General Chemistry: Atoms First (McMurry/Fay/Pribush)Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium

16.1 Multiple Choice Questions

1) Which of the following statements is not true?A) The reverse of a spontaneous reaction is always nonspontaneous.B) A spontaneous process always moves toward equilibrium.C) A nonspontaneous process cannot be caused to occur.D) A highly spontaneous process need not occur rapidly.Answer: CTopic: Section 16.1 Spontaneous Processes

2) Which forward reaction is a nonspontaneous process?A) the expansion of a gas into a vacuumB) N2(g) + 3 H2(g) 2 NH⇌ 3(g) if PH₂ = PN₂ = 1 atm, PNH₃ = 0, and Kp = 4 × 105

C) 2 NH3(g) N⇌ 2(g) + 3 H2(g) if PNH₃ = 1 atm, PH₂ = PN₂ = 0, and Kp = 2 × 10-6

D) none of the aboveAnswer: DTopic: Section 16.1 Spontaneous Processes

3) The chemical system shown below is at equilibrium. Which change in conditions will not result in a spontaneous forward reaction?

N2(g) + 3 H2(g) 2 NH⇌ 3(g) Kp = 4 × 105

A) adding a catalystB) adding more H2C) adding more N2D) reducing the volumeAnswer: ATopic: Section 16.1 Spontaneous Processes

4) Classify each of the following processes as spontaneous or nonspontaneous.I. H2O(l) → H2O(g) T = 25°C, vessel open to atmosphere with 50% relative

humidityII. H2O(s) → H2O(l) T = 25°C, P = 1 atm

A) I and II are both spontaneous.B) I is spontaneous and II is nonspontaneous.C) I is nonspontaneous and II is spontaneous.D) I and II are both nonspontaneous.Answer: ATopic: Section 16.1 Spontaneous Processes

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5) The reaction A(g) → B(g) is spontaneous under standard conditions. Which of the following statements must be true?

I. The reaction B(g) → A(g) is nonspontaneous under standard conditions.II. A(g) will be completely converted to B(g) if sufficient time is allowed.III. A(g) will be completely converted to B(g) rapidly.

A) none of theseB) I C) I and IID) I, II, and IIIAnswer: BTopic: Section 16.1 Spontaneous Processes

6) Which of the following processes are spontaneous?I. dissolving more solute in an unsaturated solutionII. dissolving more solute in a saturated solutionIII. dissolving more solute in a supersaturated solution

A) none of theseB) IC) I and IID) I, II, and IIIAnswer: BTopic: Section 16.1 Spontaneous Processes

7) Which of the following processes is spontaneous?A) a mixture of two gases separating into pure compoundsB) reaction of sodium with oxygenC) precipitation of solute from a saturated solutionD) water flowing uphillAnswer: BTopic: Section 16.1 Spontaneous Processes

8) Entropy is a measure ofA) free energy.B) the heat of a reaction.C) molecular randomness.D) the rate of a reaction.Answer: CTopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

9) For which of the following will the entropy of the system increase?A) condensation of steamB) reaction of magnesium with oxygen to form magnesium oxideC) reaction of nitrogen and hydrogen to form ammoniaD) sublimation of dry iceAnswer: DTopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

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10) For which process is the sign of △S negative in the system?A) 2 H2(g) + O2(g) → 2 H2O(g)

B) 2 H2O(l) + 2 K(s) → 2 K+(aq) +2 OH–(aq) + H2(g)C) H2O(s) → H2O(g)D) H2O(l) → H2O(g)Answer: ATopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

11) Predict the sign of ΔS of the system for both of the following.I. 2 C(graphite) + O2(g) → 2 CO(g)II. C4H10(g) → C4H10(l)

A) ΔS should be negative for I and negative for II.B) ΔS should be negative for I and positive for II.C) ΔS should be positive for I and negative for II.D) ΔS should be positive for I and positive for II.Answer: CTopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

12) Sodium reacts violently with water according to the equation:2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

The resulting solution has a higher temperature than the water prior to the addition of sodium. What are the signs of and for this reaction?A) ΔH° is negative and ΔS° is negative.B) ΔH° is negative and ΔS° is positive.C) ΔH° is positive and ΔS° is negative.D) ΔH° is positive and ΔS° is positive.Answer: BTopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

13) The brown color associated with photochemical smog is due to NO2(g), which is involved in an equilibrium with N2O4(g) in the atmosphere.

2 NO2(g) N⇌ 2O4(g)Predict the signs of the enthalpy and entropy change for the forward reaction.A) The enthalpy change is negative and the entropy change is negative.B) The enthalpy change is negative and the entropy change is positive.C) The enthalpy change is positive and the entropy change is negative.D) The enthalpy change is positive and the entropy change is positive.Answer: ATopic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

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14) What is W in Boltzmann's formula, S = k ln W?A) a fraction indicating the probability of obtaining a resultB) a random numberC) the number of ways of obtaining the stateD) the work times Avogadro's numberAnswer: CTopic: Section 16.3 Entropy and Probability

15) Which electron on an atom of copper would have the highest value of W in the Boltzmann formula?A) 3sB) 3dC) 4sD) 4pAnswer: BTopic: Section 16.3 Entropy and Probability

16) An electron in an oxygen p orbital on which of the following would have the highest entropy?A) CH3CH2OH

