Thermodynomics

HOW OFTEN DOES THERMODYNAIJIICS APPEAR ON IHEEXAT'I?In the multiple-choice section, this topic appears in about 5 out of 75 questions. In the free-responsesection, this topic appears every year.

THE FIRST AND SECOND IAWS OF THERMODYNAT|JIICSThe first law of thennodynamics says that the energy of the universe is constant. Energy can beneither created nor deshoyed, so while energy can be conaerted in a chemical process, the total energyremains constant.

The second law of thermodynamics says that if a process is spontaneous in one direction, then itcan't be spontaneous in the reverse direction, and that the entropy of the universe always increasesduring spontaneous reactions.

il3

SIATE FUNCTIONSEnthalpy change (AlI), enhopy change (AS), and free-energy change (AG) are state functions. Thatmeans they all depend only on the change between the initial and final states of a system, not on theprocess by which the change occurs. For a chemical reaction, this means that the thermodynamic statefunctions are independent of reaction pathway; for instance, the addition of a catalyst to a reactionwill have no effect on the overall energy or entropy change of the reaction.

STANDARD STATE CONDITIONS\{hen the values of thermodynamic quantities are given on the test, they are almost always given forstandard state conditions. A thermodynamic quantity under standard state conditions is indicated bythe little superscript circle, so the following is true under standard state conditions:

AH = AHo

AS = ASo

AG = AGo

Standard State Conditions. All gases are at 1 atmosphere pressure.. All liquids are pure.. All solids are pure.. All solutions are at 1-molar (1 M) concentration.. The energy of formation of an element in its normal state is defined

as zelo.o The temperature used for standard state values is almost invariably

room temperature: 25"C (298 K). Standard state values can becalculated for other temperatures, however.

ENTHALPY

ErrunrpY Cnnnar, AflThe enthalpy of a substance is a measure of the energy that is released or absorbed by the substance

when bonds are broken and formed during a reaction.

The Basic Rules of EnthalpyWhen bonds are formed, energy is released.

When bonds are broken, energy is absorbed.

The change in enthalpy, AH, that takes place over the course of a reaction can be calculated bysubtracting the enthalpy of the reactants from the enthalpy of the products.

Enthalpy ChangeNf=H -HProducts reactilts

I I4 I (RACKING THE AP CHTMISTRY T)(AM

If the products have stronger bonds than the reactants, then the products have lower enthalpythan the reactants and are more stable; in this case, energy is released by the reaction, which isexothermic.

If the products have weaker bonds than the reactants, then the products have higher enthalpythan the reactants and are less stable; in this case, energy is absorbed by the reaction, which isendothermic.

All substances like to be in the lowest possible energy state, which gives them the greatest stabil-ity. This means that, in general, exothermic processes are more likely to occur spontaneously thanendothermic processes.

Hmr or FonmrrroN, AH;Heat of formation is the change in energy that takes place when one mole of a compound is formedfrom its component pure elements under standard state conditions. Heat of formation is almost al-ways calculated at a temperature of 25oC (298 K).

Remember, AH! for a pure element is defined as zero.

. If. Nl'rfor a compound is negative, energy is released when the compound isformed from pure elements, and the product is more stable than its constituentelements. That is, the process is exothermic.

. If AH; for a compound is positive, energy is absorbed when the compound isformdd from pure elements, and the product is /ess stable than its constituentelements. That is, the process is endothermic.

If the AHivalues of the products and reactants are known, AH for a reaction can be calculated.

AHo = >,Nl"f products -DNlot reactants

Let's find AHo for the reaction below.

2 CH,OH(g) + 3 O,(g) -+ 2 CO,(g) + 4 H,O(g)

Compound AIl" (k/mol)

CH3OH(g) -207Or(g) 0

co,k) -3e4HrOk) -242

AHo = >, Ni7 products -E NI? reactants

AHo = t(2)(^Hicq)+ (+)(aHiHPlt- t(2)(^HicHPH) * (3)(^H;o,)l

AHo = l(2)(-3e4kl + g)(aazkDl- t(2X-201 kD + (3X0 kl)l

AHo = (-1,756 kD - (-402 kJ)

AHo = -1,354 kJ

THER['l0DYNAl,llCS I I l5

Bono ErrnoYBond energy is the energy required to break a bond. Because the breaking of a bond is an endothermicProcess, bond energy is always a positive number. When a bond is formed, energy equal to the bondenergy is released.

