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This print-out should have 72 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsFor the reaction

2H2(g) + O2(g)→ 2H2O(ℓ)

find the value for the work done at 300 K.

1. 2.5 kJ

2. −7.5 kJ

3. 7.5 kJ correct

4. −2.5 kJ

Explanation:

T = 300 K R =8.314 J

mol ·K∆n = −3 mol

w = −P ∆V

= −∆nRT

= −(−3 mol)

(

8.314 J

mol ·K

)

(300 K)

= 7482.6 J

002 10.0 pointsThe enthalpy of fusion of methanol (CH3OH)is 3.16 kJ/mol. How much heat would beabsorbed or released upon freezing 25.6 gramsof methanol?

1. 2.52 kJ absorbed

2. 0.253 kJ absorbed

3. 2.52 kJ released correct

4. 3.95 kJ released

5. 3.95 kJ absorbed

6. 0.253 kJ released

Explanation:

MW CH3OH = 32.042g

∆H(fusion) = −∆H(freezing)

q = 25.6 g×1mol

32.042g×

−3.16 kJ

mol= −2.525 kJ

003 10.0 pointsA 0.2 gram sample of a candy bar is com-busted in a bomb calorimeter, increasing thetemperature of the 2000 g of water from25.00◦C to 25.47◦C. What is ∆U in kJ/g?Ignore any heat loss or gain by the calorime-ter itself.

1. 19.6 kJ/g

2. -3.9 kJ/g

3. -0.08 kJ/g

4. -19.6 kJ/g correct

5. 3.9 kJ/g

6. 0.08 kJ/g

Explanation:None

004 10.0 pointsFor the combustion reaction of ethylene(C2H4)

C2H4 + 3O2 → 2CO2 + 2H2O

assume all reactants and products are gases,and calculate the ∆H0

rxn using bond energies.

1. 0 kJ/mol

2. 251 kJ/mol

3. 680 kJ/mol

4. −1300 kJ/mol correct

5. −251 kJ/mol

6. 1300 kJ/mol


7. −680 kJ/mol

Explanation:

∆H0rxn =

∑

BE reactants −

∑

BE products

=[

(C C) + 4 (C H)

+3 (O O)]

−

[

4 (C O) + 4 (H O)]

=

[(

602kJ

mol

)

+ 4

(

413kJ

mol

)

+3

(

498kJ

mol

)]

−

[

4

(

799kJ

mol

)

+ 4

(

463kJ

mol

)]

= −1300kJ

mol

005 10.0 pointsMethyl tert-butyl ether or MTBE is an oc-tane booster for gasoline. The combustionof 0.9211 grams of MTBE (C5H12O(ℓ), 88.15g/mol) is carried out in a bomb calorimeter.The calorimeter’s hardware has a heat capac-ity of 1.540 kJ/◦C and is filled with exactly2.022 L of water. The initial temperature was26.336◦C. After the combustion, the temper-ature was 29.849◦C. Analyze this calorimeterdata and determine the molar internal energyof combustion (∆U) for this octane booster.

1. -3362 kJ/mol correct

2. -3120 kJ/mol

3. -3560 kJ/mol

4. -1957 kJ/mol

5. -2286 kJ/mol

6. -2748 kJ/mol

7. -4293 kJ/mol

Explanation:∆T = 29.849− 26.336 = 3.513◦

Ccal = [2022(4.184) + 1540 ]/1000= 10.00 kJ/◦C

qcal = 10(3.513) = 35.130 kJmoles = 0.9211/88.15 = 0.01045 mol∆U = -35.130 kJ / 0.01045 mol

= -3362 kJ/mol

006 10.0 pointsCalculate the standard reaction enthalpy

for the oxidation of nitric oxide to nitrogendioxide

2NO(g) + O2(g)→ 2NO2(g)

givenN2(g) + O2 → 2NO(g)

∆H◦ = +180.5 kJ ·mol−1

2NO2(g)→ N2(g) + 2O2(g)∆H◦ = −66.4 kJ ·mol−1

1. −294.6 kJ ·mol−1

2. +114.1 kJ ·mol−1

3. −114.1 kJ ·mol−1 correct

4. +246.9 kJ ·mol−1

5. −246.9 kJ ·mol−1

Explanation:We need to reverse the reactions and add

them:

2NO(g)→ N2(g) + O2(g)∆H = −180.5 kJ/mol

NO2(g) + 2O2(g)→ 2NO2(g)∆H = +66.4 kJ/mol

2NO(g) + 2O2(g)→ O2(g) + 2NO2(g)

2NO(g) + O2(g)→ 2NO2(g)∆H = −114.1 kJ/mol

007 10.0 pointsYou have a 12 oz. can (355 mL) of beer. Youtest the temperature and see that it reads 0◦C.Now this isn’t just any beer; this is Guinnessand you’ve heard that Guinness is best atroom temperature (20◦C). If the specific heat


of Guinness is 4.186 J/g·◦C, how much heatshould you add in order to raise the tempera-ture? The density of Guinness is 1.2 g/mL.

1. 33.6 kJ

2. 83 J

3. 33.6 J

4. 35.6 kJ correct

Explanation:

008 10.0 pointsA student runs a reaction in a closed system.In the course of the reaction, 64.7 kJ of heatis released to the surroundings and 14.3 kJof work is done on the system. What isthe change in internal energy (∆U) of thereaction?

1. -79.0 kJ

2. 50.4 kJ

3. 79.0 kJ

4. -50.4 kJ correct

5. 90.4kJ

Explanation:The change in internal energy is given by

the formula: ∆U = q + w. In this reaction,q is -64.7 kJ and w is 14.3 kJ. The answer is-50.4 kJ.

009 10.0 pointsConsider the plot below for three differentsamples of pure water.

T (K)

∆H(J)

Y

X

Z

Based on the plot, which answer choicebelow is a correct statement regarding thethree samples of pure water?

1. All three samples have the same heatcapacity.

2. Sample Z has the greatest heat capacity.

3. Sample X has the smallest mass. correct

4. All three samples have different specificheat capacities.

5. Sample Y would require the least heat toraise its temperature by 1 K.

Explanation:It is given that the samples are all pure

water, which ensures that all three sampleshave the same specific heat capacity. Since thesamples have different changes in T for a giveninput of energy, we can infer that all threesamples have different heat capacities. Sincesample X experiences the largest increase inT for a given amount of energy, it must bethe least massive sample. Sample Z wouldactually have a heat capacity intermediatebetween that of X and Y. Sample Y requriesthe most heat to raise its temperature by anygiven amount.

