I 86 | Cracking the AP Chemistry Exam

CHAPTER 7 OUESTIONS

Multiple-Choice 0uestions

Use the following information to answer questions 1-5.

Reaction 1: NrHo(D + Hr(S) -+ 2NH,(S)

Reaction 2: NrHo(I) + CHuO(f -+ CHrO(g) + Nr(S) + 3Hr(S)

Reaction 3: Nr(s) + 3Hr(s) -+ 2NHr(s)Reaction a: CH.O(f + CHrO(g) + Hr(S)

t. What is the enthalpy change for reaction [ ?

(A) -148 kJ/molon(B) -56 kJ/molon

I

(C) -18 kJlmol*(D) +148 kJ/mol*"

NI =?L,H =-3'1 kJ/mol dnLH =46 kJ/mol ftnLH =45 kJ./mol m

2. If reaction 2 were repeated at a higher temperature, how would the reaction's value for

AG be affected?

(A) It would become more negative because entropy is a driving force behind this

reaction.(B) It would become more positive because the reactant molecules would collide more

often.(C) It would become more negative because the gases will be at a higher pressure-

(D) It will stay the same; temperature does not affect the value for AG.

3. Under what conditions would reaction 3 be thermodynamically favored?

(A) It is always favored.

(B) It is never favored.(C) It is only favored at low temperatures.

(D) It is only favored at high temperatures.

4. It 64 g of CH.O were to decompose via reaction 4, approximately how much energy

would be released or absorbed?

(A) 65 kJ of energy will be absorbed.

(B) 65 kJ of energy will be released.

(C) 130 kJ ofenergy will be absorbed.

(D) 130 kJ of energy will be released.

5. Which of the following is true for all of the reactions?

(A) The enfopy increases.

(B) If completed inside a sealed container, the pressure will decrease as the reactions

go to completion.(C) The activation energy for all four reactions will decrease with increasing

temperature.

@) More energy is released when the bonds in the products form than is necessary tobreak down the bonds of the reactants.

7.

2A1(s) + 3Cl,(g) -+ 2AlClr(s)

The reaction above is not thermodynamically favored under standard conditions, butit becomes thermodynamically favored as the temperature decreases toward absolute

zero. Which of the following is true at standard conditions?

(A) AS and AIl are both negative.(B) AS and NI arc both positive.(C) AS is negative, and AFl is positive.(D) AS is positive, and AFI is negative.

Which of the following is true of the reaction shown in the diagram above?

(A) The reaction is endothermic because the reactants arc at a higher energy level than

the products.(B) The reaction is endothermic because the reactants are at a lower energy level than

the products.(C) The reaction is exothermic because the reactants are at a higher energy level than

the products.(D) The reaction is exothermic because the reactants are at a lower energy lelel than

the products.

How does the addition of a catalyst affect the overall energy change of a reaction?

(A) Catalysts make reactions more endothermic as they help break the bonds of the

reactants. i(B) Catalysts make reactions more exothermic as they add energy to the system.

(C) Catalysts reduce the energy change in the reaction to zero.

(D) Catalysts have no effect on the overall energy change of a reaction.

XbolrO)

F]6

0)

oF<

Reaction Coordinate

Big ldea #5: Laws of Thermodynamics and Changes in trlatter I t e I

1gg .'l Cracking:the,AP CtremistrrT Exam

9. C(s) +2wr(il -+ CH+(g)

C(s) + Or(S) -+ COz(g)

H,(s)+ 1or{s)+Hro(l)2

11.

10.

Bised on the information given above, what is A.I/" for the following reaction?

Ctt(s) +ZO.,(d -+ CO,(s) +2H,O(t)

(A) x+y+z(B) x+y-z(C) z+y-?s,(D) 2z+y-x

Is AS positive or negative for the above reaction, and why?,

(A) Positive; the productq are more disordered than the reactants

(B) Positive; energy is absorbed(C) Negative; the products ar€ more disordered than the reactants

(D) Negative; energy is released

ZHr(d+ Or(s) -+ 2HrO(g)

Based oii the infoiination given in'the table below, what is Af1" for the above reaction?

L,H" = x

M:1" =!