B) CH3CH2O–

C) CH3CO2OH

D) CH3CO2–

Answer: DTopic: Section 16.3 Entropy and Probability

17) What is k in Boltzmann's formula, S = k ln W?A) the degeneracy of the stateB) the equilibrium constant for the processC) the universal gas constant divided by Avogadro's numberD) the universal gas constant times Avogadro's numberAnswer: CTopic: Section 16.3 Entropy and Probability

18) The entropy change associated with the expansion of one mole of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature is given by the equation, ΔS = R ln (Vf/Vi). What is the entropy change associated with the expansion of three moles of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature?A) ΔS = R ln (Vf/Vi)B) ΔS = 3 mol × R ln (Vf/Vi)

C) ΔS = R ln (Vf × 23/Vi)D) ΔS = R ln (Vf × 3!/Vi)Answer: BTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

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19) What is the entropy change associated with the expansion of one mole of an ideal gas from an initial volume of V to a final volume of V of 2.50V at constant temperature?A) ΔS = 2.50 R ln (Vf/Vi)B) ΔS = -2.50 R ln (Vf/Vi)C) ΔS = R ln 2.50D) ΔS = -R ln 2.50Answer: CTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

20) Predict the sign of ΔS for each of the following processes, which occur at constant temperature.

I. The volume of 2.0 moles of O2(g) increases from 44 L to 52 L.II. The pressure of 2.0 moles of O2(g) increases from 1.0 atm to 1.2 atm.

A) I: ΔS= negative; II: ΔS= negativeB) I: ΔS= negative; II: ΔS= positiveC) I: ΔS= positive; II: ΔS= negativeD) I: ΔS= positive; II: ΔS= positiveAnswer: CTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

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21) Assume a heteronuclear diatomic molecule, AB, forms a one-dimensional crystal by lining up along the x-axis. Also assume that each molecule can only have one of six possible orientations, corresponding to atom A facing in either the positive or negative direction along the x-, y-, or z-axis. If the molecules are arranged randomly in the six directions, the molar entropy at absolute zero should beA) R ln 6.B) R ln 66.C) R ln 6!D) 0.Answer: ATopic: Section 16.3 Entropy and Probability

22) The Boltzmann formula is S = k ln W. A perfect crystal has a molar entropy of 0 at absolute zero becauseA) W = 0.B) W = 1.C) W = NA.D) k = 1.Answer: BTopic: Section 16.3 Entropy and Probability

23) What is the sign of △S for each of the following processes?I. The separation of gaseous molecules of UF6, into 238UF6 and 235UF6 at constant

temperature and pressure.II. The dissolving of I2(s) in CCl4(l).

A) ΔS is negative for I and negative for II.B) ΔS is negative for I and positive for II.C) ΔS is positive for I and negative for II.D) ΔS is positive for I and positive for II.Answer: BTopic: Section 16.3 Entropy and Probability

24) Which has the lowest entropy?A) CH3OH(s, –25°C)B) CH3OH(s, –15°C)C) CH3OH(l, 15°C)D) CH3OH(l, 25°C)Answer: ATopic: Section 16.4 Entropy and Temperature

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25) Which has the highest entropy in each set?I. H2O(s), H2O(l), H2O(g) at 0.1°C, 4.58 atmII. H2O(l) at 0°C, H2O(l) at 25°C, H2O(l) at 100°C (all at 1.0 atm pressure)

A) H2O(l) in set I and H2O(l) at 0°C in set IIB) H2O(s) in set I and H2O(l) at 100°C in set IIC) H2O(g) in set I and H2O(l) at 0°C in set IID) H2O(g) in set I and H2O(l) at 100°C in set IIAnswer: DTopic: Section 16.4 Entropy and Temperature

26) Which provides the greatest increase in entropy?A) H2O (s, 0°C) → H2O (l, 0°C)B) H2O (l, 0°C) → H2O (l, 25°C)C) H2O (g, 0.1°C) → H2O (s, 0.1°C)D) H2O (l, 100°C) → H2O (g, 100°C)Answer: DTopic: Section 16.4 Entropy and Temperature

27) According to the third law of thermodynamics,A) energy is conserved in any transformation of matter.B) the entropy increases for any spontaneous process.C) the entropy of a perfectly ordered, crystalline substance is zero at 0 Kelvin.D) the entropy of the universe increases for any spontaneous process.Answer: CTopic: Section 16.4 Entropy and Temperature

28) Which of the following statements must be true for the entropy of a pure solid to be zero?I. The temperature must be 0 K.II. The solid must be crystalline, not amorphous.III. The solid must be perfectly ordered.IV. The solid must be an element.