010 10.0 pointsWhat is the total heat flow when 12 grams ofice at -40◦C are heated to become water at25◦C?


1. 0.97 kJ

2. 2.26 kJ

3. 29.39 kJ

4. 4.01 kJ

5. 27.12 kJ

6. 6.27 kJ correct

Explanation:12 g

H2O(s)−40◦C

step 1−→

12 g

H2O(s)0◦C

step 2−→

12 g

H2O(ℓ)0◦C

step 3−→

12 g

H2O(ℓ)25◦C

Step 1:2.09 J

g·◦C· (12 g) · (0−− 40)◦C

= 1,003.2 J

Step 2:334 J

g· (12 g)

= 4,008 J

Step 3:4.184 J

g ·◦ C· (12 g) · (25− 0)◦C

= 1,255.2 JTotal = 1, 003.2 J + 4, 008 J + 1, 255.2 J

= 6,266.4 J = 6.27 kJ

011 10.0 pointsA CD player and its battery together do 500kJ of work, and the battery also releases 250kJ of energy as heat and the CD player re-leases 50 kJ as heat due to friction from spin-ning. What is the change in internal energy ofthe system, with the system regarded as thebattery and CD player together?

1. +200 kJ

2. −700 kJ

3. −750 kJ

4. −200 kJ

5. −800 kJ correct

Explanation:

Heat from the CD player is −50 kJ.Heat from the battery is −500 kJ.Work from both together on the surround-

ings is −250 kJ.This question is testing your ability to see

what the system is, and then look at ONLYthe energy flow for the system. Here the sys-tem is the battery and the CD player together.

∆U = q + w

= [−50 kJ + (−250 kJ)] + (−500 kJ)

= −800 kJ

012 10.0 points3 g of a hydrocarbon fuel is burned in a bombcalorimeter that contains 200 grams of waterinitially at 25.00◦C. After the combustion re-action, the temperature is 27.00◦C. Howmuchheat is evolved per gram of fuel burned? Theheat capacity of the calorimeter (hardwareonly) is 150 J/◦C.

1. 21220 J/g

2. 1673 J/g

3. 1973 J/g

4. 557 J/g

5. 7505 J/g

6. 7073 J/g

7. 300 J/g

8. 657 J/g correct

Explanation:mfuel = 3 g mwater = 200 g∆T = 27.00◦C− 25.00◦C = 2.0 ◦C

The amount of heat evolved by the reactionis equal to the amount of heat gained by thewater plus the amount of heat gained by thecalorimeter.The specific heat of water is 4.184 J/g·◦C,

so we have to multiply by grams and temper-ature change in order to obtain Joules:

∆H of water = (cH2O)(mwater)(∆T )


= (4.184 J/g ·◦ C) (200 g)

× (2.00◦C)

= 1673 J

The heat capacity of the calorimeter is 150J/◦C. This is not per gram, so we just haveto multiply by the temperature change to getJoules:

∆H of calorimeter = (ccal)(∆T )

= (150 J/◦C)(2.00◦C)

= 300 J

∆H of rxn = ∆H of water

+ ∆H of calorimeter

= 1673 J + 300 J = 1973 J

This is the total amount of heat evolved bythe combustion of the 3 g of fuel. To get theamount per gram of fuel burned, we divide∆H of the reaction by the amount of fuelburned (3 grams):

1973 J

3 g= 657 J/g

013 10.0 pointsThe specific heat of water is 1.00 cal/g·◦C, theheat of vaporization of water is 540 cal/g, andthe heat of fusion of water is 80 cal/g. Howmuch heat would be required to convert 10grams of ice at 0◦C to 10 grams of water at75◦C?

1. 15.5 cal

2. 6150 cal

3. 155 cal

4. 1.55 kcal correct

5. 61.5 kcal

Explanation:10 g

H2O(s)0◦C

step 1−→

10 g

H2O(ℓ)0◦C

step 2−→

10 g

H2O(ℓ)75◦C

Step 1:80 cal

g(10 g) = 800 cal

Step 2:1 cal

g ·◦ C(10 g)(75− 0)◦C = 750 cal

Total = 800 cal + 750 cal

= 1550 cal

= 1.55 kcal

014 10.0 points1 g of cake is combusted in a bomb calorime-ter. The heat capacity of the calorimeterhardware is 12 calories ·K−1. The calorime-ter contains 4 L of water; the specific heatcapacity of water is 1 calorie · g−1

·K−1 andthe density of water is 1 g ·mL−1. You det-onate the cake and the temperature of thewater increases by 1.2 K. Calculate the calo-ries in the one-gram sample of cake, ∆U .

1. 4814.4 calories correct

2. 1150.7 calories

3. 20083.2 calories

4. 20143.4 calories

5. 4800.0 calories

6. 1147.2 calories

Explanation:1calorie · g−1

·K−1× 4000ml× 1g ·mL−1

× 1.2K

015 10.0 pointsReaction of tertiary butyl alcohol with hydro-bromic acid produces tertiary butyl bromideby the following reaction. Use bond energies(provided in preamble) to estimate the changein enthalpy, ∆H, for this reaction.

+ HBr

OH Br

+ H2O

1. +105 kJ/mol

2. +186 kJ/mol

3. +24 kJ/mol


4. −105 kJ/mol


6. −186 kJ/mol

Explanation:In this reaction, you break two bonds in thereactants: a C-O bond and a H-Br bond. Youform new bonds in the products: a C-Br bondand a H-O bond.The change in enthalpy is the energy in forbreaking the bonds combined with the energyout gained from forming the new bonds.

∆Hrxn = B.E.C−O + B.E.H−Br

−B.E.C−Br − B.E.H−O

∆Hrxn = 358 + 366− 285− 463

∆Hrxn = −24 kJmol−1

016 10.0 points

Estimate the heat released when 1-butene(CH3CH2CH CH2) reacts with bromine togive CH3CH2CHBrCH2Br. Bond enthalpiesareC H : 412 kJ/mol; C C : 348 kJ/mol;C C : 612 kJ/mol; C Br : 276 kJ/mol;Br Br : 193 kJ/mol.