AHo = E Bond energies of bonds broken - E Bond energies of bonds formed

The bonds broken will be the reactant bonds, and the bonds formed will be the product bonds.Let's find AHo for the reaction below.

z}{,r$ + qk) -+ 2 HPk)

H-H 436

O=O 499

o-H 463

AHI = X Bond energies of bonds broken - E Bond energies of bonds formedAHo = I(2XH-H) + (1)(o=o)l- t(4xo-H)lAHo = l(2)(436 kl) + (1)(4ee kDI- (4X463 kDI

AHo = (1,,37tkD _ (1,952 kJ)

AHo = -481 kJ

Hrss's LrwHess's law says that if a reaction can be described as a series of steps, then AH for the overall reactionis simply the sum of the AH values for all the steps.

For example, let's say you wanted to calculate the enthalpy change for the following reaction:

c,[tgl + Hp(/) + C,H,OH(/) AHo = ?

And let's say that you know the enthalpy changes for the following two reactions:

C,rLk) +3O,(g) +2CQk) +2lto(/) AFI. = -1,AlLkl

2Cor(g) + 3 Hp(l) + CfIuOH(l) + 3 Qk) AHo = +1,358 kJ

If you add the two reactions whose enthalpy changes you know (as though they were simultane-ous equations) and cancel the substances that appear on both sides, you'Il get the reaction that you'relooking for. This means that you can add the enthalpy changes for the reactions that you know to get

the enthalpy change that you're looking for.

-L,41L k] + 1,368 kl = -43 k]

It5 I (RACI(It{G IHE AP CHEIIISIRY EXAI'I

Hrnr Cnplcrrv nno Sprclnc HrlrHeat capacity, Cn, is a measure of how much the temperature of an object is raised when it absorbs heat.

1-=M'*- LTCp = heat capacity

AH= heat added (J or cal)

AT = temperature change (K or "C)

An object with a large heat capacity can absorb a lot of heat without undergoing much of a changein temperature, whereas an object with a small heat capacity shows a large increase in temperatureeven if only a small amount of heat is absorbed.

Specific heat is the amount of heat required to raise the temperature of one gram of a substance

one degree Celsius.

4 = mcLTheat added (J or cal)

mass of the substance (g or kg)specific heat

temperature change (K or "C)

ENTROPYThe entropy, S, of a system is a measure of the randomness or disorder of the system; the greater thedisorder of a system, the greater its entropy. Because zero entropy is defined as a solid crystal at 0 K,all substances that we encounter will have some positive value for entropy. Standard entropies, So,

are calculated at 25"C (298 K).You should be familiar with several simple rules concerning entropies.

. Liquids have higher entropy values than solids.

. Gases have higher entropy values than liquids.

. Particles in solution have higher entropy values than solids.

. Two moles of a substance have higher entropy value than one mole.

Erurnopv Cnlnor, A5"The standard entropy change, AS? that has taken place at the completion of a reaction is the differencebetween the standard enhopies of the products and the standard entropies of the reactants.

Specific Heat ,

ASo=ISo -I SoProducts reactants

THERMODYNAMICS I II7

GIBBS FREE ENERGYThe Gibbs free energy, or simply free energy, G, of. a process is a measure of the spontaneity of theProcess.

For a given reaction

o if AG is negative, the reaction is spontaneous

o if AG is positive, the reaction is not spontaneous

r if AG = 0, the reaction is at equilibrium

Fnrr Errncv (nmcE, LGThe standard free energy change, AG, for a reaction can be calculated from the standard free energiesof formation, LG,, of its products and reactants in the same way that ASo was calculated.

AGo=EAG:. -EAG-o/ PftXtucts , reactants

LG, LH, AilD ASIn general, nature likes to move toward two different and seemingly contradictory states-low energyand high disorder, so spontaneous processes must result in decreasing enthalpy or increasing entropyor both.