A) IB) I and IIC) I, II, and IIID) I, II, III, and IVAnswer: CTopic: Section 16.4 Entropy and Temperature

29) Under which of the following conditions would one mole of Ne have the highest entropy, S?A) 27°C and 25 LB) 137°C and 25 LC) 27°C and 35 LD) 137°C and 35 LAnswer: DTopic: Section 16.4 Entropy and TemperatureAlgo. Option: algorithmic

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30) Which has the highest standard molar entropy at 25°C? A) Al(s)B) Al(l)C) Al(g)D) All three should have a standard molar entropy of zero.Answer: CTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

31) Which has the highest standard molar entropy at 25°C? A) F2(g)B) Cl2(g)C) Br2(g)D) I2(g)Answer: DTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

32) Which of the following gas molecules has the greatest standard molar entropy at 25°C?A) C2H2B) CH2CH2C) CH3CH3D) All have the same entropy.Answer: CTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

33) Which substance has the highest standard molar entropy at 25°C ?A) C(graphite)B) C2H4(g)C) CH3OH(l)D) MgCO3(s)Answer: BTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

34) Which one of the following has the lowest standard molar entropy, S°, at 25°C?"A) C8H18(s)B) C8H18(l)C) C12H26(s)D) C12H26(l)Answer: ATopic: Section 16.5 Standard Molar Entropies and Standard Entropies of ReactionAlgo. Option: algorithmic

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35) Calculate ΔS° for the following reaction.N2(g) + 2 O2(g) → 2 NO2(g)

A) -156.5 J/KB) -121.5 J/KC) 15.5 J/KD) 636.5 J/KAnswer: BTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

36) ΔS° = – 198.7 J/K for the reaction shown below. Calculate S° for NH3(g).N2(g) + 3 H2(g) → 2 NH3(g)

A) 61.7 J/K∙molB) 123.4 J/K∙molC) 192.3 J/K∙molD) 384.6 J/K∙molAnswer: CTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

37) Calculate ΔS° for the formation of one mole of solid sodium bromide from the elements at 25°C.

A) -116.7 J/KB) -81.2 J/KC) -40.5 J/KD) 86.8 J/KAnswer: CTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

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38) The standard molar entropy for Br2(g) is 245.46 J/(mol ∙ K) at 25°C. Given that ΔS° = 104.58 J/K for the dissociation of one mole of Br2(g) into Br(g) at 25°C, find the standard molar entropy for Br(g) at 25°C.A) 70.44 J/(mol ∙ K)B) 140.08 J/(mol ∙ K)C) 175.02 J/(mol ∙ K)D) 350.04 J/(mol ∙ K)Answer: CTopic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

39) Which of the three laws of thermodynamics provides a criterion for spontaneity?A) the first law of thermodynamicsB) the second law of thermodynamicsC) the third law of thermodynamicsD) both the second and third laws of thermodynamicsAnswer: BTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

40) Which of the following is a criterion for spontaneity that holds for any process?A) ΔG < 0B) ΔG > 0C) ΔStotal < 0D) ΔStotal > 0Answer: DTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

41) According to the second law of thermodynamics, all reactions proceed spontaneously in the direction that increases the entropy of theA) surroundings.B) system.C) system – surroundingsD) system + surroundingsAnswer: DTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

42) For a process to be at equilibrium, it is necessary thatA) ΔSsys = ΔSsurr.B) ΔSsys = - ΔSsurr.C) ΔSsys = 0.D) ΔSsys = 0 and ΔSsurr = 0.Answer: BTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

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43) For a spontaneous processA) energy and entropy are conserved.B) energy is conserved and the entropy of the system and surroundings increases.C) the energy of the system and the surroundings decreases and the entropy of the system and surroundings increases.D) both the energy and the entropy of the system and surroundings decrease.Answer: BTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

44) For the processCaCO3(calcite) → CaCO3(aragonite) ΔH° = -0.21 kJ, ΔS° = -4.2 J/K

Assuming that the surroundings can be considered a large heat reservoir at 25°C, calculate ΔSsurr and ΔStotal for the process at 25°C and 1 atm pressure. Is the process spontaneous at 25°C and 1 atm pressure?A) ΔSsurr = 4.2 J/K, Δtotal = 0, not spontaneousB) ΔSsurr = 0.7 J/K, ΔStotal = -3.5 J/K, not spontaneousC) ΔSsurr = -0.7 J/K, ΔStotal = -4.9 J/K, spontaneousD) ΔSsurr = -0.7 J/K, ΔStotal = -4.9 J/K, not spontaneousAnswer: BTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

45) During perspiration,A) the entropy of the water evaporated decreases and the entropy of the body decreases.B) the entropy of the water evaporated decreases and the entropy of the body increases.C) the entropy of the water evaporated increases and the entropy of the body decreases.D) the entropy of the water evaporated increases and the entropy of the body increases.Answer: CTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

46) A hot penny is dropped into cold water inside a polystyrene foam cup. Assuming negligible heat loss to the atmosphere and the cup,A) the decrease in entropy of the penny is equal to the increase in entropy of the water.| ΔSpenny | = | ΔSwater |B) the decrease in entropy of the penny is less than the increase in entropy of the water.| ΔSpenny | < | ΔSwater |C) the decrease in entropy of the penny is more than the increase in entropy of the water.| ΔSpenny | > | ΔSwater |D) the entropy of both the penny and the water increases.Answer: BTopic: Section 16.6 Entropy and the Second Law of Thermodynamics

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47) At constant pressure and temperature, which statement is true?A) All reactions for which ∆H < 0 are spontaneous.B) All reactions for which ∆S < 0 are spontaneous.C) All reactions for which ∆G < 0 are spontaneous.D) All reactions for which K < 1 are spontaneous.Answer: CTopic: Section 16.7 Free Energy