1. 317 kJ/mol

2. 288 kJ/mol

3. 181 kJ/mol

4. 507 kJ/mol

5. 95 kJ/mol correct

Explanation:

C

H

H

H

C

H

H

C

H

C

H

H

+ Br Br →

C

H

H

H

C

H

H

C

H

Br

C

H

H

Br

∆H =∑

Ebreak −

∑

Emake

=[

(C C) + (Br Br)]

−

[

2 (C Br) + (C C)]

= 612 kJ/mol + 193 kJ/mol

−

[

2 (276 kJ/mol) + 348 kJ/mol]

= −95 kJ/mol ,

which means 95 kJ/mol of heat was released.

017 10.0 pointsWhich of the following is/are a reason thatwater is a desirable heat sink for use incalorimeters?I) Water’s heat specific capacity is very pre-

cisely known.II) Water is readily available.III) Water has an unusually large specific

heat capacity.

1. I only

2. II and III

3. I and II

4. I, II and III correct

5. II only

6. I and III

7. III only

Explanation:Water is a good heat sink for all of the rea-

sons listed above. Moreover, its large heatcapacity, liquid state and ready availabilityenable us to easily set up a calorimeter suchthat ∆T is large enough that it can be easilymeasured and small enough that phase tran-sition temperatures are not reached.


018 10.0 pointsConsider a thermodynamic system that is si-multaneously releasing heat and doing work.The internal energy of this system will:

1. Decrease correct

2. Increase, decrease, or stay the same de-pending on the magnitudes of heat and work

3. Stay exactly the same.

4. Increase

Explanation:The change in internal energy is equal to

the sum of the heat absorbed by the systemand work done on the system based on theequation: ∆U = q + w. In this case, q andw are both negative. Therefore the internalenergy will be decreasing regardless of themagnitudes of heat and work.

019 10.0 pointsWhich of the following statements is/are true?I) For a given process, ∆H must be zero

when external pressure is zero.II) For a given process, ∆U and ∆H must

have different values.III) For a given process, ∆Usys and ∆Usurr

must have the same magnitude.

1. I, II

2. I, II, III

3. I only

4. III only correct

5. II, III

6. II only

7. I, III

Explanation:∆H is heat transferred at constant pres-

sure. For a given process ∆U and ∆H do

not always have different values; for example,they can both be zero for a given process. Be-cause ∆Uuniv must be zero according to the1st law, ∆U of a system and its surround-ings not only can but must have the samemagnitude and opposite signs.

020 10.0 pointsIf you drop a piece of potassium metal intowater you get the following exothermic reac-tion:

2K(s) + 2H2O(ℓ)→ 2KOH(aq) + H2(g)

What are the values of q and w for this reac-tion, at constant temperature and pressure?

1. Both are positive.

2. q is negative and w is positive.

3. q is positive and w is negative.

4. Both are negative. correct

Explanation:

021 10.0 pointsThe formation of chemical bonds from sepa-rated atoms

1. is never spontaneous.

2. increases entropy.

3. may be either endothermic or exother-mic.

4. is always exothermic. correct

5. is always endothermic.

Explanation:The energies to break bonds are positive

(require energy input, positive ∆H) and en-ergies to form bonds are negative (release en-ergy, negative ∆H).

022 10.0 pointsWhich of

O2(g), O2(ℓ), H2(g), H2(ℓ), H2O(g), H2O(ℓ)


have a standard enthalpy of formation equalto zero?

1. O2(g), O2(ℓ), H2(g), H2(ℓ), H2O(g),H2O(ℓ)

2. O2(g), H2(g), H2O(g)

3. O2(g), O2(ℓ), H2(g), H2(ℓ)

4. O2(g), H2(g) correct

5. All of them, but only at absolute zero

Explanation:Molecules in their native state at STP have

a standard enthalpy of formation of zero.

023 10.0 pointsWhen 1 mol of methane is burned at constantpressure, −890 kJ/mol of energy is released asheat. If a 3.64 g sample of methane is burnedat constant pressure, what will be the value of∆H? (Hint: Convert the grams of methaneto moles. Also make sure your answer has thecorrect sign for an exothermic process.)1. -61.18752. -202.4753. -176.8884. -268.6695. -257.5446. -233.6257. -132.3878. -140.7319. -264.21910. -115.144

Correct answer: −202.475 kJ.

Explanation:CH4 : ∆Hc = −890 kJ/mol m = 3.64 g∆H = ?

nCH4= (3.64 gCH4)

(

mol CH4

16 g CH4

)

= 0.2275 mol CH4

∆H = (0.2275mol)(−890 kJ/mol) = −202.475 kJ

024 10.0 pointsCalculate the quantity of energy required tochange 3.00 mol of liquid water at 100◦C tosteam at 100◦C. The molar heat of vaporiza-tion of water is 40.6 kJ/mol.

1. 300 kJ

2. 122 kJ correct

3. None of these

4. 40.6 kJ

5. 13.5 kJ

Explanation:n = 3.0 mol

40.6 kJ

mol· 3 mol = 121.8 kJ

025 10.0 pointsCalculate the heat of formation for 2.6 molof sulfur dioxide (SO2) from its elements, sul-fur and oxygen. Use the balanced chemicalequation and the following information.

S(s) +3

2O2(g) −→ SO3(g)

∆H0c = −395.2 kJ/mol

2 SO2(g) + O2(g) −→ 2 SO3(g)∆H0 = −198.2 kJ/mol

1. -414.542. -562.593. -384.934. -769.865. -592.26. -503.377. -651.428. -710.649. -621.8110. -532.98

Correct answer: −769.86 kJ.

Explanation:∆Hf of SO2 from S and O = ?The balanced equation is

S(s) + O2(g) −→ SO2(g)


Reaction ∆H0f

(kJ/mol)

S(s) +3

2O2(g)→ SO3(g) −395.2

1

2

[

2 SO3(g)→ 2 SO2(g) + O2

]

1

2(198.2) = 99.1

S(s) + O2(g)→ SO2(g) −296.1

For 2.6 mol sulfur dioxide, the energy re-quired is

(−296.1 kJ/mol)(2.6 mol) = −769.86 kJ

026 10.0 points? heat capacity is the amount of heat

required to raise the temperature of one ?of an object by 1◦C. It is an ? property.

1. Specific; gram; extensive

2. Molar; gram; intensive

3. Specific; mole; extensive

4. Molar; gram; extensive

5. Specific; gram; intensive correct

6. Molar; mole; extensive

Explanation:None

027 (part 1 of 3) 10.0 pointsConsider the following chemical and physicalchanges:

A. H2O(l) −→ H2O(g)

B. H2O(l) −→ H2O(s)

C. 1

2H2(g) + 1

2F2(g)→ HF(g)

∆H = −271.1 kJ/mol

D. C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(g)

E. N2(g) + O2(g)→ 2NO(g)∆H = 180.50 kJ/mol

Which change(s) are endothermic?