There is an important equation that relates spontaneity (AG), enthalpy (AtI), and enhopy (AS) toone another.

LGo = AIf- fASoT = absolute temperature (K)

The chart below shows how different values of enthalpy and entropy affect spontaneity.

+

+

+

+

LowHigh

LowFIigh

LowFIigh

LowHigh

-)Always spontaneous

l)*.".r spontaneous

+ Not spontaneous at low temperatureSpontaneous at high temperature

Spontaneous at low temperature+ Not spontaneous at high temperature

You should note that at low temperature, enthalpy is dominant, while at high temperature, enhopyis dominant

I l8 I (RA(KIllG IHE AP (HEIIISIRY ExAl,l

AG lro LG"The standard free energy change, AGo, gives the spontaneity of a reaction when all the concentrations

of reactants and products are in their standard state concentrations (at L molar). The free energy

change, and thus the spontaneity, of a reaction will be different from the standard free energy change

if the initial concentrations of reactants and products are not 1 molar.The standard free energy change, AGo, can be related to AG for other conditions by the following

equations:

AG= AGo+ RTln Qor

AG= AGo+ 2.303RTlog QAGo= standard free energy change (J)

AG = free energy change under given initial conditions (|)

R = the gas constant, 8.31J/mol-KT = absolute temperature (K)

Q = the reaction quotient for the given initial conditions

Smuonno Fnrr Errncy CHANGT Ar{D THr EouruBRrun ConsmuLet's look at the equation relating AG and AG'.

AG=AGo+RTlnQAt equilibrium, AG = 0 and Q= K.

Knowing this, we can derive an equation relating the standard free energy change, AG? and theequilibrium constant, K.

AGo= -RTln Kor

AGo= -2.303RTlog K

AGo= the gas constant,8.31J/mol-KT = absolute temperature (K)

K = the equilibrium constant

Notice that if AGo is negative, K must be greater than 1, and products will be favored at equilibrium.Alternatively, if AG"is positive, K must be less than 1, and reactants will be favored at equilibrium.

THERM()DYilAMI(S I II9

ENERGY DIAGRAI{IS

Energy

EXOTHERMIC REACTION

The diagram above shows the energy change that takes place during an exothermic reaction. Thereactants start with a certain amount of energy (read the $aph from left to right). For the reaction toproceed, the reactants must have enough energy to reach the transition state, where they are part ofan activated complex. This is the highest point on the graph above. The amount of energy needed toreach this point is called the activation energy, E". At this point, all reactant bonds have been broken,but no product bonds have been formed, so this is the point in the reaction with the highest energy

and lowest stability. The energy needed for the reverse reaction is shown as line E,'.

Moving to the right past the activated complex, product bonds start to form, and we eventuallyreach the energy level of the products.

This diagram represents an exothermic reaction, so the products are at a lower energy level thanthe reactants and AH is negative.

Activated Complex

Reaction Coordinate

120 I CRACKIl{G THE AP (HtLllSTRY EXAM

The diagram below shows an endothermic reaction.

ENDOTHERMIC REACTION

h this diagram, the energy of the producb is greater than the mergy of the reactanb, so AH is positive.Reaction diagrams can be read in both directions, so the reverse reaction for an exothermic reac-

tion is endothermic and vice versa.

Reaction Coordinate

THERil0DYI|AilICS I r2r

Crmrvsrs mro Errnev Dncnuns

Energy

Uncatalyzed

Catalyzed

FProducts

Reactants

Reaction Coordinate

A catalyst speeds up a reaction by providing the reactanb with an altemate pathway that has a loweractivation energy, as shown in the diagram above.

Notice that the only difference between the catalyzed reaction and the uncatalyzed reaction is thatthe energy of the activated complex is lower for the catalyzedreaction. A catalyst lowers the activa-tion energy, but it has no effect on the energy of the reactants, the energy of the products, or AH forthe reaction.

Also note that a catalyst lowers the activation energy forboth the forward and the reverse reactior!so it has no effect on the equilibrium conditions.