48) Why is the sign of ΔG rather than the sign of ΔStotal generally used to determine the spontaneity of a chemical reaction?A) ΔG can be used for processes that occur under any conditions.B) ΔG involves thermodynamic functions of the system only.C) Free energy is easier to understand than entropy.D) Entropy is based on probability and is therefore less reliable.Answer: BTopic: Section 16.7 Free Energy

49) Other than only PV work, what reaction conditions must be satisfied for the sign of ΔG to be used as a criterion for spontaneity?A) constant volume and pressureB) constant temperature and pressureC) constant temperature and volumeD) constant volume onlyAnswer: BTopic: Section 16.7 Free Energy

50) For the reaction 3 C2H2(g) → C6H6(l) at 25°C, the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K. Calculate the standard free energy change at 25°C.A) 948 kJB) -503 kJC) -618 kJD) -1061 kJAnswer: BTopic: Section 16.7 Free Energy

51) For a particular process ΔG is less than ΔH. ThereforeA) ΔS is positive.B) ΔS is negative.C) ΔS is zero.D) ΔS is negative if ΔH is positive and ΔS is positive if ΔH is negative.Answer: ATopic: Section 16.7 Free Energy

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52) For a particular process, ΔG = ΔH at a given temperature and pressure. Therefore,A) ΔS is positive if ΔH is positive and negative is ΔH is negative.B) ΔS is negative if ΔH is positive and positive if ΔH is negative.C) ΔS is zero.D) ΔS = ΔG/T.Answer: CTopic: Section 16.7 Free Energy

53) The solubility of manganese(II) fluoride in water is 6.6 g/mL at 40°C and 4.8 g/L at 100°C. Based on these data, what is the sign of ΔH° and ΔS° for the process below?

MnF2(s) Mn⇌ 2+(aq) + 2 F-(aq)A) ΔH° is negative but the sign of ΔS° cannot be determined from this information.B) ΔH° is negative and ΔS° is definitely negative.C) ΔH° is positive but the sign of ΔS° cannot be determined from this information.D) ΔH° is positive and ΔS° is definitely negative.Answer: ATopic: Section 16.7 Free Energy

54) At 25°C, ΔH° = 1.895 kJ and ΔS° = -3.363 J/K for the transitionC(graphite) → C(diamond)

Based on these dataA) graphite cannot be converted to diamond at 1 atm pressure.B) diamond is more stable than graphite at all temperatures at 1 atm.C) diamond is more stable than graphite below 290°C and graphite is more stable than diamond above 290°C.D) graphite is more stable than diamond below 290°C and diamond is more stable than graphite above 290°C.Answer: ATopic: Section 16.7 Free Energy

55) For bromine, ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK-1mol-1 at 25°C. What is the normal boiling point for bromine?A) 25°CB) 58°CC) 124°CD) 332°CAnswer: BTopic: Section 16.7 Free Energy

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56) Consider the reaction:N2(g) + 3 F2(g) → 2 NF3(g) ΔH° = -249 kJ and ΔS° = -278 J/K at 25°C

Calculate ΔG° and state whether the equilibrium composition should favor reactants or products at standard conditions.A) ΔG° = -332 kJ; the equilibrium composition should favor products.B) ΔG° = -332 kJ; the equilibrium composition should favor reactants.C) ΔG° = -166 kJ; the equilibrium composition should favor products.D) ΔG° = -166 kJ; the equilibrium composition should favor reactants.Answer: CTopic: Section 16.7 Free Energy

57) Which statement is true about the formation of CaCO3(s) from CaO(s) and CO2(g) at 1.00 atm?

CaO(s) + CO2(g) → CaCO3(s) ΔH° = -178.7 kJ and ΔS° = -150.4 J/KA) The reaction is spontaneous at all temperatures.B) The reaction is spontaneous at high temperatures.C) The reaction is spontaneous at low temperatures.D) The reaction is not spontaneous at any temperature.Answer: CTopic: Section 16.7 Free Energy

58) The signs of ΔG, ΔH, and ΔS at 25°C are shown below for three reactions.

Which reaction could go in the reverse direction at high temperature?A) IB) IIC) IIID) I and IIAnswer: CTopic: Section 16.7 Free Energy

59) For the evaporation of water during perspiration on a hot, dry day,A) ΔH is positive and TΔS = ΔH.B) ΔH is positive and TΔS > ΔH.C) ΔH is positive and TΔS < ΔH.D) ΔH is negative and TΔS is positive.Answer: BTopic: Section 16.7 Free Energy

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60) For the reaction below ∆G° = + 33.0 kJ, ∆H° = + 92.2 kJ, and ∆S° = + 198.7 J/K. Estimate the temperature at which this reaction becomes spontaneous.

2 NH3(g) → N2(g) + 3 H2(g)A) 0.464 KB) 166 KC) 298 KD) 464 K Answer: DTopic: Section 16.7 Free Energy

61) Calculate the standard free energy change at 25°C for the reaction2 NO(g) + O2(g) → 2 NO2(g).