1. C and D only

2. B and E only

3. C only

4. A, D, and E only

5. A, C, and D only

6. A and E only correct

Explanation:None

028 (part 2 of 3) 10.0 pointsFor which change(s) would ∆H = ∆U ?

1. B and E only

2. C and D only

3. A and B only

4. A and D only

5. B, C, and E only correct

Explanation:None

029 (part 3 of 3) 10.0 pointsFor which change(s) would ∆Hrxn = ∆Hf ofthe product?

1. A, B, and C only

2. C only correct

3. A and C only

4. A, B, C, and E only

5. C and E only

Explanation:None

030 10.0 points


The standard enthalpy of formation of Br2(ℓ)is

1. negative.

2. zero. correct

3. positive.

Explanation:Since bromine is a liquid at 25◦C and 1 atm

pressure, it is already in its standard state, sothe standard enthalpy of formation is zero.

031 10.0 pointsConsider the combustion reaction below.

2CH3OH(g) + 3O2(g)→ 2CO2(g) + 4H2O(ℓ)

If this reaction took place in a closed,rigid container, work would be (posi-tive/negative/zero) and heat would be (posi-tive/negative/zero).

1. positive, zero

2. positive, negative

3. zero, positive

4. negative, positive

5. negative, zero

6. zero, negative correct

Explanation:Despite the fact that the reaction results in

a net decrease in moles of gas, which wouldnormally result in positive work (work doneon the system), the closed, rigid container pre-cludes a change in volume and results in zerowork. Like other combustion reactions, heatis released, and so heat would be negative.

032 10.0 pointsWhich of the reactions below is a formationreaction?

1. 2Fe(s) + 3O(g)→ Fe2O3(s)

2. B2(s) + 2 I2(ℓ) + Cl2(g)→ 2BI2Cl(g)

3. Cdiamond(s) +1

2O2(g)→ CO(g)

4. N2(g) + 2H2(g) +1

2O2(g)→ N2H4O(g)

correct

Explanation:A formation reaction produces exactly one

mole of one product from elements in theirstandard states.

033 10.0 pointsEnergy in the amount of 455 J is added to a67.0 g sample of water at a temperature of7.00◦C. What will be the final temperature ofthe water?1. 26.70392. 8.624653. 15.16164. 27.10925. 17.85016. 30.4047. 29.70168. 13.53279. 15.654510. 3.29054

Correct answer: 8.62465◦C.

Explanation:q = 455 J Ti = 7◦Cm = 67 g Tf = ?SHH2O = 4.18 J/g ·K

SHH2O =q

m∆T

∆T =q

mSHH2O

=455 J

(67 g) (4.18 J/g ·K)

= 1.62465 K

= 1.62465◦C

∆T = Tf − Ti

Tf = ∆T + Ti

= 1.62465◦C+ 7◦C

= 8.62465◦C


034 10.0 pointsA system did 150 kJ of work and its internalenergy increased by 60 kJ. How much energydid the system gain or lose as heat?

1. The system gained 60 kJ of energy asheat.

2. The system gained 90 kJ of energy asheat.

3.The system lost 210 kJ of energy as heat.

4. The system lost 90 kJ of energy as heat.

5. The system gained 210 kJ of energy asheat. correct

Explanation:

035 10.0 pointsAn important reaction that takes place in theatmosphere is

NO2(g) −→ NO(g) + O(g)

which is brought about by sunlight. Calculatethe standard enthalpy of the reaction from thefollowing information

reaction ∆H◦ (kJ)

O2(g)→ 2O(g) +498.4

NO(g) + O3(g) −→ NO2(g) + O2(g) −200.0

3

2O2(g) −→ O3(g) +142.7

1. 306.5 kJ correct

2. 820.5 kJ

3. 320.2 kJ

4. 555.7 kJ

5. 963.8 kJ

6. 106.5 kJ

7. 449.2 kJ

Explanation:Using Hess’ Law we add the reverse (flip) ofreaction 2; the reverse (flip) of reaction 3; andone half of reaction 1:

NO2(g) + O2(g) −→ NO(g) + O3(g)+200

O3(g) −→3

2O2(g) −142.7

1

2O2(g) −→ O(g) +249.2

NO2(g) −→ NO(g) + O(g)+306.5

036 10.0 pointsWhat is the value of work when an externalpressure of 2 atm compresses a piston from aninitial volume of 11.2 liters to a final volumeof 2 liters.

1. −18.4 kJ

2. 18.4 kJ

3. 1.86 kJ correct

4. −1.86 kJ

Explanation:For expansion against a constant external

pressure,

w = −Pext∆V

= (−2 atm)(2 L− 11.2 L)

× (101.325 J · L−1· atm−1)

= 1.86 kJ .

037 10.0 pointsA bomb calorimeter with a heat capacity of30 J/C contains 1000 g of water with an initialtemperature of 25◦C . A 0.5 g sample of acandy bar is placed in a bomb calorimeter andignited, resulting in a new water temperatureof 30◦C . What is ∆E for this reaction?

1. −42 kJ/g correct

2. 0 kJ/g


3. +21 kJ/g

4. −300 kJ/g

5. +300 kJ/g

6. −21 kJ/g

7. +42 kJ/g

Explanation:

038 10.0 pointsFor the combustion reaction of ethylene(C2H4)

C2H4 + 3O2 → 2CO2 + 2H2O

assume all reactants and products are gases,and calculate the ∆H0

rxn using bond energiesfrom the list below.C H : 413 kJ/mol; H O : 463 kJ/mol;O O : 146 kJ/mol; O O : 498 kJ/mol.C C : 346 kJ/mol; C C : 602 kJ/mol.C O : 358 kJ/mol; C O : 799 kJ/mol.

1. 0 kJ/mol

2. −251 kJ/mol

3. −680 kJ/mol


5. 1300 kJ/mol

6. 680 kJ/mol

7. 251 kJ/mol

Explanation:

∆H0rxn =

∑

BE reactants −

∑

BE products

=[

(C C) + 4 (C H)

+3 (O O)]

−

[

4 (C O) + 4 (H O)]

=

[(

602kJ

mol

)

+ 4

(

413kJ

mol

)

+3

(

498kJ

mol

)]

−

[

4

(

799kJ

mol

)

+ 4

(

463kJ

mol

)]

= −1300kJ

mol

039 10.0 points2.26 g of liquid water at 23.5 ◦C was com-pletely converted to ice at 0 ◦C. How muchheat was (absorbed/released) by the systemduring this process?

