Test Bank Answers 8th Edition Chapter.18 Thermodynamics

Chemistry: Molecular Nature of Matter, 8e (Jespersen)

Chapter 18 Thermodynamics

1) The mathematical equation that expresses the first law of thermodynamics is

A) ΔH = ΔE + pΔV.

B) ΔH = ΔE - pΔV.

C) ΔH = q + w.

D) ΔE = q + w.

E) ΔH = q + ΔE.

Diff: 1

Section: 18.1

2) The standard enthalpy of reaction, ΔH°rxn for

C2H2(g) + 2 H2(g) → C2H6(g)

is −311.5 kJ mol−1. Determine the value of ΔE°rxn for this reaction.

A) −306.5 kJ mol−1

B) −309.0 kJ mol−1

C) −314.0 kJ mol−1

D) −316.46 kJ mol−1

E) +4646 kJ mol−1

Diff: 1

Section: 18.1

3) The standard enthalpy of reaction, ΔH° at 25 C°

NH3(g) + HCl(g) → NH4Cl(s)

is −175.9 kJ mol−1. Determine the value of ΔE°rxn for this reaction.

A) −164.8 kJ mol−1

B) −170.9 kJ mol−1

C) −173.4 kJ mol−1

D) −180.9 kJ mol−1

E) +5134 kJ mol−1

Diff: 2

Section: 18.1

4) The standard enthalpy of reaction, ΔH°rxn, for the reaction,

C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(l)

is −2877 kJ mol−1. Determine the value of ΔE°rxn for this reaction.

A) -2868 kJ mol−1

B) -2871 kJ mol−1

C) -2880 kJ mol−1

D) -2886 kJ mol−1

E) +2886 kJ mol−1

Diff: 2

Section: 18.1

5) The standard enthalpy of reaction, ΔH°rxn, for the reaction,

CaO(s) + SO3(g) → CaSO4(s)

is -401.5 kJ mol−1. Determine the value of ΔE°rxn for this reaction.

A) -362.2 kJ mol−1

B) -399.0 kJ mol−1

C) -404.0 kJ mol−1

D) -2880 kJ mol−1

E) +2077 kJ mol−1

Diff: 2

Section: 18.1

6) Which statement is true?

A) Spontaneous changes are always accompanied by an increase in the entropy of the system.

B) Spontaneous changes are always accompanied by a decrease in the entropy of the system.

C) Spontaneous changes are always accompanied by an increase in the enthalpy of the system.

D) Spontaneous changes are always accompanied by a decrease in the enthalpy of the system.

E) Most highly exothermic chemical reactions are also spontaneous chemical reactions.

Diff: 2

Section: 18.2

7) Which process is accompanied by an increase in entropy?

A) setting up a stack of dominos

B) setting up decorations on a Christmas tree

C) filing correspondence in file folders and placing them in hanging file folders

D) dropping a glass pane on the front walk of your residence

E) restocking a canned goods shelf display in a supermarket

Diff: 1

Section: 18.3

8) Which reaction is accompanied by an increase in entropy?

A) ZnS(s) + 3/2 O2(g) → ZnO(s) + SO2(g)

B) CH4(g) + H2O(g) → CO(g) + 3H2(g)

C) BaO(s) + CO2(g) → BaCO3(s)

D) Na2CO3(s) + CO2(g) + H2O(g) → 2 NaHCO3(s)

E) N2(g) + 3 H2(g) → 2 NH3(g)

Diff: 1

Section: 18.3

9) Which reaction is accompanied by an increase in entropy?

A) 2 H(g) → H2(g)

B) NiCl2(s) + 6 NH3(g) → NiCl2∙6NH3(s)

C) I2(g) → 2 I(g)

D) ZnO(s) + CO2(g) → ZnCO3(s)

E) C2H4(g) + Cl2(g) → C2H4Cl2(l)

Diff: 1

Section: 18.3

10) Which reaction is accompanied by an increase in entropy?

A) C12H20(l) + 17 O2(g) → 12 CO2(g) + 10 H2O(l)

B) NH4Cl(s) → NH3(g) + HCl(g)

C) 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(s)

D) Ba(OH)2(s) + 2 HCl(g) → BaCl2∙2H2O(s)

E) (CH3)2CO(l) + 4 O2(g) → 3 CO2(g) + 3 H2O(l)

Diff: 1

Section: 18.3

11) Which reaction is accompanied by an increase in entropy?

A) C8H16(l) + 12 O2(g) → 8 CO2(g) + 8 H2O(l)

B) N2(g) + 3 H2(g) → 2 NH3(g)

C) 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(s)

D) Ba(OH)2(s) + CO2(g) → BaCO3(s) + H2O(l)

E) NH4NO2(s) → N2(g) + 2 H2O(l)

Diff: 1

Section: 18.3

12) Which process is accompanied by a decrease in the entropy of the system?

A) the mixing of one liter of water with one liter of ethylene glycol to produce one liter of solution

B) the breaking of a large rock into very many smaller pieces of crushed gravel

C) the thawing of the frozen orange juice concentrate that was left in the car

D) the spontaneous chemical reaction of TNT (a solid chemical compound) wherein it decomposes into several simple compounds, some of which are gaseous

E) the absorption of odorous gaseous compounds by the charcoal filter in your home central air cleaning unit

Diff: 1

Section: 18.3

13) Which process is accompanied by an increase in the entropy of the system?