A) -4.7 kJB) -72.6 kJC) -157.8 kJD) -532.6 kJAnswer: BTopic: Section 16.8 Standard Free-Energy Changes for Reactions

62) For any thermodynamic function Y, ΔY° for a reaction refers to the change in Y for the process in whichA) the mixed reactants at 1 atm go to equilibrium at 1 atm.B) the separate reactants at 1 atm go to equilibrium at 1 atm.C) the separate reactants in their standard states are completely converted to separate products in their standard states.D) the spontaneous reaction occurs.Answer: CTopic: Section 16.8 Standard Free-Energy Changes for Reactions

63) Which of the following is true?A) As a reaction at constant temperature and pressure goes to equilibrium, |ΔG| decreases.B) The larger ΔG°, the faster the reaction.C) The standard state for solutes is the pure solute at 1 atm.D) When a reaction reaches equilibrium, ΔG° = 0.Answer: ATopic: Section 16.8 Standard Free-Energy Changes for Reactions

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64) Which statement is true concerning the standard states of F2(g) and C6H12O6(aq)?A) The standard state for F2(g) is the pure gas at 1 atm and for C6H12O6(aq) is the pure solid at 1 atm.B) The standard state for F2(g) is the pure gas at 1 mol/L and for C6H12O6(aq) is the pure solid at 1 atm.C) The standard state for F2(g) is the pure gas at 1 atm and for C6H12O6(aq) is the solution at a concentration of 1 mol/L.D) The standard state for F2(g) is the pure gas at 1 mol/L and for C6H12O6(aq) is the solution at a concentration of 1 mol/L.Answer: CTopic: Section 16.8 Standard Free-Energy Changes for Reactions

65) Calculate the standard free energy for the reaction given.2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l)

A) -465.2 kJB) -797.8 kJC) -1404.8 kJD) -2069.8 kJAnswer: CTopic: Section 16.9 Standard Free Energies of Formation

66) Which is the lowest at 25°C?A) ΔG°f for H2O (s)B) ΔG°f for H2O (l)C) ΔG°f for H2O (g)D) 1/2ΔG°f for O2 (g) plus ΔG°f for H2O (g)Answer: BTopic: Section 16.9 Standard Free Energies of Formation

67) Which of the following is zero at 25°C?A) ΔG°f for N2(g)B) ΔG°f for H2O (l)C) S° for N2 (g)D) S° for H2O (l)Answer: ATopic: Section 16.9 Standard Free Energies of Formation

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68) A positive value of ΔG°f for a solid compound at 25°C means theA) compound cannot exist at 25°C and 1 atm.B) compound must be a liquid or a gas at 25°C and 1 atm.C) process of forming the compound from the elements is exothermic.D) process of forming the compound from the stable elements at 25°C and 1 atm is nonspontaneous.Answer: DTopic: Section 16.9 Standard Free Energies of Formation

69) At 25°C, ΔG°f is -620 kJ/mol for SiCl4(g) and -592 kJ/mol for MgCl2(s). Calculate ΔG° for

the reaction, and determine if the reaction is spontaneous at 25°C if the pressure of SiCl4(g) is 1 atm.A) ΔG° = 28 kJ; the process is spontaneous.B) ΔG° = 28 kJ; the process is nonspontaneous.C) ΔG° = -564 kJ; the process is spontaneous.D) ΔG° = -564 kJ; the process is nonspontaneous.Answer: CTopic: Section 16.9 Standard Free Energies of Formation

70) Which of the following are unstable with respect to their constituent elements at 25°C?

A) C8H18(l), CH3OH(l)B) C8H18(l), C2H2(g)C) C2H2(g)D) CH3OH(l)Answer: CTopic: Section 16.9 Standard Free Energies of Formation

71) In general, as a reaction goes to equilibriumA) ΔG decreases.B) ΔG°f decreases.C) ΔG goes to zero.D) ΔG° decreases.Answer: CTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

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72) At 25°C, ΔG° = -198 kJ for the reaction, NO(g) + O3(g) NO⇌ 2(g) + O2(g). Calculate ΔG under the following conditions:

A) -159 kJB) -167 kJC) -198 kJD) -236 kJAnswer: ATopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

73) For a reaction at constant temperature, as Q increasesA) ΔG and ΔG° increase.B) ΔG and ΔG° decrease.C) ΔG increases, but ΔG° remains constant.D) ΔG decreases, but ΔG° remains constant.Answer: CTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

74) At high temperatures boron carbide vaporizes according to the equationB4C(s) 4 B(⇌ g) + C(s)

Which equation describes the relationship between ΔG° and ΔG for this reaction?A) ΔG = ΔG° + R T ln (pB ∙ [C]/[B4C])B) ΔG = ΔG° + R T ln pBC) ΔG = ΔG° + 4 R T ln pBD) ΔG = ΔG° - 4 R T ln pBAnswer: CTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

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75) At 2600 K, ΔG° = 775 kJ for the vaporization of boron carbide:B4C(s) 4 B(⇌ g) + C(s)

Find ΔG and determine if the process is spontaneous if the reaction vessel contains 4.00 mol B4C(s), 0.400 mol of C(s), and B(g) at a partial pressure of 1.0 × 10-5 atm. At this temperature, R T = 21.6 kJ.A) ΔG = -270 kJ; spontaneous.B) ΔG = -270 kJ; nonspontaneous.C) ΔG = -220 kJ; spontaneous.D) ΔG = -220 kJ; nonspontaneous.Answer: CTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

76) ΔG = ΔG° for a reactionA) if Q = K.B) if Q = 1.C) at STP.D) at the start of the reaction.Answer: BTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

77) What is the relationship between ΔG and the ΔG°F for the reaction below?

