)

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Measuring Heats of ReactionCALORIMETRYCALORIMETRY

Constant Volume “Bomb” Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for reaction.

58CalorimetryCalorimetry

Some heat from reaction warms waterqwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

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Calculate heat of combustion of octane. Calculate heat of combustion of octane. CC88HH1818 + 25/2 O + 25/2 O22 --> 8 CO --> 8 CO22 + 9 H + 9 H22OO

•• Burn 1.00 g of octaneBurn 1.00 g of octane• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC• Calorimeter contains 1200 g waterCalorimeter contains 1200 g water• Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

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Step 1Step 1 Calc. heat transferred from reaction to water.Calc. heat transferred from reaction to water.q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J

Step 2Step 2 Calc. heat transferred from reaction to bomb.Calc. heat transferred from reaction to bomb.q = (bomb heat capacity)(∆T)q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J

Step 3Step 3 Total heat evolvedTotal heat evolved 41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 JHeat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

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