1. 1478 J; absorbed

2. 755 J; absorbed

3. 1478 J; released

4. 977 J; absorbed

5. 977 J; released correct

6. 755 J; released

Explanation:for 1 gram (cooling + freezing):23.5(4.184) + 334 = 432.324 J/gscale up to 2.26 g :432.324(2.26) = 977.052 J= 977 J released

040 10.0 pointsWhich of the following reactions is an en-thalpy of formation reaction?

1.1

2N2(ℓ) +

3

2H2(g)→ NH3(g)

2. NaOH(aq) + HCl(g)→H2O(ℓ) + NaCl(aq)

3. 2Fe(s) +3

2O2(g)→ Fe2O3(s) correct

4. CH4(g)→ Cgraphite + 2H2(g)

Explanation:Formation reactions describe production of

exactly one mole of one product from stoichio-metric quantities of elements in their standardstates.


041 10.0 pointsConsider a system where 2.50 L of ideal gasexpands to 6.25 L against a constant externalpressure of 330 torr. Calculate the work (w)for this system.

1. −1238 J

2. +1238 J

3. −1.63 J

4. +165 J

5. +1.63 J

6. −165 J correct

Explanation:Convert torr to atm, and then convert an-

swer in L·atm to joules. The answer will benegative due to expansion of the gas.w = −P∆V = −(330/760)(3.75L)w = −1.628 L atm×101.325 J/(L atm) = −165 J


for the reaction

C2H5OH(ℓ) + 3O2(g)→ 2CO2(g) + 3H2O(ℓ)

givenC2H5OH(ℓ)→ 2Cgraphite(s) + 3H2(g) + 1/2O2(g)

∆H◦ = 228 kJ ·mol−1

CO2(g)→ Cgraphite(s) + O2(g)

∆H◦ = 394 kJ ·mol−1

H2(g) + 1/2O2(g)→ H2O(ℓ)∆H◦ = −286 kJ ·mol−1

1. 730 kJ ·mol−1

2. −846 kJ ·mol−1

3. −1, 418 kJ ·mol−1 correct

4. 336 kJ ·mol−1

5. −452 kJ ·mol−1

Explanation:We need to reverse and double the second

reaction as well as triple the third reactionand then add all three reactions.

043 10.0 pointsCalculate the standard reaction enthalpy(∆H◦

rxn) for the final stage in the productionof nitric acid, when nitrogen dioxide dissolvesin and reacts with water:

3NO2(g) + H2O(ℓ)→ 2HNO3(aq) + NO(g)

1. −370 kJ

2. +70 kJ

3. −104 kJ

4. +136 kJ

5. −304 kJ


Explanation:Values for ∆H◦

f from external table are inorder (from reaction) +33, -286, -207, and+90

∆H◦

rxn =(

∑

n∆H◦

j

)

products

−

(

∑

n∆H◦

j

)

reactants

=[

2∆H◦

f, HNO3(aq)+∆H◦

f, O(g)

]

−

[

3∆H◦

f, NO2(g)+∆H◦

f, H2O(ℓ)

]

=[

2 (−207) + 90]

−

[

3 (33) + (−286)]

= −137 kJ

044 10.0 pointsThe molar heat capacity of C6H6(ℓ) is 136J/mol ·◦C and of C6H6(g) is 81.6 J/mol ·◦C.The molar heat of fusion for benzene is 9.92kJ/mol and its molar heat of vaporization is


30.8 kJ/mol. The melting point of benzeneis 5.5◦C, its boiling point is 80.1◦C, and itsmolecular weight 78.0 g/mol. How much heatwould be required to convert 234 g of solidbenzene (C6H6(s)) at 5.5

◦C into benzene va-por (C6H6(g)) at 100.0

◦C?

1. 97.2715 kJ

2. 157.468 kJ correct

3. 4931.72 kJ

4. 60.1968 kJ

5. 152.597 kJ

Explanation:mbenzene = 234 g T1 = 5.5◦CT2 = 100.0◦C

234 g ×mol

78.0 g= 3mol

C6H6(s)5.5◦C

step 1−→ C6H6(ℓ)

5.5◦C

step 2−→

C6H6(ℓ)80.1◦C

step 3−→ C6H6(g)

80.1◦C

step 4−→ C6H6(g)

100.0◦C

Step 1 :9.92 kJ

mol× 3mol = 29.76 kJ

Step 2 :136 J

mol ·◦ C× (3 mol)×(80.1− 5.5)◦C

= 30436.8 J = 30.4368kJ

Step 3 :30.8 kJ

mol× (3mol) = 92.4 kJ

Step 4 :81.6 J

mol ·◦ C× (3 mol)

×(100.0− 80.1)◦C

= 4871.52 J = 4.87152kJ

Total = 29.76 kJ + 30.4368 kJ

+92.4 kJ + 4.87152 kJ

= 157.468 kJ

045 10.0 pointsCalculate the standard reaction enthalpy forthe reaction

NO2(g)→ NO(g) + O(g)

given +142.7 kJ/mol for the standard en-thalpy of formation of ozone andO2(g)→ 2O(g) ∆H◦ = +498.4 kJ/molNO(g) + O3(g)→ NO2(g) + O2(g)

∆H◦ = −200 kJ/molRemember the definition of the standard en-thalpy of formation of a substance.

1. +306 kJ/mol correct

2. +355 kJ/mol

3. +192 kJ/mol

4. +592 kJ/mol

5. +555 kJ/mol

Explanation:

O2(g)→ 2O(g)

∆H◦ = +498.4 kJ/mol

NO(g) + O3(g)→ NO2(g) + O2(g)

∆H◦ = −200 kJ/mol

The standard formation of ozone is

3

2O2(g)→ O3(g)

∆H = +142.7 kJ/mol

We calculate the ∆Hrxn using Hess’ Law:To combine the reactions and get the de-

sired reaction, reverse the second and thirdequations and add half of the first one:

NO2(g) + O2(g)→ NO(g) + O3(g)

∆H = +200 kJ/mol

O3(g)→3

2O2(g)

∆H = −142.7 kJ/mol1

2O2(g)→ O(g)

∆H =1

2(498.4 kJ/mol)

NO2(g)→ NO(g) + O(g)

∆Hrxn = 306.5 kJ/mol


046 10.0 pointsA coffee cup calorimeter measures the heatat constant ? whereas a bomb calorimetermeasures the heat at constant ?

