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: C Topic: Section 16.1 Spontaneous Processes 2) Which forward reaction is a nonspontaneous process? A) the expansion of a gas into a vacuum B) N2(g) + 3 H2(g) ⇌ 2 NH3(g) if PH₂ = PN₂ = 1 atm, PNH₃ = 0, and Kp = 4 × 105 C) 2 NH3(g) ⇌ N2(g) + 3 H2(g) if PNH₃ = 1 atm, PH₂ = PN₂ = 0, and Kp = 2 × 10-6 D) none of the above Answer: D Topic: 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 NH3(g) Kp = 4 × 105 A) adding a catalyst B) adding more H2 C) adding more N2 D) reducing the volume Answer: A Topic: 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 humidity II. 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: A Topic: Section 16.1 Spontaneous Processes 1 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 these B) I C) I and II D) I, II, and III Answer: B Topic: Section 16.1 Spontaneous Processes 6) Which of the following processes are spontaneous? I. dissolving more solute in an unsaturated solution II. dissolving more solute in a saturated solution III. dissolving more solute in a supersaturated solution A) none of these B) I C) I and II D) I, II, and III Answer: B Topic: Section 16.1 Spontaneous Processes 7) Which of the following processes is spontaneous? A) a mixture of two gases separating into pure compounds B) reaction of sodium with oxygen C) precipitation of solute from a saturated solution D) water flowing uphill Answer: B Topic: Section 16.1 Spontaneous Processes 8) Entropy is a measure of A) free energy. B) the heat of a reaction. C) molecular randomness. D) the rate of a reaction. Answer: C Topic: 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 steam B) reaction of magnesium with oxygen to form magnesium oxide C) reaction of nitrogen and hydrogen to form ammonia D) sublimation of dry ice Answer: D Topic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review 2 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: A Topic: 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: C Topic: 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: B Topic: 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) ⇌ N2O4(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: A Topic: Section 16.2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review 3 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.3 Entropy and Probability 15) Which electron on an atom of copper would have the highest value of W in the Boltzmann formula? A) 3s B) 3d C) 4s D) 4p Answer: B Topic: Section 16.3 Entropy and Probability Algo. Option: algorithmic 4 . S = k ln W? A) a fraction indicating the probability of obtaining a result B) a random number C) the number of ways of obtaining the state D) the work times Avogadro's number Answer: C Topic: Section 16. S = k ln W? A) the degeneracy of the state B) the equilibrium constant for the process C) the universal gas constant divided by Avogadro's number D) the universal gas constant times Avogadro's number Answer: C Topic: 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: D Topic: Section 16.14) What is W in Boltzmann's formula. 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: B Topic: Section 16.3 Entropy and Probability 17) What is k in Boltzmann's formula. ΔS = R ln (Vf/Vi). II.50 Answer: C Topic: Section 16.2 atm. II: ΔS= positive C) I: ΔS= positive. The pressure of 2. A) I: ΔS= negative.0 atm to 1. II: ΔS= negative B) I: ΔS= negative.3 Entropy and Probability Algo.50 D) ΔS = -R ln 2. II: ΔS= negative D) I: ΔS= positive. Option: algorithmic 20) Predict the sign of ΔS for each of the following processes.50 R ln (Vf/Vi) B) ΔS = -2. Option: algorithmic 5 . which occur at constant temperature. II: ΔS= positive Answer: C Topic: Section 16. I.3 Entropy and Probability