122 a CRACI(Il{G THE AP CHEI,IISTRY EXAII

CHAPTER 8 OUESIIONS

Muulrlr-Cxolcr Qursrlous

Ouestions 1-4

(A) Free energy change (AG)

(B) Entropy change (AS)

(C) Heat of vaPorization(D) Heat of fusion(E) Heat caPacitY

1. If this has a negative value for a process,

then the Process occurs spontaneously.

2. This is a measure of how the disorder of a

system is changing'

3. This is the energy given off when a sub-

stance condenses.

4. This is the energy taken in by a substance

when it melts.

5. 2AlG)+ s Clr(g) -+ 2AlCI,(s)

The reaction above is not spontaneousunder standard conditions, but becomes

spontaneous as the temperature decreases

toward absolute zero. Which of thefollowing is true at standard conditions?

(A) AS and AH are both negative.(B) AS and AH are both Positive.(C) AS is negative, and AH is positive.(D) AS is positive, and AH is negative.(E) AS and AH are both equal to zero.

Reaction Coordinate

\Atrhich of the following is true of the reaction

shown in the diagram above?

(A) The reaction is endothermic because

the reactants are at a higher energylevel than the products.

(B) The reaction is endothermic because

the reactants are at a lower energy levelihan the products.

(C) The reaction is exothermic because thereactants are at a higher energy levelthan the products.

(D) The reaction is exothermic because the

reactants are at a lower energy levelthan the products.

(E) The reaction is endothermic because

the reactants are at the same energy

level as the products.

>.bo!0,

rI](n

o

THERMODYNAMI(S I I23

2 H,(g) + o,(8) -+ 2 Hp(g)

Based on the information given in the tablebelow what is AHo for the above reaction?

Bond Average bond energy (kllmol)

H-H 500

O=O 500

o-H 500

(A) -2,000 kl(B) -1,500 kJ(c) -soo kl(D) +1,000 kJ(E) +2,000 kJ

lA/hich of the following is true of a reactionthat is spontaneous at298K but becomesnonspontaneous at a higher temperature?

(A) AS" and A.FI. are both negative.(B) ASo and AHo are both positive.(C) A5'is negative, and AH'is positive.(D) AS'is positive, and A-FI" is negative.(E) AS" and AHo are both equal to zero.

Which of the following will be true whena pure substance in liquid phase freezesspontaneously?

(A) LG, Nf , and AS are all positive.(B) LG, M.l, and AS are all negative.(C) AG and AH are negative, but AS is

positive.(D) AG and AS are negative, but AH is

positive.(E) AS and AH are negative, but AG is

positive.

\Atrhich point on the graph shown abovecorresponds to activated complex ortransition state?

(A) 1

(B) 2(c) 3(D) 4(E) 5

11. C(s) + q(g) -+ COr(g) AHo = -390 kllmol

H,(g) + )o;tl+ H,o(/) aH" = -2eokllmol

2C(s) +Hr(8) -qHrk) LH" = +23}kJlmol

Based on the information given above, r,r,hatis AH" for the following reaction?

q

CrHr(g) + ioz@)

-+2COr(s)+HrO(/)

(A) -1,300 kJ(B) -1.,070kI(c) -840 kl(D) -780 ki(E) -680 kI

10.7.

bok0)atq(c

OJ

E"

8.

9.

Reaction Coordinate

I24 I (RACKI]'IG THE AP (HTMISTRY EXAM

15.t2. If an endothermic reaction is spontaneous at298K, which of the following must be truefor the reaction?

I. AG is greater than zero.II. AH is greater than zero.m. AS is greater than zero.

(A) I only(B) II only(C) I and II only(D) II and III only(E) I,II, and III

The addition of a catalyst will have which ofthe following effects on a chemical reaction?

I. The enthalpy change will decrease.II. The entropy change will decrease.III. The activation energy will

decrease.

(A) I only(B) II only(C) III only(D) I and II only(E) II and III only

C(s) + 2 4k) --, CH,(g) LHo = x

C(s) + Qk) -; co,(s) NI" =A

HJs) + *o,rrr-+

H,og) LIro =z

Based on the information given above, whatis AHo for the following reaction?

CHnk) +2Or(g) + COr(S) +ZHrO(l)

(A) x+y+z(B) x+y-z(C) z+y-Zx(D) 2z+y-x(E) 2z+y-2x

For which of the following processes will AS

be positive?