MgF2(s) → Mg2+(aq) + 2 F-(aq)

A) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)] - ΔG°f [MgF2 (s)]} + RT ln ([Mg2+]

[F-]2/[MgF2])

B) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)] - ΔG°f [MgF2 (s)]} + RT ln ([Mg2+] [F-])2)

C) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)]} + RT ln ([Mg2+] [F-]2)

D) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F- (aq)] - ΔG°f [MgF2 (s)]} + RT ln KspAnswer: BTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

78) If Q increasesA) ΔG increases and the reaction becomes more spontaneous.B) ΔG increases and the reaction becomes less spontaneous.C) ΔG decreases and the reaction becomes more spontaneous.D) ΔG decreases and the reaction becomes less spontaneous.Answer: BTopic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture

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79) What is the relationship between ΔG, Qp, and Kp for a reaction involving gases?A) ΔG = Qp/KpB) ΔG = Kp/QpC) ΔG = RTln(Qp/Kp)D) ΔG = RTln(Kp/Qp)Answer: CTopic: Section 16.11 Free Energy and Chemical Equilibrium

80) When equilibrium is reached at constant temperature and pressure,A) Q = 1.B) ΔG° = 0.C) S is maximized.D) G is minimized.Answer: DTopic: Section 16.11 Free Energy and Chemical Equilibrium

81) Calculate Ksp for PbI2 at 25°C based on the following data:

A) 4 × 10-31

B) 8 × 10-18

C) 9 × 10-9

D) 5 × 10-5Answer: CTopic: Section 16.11 Free Energy and Chemical Equilibrium

82) At high temperatures, boron carbide vaporizes according toB4C(s) 4 B(⇌ g) + C(s)

At 2500 K, the equilibrium pressure of B(g) is 0.0342 mm Hg over a mixture of 0.300 mol B4C(s) and 0.500 mol C(s). Calculate for this process.A) 832 kJB) 799 kJC) 281 kJD) 247 kJAnswer: ATopic: Section 16.11 Free Energy and Chemical Equilibrium

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83) Solid NaHCO3 is heated to 90°C. At equilibrium the total pressure of the gases produced is 0.545 atm. Calculate ΔG° at 90°C for the reaction

2 NaHCO3(s) Na⇌ 2CO3(s) + H2O(g) + CO2(g).A) -7.85 kJB) -3.67 kJC) +3.67 kJD) +7.85 kJAnswer: DTopic: Section 16.11 Free Energy and Chemical Equilibrium

84) What is K if ΔG° = -18.0 kJ for a reaction at 25°?A) 1.4 × 103

B) 1.2 × 102

C) 8.1 × 10-3

D) 7.3 × 10-4Answer: ATopic: Section 16.11 Free Energy and Chemical Equilibrium

85) If ΔG° is negative for a reaction,A) K < 0.B) K = 0.C) K is between 0 and 1.D) K > 1.Answer: DTopic: Section 16.11 Free Energy and Chemical Equilibrium

86) If ΔG° is positive for a reaction,A) K < 0.B) K = 0.C) K is between 0 and 1.D) K > 1.Answer: CTopic: Section 16.11 Free Energy and Chemical Equilibrium

87) For the following reaction find Kp at 25°C and indicate whether Kp should increase or decrease as the temperature rises.

NH4HS(s) H⇌ 2S(g) + NH3(g)ΔH° = 83.47 kJ and ΔG° = 17.5 kJ at 25°C.

A) Kp = 8.6 × 10-4 and Kp should increase as the temperature rises.

B) Kp = 8.6 × 10-4 and Kp should decrease as the temperature rises.

C) Kp = 1.2 × 103 and Kp should increase as the temperature rises.

D) Kp = 1.2 × 103 and Kp should decrease as the temperature rises.Answer: ATopic: Section 16.11 Free Energy and Chemical Equilibrium

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88) If ΔG is small and positive,A) the forward reaction is spontaneous and the system is far from equilibrium.B) the forward reaction is spontaneous and the system is near equilibrium.C) the reverse reaction is spontaneous and the system is far from equilibrium.D) the reverse reaction is spontaneous and the system is near equilibrium.Answer: DTopic: Section 16.11 Free Energy and Chemical Equilibrium

89) In figure (1) below argon atoms, represented by unshaded spheres, and neon atoms, represented by shaded spheres, are in separate compartments. Figure (2) shows the equilibrium state of the system after the stopcock separating the two compartments is opened. Assuming that argon and neon behave as ideal gases, what are the signs (+, -, or 0) of ΔH, ΔS, and ΔG for this process?

A) ΔH = +, ΔS = -, ΔG = +B) ΔH = 0, ΔS = +, ΔG = -C) ΔH = 0, ΔS = -, ΔG = +D) ΔH = -, ΔS = +, ΔG = -Answer: BTopic: Key Concept Problems

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90) In figure (1) below oxygen molecules, represented by unshaded spheres, and chlorine molecules, represented by shaded spheres, are in separate compartments. Figure (2) shows the equilibrium state of the system after the stopcock separating the two compartments is opened. Assuming the oxygen and the chlorine behave as ideal gases, what are the signs (+, -, or 0) of ΔH, ΔS, and ΔG for this process?