Algo.50 R ln (Vf/Vi) C) ΔS = R ln 2.19) 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 = 2.0 moles of O2(g) increases from 44 L to 52 L. The volume of 2.0 moles of O2(g) increases from 1. The separation of gaseous molecules of UF6. corresponding to atom A facing in either the positive or negative direction along the x-. B) W = 1. C) R ln 6! D) 0. –25°C) B) CH3OH(s. Answer: B Topic: Section 16.4 Entropy and Temperature 6 . D) k = 1. Also assume that each molecule can only have one of six possible orientations. y-. the molar entropy at absolute zero should be A) R ln 6. 25°C) Answer: A Topic: Section 16. AB. A perfect crystal has a molar entropy of 0 at absolute zero because A) W = 0.3 Entropy and Probability 22) The Boltzmann formula is S = k ln W.21) Assume a heteronuclear diatomic molecule. A) ΔS is negative for I and negative for II. The dissolving of I2(s) in CCl4(l). D) ΔS is positive for I and positive for II. forms a one-dimensional crystal by lining up along the x-axis. B) R ln 66.3 Entropy and Probability 23) What is the sign of △S for each of the following processes? I. Answer: B Topic: Section 16. into 238UF6 and 235UF6 at constant temperature and pressure. If the molecules are arranged randomly in the six directions. C) W = NA. B) ΔS is negative for I and positive for II. II. or z-axis. 15°C) D) CH3OH(l. C) ΔS is positive for I and negative for II.3 Entropy and Probability 24) Which has the lowest entropy? A) CH3OH(s. –15°C) C) CH3OH(l. Answer: A Topic: Section 16. and III D) I. IV.4 Entropy and Temperature 28) Which of the following statements must be true for the entropy of a pure solid to be zero? I. 0. II. 0°C) → H2O (l. 25°C) C) H2O (g. The solid must be crystalline. 100°C) → H2O (g. H2O(l) at 0°C.25) Which has the highest entropy in each set? I. The solid must be perfectly ordered.58 atm II. Answer: C Topic: Section 16. III.4 Entropy and Temperature 26) Which provides the greatest increase in entropy? A) H2O (s. 0. H2O(l). and IV Answer: C Topic: Section 16. The temperature must be 0 K. D) the entropy of the universe increases for any spontaneous process.1°C) D) H2O (l. II. 0°C) B) H2O (l. C) the entropy of a perfectly ordered. 100°C) Answer: D Topic: Section 16.4 Entropy and Temperature 27) According to the third law of thermodynamics. A) energy is conserved in any transformation of matter. H2O(l) at 100°C (all at 1. H2O(g) at 0. H2O(s). H2O(l) at 25°C. 4. S? A) 27°C and 25 L B) 137°C and 25 L C) 27°C and 35 L D) 137°C and 35 L Answer: D Topic: Section 16. II. crystalline substance is zero at 0 Kelvin. 0°C) → H2O (l.4 Entropy and Temperature Algo.0 atm pressure) A) H2O(l) in set I and H2O(l) at 0°C in set II B) H2O(s) in set I and H2O(l) at 100°C in set II C) H2O(g) in set I and H2O(l) at 0°C in set II D) H2O(g) in set I and H2O(l) at 100°C in set II Answer: D Topic: Section 16.4 Entropy and Temperature 29) Under which of the following conditions would one mole of Ne have the highest entropy. not amorphous. The solid must be an element. B) the entropy increases for any spontaneous process. Option: algorithmic 7 .1°C) → H2O (s. A) I B) I and II C) I.1°C. III. Option: algorithmic 8 . Answer: C Topic: Section 16. at 25°C?" A) C8H18(s) B) C8H18(l) C) C12H26(s) D) C12H26(l) Answer: A Topic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction 34) Which one of the following has the lowest standard molar entropy.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: D Topic: Section 16.5 Standard Molar Entropies and Standard Entropies of Reaction Algo. Answer: C Topic: 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) C2H2 B) CH2CH2 C) CH3CH3 D) All have the same entropy.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.