I. NaCl(s) --> Na.(aq) + Cl-(aq)II. 2 Hr(g) + or(8) -+ 2 Hro(g)III. CaCO.(s) + CaO(s) + COr(g)

(A) I only(B) II only(C) I and II only(D) I and III only(E) I,II, and III

In which of the following reactions isentropy increasing?

(A) 2 Sor(g) * oJS) -+ z So,(8)(B) CO(g) + H,O(g) -+ H,(B) + CO,(S)(C) H,(g) + Cl,(g) -+ 2 HCI(g)(D) 2 NO,(g) -+ z No(g) * o,(8)(E) 2HrS(g) +3Or(g) -+2HzO(8) *2SOr(8)

When pure sodium is placed in anatmosphere of chlorine gas, the followingspontaneous reaction occurs.

2 Na(s) + Clr(g) + 2 NaCl(s)

Which of the following statements is trueabout the reaction?

AS>OAH<OAG>O

I onlyII onlyI and II onlyII and III onlyI, II, and III

t5.13.

t7.

14.

I.il.

m.

(A)(B)(c)(D)(E)

IHERI'40DYNAftll(S I I 25

18. HrQ)+F,G)-2HF(8) 20. 2S(s)+3O,(s)-+2SQk)

Gaseous hydrogen and fluorine combine in AH = +800 kllmot

the reaction aboveto form hydrogen fluo.r]le 2 SO.(g) _+ 2 Sek) + ek)lvith an enthalpy change of -540 kJ. What.is m J jOOkJ/rilthe value of the heat of formation of HF(g)?

(A) -1,0g0 kIlmol Based on the information given above, what

(B) -540 kJlmol is AH for the following reaction?

(C) )70k1/mol S(s)+Qk)+Sor(S)(D) 270kJ/mot(E) 540 kllmol (A) 300 kl

(B) s00 kl

D. HPttl -rH,o(f (C) 600k1(D) 1,000 kI

Which of the following is true of the reaction (E) 1,200 kIshown above at room temperature?

I. AG is greater than zero.[. AH is greater than zero.m. AS is greater than zero.

(A) II only(B) III only(C) I and II only(D) I and III only(E) II and III only

I26 I CRACKING THE AP CHEilISTRY EXAM

Pnorrrms

C.HrrO.(s)

o, (g)

COr(8)

rLO (D

212.L3

205

213.6

69.9

180

32

MI8

Energy is released when glucose is oxidized in the following reaction, which is a metabolismreaction that takes place in the body.

C.H,pr(s) + 5 qk) -+ 5 CQk) + 6 Hp(I)

The standard enthalpy change, AH", for the reaction is -2,801. k] at 298 K.

(a) Calculate the standard entropy drange, ASo, for the oxidation of glucose.

(b) Calculate the standard free energy change, AG", for the reaction at 298 K.

(c) Whatisthevalue of Krfor thereaction?

(d) How mudr energy is grven off by the oxidation of 1.00 gram of glucose?

c-HO=O

C=O

G-H

415

495

799

463

CH.(8) + 2 o,(g) + CQk)+ 2 rLoG)

The standard free energy change, AGo, for the reaction above is +01 kI at 298 K.

(a) Use the table of bond dissociation energies to find AHo for the reaction above.

(b) Whatis the value of Knforthe reaction?

(c) What is the value of ASo for the reaction at 298 K?

(d) Give an explanation for the size of the entropy change found in (c).

2.

THERT()DYlIAIII(S I I27

3. Nr(g) + 3 Hr(s) <--s 2 Nu.(s)

Theheatof formation,AHi, of NHr(ilis-46.ZkJlmol.Thefreeenergyof formatiory LGo,, ofNH,(S) is -1.6.7 kllmol.

(a) What are the values of AH" and AGo for the reaction?

(b) What is the value of the entropy change, AS? for the reaction above at 29BK?

(c) As the temperature is increased, what is the effect on AG for the reaction? How does thisaffect the spontaneity of the reaction?

(d) At what temperature can \, H, and NH, gases be maintained together in equilibrium, eachwith a partial pressure of 1 atm?