A) solid gold melting

B) the condensation of water on cold surface

C) the freezing of a popsicle

D) sewing a quilt

E) a cup of coffee cooling in a mug

Diff: 1

Section: 18.3

14) Which process is accompanied by a decrease in the entropy of the system?

A) a cup of juice spreading across the floor after being spilled

B) soaking up a chemical spill with an absorbent material

C) the sublimation of iodine crystals

D) water boiling in a steam kettle

E) butter melting for a cake

Diff: 1

Section: 18.3

15) Which of these species has the highest entropy (S°) at 25°C?

A) H2O (s)

B) H2O (l)

C) H2O (g)

Diff: 1

Section: 18.3

16) Which of these species has the highest entropy (S°) at 25°C?

A) CO2 (s)

B) CO2 (l)

C) CO2 (g)

Diff: 1

Section: 18.3

17) Which statement below is always true for a spontaneous chemical reaction?

A) ΔSsys + ΔSsurr = 0

B) ΔSsys + ΔSsurr < 0

C) ΔSsys + ΔSsurr > 0

D) ΔSsys − ΔSsurr = 0

E) ΔSsys − ΔSsurr < 0

Diff: 1

Section: 18.3

18) Of the species listed, which should possess the highest standard entropy (S°) for one mole of substance?

A) (CH3)2CO(l)

B) C4H10(g)

C) K2SO4(s)

D) H2O(l)

E) Br2(l)

Diff: 1

Section: 18.3

19) Which species has the highest standard entropy (S°) for one mole of substance?

A) Au(s)

B) Cd(s)

C) Hg(l)

D) Ni(s)

E) K(s)

Diff: 1

Section: 18.3

20) Which species has the greatest standard entropy (S°) for one mole of substance?

A) I2(s)

B) Br2(l)

C) N2(l)

D) Cl2(g)

E) He(l)

Diff: 1

Section: 18.3

21) Which species should possess the lowest standard entropy (S°)?

A) CH4(g)

B) CO2(s)

C) NH3(l)

D) H2O(l)

E) Ar(g)

Diff: 1

Section: 18.3

22) Which set has the species listed in order of increasing standard entropy, S°?

A) Au(s) < CaCO3(s) < H2O(l)

B) CaCO3(s) < H2O(l) < Au(s)

C) Au(s) < H2O(l) < CaCO3(s)

D) CaCO3(s) < Au(s) < H2O(l)

Diff: 2

Section: 18.3

23) Which set below has the species listed in order of increasing standard entropy, S°?

A) NaHCO3(aq) < C2H5OH(l) < Cr(s) < N2(g)

B) Cr(s) < N2(g) < NaHCO3(aq) < C2H5OH(l)

C) Cr(s) < C2H5OH(l) < NaHCO3(aq) < N2(g)

D) Cr(s) < NaHCO3(aq) < C2H5OH(l) < N2(g)

E) N2(g) < NaHCO3(aq) < Cr(s) < C2H5OH(l)

Diff: 2

Section: 18.3

24) Which set below has the species listed in order of increasing standard entropy, S°?

A) CaSO4(s) < C2H5OH(l) < Ar(g)

B) CH3CH2—O—H(l) < Ar(g) < CaSO4(s)

C) CaSO4(s) < Ar(g) < C2H5OH(l)

D) C2H5OH(l) < CaSO4(s) < Ar(g)

Diff: 2

Section: 18.3

25) For a certain chemical reaction, ΔH is < 0 and Δ S is < 0. This means that

A) we conclude the reaction must be spontaneous regardless of temperature and becomes even more so at higher temperatures.

B) we conclude the reaction must be spontaneous regardless of temperature and becomes even more so at lower temperatures.

C) we conclude the reaction may or may not be spontaneous, but spontaneity is favored by low temperatures.

D) we conclude the reaction may or may not be spontaneous, but spontaneity is favored by high temperatures.

E) we cannot make any conclusion about spontaneity or even tendencies from the limited information presented.

Diff: 2

Section: 18.4

26) For a certain chemical reaction, ΔH is > 0 and ΔS is < 0. This means that

A) we conclude the reaction must be nonspontaneous regardless of temperature.

B) we conclude the reaction must be spontaneous regardless of temperature and becomes even more so at lower temperatures.

C) we conclude the reaction may or may not be spontaneous, but spontaneity is favored by low temperatures.

D) we conclude the reaction may or may not be spontaneous, but spontaneity is favored by high temperatures.

E) we cannot make any conclusion about spontaneity or even tendencies from the limited information presented.

Diff: 2

Section: 18.4

27) The requirement for a spontaneous chemical reaction is

A) ΔG = 0.

B) ΔH > 0.

C) ΔS = 0.

D) ΔE > 0.

E) ΔG < 0.

Diff: 1

Section: 18.4

28) The normal melting point of benzoic acid is 122.4°C. Predict the signs of ΔH, ΔS, and ΔG for the process in which liquid benzoic acid freezes at 120°C and 1 atm: C7H6O2(l) → C7H6O2(s)

A) A

B) B

C) C

D) D

E) E

Diff: 2

Section: 18.4

29) The normal melting point of carbon dioxide is -78°C. Predict the signs of ΔH, ΔS, and ΔG for the process in which solid carbon dioxide sublimes at -50°C and 1 atm: CO2(s) → CO2(g)

A) A

B) B

C) C

D) D

E) E

Diff: 2

Section: 18.4

30) The normal melting point of naphthalene is 80.3°C. Predict the signs of ΔH, ΔS, and ΔG for the process in which solid naphthalene melts at 82.5°C and 1 atm: C10H8(s) C10H8(l)

A) A

B) B

C) C

D) D

E) E

Diff: 2

Section: 18.4

31) Which property associated with a chemical reaction is dependent on how the reaction is carried out, not on just its initial and final states?