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: B Topic: Section 16. S°. 5 J/K D) 636.5 J/K C) 15.5 J/K Answer: B Topic: Section 16.35) Calculate ΔS° for the following reaction.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.6 J/K∙mol Answer: C Topic: Section 16. N2(g) + 2 O2(g) → 2 NO2(g) A) -156.3 J/K∙mol D) 384.2 J/K C) -40. N2(g) + 3 H2(g) → 2 NH3(g) A) 61. A) -116.5 Standard Molar Entropies and Standard Entropies of Reaction 36) ΔS° = – 198.5 J/K B) -121.7 J/K for the reaction shown below.7 J/K∙mol B) 123.5 Standard Molar Entropies and Standard Entropies of Reaction 9 .8 J/K Answer: C Topic: Section 16.7 J/K B) -81.4 J/K∙mol C) 192. Calculate S° for NH3(g).5 J/K D) 86. 02 J/(mol ∙ K) D) 350. Given that ΔS° = 104.6 Entropy and the Second Law of Thermodynamics 40) Which of the following is a criterion for spontaneity that holds for any process? A) ΔG < 0 B) ΔG > 0 C) ΔStotal < 0 D) ΔStotal > 0 Answer: D Topic: 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 thermodynamics B) the second law of thermodynamics C) the third law of thermodynamics D) both the second and third laws of thermodynamics Answer: B Topic: Section 16. Answer: B Topic: Section 16. A) 70.58 J/K for the dissociation of one mole of Br2(g) into Br(g) at 25°C.08 J/(mol ∙ K) C) 175.44 J/(mol ∙ K) B) 140.6 Entropy and the Second Law of Thermodynamics 41) According to the second law of thermodynamics. C) ΔSsys = 0. D) ΔSsys = 0 and ΔSsurr = 0. it is necessary that A) ΔSsys = ΔSsurr. find the standard molar entropy for Br(g) at 25°C.46 J/(mol ∙ K) at 25°C. all reactions proceed spontaneously in the direction that increases the entropy of the A) surroundings. B) ΔSsys = .04 J/(mol ∙ K) Answer: C Topic: Section 16.38) The standard molar entropy for Br2(g) is 245. B) system. C) system – surroundings D) system + surroundings Answer: D Topic: Section 16.6 Entropy and the Second Law of Thermodynamics 10 .6 Entropy and the Second Law of Thermodynamics 42) For a process to be at equilibrium.ΔSsurr. D) both the energy and the entropy of the system and surroundings decrease. ΔStotal = -4. Answer: C Topic: Section 16. B) energy is conserved and the entropy of the system and surroundings increases.2 J/K Assuming that the surroundings can be considered a large heat reservoir at 25°C. ΔS° = -4.2 J/K. | ΔSpenny | > | ΔSwater | D) the entropy of both the penny and the water increases. Answer: B Topic: Section 16. not spontaneous C) ΔSsurr = -0. not spontaneous B) ΔSsurr = 0.21 kJ.43) For a spontaneous process A) energy and entropy are conserved. A) the entropy of the water evaporated decreases and the entropy of the body decreases.7 J/K.9 J/K. spontaneous D) ΔSsurr = -0.7 J/K. D) the entropy of the water evaporated increases and the entropy of the body increases. not spontaneous Answer: B Topic: Section 16. calculate ΔSsurr and ΔStotal for the process at 25°C and 1 atm pressure. B) the entropy of the water evaporated decreases and the entropy of the body increases. Δtotal = 0.6 Entropy and the Second Law of Thermodynamics 44) For the process CaCO3(calcite) → CaCO3(aragonite) ΔH° = -0.6 Entropy and the Second Law of Thermodynamics 11 .6 Entropy and the Second Law of Thermodynamics 45) During perspiration. | ΔSpenny | < | ΔSwater | C) the decrease in entropy of the penny is more than the increase in entropy of the water. C) the entropy of the water evaporated increases and the entropy of the body decreases. ΔStotal = -4.5 J/K.9 J/K. A) the decrease in entropy of the penny is equal to the increase in entropy of the water. Is the process spontaneous at 25°C and 1 atm pressure? A) ΔSsurr = 4. Answer: B Topic: Section 16. ΔStotal = -3.7 J/K. | ΔSpenny | = | ΔSwater | B) the decrease in entropy of the penny is less than the increase in entropy