A) ΔH = +, ΔS = -, ΔG = +B) ΔH = 0, ΔS = +, ΔG = -C) ΔH = 0, ΔS = -, ΔG = +D) ΔH = -, ΔS = +, ΔG = -Answer: BTopic: Key Concept Problems

91) The figure represents the spontaneous deposition of iodine in which iodine vapor, I2(g), becomes crystalline iodine solid I2(s): I2(g): → I2(s). What are the signs (+ or -) of ΔH, ΔS, and ΔG for this process?

A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -, ΔG = -Answer: DTopic: Key Concept Problems

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92) The figure represents the spontaneous evaporation of nitrogen in which liquid nitrogen, N2(l), becomes gaseous nitrogen, N2(g): N2(l) → N2(g). What are the signs (+ or -) of ΔH, ΔS, and ΔG for this process?

A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -, ΔG = -Answer: BTopic: Key Concept Problems

93) An ideal gas is expanded at constant temperature. What are the signs (+, -, or 0) of ΔH, ΔS, and ΔG for this system?

A) ΔH = +, ΔS = -, ΔG = +B) ΔH = 0, ΔS = +, ΔG = -C) ΔH = 0, ΔS = -, ΔG = +D) ΔH = -, ΔS = +, ΔG = -Answer: BTopic: Key Concept Problems

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94) The figure above represents the nonspontaneous reaction O2(g) → 2O(g). What are the signs (+ or -) of ΔH, ΔS, and ΔG for this process?A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -, ΔG = -Answer: ATopic: Key Concept Problems

95) The figure above represents the reaction O2(g) → 2O(g), which is nonspontaneous at 25°C. How will the spontaneity of this reaction vary with temperature? This reaction is A) nonspontaneous at all temperatures.B) nonspontaneous at high temperatures and spontaneous at low temperatures.C) spontaneous at high temperatures and nonspontaneous at low temperatures.D) spontaneous at all temperatures.Answer: CTopic: Key Concept Problems

The figure below represents the spontaneous reaction of H2 (shaded spheres) with O2 (unshaded spheres) to produce gaseous H2O.

96) What are the signs (+, or -) of ΔH, ΔS, and ΔG for this process?A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -,ΔG = -Answer: DTopic: Key Concept Problems

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97) How will the spontaneity of this reaction vary with temperature? This reaction is A) nonspontaneous at all temperatures.B) nonspontaneous at high temperatures and spontaneous at low temperatures.C) spontaneous at high temperatures and nonspontaneous at low temperatures.D) spontaneous at all temperatures.Answer: BTopic: Key Concept Problems

Consider the reaction 2A(g) A⇌ 2(g). The following pictures represent two possible initial states and the equilibrium state of the system.

98) For initial state 1 what is the relationship between the reaction quotient, Qp, and the equilibrium constant, Kp?A) Qp < KpB) Qp = Kp = 1C) Qp = Kp ≠ 1D) Qp > KpAnswer: ATopic: Key Concept Problems

99) For initial state 2 what is the relationship between the reaction quotient, Qp, and the equilibrium constant, Kp?A) Qp < KpB) Qp = Kp = 1C) Qp = Kp ≠ 1D) Qp > KpAnswer: DTopic: Key Concept Problems

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100) What are the signs (+ or -) of ΔH, ΔS, and ΔG when the system spontaneously goes from initial state 1 to the equilibrium state?A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -, ΔG = -Answer: DTopic: Key Concept Problems

101) What are the signs (+ or -) of ΔH, ΔS, and ΔG when the system spontaneously goes from initial state 2 to the equilibrium state?A) ΔH = +, ΔS = +, ΔG = +B) ΔH = +, ΔS = +, ΔG = -C) ΔH = -, ΔS = -, ΔG = +D) ΔH = -, ΔS = -, ΔG = -Answer: BTopic: Key Concept Problems

Consider the following gas-phase reaction of A2 (shaded spheres) and B2 (unshaded spheres):A2(g) + B2(g) 2 AB(⇌ g) ΔG ° = +25 kJ

102) Which of the above reaction mixtures has the least spontaneous forward reaction?A) (1)B) (2)C) (3)D) (4)Answer: DTopic: Key Concept Problems

103) Which of the above reaction mixtures has the most spontaneous forward reaction?A) (1)B) (2)C) (3)D) (4)Answer: ATopic: Key Concept Problems

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104) Which of the above reaction mixtures is ΔG of reaction = ΔG ° ?A) (1)B) (2)C) (3)D) (4)Answer: CTopic: Key Concept Problems

105) According to the diagram above, the forward reaction isA) nonspontaneous at d and e, and spontaneous at f.B) nonspontaneous at d, at equilibrium at e, and spontaneous at f.C) spontaneous at d, at equilibrium at e, and nonspontaneous at f.D) spontaneous at d, e, and f.Answer: CTopic: Key Concept Problems

106) According to the diagram above,A) ΔG° is positive and the equilibrium composition is rich in products.B) ΔG° is positive and the equilibrium composition is rich in reactants.C) ΔG° is negative and the equilibrium composition is rich in products.D) ΔG° is negative and the equilibrium composition is rich is reactants.Answer: BTopic: Key Concept Problems

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107) According to this diagram,A) ΔG° is positive and is equal to a - b.B) ΔG° is positive and is equal to b - c.C) ΔG° is negative and is equal to b - c.D) ΔG° is negative and is equal to a - c.Answer: ATopic: Key Concept Problems

108) The following pictures represent three equilibrium mixtures for the interconversion of A, B, and C molecules (unshaded spheres) into X, Y, and Z molecules (shaded spheres), respectively. What is the sign of ΔG ° for each of the three reactions?