A) ΔS

B) ΔH

C) ΔE

D) w

E) ΔG

Diff: 1

Section: 18.4

32) A negative sign for ΔG indicates that, at constant temperature and pressure,

A) the reaction is spontaneous.

B) ΔS must be greater than zero.

C) the reaction must be exothermic.

D) the reaction must be fast.

E) the reaction must be endothermic.

Diff: 1

Section: 18.4

33) For the reaction 2NO(g) + O2(g) → 2NO2(g), ΔH° = −113.1 kJ/mol and ΔS° = -145.3 J/K mol.

Which of these statements is true?

A) The reaction is spontaneous at all temperatures.

B) The reaction is only spontaneous at low temperatures.

C) The reaction is only spontaneous at high temperatures.

D) The reaction is at equilibrium at 25°C under standard conditions.

E) ΔG° becomes more favorable as temperature increases.

Diff: 1

Section: 18.4

34) According to the second law of thermodynamics, a spontaneous reaction must result in

A) no change in free energy for the reaction.

B) the enthalpy of the universe increasing.

C) the entropy of the universe decreasing.

D) the temperature of the surrounding increasing.

E) the entropy of the universe increasing.

Diff: 1

Section: 18.4

35) Using the standard entropy values:

H2(g), S° = +130.6 J mol−1 K−1

I2(s), S° = +116.12 J mol−1 K−1

HI(g), S° = +206.0 J mol−1 K−1

calculate the standard entropy change, S°, for the reaction:

H2(g) + I2(s) → 2 HI(g)

A) −40.8 J mol−1 K−1

B) +40.8 J mol−1 K−1

C) −165.3 J mol−1 K−1

D) +165.3 J mol−1 K−1

E) +206.0 J mol−1 K−1

Diff: 2

Section: 18.5

36) Using the standard entropy values:

NO(g), S° = +210.6 J mol−1 K−1

O2(g), S° = +205.0 J mol−1 K−1

NO2(g), S° = +240.5 J mol−1 K−1

calculate the standard entropy change, S°, for the reaction:

NO(g) + ½ O2(g) → NO2(g)

A) −72.6 J mol−1 K−1

B) −175.1 J mol−1 K−1

C) +72.6 J mol−1 K−1

D) +175.1 J mol−1 K−1

E) −656.1 J mol−1 K−1

Diff: 2

Section: 18.5

37) Using the standard entropy values:

C2H4(g), S° = +219.8 J mol−1 K−1

H2(g), S° = +130.6 J mol−1 K−1

C2H6(g), S° = +229.5 J mol−1 K−1

calculate the standard entropy change, S°, for the reaction:

C2H4(g) + H2(g) → C2H6(g)

A) +120.9 J mol−1 K−1

B) −98.9 J mol−1 K−1

C) −120.9 J mol−1 K−1

D) +140.3 J mol−1 K−1

E) +579.9 J mol−1 K−1

Diff: 2

Section: 18.5

38) Using the standard entropy values:

SO2(g), S° = +248.0 J mol−1 K−1

SO3(g), S° = +256.0 J mol−1 K−1

NO(g), S° = +210.6 J mol−1 K−1

NO2(g), S° = +240.5 J mol−1 K−1

Calculate the standard entropy change, ΔS°, for the reaction:

NO2(g) + SO2(g) → NO(g) + SO3(g)

A) -6.2 J mol−1 K−1

B) +21.9 J mol−1 K−1

C) -21.9 J mol−1 K−1

D) -37.9 J mol−1 K−1

E) +52.9 J mol−1 K−1

Diff: 2

Section: 18.5

39) Using the standard free energies of formation:

BaCO3(s), ΔG°f = -1139.0 kJ mol−1

BaSO4(s), ΔG°f = -1353.0 kJ mol−1

CO2(g), ΔG°f = -394.4 kJ mol−1

SO3(g), ΔG°f = -370.0 kJ mol−1

Calculate the standard free energy change, ΔG°, for the reaction:

BaCO3(s) + SO3(g) → BaSO4(s) + CO2(g)

A) +189.6 kJ mol−1

B) -238.4 kJ mol−1

C) +472.4 kJ mol−1

D) +238.4 kJ mol−1

E) -2516.4 kJ mol−1

Diff: 2

Section: 18.6

40) Using the standard free energies of formation:

NO2(g), ΔG°f = +51.84 kJ mol−-1

NO(g), ΔG°f = +86.69 kJ mol−1

SO2(g), ΔG°f = -300.0 kJ mol−1

SO3(g), ΔG°f = -370.0 kJ mol−1

Calculate the standard free energy change, ΔG°, for the reaction:

NO2(g) + SO2(g) → NO(g) + SO3(g)

A) -35.15 kJ mol−1

B) -104.9 kJ mol−1

C) -429.2 kJ mol−1

D) -619.6 kJ mol−1

E) +35.15 kJ mol−1

Diff: 2

Section: 18.6

41) Given the data:

H2(g), ΔH°f = 0 kJ mol−1, S° = +130.6 J mol−1 K−1

I2(s), ΔH°f = 0 kJ mol−1, S° = +116.12 J mol−1 K−1

HI(g), ΔH°f = +26 kJ mol−1, S° = +206 J mol−1 K−1

Calculate the standard free energy change, ΔG°, for the reaction:

H2(g) + I2(s) → 2 HI(g)

A) +2.7 kJ mol−1

B) -46.5 kJ mol−1

C) -2.7 kJ mol−1

D) +128.2 kJ mol−1

E) -165.3 kJ mol−1

Diff: 2

Section: 18.6

42) Given the data:

Ag2O(s), ΔH°f = -31.1 kJ mol−1, S° = +121.3 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = +205 J mol−1 K−1

Ag(s), ΔH°f = 0.00 kJ mol−1, S° = +42.55 J mol−1 K−1

Calculate the standard free energy change ΔG°, for the reaction:

Ag2O(s) → 2 Ag(s) + ½ O2(g)

A) +11.3 kJ mol−1

B) -24.0 kJ mol−1

C) -38.2 kJ mol−1

D) -50.4 kJ mol−1

E) -1.3 kJ mol−1

Diff: 2

Section: 18.6

43) Given the data:

N2(g), ΔH°f = 0.00 kJ mol−1, S° = +191.5 J mol−1 K−1

H2(g), ΔH°f = 0.00 kJ mol−1, S° = +130.6 J mol−1 K−1

NH3(g), ΔH°f = -46.0 kJ mol−1, S° = +192.5 J mol−1 K−1

Calculate the standard free energy change, ΔG° for the reaction:

N2(g) + 3 H2(g) → 2 NH3(g)

A) -7.4 kJ mol−1

B) -32.9 kJ mol−1

C) -84.6 kJ mol−1

D) +112.3 kJ mol−1

E) +32.9 kJ mol−1

Diff: 2

Section: 18.6

44) Given the data:

NH3(g), ΔH°f = -46.0 kJ mol−1, S° = +192.5 J mol−1 K−1

NO(g), ΔH°f = +90.4 kJ mol−1, S° = +210.6 J mol−1 K−1

H2O(l), ΔH°f = -286 kJ mol−1, S° = +69.96 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = +205 J mol−1 K−1

Calculate the standard free energy change, ΔG°, for the reaction:

2 NH3(g) + 5/2 O2(g) → 2 NO(g) + 3 H2O(l)

A) -100.8 kJ mol−1

B) -206.7 kJ mol−1

C) -276.5 kJ mol−1

D) -505.8 kJ mol−1

E) +664.3 kJ mol−1

Diff: 2

Section: 18.6

45) Given the data:

PbO(s), ΔH°f = -217.3 kJ mol−1, S° = +68.7 J mol−1 K−1

PbO2(s), ΔH°f = -277.0 kJ mol−1, S° = +68.6 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = +205 J mol−1 K−1

Calculate the standard free energy change, ΔG°, for the reaction:

PbO(s) + ½ O2(g) PbO2(s)

A) +1.45 kJ mol−1

B) −29.1 kJ mol−1

C) −68.3 kJ mol−1

D) −90.3 kJ mol−1

E) +29.1 kJ mol−1

Diff: 2

Section: 18.6

46) Using the data:

C2H4(g), ΔH°f = +51.9 kJ mol−1, S° = 219.8 J mol−1 K−1

CO2(g), ΔH°f = -394 kJ mol−1, S° = 213.6 J mol−1 K−1

H2O(l), ΔH°f = -286.0 kJ mol−1, S° = 69.96 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = 205 J mol−1 K−1

Calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process:

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)

A) 1332 kJ mol−1

B) 1380 kJ mol−1

C) 1451 kJ mol−1

D) 1492 kJ mol−1

E) 2422 kJ mol−1

Diff: 2

Section: 18.7

47) Which of the following is the best example of a thermodynamically reversible process?

A) water dripping out of a leak in a pail

B) the conversion of ice to water at 0°C

C) the popping of a balloon

D) the dissolving of sugar into water

E) the cracking of a block of ice with a hammer

Diff: 2

Section: 18.7

48) Which of the following is the best example of a thermodynamically reversible process?

A) a nail rusting in a deck

B) the dispersion of food coloring when added to a glass of water

C) the buring of a piece of paper

D) the conversion of water to ice in a sealed at 0°C

E) the conversion of dry ice to CO2 gas on a counter at room temperature

Diff: 2

Section: 18.7

49) In order to get the most efficient work out of a thermochemical process, what needs to be done?

A) The process must take place at high temperature.

B) The process must be endothermic.

C) The process must be exothermic.

D) The process must be done reversibly.

E) The process must cause an increase in entropy for the system.

Diff: 2

Section: 18.7

50) Using the data:

C2H6(g), ΔH°f = -84.5 kJ mol−1, S° = +229.5 J mol−1 K−1

CO2(g), ΔH°f = -394.0 kJ mol−1, S° = +213.6 J mol−1 K−1

H2O(l), ΔH°f = -286.0 kJ mol−1, S° = +69.96 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = +205 J mol−1 K−1

Calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process:

C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l), per mole of C2H6

A) 1426 kJ mol−1

B) 1469 kJ mol−1

C) 1654 kJ mol−1

D) 2938 kJ mol−1

E) 3029 kJ mol−1

Diff: 2

Section: 18.7

51) Assuming that, since the physical states do not change, the values of ΔH and sS do not change as we raise the temperature in the reaction below. Using the following values at 298 K,