Esslys

4. 2Hr(g) * oJS) -+zt\o(t)

The reaction above proceeds spontaneously from standard conditions at298K.

(a) Predict the sign of the entropy change, AS",for the reaction. Explain.

(b) How would the value of AS" for the reaction change if the product of the reaction wasH'o(g)?

(c) Whatisthesignof AGo at298K? Explain.

(d) Whatisthesignof AHo at298K? Explain.

5. CaO(s) + CQk) -+ CaCOr(s)

The reaction above is spontaneous at298K, and the heat of reaction, AHo, is -778k|.

(a) Predict the sign of the entropy change, AS7 for the reaction. Explain.

O) What is the sign of AGoat 298K? Explain.

(c) Vy'hat change , if. any, occurs to the value of AG" as the temperature is increased from 298 K?

(d) As the reaction takes place in a closed container, what changes will occur in the concentrationof CO, and the temperature?

H,o(r) * HPk)At298 K, the value of the equilibrium constant, K, for the reaction above is 0.036.

(a) What is the sign of ASo for the reaction above at 298K?

(b) Whatis the sign of AHo for the reaction above at 2gBK?

(c) What is the sign of AGo for the reaction above at 298K?

(d) At approximately what temperature will AG for the reaction be equal to zero?

I 2E I CRA(KING THE AP (HEMISTRY EXAM

CHAPTER 8 ANSWERS AND EXPTANATIONS

Mumrrrt-Cxolcr Qurslors1. A A negative value for AG means that the process is spontaneous. A positive value for AG means

that the process is nonspontaneous.

2. B Entropy is a measure of the disorder of a system. A positive value for AS means that a system

has become more disordered. A negative value for AS means that a system has become more

orderly.

3. C The heat of vaporization is the heat given off when a substance condenses. It is also the heat

that must be put into a substance to make it vaporize.

4. D The heat of fusion is the heat that must be put into a substance to melt it. It is also the heat

given off when a substance freezes.

By the way, choice (E), heat capacity, is a measure of how much heat must be added to a givenobject to raise its temperature 1oC.

5. A Remember AG = AH- TAS.

If the reaction is spontaneous only when the temperature is very low, then AG is only negativewhen T is very small. This can only happen when AH is negative (which favors spontaneity)and AS is negative (which favors nonspontaneity). A very small value for 7 will eliminate theinfluence of AS.

6. C In an exothermic reaction, energy is given off as the products are created because the productshave less potential energy than the reactants.

7. C The bond energy is the energy that must be put into a bond to break it. First let's figure out howmuch energy must be put into the reactants to break their bonds.

To break 2 moles of H-H bonds, it takes (2)(500) kI = 1,000 kI

To break 1 mole of O=O bonds, it takes 500 kj.

So to break up the reactants, it takes +1,500 kJ.

Energy is given off when a bond is formed; that's the negative of the bond energy. Now let's see

how much energy is given off when 2 moles of HrO are formed.

2 moles of HrO molecules contain 4 moles of O-H bonds, so (4)(-500) kI = -2,000 k] are given off.

So the value of AHo for the reaction is

(-2,000 k], the energy given off) + (1,500 kJ, the energy put in) = -500 kI.

8. A Remember,AG= Nl-T AS. If thereactionisspontaneousatstandardtemperaturebut becomes nonspontaneous at higher temperatures, then AG is negative only at lowertemperatures. This can only happen when AH is negative (which favors spontaneity) and AS

is negative (which favors nonspontaneity). As the value of T increases, the influence of AS

increases, eventually making the reaction nonspontaneous.

9. B The process is spontaneous, so AG must be negative. The intermolecular forces become strongerand the substance moves to a lower energy level when it freezes, so AH must be negative. Thesubstance becomes more orderly when it freezes, so AS must be negative.

THERMODYNAMICS I I29

10. C Point 3 represents the activated complex, which is the point of highest energy. This point is thetransition state between the reactants and the products.

11. A The equations given on top give the heats of formation of all the reactants and products(remember, the heat of formation of O, an element in its most stable form, is zero).

AHo for a reaction = (AHo for the products) - (lH. for the reactants).