L,Ho = z

Average bond energy (kJ/mol)Bond

H-.HO=OO=H

,.: .,t: " ,500

500'.:. ,,,..',. '' 500

t2.

(A) -2,000 kJ(B) -500 kJ' :-.: 1j::.. :.:l(C).+1r0Q0..9,,,,., , ,,: ,i.,;, .:

(D) +2,ooo kJ

' , ' H2O(D -+ HrO(s) :

:l

Which of the fqllowing is true for the above reaction?

(A) The value for AS is positive.

(B) The value for AG is zero.

(C) The value for AFl is negative.

@) The reaction is favored at 1.0 atm and 298 K.

botr0)d

rI](g

tr0,)

o

Reaction Coordinate: 'l: ' '

lVhich pqlnt on lhc graph shown lbove correspo4ds b Tl^l:d, comntex or transitio:t state?

(A) I(B) 2

(c)3 :

(D) 4

C(s)+Or(s)+COr(s),

H"(e) + I o"1g; -r lt,og;'22C(s) + H,(e) -+ crHr(s)

Based on the information given above, what is AIl" for thq following reaction?

C,H"(8) + ' O,t8) -+ 2COr(d +'Hro(l)a-L

'(A) -1,300kJ

(B) -1,070kJ(c) -840 kJ(D) -780 kJ

15. In which of the following reactions is entropy incrllsine?

(A) 2SOr(8) +O,(s) -+ 2SOrG)'(B) CO(s) + H,O(s) -+ H,(s) + COr(s)(C) 4(g) + Clr(s) -+ 2HCI(g)(D) 2No,(8) + 2No(g) + or(s)

AIl' = -390 kJ/mol

AI1" = -290 kJ/mol

Af1'= +230kJ/mol

Big ldea #5: Laws of Thermodynamics and Changes in Matter I 1 I I

190 | Cracking the AP Chemistry Exam

16. When pure sodium is placed in an atmosphere of chlorine gas, the following reaction

occurs without the addition of additional energy:

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

Which of the following correctly identifies the values for AG, AS, and AI1?

(A) G>0,S>0,H<0(B) G<0,S<0,H<0(c) G<0,s<0,1{>0(D) C>0,S<0,H<0

17. Hr(s) + F2(8) + 2HF(g)

Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen

fluoride with an enthalpy change df -S+O kJ. What is the value of the heat of formation

of HF(sX

(A) -1,080 kJ/mol(B) -270 kJ/mol(C) zl\kJlmol(D) 540 kJ/mol

18. ., HrO(s) -+ HrO(l)

At room temperature, which of the following is true?

(A) AG and Af1 are positive; AS is negative.

(B) AG is positive; Af/ and AS are negative.

(C) AG is negative; AI1 and AS are positive.

(D) AG and AS are negative; AIl is positive.

t9. 2S(s) + 3or(g) -+ 2SO3(s)

AI1= +800 kJ/mol

ZSOr(s) -+ ZSOr(s) + Or(s)AI1 = -200 kJ/mol

Based on the information given above, what is Afl for the following reaction?

S(s) + Or(s) -+ SOr(s)

(A) 300 kJ(B) soO kJ(c) 600 kJ(D) 1,000 kJ

Free-Response 0uestions

l.

Energy is released when glucose is oxidized in the following reaction, which is a

metabolism reaction that takes place in the body.

CuH,,Ou(s) + 6Or(s) -+ 6COr(s) + 6HrO(l)

The standard enthalpy change, A-Flo, for the reaction is -2,801 kJ/moloo at 298 K.

(a) Calculate the standard entropy change, A,So, for the oxidation of glucose.

(b) Calculate the standard free energy change, AGo, for lhe reaction at 298 K.(c) Using the zxis below, draw an energy profile for the reaction.

Reaction Progress

(d) How much energy is given off by the oxidation,of 1.00 gram of glucose?

>,boHo

212.13

205

213.6

69.9

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I 92 | Cracking the AP Chemistry Exam

415

495

799

463

cHo(s) +2or(d -+ cO'(s) + 2Hro(6)

The standard free energy change, AGo, for the reaction above is -801 kJ/mol*" at 298 K'

(a) Use the table of bond dissociation energies to find Alf for the reaction above'

itt How many grams of methane must feact with excess oxygen in order to release

1500 kJ of heat?