CdO(s), ΔH°f = -258.2 kJ mol−1, S° = +54.8 J mol−1 K−1

CdSO4(s), ΔH°f = -933.5 kJ mol−1, S° = +123 J mol−1 K−1

SO3(g), ΔH°f = -396 kJ mol−1, S° = +256 J mol−1 K−1

calculate a value for the free energy change, ΔG°T for the reaction,

CdSO4(s) → CdO(s) + SO3(g) at 750 K

A) +223.3 kJ mol−1

B) +138.5 kJ mol−1

C) -140.6 kJ mol−1

D) +420.5 kJ mol−1

E) -420.5 kJ mol−1

Diff: 2

Section: 18.8

52) Assuming that, since the physical states do not change, the values of ΔH and ΔS do not change as we raise the temperature in the reaction below. Using the following values at 25 °C,

CaO(s), ΔH°f = -635.5 kJ mol−1, S° = +40.0 J mol−1 K−1

CaCO3(s), ΔH°f = -1207 kJ mol−1, S° = +92.9 J mol−1 K−1

CO2(g), ΔH°f = -394 kJ mol−1, S° = +213.6 J mol−1 K−1

calculate a value for the free energy change, ΔG°T for the reaction,

CaCO3(s) → CaO(s) + CO2(g) at 815 °C

A) +2.6 kJ mol−1

B) +46.5 kJ mol−1

C) -174.7 kJ mol−1

D) +308.5 kJ mol−1

E) -352.4 kJ mol−1

Diff: 2

Section: 18.8

53) Naphthalene (C10H8) is a solid at room temperature. It sublimes by the following process. Using the data below determine the best estimate of the sublimation temperature for naphthalene (the temperature at which this reaction will become spontaneous under standard state conditions.

C10H8(s) → C10H8(g)

ΔH° = 72.1 kJ/mol

ΔS° = 196.55 J/K mol

A) 2.73 K

B) 421 K

C) 561 K

D) 367 K

E) 315 K

Diff: 2

Section: 18.8

54) Assuming that, since the physical states do not change, the values of ΔH and ΔS do not change as we raise the temperature, and using,

N2(g), ΔH°f = 0.00 kJ mol−1, S° = +191.5 J mol−1 K−1

H2(g), ΔH°f = 0.00 kJ mol−1, S° = +130.6 J mol−1 K−1

NH3(g), ΔH°f = -46.0 kJ mol−1, S° = +192.5 J mol−1 K−1

calculate a value for the free energy change, ΔG°T for the reaction below, at 500 °C

N2(g) + 3 H2(g) → 2 NH3(g)

A) +7.2 kJ mol−1

B) +245.3 kJ mol−1

C) +99.1 kJ mol−1

D) +61.3 kJ mol−1

E) +153.2 kJ mol−1

Diff: 2

Section: 18.8

55) Over a wide temperature range, the reaction, M2O3(s) + C(s) → M(s) + CO2(g), is spontaneous at low temperatures but non-spontaneous at high temperatures. If we assume that, since the physical states do not change, the values of ΔG°T and ΔS°T are constant over this temperature range, we can then deduce that

A) ΔH < 0 and ΔS > 0.

B) ΔH < 0 and ΔS < 0.

C) ΔH > 0 and ΔS < 0.

D) ΔH > 0 and ΔS > 0.

E) The information is insufficient to make any judgment as to the signs of ΔH and ΔS.

Diff: 2

Section: 18.8

56) Given the data:

Ag2O(s), ΔH°f = -31.1 kJ mol−1, S° = +121.3 J mol−1 K−1

Ag(s), ΔH°f = 0.00 kJ mol−1, S° = +42.55 J mol−1 K−1

O2(g), ΔH°f = 0.00 kJ mol−1, S° = +205.0 J mol−1 K−1

Calculate the temperature at which ΔG°T = 0 for the reaction, Ag2O(s) → 2 Ag(s) + ½ O2(g). Assume that, since the physical states do not change, ΔH° and ΔS° are independent of temperature between -50.0 °C and 950.0 °C.

A) +196 °C

B) +246 °C

C) +423 °C

D) +610 °C

E) +818 °C

Diff: 2

Section: 18.8

57) The thermochemical equation representing one process used for the synthesis of ammonia involves the equilibrium shown,

N2(g) + 3 H2(g) 2 NH3(g)

or this reaction, ΔH° = -92.2 kJ, ΔG° = -33.4 kJ. All things considered, one concludes that

A) the coefficients give us the mole ratios but not the volume ratios of the reacting species.

B) an increase in pressure favors an increase in the yield of ammonia.

C) carrying out the reaction at a higher temperature shifts the position of equilibrium to the right, thus favoring an increase in the yield of ammonia.

D) cooling the mixture to remove ammonia from the reaction mixture unfortunately also decreases the overall yield of ammonia.

E) the mixture does not need heating, in fact, cooling the mixture assists the rapid establishment of equilibrium.

Diff: 2

Section: 18.8

58) Determine the equilibrium constant Kp at 25°C for the reaction:

N2(g) + 3H2(g) 2NH3(g)

(ΔG°f (NH3(g)) = -16.6 kJ/mol)

A) 1.52 × 10−6

B) 2.60

C) 8.28 × 10−2

D) 13.4

E) 6.60 × 105

Diff: 1

Section: 18.9

59) Determine the equilibrium constant Kp at 25°C for the reaction:

2NO(g) + O2(g) 2NO2(g)

(ΔG°rxn = -69.7 kJ/mol)

A) 1.65 × 1012

B) 8.28 × 10−2

C) 2.60

D) 13.4

E) 6.07 × 10−13

Diff: 2

Section: 18.9

60) The equilibrium constant at 427°C for the reaction

N2(g) + 3H2(g)2NH3(g)

is Kp = 9.4 × 10−5. Calculate the value of ΔG° for the reaction at this temperature.