A) ΔG °(1) = -; ΔG °(2) = +; ΔG °(3) = 0B) ΔG °(1) = -; ΔG °(2) = 0; ΔG °(3) = +C) ΔG °(1) = 0; ΔG °(2) = -; ΔG °(3) = +D) ΔG °(1) = +; ΔG °(2) = 0; ΔG °(3) = -Answer: BTopic: Key Concept Problems

16.2 Algorithmic Questions

1) The entropy change associated with the expansion of one mole of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature is given by the equation, ΔS = R ln (Vf/Vi). What is the entropy change associated with the expansion of three moles of an ideal gas from an initial volume of Vi to a final volume of Vf at constant temperature? A) ΔS = R ln (Vf/Vi)B) ΔS = 3 mol × R ln (Vf/Vi)

C) ΔS = R ln (Vf × 23/Vi)D) ΔS = R ln (Vf × 3!/Vi)Answer: BTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

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2) What is the entropy change associated with the expansion of one mole of an ideal gas from an initial volume of V to a final volume of 4.50 V at constant temperature? A) ΔS = 4.50 R ln (Vf/Vi)B) ΔS = -4.50 R ln (Vf/Vi)C) ΔS = R ln 4.50D) ΔS = -R ln 4.50Answer: CTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

3) Predict the sign of ΔS for each of the following systems, which occur at constant temperatureI. The volume of 2.0 moles of O2(g) increases from 44 L to 52 L.II. The pressure of 2.0 moles of O2(g) increases from 1.0 atm to 1.2 atm. A) I: ΔS = negative; II: ΔS = negativeB) I: ΔS = negative; II: ΔS = positiveC) I: ΔS = positive; II: ΔS = negativeD) I: ΔS = positive; II: ΔS = positiveAnswer: CTopic: Section 16.3 Entropy and ProbabilityAlgo. Option: algorithmic

4) Under which of the following conditions would one mole of He have the highest entropy, S? A) 17°C and 15 LB) 127°C and 15 LC) 17°C and 25 LD) 127°C and 25 LAnswer: DTopic: Section 16.4 Entropy and TemperatureAlgo. Option: algorithmic

5) Which one of the following would be expected to have the lowest standard molar entropy, S°, at 25°C?A) C10H22(s)B) C10H22(l)C) C14H30(s)

D) C14 OH(l)Answer: ATopic: Section 16.5 Standard Molar Entropies and Standard Entropies of ReactionAlgo. Option: algorithmic

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16.3 Short Answer Questions

1) Chemical and physical changes can be classified as spontaneous or nonspontaneous. At 25°C and 1 atm pressure the decomposition of water into hydrogen and oxygen is classified as ________, and the melting of ice is classified as ________.Answer: nonspontaneous, spontaneousTopic: Section 16.1 Spontaneous Processes

2) The sign (+ or –) of ∆H is ________ and the sign (+ or –) of ∆S is ________ for the evaporation of water.Answer: +, +Topic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review

3) A 1.0 mole sample of gas at STP has a ________ entropy than 1.0 mole of gas at 273 K and 835 mm Hg.Answer: higherTopic: Section 16.3 Entropy and Probability

4) The entropy of water at 25° is ________ than the entropy of water at 35°C.Answer: less thanTopic: Section 16.4 Entropy and Temperature

5) Standard molar entropies, S°, in J/K∙mol, are given below each reactant and product in the reaction shown below. The standard entropy of reaction, ∆S°, for this reaction is ________ J.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)186.2 205.0 213.6 69.9

Answer: – 242.8Topic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction

6) A reaction for which ∆H° = + 98.8 kJ and ∆S° = + 141.5 J/K is ________ (spontaneous or nonspontaneous) at low temperatures and ________ (spontaneous or nonspontaneous) at high temperatures.Answer: nonspontaneous, spontaneousTopic: Section 16.7 Free Energy

7) A reaction has ∆G° = + 21.5 kJ/mol, ∆H° = + 25.0 kJ/mol, and ∆S° = + 15.0 J/mol∙K can become spontaneous at a temperature of ________ K.Answer: 1670Topic: Section 16.7 Free Energy

8) Acetylene, C2H2, has a standard enthalpy of formation, ∆H° = 226.7 kJ/mol, and a standard entropy change for its formation from its elements, ∆S° = 58.8 J/K∙mol. The standard free energy of formation of acetylene is ________ kJ/mol.Answer: + 209.2Topic: Section 16.8 Standard Free-Energy Changes for Reactions

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9) Standard free energies of formation, ∆G°, in kJ/mol, are given below each reactant and product in the reaction shown below. The standard free energy of reaction, ∆G°, for this reaction is ________ kJ.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)– 50.8 0 – 394.4 – 237.2

Answer: – 818.0Topic: Section 16.9 Standard Free Energies of Formation

10) The standard free energy for a reaction is ∆G° = – 33.0 kJ. At 25°C the equilibrium constant for this reaction , Kp = ________.

Answer: 6.09 × 105Topic: Section 16.11 Free Energy and Chemical Equilibrium

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