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

iO Give an explanation for the size of the entropy change found in (c)'

3. Nr(s) +: Hr(s) -r z NH,(s)

The heat of formation, A1l"r, of NHr(g) is - 46.2kJlmol. The free engrgy of formation,

AG"r, of NHr(g) is -16.7 ki/mol'

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

o)Whatisthevalueoftheentropychange,AS,forthe'reactionaboveat29SK?(c) As the temperature is increased, what is the effect on AG for the reaction? How

does this affect the thermodynamic favorability of the reaction

(d) At what temperahre can N? Hr, and NI{, g4sesbe maintained together in

equilibrium, each with a partial pressure of 1 atm?

2Hr(il + Or(8r) -+ 2r\O(I)

The reaction above is thermodynamically favored at standard conditions at 298 K.

(a) Predict the sign of the enhopy change, ASo, for tbe reaction. Explain.

O) How would the value of A.f for &e reaction change if the product of the reactionwas HrO(g)?

(c) What ii the sign of AGo at 298 K? Explain.(d) What is the sign of AIf at 298 K? Explain.

5. CaO(s) + CO'(S/ -+ CaCOr(s)

The reaction above is thermodynamically favored at2g8 K', and the heat of reaction,AHo, is -178 kJ/mol*".

(a) hedict the sign of the entropy change, ASo, for the reaction. Explain.(b) What is the sign of AGo at 298 K? Explain.(c) What change, if any, occurs to the va.lue of AGo as the temperature is increased

from 298 K?(d) As the reaction takes place in a closed container, what changes will occur in the

concenhation of CO, and the temperature?

6. A student designs an experiment to determine the specific heat of aluminum. The studentheats a piece of aluminum with a mass of 5.86 g to various temperatures, then drops itinto a calorimeter containing 25.0 nL of water. The following data is gathered dwingoneofthefrials:j

109.1 23.2 26.8

(a) Gjven that the specific heat of water is 4.18 J/g."C and assuming its density isexactly 1.00 g/ml, calculate the heat gained by the water.

(b) Calculate the specific heat of aluminum from the experimental data given.(c) Calculate the enthalpy change for the cooling of aluminum in water in

kJ/mol.(d) If the accepted specific heat of aluminum is 0.900 J/g."C, calculate the percent

eITOr.

(e) Suggest two potential sources of error that would lead the student's experimentalvalue to be different from the actual value. Be specific in your reasoning and makesure any identified error can be quantitatively tied to the student's results.

Big idea #5: Laws of Thermodynamics and Changes ln watter I t Sl

CHAPTER 7 ANSWERS AND EXPLANATIONS

Multiple-Choice1. C To ger reacrion 1, all that is needed is ro flip reaction 4 and then add

it to reactions 2 and 3. Doing so will yield reaction 1:

NrH4(4 + CH*O(/) -+ CHrO(g) * Nr(g) + 3H,(g)

Nr(g) * aHr(g) -+ 2NHr(g)

CH,O(g) * H,(f) -+ CH.O(/)

C37 kJlmol',*^) * (-46 kJ/mol..") + (65 kJ/mol,*") = -18 kJlmol*.

2. A Entropy is posirive dl.rring reacrion 2, as creating gas molecules out

of liquid molecules demonstrates an increase in disorder. Using

LG = LH - 7AS, if the temperature increases, the 7A.S term will

also increase. As the overall value for that term is negative, increasing

7AS makes AG more negative.

3. C Use che LG = AH - fAS equation. To do that, determine the signs

oFboth LH andthe IAS term.

'We know that AF1 is negative. AS is negative as well' because the

reaction is going from four molecules to two molecules, meaning

it is becoming more ordered. That, in turn' means the 7A.S term is

positive.

Reactions are only favored when AG is negative. As temperature

increases, the IAS term becomes larger and more likely to be greater

in magnitude than the Aflterm. If the temperature remains low, the

7AS value is much more likely to be smaller in magnitude than the

Aflvalue, meaning AG is more likely to be negative and the reaction

will be favored.