First, the products:

From CO, we get (2X-390 kD = -780 klFrom HrO, we get 190kJSo AHo for the products = (-780 k|) + (-290 kD = -1,070 kJ

Now the reactants:

From CrH, we get +230 k]. The heat of formation of Q is defined as zero, so that's it for thereactants.

AHo for the reaction - (-1,070 k]) - (+230 kD = -1,300 kI.

12. D The reaction is spontaneous, so AG must be less than zeto, so (I) is not true. The reaction isendothermic, so A.Ff must be greater than zero; therefore, (II) is true.

The only way that an endothermic reaction can be spontaneous is if the entropy is increasing, soAS is greater than zero, and (II! is true.

13. C The addition of a catalyst speeds up a reaction by lowering the activation energy. A catalyst hasno effect on the entropy or enthalpy change of a reaction.

14. D The equations given above the question give the heats of formation of all the reactants andproducts (remember, the heat of formation of O, an element in its most stable form, is zero).

AHo for a reaction = (AHo for the products) - (AH" for the reactants).

First, the products:

From 2 HrO,we get2z

From COr, we get y

So AHo for the products = h + y

Now the reactants:

From CH' we get r. The heat of formation of Q is defined to be zero, so that's it for the reactants.

AHo for the reaction = (22 + y) - (x) = 2z + y - x.

15. D In (I), NaCl goes from a solid to aqueous particles. Aqueous particles are more disorderly than asolid, so AS will be positive.

In (II), we go from 3 moles of gas to 2 moles of gas. Fewer moles and less gas means less entropy,so AS will be negative.

In (III), we go from a solid to a solid and a gas. More moles and the production of more gasmeans increasing entropy, so AS will be positive.

16. D Choice (D) is the only reaction where the number of moles of gas is increasing, going from 2moles of gas on the reactant side to 3 moles of gas on the product side. In all the other choices,the number of moles of gas either decreases or remains constant.

I30 : CRA(KING IHE AP (HTMISTRY EXAM

17. B In the reaction, a solid combines with a gas to produce fewer moles of solid, so the entropychange is negative and (I) is not true. The reaction is spontaneous, so the free energy change isnegative and (IIf is not true. For a reaction with decreasing enhopy to be spontaneous, it mustbe exothermic. The enthalpy change is negative for an exothermic reaction and (II) is true.

18. C The reaction that forms 2 moles of HF(g) from its constituent elements has an enthalpy changeof -540 kj. The heat of formation is given by the reaction that forms L mole from these elements,so you can just divide -540 kI by 2 to get -270 kJ.

19. E Ice melts spontaneously at room temperature, so AG is less than zero; therefore, (I) is not true.

Lr this reactiory ice melts, so heat is absorbed to break up the intermolecular forces, and AH isgreater than zero; therefore, (II) is true.

Liquid water is less orderly than ice, so the entropy change when ice melts is positive; therefore,AS is greater than zero, and (III) is true.

20. A You can use Hess's law. Add the two reactions together, and cancel things that appear on bothsides.

2 S(s) + 3 OJS) -+ 2 SOr(8) AiT = +800 kJ

2 SO3G) -+ Z SOr(8) + Or(8) AH = -200 kI

2 S(s) + 3 Qk) + z So.(S) -+ 2 Solg) + 2 So.(S) + Qk)

This reduces to

2 S(s) + 2O2@) -+ 2 SQk) AH = +600 kl

Now we can cut everything in half to get the equation we want.

S(s) + O,(g) --> SOr(g) AII = +300 kJ

THERII0DYl{Al,llCS I I 3l

Pnosrrms

1. (a) Use the entropy values in the table.

ASo=IsJ.o*o-IsL",*o

As'= [(6)(213.5) + (5)(6e. s)l-l(212.1s) + (6X20s)] I/KLS" =259J/K

(b) Use the equation below Remember that enthalpy values are given in k] and entropy valuesare given in].

AGo = AHo- TASo

tr6'= (-2,801 kD - Q98)(0.259 kD = -2,880 kI

(c) Use the equation below. Remember that the gas constant is grven in terms of ].AGOlosK=" -2.303RT

(-2.s8o.ooo)rogK= (_rsom;60 =sos

K = 16sos

(d) The enthalpy change of the reaction , H', is a measure of the energy given off by 1 mole ofglucose.