1. pressure (qp = ∆H); volume (qv = ∆U)correct

2. pressure (qp = ∆U); volume (qv = ∆H)

3. volume (qv = ∆H); pressure (qp = ∆U)

4. volume (qv = ∆U); pressure (qp = ∆H)

Explanation:None

047 10.0 pointsYou have two liquids of identical mass, andboth with initial temperatures of 15◦C. Oneis ethanol, C2H5OH, with a specific heat of2.46 J/g◦C and the other is benzene, C6H6,with a specific heat of 1.74 J/g◦C. If bothliquids absorb the same amount of heat, whichone will have the highest final temperature?Assume that neither liquid reaches its boilingpoint.

1. Cannot tell without more informationgiven.

2. ethanol

3. Both liquids will have the same final tem-perature.

4. benzene correct

Explanation:Temperature rise (∆T ) is inversely propor-

tional to the heat capacity.

∆T =q

mCs

Therefore, because benzene has a smallerheat capacity, Cs, it will have the larger tem-perature rise.

048 10.0 points

1-bromo-isobutane will undergo and elimina-tion reaction to yield isobutene and hydrogenbromide as shown in the reaction below. Usebond energies (provided in preamble) to esti-mate the change in enthalpy, ∆H, for this gasphase reaction.

H3C

H3CCH CH2 Br

H3C

H3CC CH2 + HBr

1. −270 kJ/mol

2. +270 kJ/mol

3. −76 kJ/mol

4. +337 kJ/mol

5. +76 kJ/mol correct

6. −337 kJ/mol

Explanation:In this reaction, you break three bonds in thereactants: a C-C, a C-H, and a C-Br bond.You form new bonds in the products: a H-Brbond and a C=C bond.The change in enthalpy is the energy in forbreaking the bonds combined with the energyout gained from forming the new bonds.

∆Hrxn = 346 + 413 + 285− 366− 602

∆Hrxn = +76 kJmol−1

049 10.0 pointsWhich is true, considering the first law ofthermodynamics?

1. ∆U = q − w, where heat and work canboth be positive for the same process

2. ∆U = q + w, where heat and work cannever both be positive for the same process

3. ∆U = q − w, where heat and work cannever both be positive for the same process

4. ∆U = q + w, where heat and work canboth be positive for the same process. cor-


rect

Explanation:none

050 10.0 pointsFor which of the following chemical equationswould ∆H◦

rxn = ∆H◦

f ?

1. O2(g) + H2(g)→ H2O2(ℓ) correct

2. C(s, graphite) + 32O2(g) + H2(g) →

CO2(g) + H2O(g)

3. CO(g) + 12O2(g)→ CO2(g)

4. N2(ℓ) + 3 F2(g)→ 2 NF3(ℓ)

Explanation:For O2(g) + H2(g) → H2O2(ℓ), ∆H◦

f ofO2(g) and H2(g) are 0. Therefore, ∆H◦

rxn =∆H◦

f (H2O2(ℓ))

051 10.0 pointsThe combustion of methane gas (CH4) formsCO2(g)+ H2O(ℓ). Calculate the heat pro-duced by burning 1.98 mol of the methanegas. Use these ∆H0

f data to help:CH4(g)= -74.9 kJ/molCO2(g)= -393.5kJ/molH2O(ℓ)= -285.8kJ/mol.1. 1566.752. 1513.343. 1290.794. 1459.935. 1726.996. 1175.067. 1424.328. 1201.779. 1121.6510. 1762.6

Correct answer: 1762.6 kJ.

Explanation:nCH4

= 1.98 mol

CH4(g) + 2O2 −→ CO2(g) + H2O(ℓ)

∆H0 for combustion of CH4 = ?

Reaction ∆H0f

(kJ/mol)

CH4(g)→ C(s) + 2H2(g) 74.9

C(s) + O2(g)→ CO2(g) −393.5

2H2(g) + O2 → 2H2O(ℓ)2(−285.8) = −571.6

CH4(g) + 2O2 → CO2(g) + H2(ℓ) −890.2

∆H0 = (890.2 kJ/mol) (1.98mol) = 1762.6 kJ

052 10.0 pointsCalculate the standard reaction enthalpy forthe reaction.

CH4(g) + H2O(g)→ CO(g) + 3H2(g)

given2H2(g) + CO(g)→ CH3OH(ℓ)

∆H◦ = −128.3 kJ ·mol−1

2CH4(g) + O2(g)→ 2CH3OH(ℓ)∆H◦ = −328.1 kJ ·mol−1

2H2(g) + O2(g)→ 2H2O(g)∆H◦ = −483.6 kJ ·mol−1

1. +155.5 kJ ·mol−1

2. +206.1 kJ ·mol−1 correct

3. +216 kJ ·mol−1

4. +412.1 kJ ·mol−1

5. +42.0 kJ ·mol−1

Explanation:We need to reverse the first reaction, halve

the second, halve and reverse the third andadd the results:

CH3OH(ℓ) → 2H2(g) + CO(g)

∆H = 128.3 kJ/molCH4(g) + 0.5O2(g) → CH3OH(ℓ)

∆H = −164.05 kJ/molH2O(g)→ H2(g) + 0.5O2(g)

∆H = +241.8 kJ/mol

CH4(g) + H2O(g)→ CO(g) + 3H2(g)

∆H = 206.05 kJ/mol


053 10.0 pointsA system absorbs 237 J of heat while it per-forms 435 J of work. What is the change inthe internal energy of the system?

1. 672 J

2. 198 J

3. −198 J correct

4. −672 J

Explanation:

∆U = q + w = 237 J +− 435 J = −198 J

054 10.0 pointsCalculate the enthalpy change that occurswhen 1.00 kg of acetone condenses at its boil-ing point (329.4 K). The standard enthalpy ofvaporization of acetone is 29.1 kJ ·mol−1.

1. −29.1 kJ

2. +29.1 kJ

3. +501 kJ


5. −2.91× 104 kJ

Explanation:Acetone is CH3COCH3

MM = 58.0798 g/mol

n =(1000 g)(1 mol acetone)

58.0798 g/mol= 17.2177 mol

Condensation is the opposite of vaporization:

∆H = q = n∆Hcond

= n(−∆Hvap)

= (17.2177 mol)(−29.1 kJ/mol)

= −501.035 kJ

055 10.0 pointsFor which of the following reactions at roomtemperature (25◦C) would there be 5.0 kJ ofwork done on the system?