4. D The molar mass of CH4O is 32 glmol, so 64 g of it represent two

moles. For every one mole oF CH.O that decomposes, 65 kJ oF

energy is released-this is indicated by the negative sign on the Halue. If two moles decomposes, twice thac amount-I30 kJ-will

be released.

I 9 4 | Cr:acking the AP Chemistry Exam

D5 All four reactions have a negative value for enthalpy. Energy is

absorbed (positive A,Ef when bonds break and released (negative

A-If when bonds form, so'in order for the overall Allvalue to be

negative, more energy must be released than is absorbed during each

feactlon.

Remember that AG = LH - TLS.

If the reaction is thermodynamically favored only when the temper-

ature is very low, then AG is negative only when 7is very small. This

can happen only when Al1. is negative and AS is negative. A very

small value for Twill eliminate the influence of AS'

In an exothermic reaction, €nergy is given off as the products are

created because the products have less potential energy than the

reactants.

D A catalyst lowers the activation energy of a reaction and makes it

proceed faster. However, it has no effect on the amount of bond

energy in either the reactants or products.

The equations given above the question give the heats of formation

of all the reactants and products (remember: the heat of formation of

Or, an element in its most stable form, is zero).

A,Ho for a reaction = (AlJ" for the products) - (41/" for the reactants).

First, rhe products:

From 2HrO , we get 22

From COr, we get !So A11o for the products -- Ztt t

Now the reactants:

From CHn, we get x. The heat of formation of O, is defined to

be zero, so that's it for the reactants.

LHo for the reaction = (22 + t - @ = 2z + Jt - x.

A solid is changing to a gas, that means the products are less ordered

than the reactants. Increasing disorder is represented by a positive

value for A.t.

6.

'7

A

C

8.

D9.

A10.

Big lclea #5: Laws of Thermodynamics and Changes in Matter I t S S

I I I I Cracking the AP Chemistry Exam

C18. The reaction occurs without additional energy being added at room

temperature, therefore it is favored, meaning AG is negative' A liquid

is more disordered than a solid, meaning AS is positive. Finally, heat

is absorbed by the ice as it melts, meaning AIl is also positive'

you can use Hess's law. Add the two reactions together, and cancel

rhings that appear on both sides.

2SG) + 3Ork) -+ 2so3k) ^H

=+8oo kJ

2so3k) -+ 2sor(g) * Ork) , * = -2oo kJ

2S(s) + 3),r(g) + 2SOr(3) -+ 2SOr(g) + 2SOr(8) * Ork)

This reduces to ! '

2SG) + }Or(d -+ 2SOr(g) ^H

= +600 kJ:

phing in half to get the equation we want'Now we can cut ever

S(s) + Or(g) -> SOr(g) Al1= +300ld

19. A

Free-ResponseG) Use the €ntropy values in the table.

AS"=ISo -t.s'- "prcducs - uectmr

Aso = t(6X213.6) + (6X69.9)l -[Qr2.I3) +. (qx20t]J/rnol.K:

LS" =259Jlmol'K

(b) Use the equp.tion below. Remember that enthalpy values are given in

kJ arid entropy values are given in J.

LGo=AHo-?a.9"

AGo = (J,801 kJ/mol*") - Qgg KX0.259 kJ/mol,,".K) =

-2,880 kJ/mol, "

:

Reaction Progress

...-.The reaction is exothermic; therefore the reactants must have more

energy than the pioducts, as'indicated. The difference in energy is

equal to the Allfor this reaction.

iThe enthalpy change of the reaction, Ho, is a measure of the energy

given offby 1 mole of glucose.

Moles = grams

M$r

M"l.r.f ;r;.or. = . (t'oo g) . = 0.00556 moles" (180 g/mol)

(0.00556 molX2,80l k]/mol) = 15.6 kl

1.

k)

(d)

Big ldea #5. Laws of Thermodynamics and Changes in tvtatter I t S S

lo o I Cracf ing:the:AP Chemistry:Exam

2. G)

3. (")

(b)

k)

Use the relationship below.

A,H" = I Energies of the bonds broken - I Energies of the bonds formed

L,H" = t4)@t, + (2)(4e5)l - lQ)Qee) + (4)(463)l kJ/mol

LH" =-800 kJ/mot

r,5ookj. T*+ = r.e mor cHo x H# = 3o g cHn

Use AGo = LH" - fA.S"

Remember that enthalpy values are given in kJ and entropy values

are given in J.