A) 56 kJ/mol

B) -56 kJ/mol

C) -33 kJ/mol

D) 33 kJ/mol

E) 1.3 J/mol

Diff: 2

Section: 18.9

61) The equilibrium constant at 25°C for the reaction

2NO(g) + O2(g) 2NO2(g)

is Kp = 1.65 × 1012. Calculate the value of ΔG° for the reaction at this temperature.

A) -4.09 kJ/mol

B) -5.85 kJ/mol

C) +5.85 kJ/mol

D) -69.7 kJ/mol

E) 1.65 kJ/mol

Diff: 2

Section: 18.9

62) Using these bond energies, ΔH°:

C—C: 348 kJ/mol C═C: 612 kJ/mol C≡C: 960 kJ/mol

C—H: 412 kJ/mol H—H: 436 kJ/mol

Calculate the value of ΔH° of reaction for

H2C═CH2(g) + H2(g) → CH3—CH3(g)

A) -560 kJ/mol

B) -124 kJ/mol

C) -388 kJ/mol

D) +224 kJ/mol

E) -212 kJ/mol

Diff: 2

Section: 18.10

63) Using these bond energies, ΔH°:

C—C: 348 kJ/mol C═C: 612 kJ/mol C≡C: 960 kJ/mol C—H: 412 kJ/mol

C—O: 360 kJ/mol C═O: 743 kJ/mol H—H: 436 kJ/mol H—O: 463 kJ/mol

Calculate the value of ΔH° of reaction for

O═C═O(g) + 3 H2(g) → CH3—O—H(g) + H—O—H(g)

A) -191 kJ/mol

B) +272 kJ/mol

C) -272 kJ/mol

D) -5779 kJ/mol

E) +5779 kJ/mol

Diff: 2

Section: 18.10

64) Using these bond energies, ΔH°:

C—C: 348 kJ/mol C═C: 612 kJ/mol C—H: 412 kJ/mol C—O: 360 kJ

C═O: 743 kJ/mol H—H: 436 kJ/mol H—O: 463 kJ/mol O═O: 498 kJ

Calculate the value of ΔH° of reaction for

H2C═CH2(g) + H—O—H(g) → CH3—CH2—O—H(g)

A) -13 kJ/mol

B) -45 kJ/mol

C) -124 kJ/mol

D) +224 kJ/mol

E) -508 kJ/mol

Diff: 2

Section: 18.10

65) According to the first law of thermodynamics the internal energy of an isolated system must ________.

Diff: 1

Section: 18.1

66) According to the first law of thermodynamics if the energy of a system decreases ________ amount of energy must be transferred to the surroundings.

Diff: 2

Section: 18.1

67) The ________ phase of a substance will always have more entropy then the liquid phase.

Diff: 1

Section: 18.3

68) The reaction N2(g) + 3H2(g) → 2NH3(g) would result in a(n) ________ in entropy of the system.

Diff: 1

Section: 18.3

69) Consider the following chemical reaction:

Cr3+(aq) + 6H2O(l) → [Cr(H2O)6]3+(aq)

Would this reaction have a positive or negative entropy of formation?

Diff: 1

Section: 18.3

70) Which of the following processes are accompanied by an increase in entropy. One or more of the following answers may be selected.

1. 2SO2(g) + O2(g) → 2SO3(g)

2. H2O(l) → H2O(s)

3. Br2(l) → Br2(g)

4. H2O2(l) → H2O(l) + (1/2)O2(g)

Diff: 1

Section: 18.3

71) According to the second law of thermodynamics, the entropy of the universe is always ________.

Diff: 1

Section: 18.4

72) For the reaction H2(g) + S(s) → H2S(g), ΔH° = -20.2 kJ/mol and ΔS° = +43.1 J/K·mol, at which temperatures would the reaction be spontaneous?

Diff: 2

Section: 18.4

73) For the reaction 2NO(g) + O2(g) → 2NO2(g), ΔH° = −113.1 kJ/mol and ΔS° = -145.3 J/K·mol. Under which temperature conditions would the reaction be spontaneous?

Diff: 2

Section: 18.4

74) Consider a reaction that is both exothermic and increases entropy. How will temperature affect the spontaneity of the process?

Diff: 1

Section: 18.4

75) Consider a reaction that is both endothermic and increases entropy. How will temperature affect the spontaneity of the process?

Diff: 2

Section: 18.4

76) Consider a reaction that is both exothermic and decreases order. How will temperature affect the spontaneity of the process?

Diff: 2

Section: 18.4

77) Consider a reaction that is both endothermic and increasingly ordered. How will temperature affect the spontaneity of the process?

Diff: 2

Section: 18.4

78) Describe, using the free energy relationship ΔG = ΔH - TΔS, the process of a hot object coming into thermal equilibrium with a cold object.

molecules are moving rapidly. The cold object has lower H and smaller S. When the objects are placed next to each other, heat is transferred from the hotter object to the colder one. At the same time, the kinetic energy of the molecules in the colder object increases, resulting in more disorder. The kinetic energy of the molecules in the hotter system, however, decreases, and the molecules become slightly more ordered. This transfer continues until both objects are at thermal equilibrium, and the values of G are equal.