Moles - grams

MW

Moles of grucose = d#fu = 0.00555 moles

(0.00555 mol)(2,801kJ/mol) = 15.5 kJ

2. (a) Use the relationship below.

47' = f Err"rgies of the bonds broken - )fnergies of the bonds formed

6s'= [(4)(41s) + (2)(aes)] -l(2)(7ee) + (a)(4ffi)l k]

AH" = -800 kI

O) Use the following equation. Remember that the gas constant is given in terms of ].

AG"IosK=" -2.303RT

logK= (-8o1,ooo)= L40

(-z.aoa)(a.ar)(2e8) -K = lgtro

I32 I CRACKI]IG THE AP CHEilISTRY EXAII

(c) Use AGo = AHo- TASo

Remember that enthalpy values are given in k] and entropy values are given inJ.

^"o_ AH-AG _ (-800 kI)-(-Aot tJ)

r (2es K)ASo = 0.003 kI/K=31/K

(d) AS'is very small, which means that the entropy change for the process is very small. Thismakes sense because the number of moles remains constant, the number of moles of gas remainsconstant, and the complexity of the molecules remains about the same.

3. (a) By definition, AHiand AGi for \(S) and Hr(g) are equal to zero.

6.Fl' = ! aHoyproducts - ) AHorreactants

lJI" = l(2 mol)(- 46.2 kJlmol)l - 0 = -92.4k1

AGo = >Acoyproducts - lAG'rreactants

AG" = [(2 mol)(-1 6.7 kI / mol)l- 0 = -33.4 kI

(b) Use AGo = A.Ff" - TAS.

Remember that enthalpy values are given in kl an{ entropy values are given inJ.

as" _ aH - ac _ (-e2.4 kI) - (-33.4 kDr (2e8 K)

ASo = - 0.198 kJlK = -798J /K

(c) Use AG = AH" - T ASo

From (b), AS" is negative, so increasing the temperature increases the value of LGo,making the

reaction less spontaneous.

(d) Use AG = AH"- TASo

At equilibrium, AG = 0

T = Allo _ (-gz'Eool)

= 467 Kaso (-1e8I / K)

THtRMODYl{AIIICS I I33

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4' (a) AS" is negative because the products are less random than the reactants. That's because gas isconverted into liquid in the reaction.

(b) The value of ASo would increase, becoming less negative because HrO(g) is more randomthan water but remaining negative because the entropy would still deciease from reactants toproducts.

(c) AG" is negative because the reaction proceeds spontaneously.

(d) AH' must be negative at 298 K. For a reaction to occur spontaneously from standard condi-tions, either AS" must be positive or A-Ff" must be negative. This reaction is spontaneous althoughAS9 is negative, so AII. must be negative.

5. (a) AS'is negative because the products are less random than the reactants. That's because twomoles of reactants are converted to one mole of products and gas is converted into solid in thereaction.

(b) AG" is negative because the reaction proceeds spontaneously.

(c) Use AG" = AHo - T ASo

AG'will become less negative because as temperature is increased, the entropy change of areaction becomes more important in determining its spontaneity. The enfropy change for thisreaction is negative, which discourages spontaneity, so increasing temperature will make thereaction less spontaneous, thus making AG" less negative.

(d) The concentration of CO, will decrease as the reaction proceeds in the forward direction andthe reactants are consumed. The temperature will increase as heat is given off by the exothermicreaction.

6. (a) AS'is positive because the product is more random than the reactant. That's because liquid isconverted into gas in the reaction.

(b) AH'is positive because HrO(S) is less stable than water. Energy must be put into water toovercome intermolecular forces and create water vapor.

(c) Use LG' = ).203RT log K

If K is less than 1, log K will be negative, making AGo positive.

(d) AG will be equal to zero at 373 K or 100"C. AG will be equal to zero at equilibrium. The pointat which water and HrO(g) are in equilibrium is the boiling point.

I34 I CRACI(ING THE AP CHEMISTRY EXAM