A.So =AH-AG

I

_ (-eoo kJlmol)-(-eot kJlmol)

(zrs r)AS" = 0.003 g/K = 3JtK

ASo is very small, which means that the entropy change for the pro-

cess is very small. This makes sense because the number of moles

remains constant, the number of moles of gas remains constant, and

the complexity of the molecules lemains about thC same.

By definition , LH"rand LG", for Nrfu) and Hr(g') are equal to zero.

LHo = XAFlo, products i IAflo; reactants

A,H" = l(2)(- 46.2 kJ/mol)l - 0 = -92.4 kJ/mol' ...,:, '. ,.::

LG" = IAG', products - tAGo/ reactants

LG" = t?)erc] kl/mol)l 'g =:4j.,4'1r1,1*o1

..:,..

Use AGo = LI/" - ZAS" -':

. : : ,. , : .

Remember that enthalpy values are given in kJ and entroPy values

are given in J.

t co - L,H -L.G (-gz.+kJ/m-ol)- (-s1.4'kltmot)(zrs r)

A.S" = - 0.198 kJ/mol,K = -i98 J/mql'K

T

(d)

(b)

k)

(d)

use AG = A,Ho _ ZAS.

From (b), ASo is negative, so increasing the temPerature increases the

value of AGo, making the reaction less thermodynamically favored.

Use AG = LHo - ZA.to

At equilibrium, AG = 0

T = LH' -(-gz,4oo Jtmol) = 467 K

a.t" (-tls l/mol.r)ASo is negative because the products are less random than the reac-

tants. That's because ga-s is converted into liquid in the reaction.

The value of AS' would increase, becoming less negative because

HrO(g) is more random than water but remaining negative because

the entropy would still decrease from reactants to products.

AGo is negative because the reaction is favored.

A-F1o must be negative at 298 K. For a reaction to occur sponta-

neously from standard conditions, either ASo must be positive or

A11o must be negative. This reaction is thermodynamically favored

although ASo is negative, so Allo must be negative.

ASo is negative because the products are less random than the reac-

tants. That's because two moles of reactants are conv€rted to one

mole of products and gas is converted into solid in the reaction.

AGo is negative because the reaction is favored.

Use AGo = LHo: ZA.to

AG" will become less negative beca0se as temPerature i-s increased,

the entropy change of a reaction becomes more important in deter-

mining its favorability. The entropy change for this reaction is nega-

tive, which discourages favorability, so increasing temperature will

make the reaction less thermodynamically favored, thus making

AG'less negative.

The concentration of CO, will decrease as the reaction proceeds in the

forward direction and the r€actants are consumed. The temPerature

will increase as heat is given offby the exothermic reaction'

4. (")

(b)

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202 | Cracking the AP Chemistry Exam

6. (") q = mcLT

q = (25.0 sX4.18 J/g''C)(3'6'C)q=376J

The heat gained by the water is the same as the heat lost by the

aluminum.

q = mcLT

-3761= (5.86 gXrX-82'3 "C)

c = 0.780 Jlg"C

lmolAl5.86s^tffi =0'217 molAl

376 Jlo.zr7 mol = 173b J/mol = 1'73 kJ/mol

lexperimenml - accePtedl

accePted

(b)

G)

x 100%(d) o/o error =

(e)

lo.zso - o'gool x 100% = t3.30/o error

0.900

Error 1: If some of the heat that was lost by the aluminum was not

absorbed by che water, that would cause the calculated heat gained

by the watet in Part (a) to be artificially low' This' in tT": *"."1d

,.dl'r.. the value of the specific heat of aluminum as calculated in

Parr (b).

Error 2; If there was more than 25'0 mL of water in the calori-::t:'

that would mean the mass in part (a) was artificially low' which

would make the calculation foi the heat gained by the water also

artificially [ow. This, in turn, would reduce the value of the specific

heat o[aluminum calculated in part (b)'

There are many Potential errors here' but as long as you can quan-

titatively follow them to the conclusion that the experimental value

would be too low, any error (within reason) can be acceptable'