Diff: 2

Section: 18.4

79) HI has a normal boiling point of -35.4°C, and its ΔHvap is 21.16 kJ/mol. Calculate the molar entropy of vaporization (ΔSvap) for HI.

Diff: 2

Section: 18.4

80) The entropy of perfectly ordered pure crystalline substance at zero Kelvin is ________.

Diff: 1

Section: 18.5

81) Using the data,

I2(g), ΔH°f = +62.4 kJ mol−1, S° = +260.7 J mol−1 K−1

I2(s), ΔH°f = 0.00 kJ mol−1, S° = +116.12 J mol−1 K−1

calculate the temperature at which solid iodine should have a vapor pressure of 1.00 atm, based on the reaction, I2(g) I2(s).

Diff: 1

Section: 18.5

82) Calculate ΔG° at 25°C for the reaction 2HNO3(l) + NO(g) → 3NO2(g) + H2O(l).

ΔG°f (kJ/mol) (at 25°C)

H2O(l) -237.2

HNO3(l) -79.9

NO(g) 86.7

NO2(g) 51.8

Diff: 2

Section: 18.6

83) Ammonia (NH3) has a normal boiling point of -33.33°C, and its ΔHvap is 23.33 kJ/mol. Calculate the molar entropy of vaporization (ΔSvap) for ammonia.

Diff: 2

Section: 18.8

84) For a chemical reaction, ΔH° = +21.16 kJ mol−1. The value of ΔS° for the same reaction was +74.50 J mol−1 K−1. If, since the physical states do not change, the value of ΔH° and ΔS° do not change with temperature over the range of interest, what should be the value of ΔG°T at

+550.0 °C?

Diff: 2

Section: 18.8

85) For a chemical reaction, ΔH° = +31.16 kJ mol−1. The value of ΔS° for the same reaction was +74.50 J mol−1 K−1. If, since the physical states do not change, the value of ΔH and ΔS do not change with temperature over the range of interest, at what temperature should ΔG°T = 0.0, in °C?

Diff: 2

Section: 18.8

86) For the system, 2 NO2(g) N2O4(g), ΔG° = −5.40 kJ. Calculate the value of the equilibrium constant, Kp, for this system.

Diff: 2

Section: 18.9

87) For the system, 2 NO2(g) N2O4(g), ΔG° = -5.40 kJ/mol. If the system was initialized using enough NO2 to create a partial pressure of 0.500 atm and enough N2O4 to create a partial pressure of 0.350 atm, calculate the value of ΔG(T= 298 K) for these non-standard conditions prevailing at the start.

Diff: 2

Section: 18.9

88) For the system, 2 NO2(g) N2O4(g), ΔG° = 113.85 kJ/mol at 1000 K. If the system was initialized using enough NO2 to create a partial pressure of 0.750 atm and enough N2O4 to create a partial pressure of 0.450 atm, calculate the value of ΔG at this temperature using these initial values.

Diff: 2

Section: 18.9

89) For the system, 2 NO2(g) N2O4(g), ΔG° = 30.38 kJ/mol at 500 K. If the system was initialized using enough NO2 to create a partial pressure of 0.85 atm and enough N2O4 to create a partial pressure of 0.150 atm, calculate the value of ΔG at this temperature using these initial values.

Diff: 2

Section: 18.9

90) Given the following values for components of a system, at 298.15 K and 1 atm:

For PCl5, ΔH°f = -374.9 kJ mol−1, S° = 364.6 J mol−1 K−1.

For PCl3, ΔH°f = -287.0 kJ mol−1, S° = 311.8 J mol−1 K−1.

For Cl2, ΔH°f = 0.0 kJ mol−1, S° = 223.0 J mol−1 K−1.

Calculate values for Kp at 298.15 K for

PCl5(g) PCl3(g) + Cl2(g)

Diff: 2

Section: 18.9

91) For the system,

PCl5(g) + 2 NO(g) PCl3(g) + 2 NOCl(g)

ΔG° = -9.720 kJ at 25 °C. If a system is charged with 0.200 atm of PCl5, 0.800 atm of PCl3, 0.800 atm of NOCl, and 0.100 atm of NO, what is the initial value of ΔG298 for the other than standard conditions given above?

Diff: 2

Section: 18.9

92) Given the following values for components of a system, at 410 K:

For PCl5, ΔH°f = -370.0 kJ mol−1, S° = 375.0 J mol−1 K−1.

For PCl3, ΔH°f = -293.0 kJ mol−1, S° = 320.5 J mol−1 K−1.

For Cl2, ΔH°f = 000.0 kJ mol−1, S° = 230.0 J mol−1 K−1.

Calculate a value for Kp at 410 K for

PCl5(g) PCl3(g) + Cl2(g)

Hint: Find ΔG, then Kp.

Diff: 3

Section: 18.9

93) At 1500°C the equilibrium constant for the reaction CO(g) + 2H2(g) CH3OH(g) has the value Kp = 1.4 × 10−7. The value of ΔG° for this reaction at 1500°C is ________.

Diff: 2

Section: 18.9

94) Any event that is accompanied by an increase in entropy of the system will always occur spontaneously.

Diff: 1

Section: 18.3

95) The standard entropy, S° for an element in its standard state is always zero.