1. N2H2(g) + CH3OH(g)→CH2O(g) + N2(g) + 2H2(g)

2. CH2O(g) + N2(g) + 2H2(g)→N2H2(g) + CH3OH(g) correct

3. 2H2O(ℓ) + O2(g)→ 2H2O2(ℓ)

4. CH4(g) + 2O2(g)→ CO2(g) + 2H2O(g)

5. 2H2O2(ℓ)→ 2H2O(ℓ) + O2(g)

6. CO2(g) + 2H2O(g)→ CH4(g) + 2O2(g)

Explanation:At room temperature (298 K), the

product of the gas constant (R =8.314 J ·mol−1

·K−1) and T is very closeto 2.5 kJ ·mol−1. Based on 5.0 kJ =−∆ngas

(

2.5 kJ ·mol−1)

, the reaction forwhich ∆ngas = −2 will be the correct an-swer.

056 10.0 pointsThe value of ∆H for the reaction

C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(ℓ)

is −2220 kJ/mol rxn. How much heat isgiven off when 33.0 g of propane gas (C3H8)is burned at constant pressure?

1. 1665 kJ correct

2. 22420 kJ

3. 2220 kJ

4. 25.96 kJ

5. 555 kJ

6. 50.5 kJ


7. 6660 kJ

Explanation:∆H = −2220 kJ/mol mC3H8

= 33.0 g

q =(

−2220 kJ

mol rxn

)( 1 mol rxn

1 mol C3H8

)

×

(1 mol C3H8

44 g C3H8

)

(33 g C3H8)

= −1665 kJ or 1665 kJ released

057 10.0 pointsThe two reactions shown below are both en-dothermic. For which reaction is ∆H < ∆U?N2(g) + O2(g)→ 2NO(g)2NO(g) + O2(g)→ 2NO2(g)

1. 2NO(g) + O2(g)→ 2NO2(g) correct

2. Neither reaction has ∆H < ∆U .

3. Both reactions have ∆H < ∆U .

4. N2(g) + O2(g)→ 2NO(g)

Explanation:.

058 10.0 pointsConsider the following reaction

H2(g) + CO2(g)→ H2O(g) + CO(g) .

∆Hf for CO2(g) is −22.5 kJ/mol;∆Hf for CO(g) is −6.3 kJ/mol;∆Hf for H2O(g) is −13.8 kJ/mol.

1. ∆H of the reaction is negative.

2. ∆H of the reaction is zero.

3. ∆H of the reaction is positive. correct

Explanation:Reactants:

∆Hf CO2(g) = −22.5 kJ/molProducts:

∆Hf CO(g) = −6.3 kJ/mol∆Hf H2O(g) = −13.8 kJ/mol

∆H0rxn =

∑

nH0f prod −

∑

nH0f rct

=

(

−13.8kJ

mol− 6.3

kJ

mol

)

−

(

−22.5kJ

mol

)

= 2.4kJ

mol

∆H is positive.

059 10.0 pointsConsider the following specific heats: copper,0.384 J/g·◦C; lead, 0.159 J/g·◦C; water, 4.18J/g·◦C; glass, 0.502 J/g·◦C. Which substance,once warmed, would be more likely to main-tain its heat and keep you warm through along football game on a cold night?

1. water correct

2. glass

3. copper

4. lead

Explanation:Water has the highest specific heat of the

substances listed, so it has the capacity toemit the largest quantity of heat with minimaltemperature loss; the emitted heat keeps youwarm. The substance continues to warm youuntil its temperature is at or below your bodytemperature.

060 10.0 pointsA block of aluminum at 25 ◦C and 1 atm isheated until it is a liquid at 700 ◦C. It is thencooled back down until it is back in the initialstate of being a solid at 25 ◦C and 1 atm. Forthis entire process (heating and cooling) ∆His...

1. positive

2. less than ∆U

3. zero correct


4. greater than ∆U

5. negative

Explanation:Since the initial and final states of the sys-

tem are identical, and since the value of astate function such as ∆H is dependent onlyon the state of the system, we can concludethat ∆H is zero.

061 10.0 pointsWhich statement about internal energy istrue?

1. The internal energy of a system is equalto w at constant volume.

2. The internal energy of a system is con-stant at constant volume.

3. The internal energy of a system is equalto w at constant pressure.

4. The internal energy of a system is equalto q at constant volume. correct

5. The internal energy of a system is equalto q at constant pressure.

6. The internal energy of a system is con-stant at constant pressure.

Explanation:

062 10.0 pointsWhen 0.100 g of graphite is burned com-pletely in a bomb calorimeter (heat capacity= 3.344 kJ/◦C), containing 3000 g of water, atemperature rise of 0.21◦C is observed. Whatis ∆E for the combustion of graphite? Thespecific heat of liquid water is 4.184 J/g·◦C.

1. ∆E = +3.34 kJ/mol

2. ∆E = −40.1 kJ/mol

3. ∆E = −285. kJ/mol

4. ∆E = −3.34 kJ/mol

5. ∆E = −401.0 kJ/mol correct

Explanation:mgraphite = 0.100 g mwater = 3000 g∆T = 0.21◦C SHwater = 4.184 J/g·◦CHC = 3.344 kJ/◦C

The amount of heat responsible for the tem-perature increase for 3000 g of water is

q =(4.184 J

g·◦C

)

(3000 g)(0.21◦C)( 1 kJ

1000 J

)

= 2.6359 kJ

The amount of heat responsible for thewarming of the calorimeter is

q =(

3.344kJ◦C

)

(0.21◦C) = 0.7022 kJ

The amount of heat released for this reactionis

2.6359 kJ + 0.7022 kJ = 3.3381 kJ

The reaction was exothermic and there were0.1 g of graphite, so

(

−3.3381 kJ

0.1 g

)(

12g

mol

)

= −401kJ

mol

063 10.0 pointsWhen a given reaction was run at a constantpressure of 1 atm, the system absorbed 5 kJof heat and the gases were consumed, causingthe volume to decrease from 3.5 L to 1.5 L.What are ∆H and ∆U , respectively?