Diff: 1

Section: 18.5

96) If a reaction is spontaneous at constant temperature and pressure, then the value of free energy of reaction, ΔGr, must be negative.

Diff: 1

Section: 18.4

97) The equation, ΔH°reaction = ΔH°f (products) - ΔH°f (reactants) is a statement of the second law of thermodynamics.

Diff: 1

Section: 18.4

98) A negative value for the standard molar entropy, S°, of a chemical substance indicates that the substance is unstable and may decompose or detonate if mishandled.

Diff: 1

Section: 18.5

99) The entropy of a solid amorphous substance at absolute zero is 0 according to the third law of thermodynamics.

Diff: 1

Section: 18.5

100) Three factors which, acting together, determine whether a reaction is spontaneous or not, are entropy, internal energy, and enthalpy.

Diff: 1

Section: 18.4

101) The entropy of a solid at absolute zero is 0 according to the third law of thermodynamics if the solid is a perfect crystal.

Diff: 1

Section: 18.5

102) The maximum amount of energy produced by a reaction that can be theoretically harnesses as work is equal to ________.

Diff: 1

Section: 18.7

103) The reaction rates of many thermodynamically spontaneous reactions are actually very slow. Which of these statements is the best explanation for this observation?

A) The activation energy of the reaction is large.

B) ΔG° for the reaction is positive.

C) KP for the reaction is less than one.

D) The entropy change is negative.

E) Such reactions are endothermic.

Diff: 2

Section: 18.2

104) The normal freezing point of ammonia is -78°C. Predict the signs of ΔH, ΔS, and ΔG for ammonia when it melts at -76°C and 1 atm: NH3(s) → NH3(l)

Liquids have more entropy than solids, so ΔS would be positive.

Because the temperature is above the normal melting point, this process would be expected to happen spontaneously, and ΔG would be negative.)

Diff: 2

Section: 18.4

105) The normal freezing point of benzoic acid is 122.4°C. Predict the signs of ΔH, ΔS, and ΔG for a process in which liquid benzoic acid freezes to solid benzoic acid at 120°C and 1 atm:

C7H6O2 (l) → C7H6O2 (s)

Liquids have more entropy than solids, so ΔS would be negative.

Because the temperature is below the normal melting point, this process would be expected to happen spontaneously, and ΔG would be negative.

Diff: 2

Section: 18.4

106) The species in the reaction,

2 KNO3(s) + 2 C(s) → K2CO3(s) + N2(g) + 3/2 CO2(g)

have the values for standard enthalpies of formation at 25 °C:

KNO3(s), ΔH°f = -492.7 kJ mol−1

CO2(g), ΔH°f = -394.0 kJ mol−1

K2CO3(s), ΔH°f = -1146 kJ mol−1

Assume that, since the physical states do not change, the values of ΔH° and ΔS1 atm = ΔS° are constant throughout a broad temperature range and use this information to determine which of the following conditions may apply.

A) The reaction is spontaneous at all temperatures.

B) The reaction is non-spontaneous at all temperatures.

C) The reaction is spontaneous at low temperatures but nonspontaneous at high

temperatures.

D) The reaction is nonspontaneous at low temperatures but spontaneous at high

temperatures.

E) It is impossible to make any judgment about spontaneity because insufficient information can be gleaned from the data and equations presented.

Diff: 2

Section: 18.8

107) The species in the reaction, KClO3(s) → KCl(s) + 3/2 O2(g) have the values for standard enthalpies of formation at 25 °C:

KClO3(s), ΔH°f = -391.2 kJ mol−1

KCl(s), ΔH°f = -436.8 kJ mol−1

Assume that, since the physical states do not change, the values of ΔH° and ΔS° are constant throughout a broad temperature range, and use this information to determine which of the following conditions may apply.

A) The reaction is spontaneous at all temperatures over a broad temperature range.

B) The reaction is nonspontaneous at all temperatures over a broad temperature range.

C) The reaction is spontaneous at low temperatures but nonspontaneous at high temperatures.

D) The reaction is nonspontaneous at low temperatures but spontaneous at high temperatures.

E) It is impossible to make any judgment about spontaneity because insufficient information can be gleaned from the data and equations presented.

Diff: 2

Section: 18.8

108) Calculate the value of ΔH° for the reaction,

C—C: 348 kJ/mol C—H: 412 kJ/mol C—O: 360 kJ/mol

C═O: 743 kJ/mol H—O: 463 kJ/mol O═O: 498 kJ/mol

+ O═O(g) → + H2O(g)

Diff: 2

Section: 18.10

109) You'll notice that the two compounds shown below have the same molecular formula, but different structural formulas. Calculate the difference in molar bond energy between the two.

C—C: 348 kJ/mol C═C: 612 kJ/mol C—H: 412 kJ/mol C—O: 360 kJ/mol

C═O: 743 kJ/mol H—H: 436 kJ/mol H—O: 463 kJ/mol O═O: 498 kJ/mol

CH3—CH2—O—H and CH3—O—CH3(g)

Diff: 2

Section: 18.10

110) Calculate the value of ΔH° for the reaction, per mole, for the conversion of isopropyl alcohol into acetone by the reaction shown below.

C—C: 348 kJ/mol C═C: 612 kJ/mol C—H: 412 kJ/mol C—O: 360 kJ/mol

C═O: 743 kJ/mol H—H: 436 kJ/mol H—O: 463 kJ/mol O═O: 498 kJ/mol

+ O2(g) → + H2O(g)

Diff: 2

Section: 18.10

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