1. +5 kJ, +0.2 kJ

2. -5 kJ, -4.8 kJ

3. +5 kJ, +5.2 kJ correct

4. +5.2 kJ, +5 kJ

5. +5 kJ, +4.8 kJ

6. -4.8 kJ, +0.2 kJ

7. -5 kJ, -5.2 kJ


8. +5 kJ, +5 kJ

9. -5 kJ, -5 kJ

Explanation:none

064 10.0 pointsJuan freezes a bottle of water to ice (500.mL)in preparation for a road trip. How much heatcan be absorbed by that ice before it is fullymelted?

1. 2090 kJ

2. 167 kJ correct

3. 0 kJ

4. 1130 kJ

5. 0.500 kJ

6. 1.50 kJ

7. 6.02 kJ

Explanation:Density of water is 1 g/mL. So 1 g = 1 mL

for water. q=(500g)x(334 J/g)=167000 J =167 kJ.

065 10.0 pointsHow much heat is absorbed in the completereaction of 3.00 grams of SiO2 with excesscarbon in the reaction below?

SiO2(g) + 3C(s)→ SiC(s) + 2CO(g)

∆H for the reaction is +624.7 kJ/mol rxn.

1. 31.2 kJ correct

2. 1.33× 104 kJ

3. 5.06 kJ

4. 366 kJ

5. 1.13× 105 kJ

Explanation:mSiO2

= 3.00 g ∆H = +624.7 kJ/mol

q =(624.7 kJ

mol rxn

)( 1 mol rxn

1 mol SiO2

)

×

(1 mol SiO2

60 g SiO2

)

(3 g SiO2)

= 31.2 kJ

066 10.0 pointsUsing bond energies, estimate the enthalpychange for the reaction between hydrogen per-oxide (H2O2) and carbon disulfide (CS2) toproduce carbon dioxide (CO2) and hydrogendisulfide (H2S2) according to the balancedequation:H2O2 +CS2 → CO2 +H2S2

1. −577 kJ/mol

2. −106 kJ/mol

3. 292 kJ/mol

4. 106 kJ/mol


6. 577 kJ/mol

Explanation:

∆Hrxn = ΣBEbroken − ΣBEformed

Total bonds broken are 2 hydrogen oxygensingle bonds (463 kJ/mol), 1 oxygen oxygensingle bond (146 kJ/mol) and 2 carbon sul-fur double bonds (577 kJ/mol). Total bondsformed are 2 hydrogen sulfur single bonds(347 kJ/mol), 1 sulfur sulfur single bond (226kJ/mol) and 2 carbon oxygen double bonds(799 kJ/mol).∆Hrxn = −292 kJ/mol

067 10.0 pointsThe following reaction occurs during the pro-duction of metallic iron:


2 Fe2O3(s) + 3 C(graphite)→4 Fe(s) + 3 CO2(g)

Calculate ∆H for this reaction at 25◦C and1 atm.∆Hf for CO2(g) = −393.51 kJ/mol, and∆Hf for Fe2O3(s) = −824.2 kJ/mol.

1.There is insufficient information to answerthis question.

2. +467.9 kJ correct

3. −430.7 kJ

4. +430.7 kJ

5. −467.9 kJ

Explanation:

∆Hrxn = ∆Hproducts −∆Hreactants

= (−393.51)(3)− (−824.2)(2)

= 467.9 kJ

Note: The enthalpy of a pure element is al-ways zero.

068 10.0 points

Based on thermodynamic table data calcu-late ∆Hrxn for

2H2O(ℓ) + 2 SO2(g)←→ 2H2S(g) + 3O2(g)

1. 560 kJ ·mol−1

2. −560 kJ ·mol−1

3. 1120 kJ ·mol−1 correct

4. −1120 kJ ·mol−1

Explanation:

∆Hrxn =∑

∆Hf, products −

∑

∆Hf, reactants

=[

2(

−21 kJ ·mol−1)

+ 3(0)]

−

[

2(

−286 kJ ·mol−1)

+ 2(

−297 kJ ·mol−1)]

= 1, 120 kJ ·mol−1


for the reaction of calcite with hydrochloricacidCaCO3(s) + 2HCl(aq) −→

CaCl2(aq) + H2O(ℓ) + CO2(g)The standard enthalpies of formation are:for CaCl2(aq) : −877.1 kJ/mol;for H2O(ℓ) : −285.83 kJ/mol;for CO2(g) : −393.51 kJ/mol;for CaCO3(s) : −1206.9 kJ/mol;and for HCl(aq) : −167.16 kJ/mol.

1. −38.2 kJ/mol

2. −98.8 kJ/mol

3. −116 kJ/mol

4. −15.2 kJ/mol correct

5. −72.7 kJ/mol

6. −165 kJ/mol

7. −215 kJ/mol

Explanation:We use Hess’ Law:

∆H◦ =∑

n∆H◦

j,prod −

∑

n∆H◦

j,reac

= ∆H◦

f, CaCl2(aq)+∆H◦

f, H2O(ℓ)

+∆H◦

f, CO2(g)−

[

∆H◦

f, CaCO3(s)

+2(

∆H◦

f, HCl(aq)

)]

= −877.1 kJ/mol + (−285.83 kJ/mol)

+ (−393.51 kJ/mol)

−

[

−1206.9 kJ/mol

+ 2 (−167.16 kJ/mol)]

= −15.22 kJ/mol .


070 10.0 pointsFor an exothermic reaction, the sum of bondenergies for the reactants are (greater/lesser)than those of the products.

1. lesser correct

2. greater

Explanation:For an exothermic reaction, the change in

enthalpy is negative.


for the reaction

H2S(g) + 2O2(g)→ SO3(g) + H2O(ℓ)

givenH2O(g)→ H2O(ℓ)

∆H◦ = −11.0 kJ ·mol−1

H2SO4(ℓ)→ H2S(g) + 2O2(g)∆H◦ = +78.5 kJ ·mol−1

H2SO4(ℓ)→ SO3(g) + H2O(g)∆H◦ = +20.5 kJ ·mol−1

1. +88.0 kJ

2. −69.0 kJ correct

3. +110.0 kJ

4. −47.0 kJ

Explanation:The overall reaction is sum of reaction 1,

reaction 3 and the reverse of reaction 2.

072 10.0 pointsWhen 17.8 g sodium is treated with excessoxygen, 160.2 kJ of heat is produced. What isthe ∆Hrxn for the below reaction?

4Na(s) + O2(g) −→ 2Na2O(s)

1. -1682 kJ/mol

2. -15.2 kJ/mol

3. -152 kJ/mol

4. -828 kJ/mol correct

5. -168.2 kJ/mol

Explanation:None

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