of the water. C) the energy of the system and the surroundings decreases and the entropy of the system and surroundings increases. Assuming negligible heat loss to the atmosphere and the cup.6 Entropy and the Second Law of Thermodynamics 46) A hot penny is dropped into cold water inside a polystyrene foam cup. 7 Free Energy 12 . D) Entropy is based on probability and is therefore less reliable. B) ΔS is negative.7 Free Energy 51) For a particular process ΔG is less than ΔH. Answer: B Topic: Section 16. B) All reactions for which ∆S < 0 are spontaneous. D) ΔS is negative if ΔH is positive and ΔS is positive if ΔH is negative. what reaction conditions must be satisfied for the sign of ΔG to be used as a criterion for spontaneity? A) constant volume and pressure B) constant temperature and pressure C) constant temperature and volume D) constant volume only Answer: B Topic: Section 16. C) All reactions for which ∆G < 0 are spontaneous. the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K.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.7 Free Energy 49) Other than only PV work. which statement is true? A) All reactions for which ∆H < 0 are spontaneous. D) All reactions for which K < 1 are spontaneous. Therefore A) ΔS is positive. Answer: A Topic: Section 16. B) ΔG involves thermodynamic functions of the system only.7 Free Energy 50) For the reaction 3 C2H2(g) → C6H6(l) at 25°C. Calculate the standard free energy change at 25°C. A) 948 kJ B) -503 kJ C) -618 kJ D) -1061 kJ Answer: B Topic: Section 16. Answer: C Topic: Section 16. C) ΔS is zero.47) At constant pressure and temperature. C) Free energy is easier to understand than entropy. 23 JK-1mol-1 at 25°C. Answer: A Topic: Section 16. D) ΔH° is positive and ΔS° is definitely negative.91 kJ/mol and ΔS°vap = 93. Answer: A Topic: Section 16.52) For a particular process.7 Free Energy 54) At 25°C. Based on these data. ΔH° = 1. C) ΔH° is positive but the sign of ΔS° cannot be determined from this information. C) diamond is more stable than graphite below 290°C and graphite is more stable than diamond above 290°C. B) ΔH° is negative and ΔS° is definitely negative.7 Free Energy 13 . D) ΔS = ΔG/T.8 g/L at 100°C.7 Free Energy 55) For bromine. B) diamond is more stable than graphite at all temperatures at 1 atm. What is the normal boiling point for bromine? A) 25°C B) 58°C C) 124°C D) 332°C Answer: B Topic: Section 16. ΔG = ΔH at a given temperature and pressure.363 J/K for the transition C(graphite) → C(diamond) Based on these data A) graphite cannot be converted to diamond at 1 atm pressure. A) ΔS is positive if ΔH is positive and negative is ΔH is negative.895 kJ and ΔS° = -3. Therefore. ΔH°vap = 30. Answer: C Topic: Section 16.7 Free Energy 53) The solubility of manganese(II) fluoride in water is 6. what is the sign of ΔH° and ΔS° for the process below? MnF2(s) ⇌ Mn2+(aq) + 2 F-(aq) A) ΔH° is negative but the sign of ΔS° cannot be determined from this information. D) graphite is more stable than diamond below 290°C and diamond is more stable than graphite above 290°C. C) ΔS is zero. B) ΔS is negative if ΔH is positive and positive if ΔH is negative.6 g/mL at 40°C and 4. 4 J/K A) The reaction is spontaneous at all temperatures. B) The reaction is spontaneous at high temperatures. B) ΔH is positive and TΔS > ΔH. C) The reaction is spontaneous at low temperatures. Which reaction could go in the reverse direction at high temperature? A) I B) II C) III D) I and II Answer: C Topic: Section 16. C) ΔH is positive and TΔS < ΔH.7 Free Energy 14 . and ΔS at 25°C are shown below for three reactions. Answer: C Topic: Section 16. C) ΔG° = -166 kJ. A) ΔH is positive and TΔS = ΔH. A) ΔG° = -332 kJ. D) ΔH is negative and TΔS is positive. the equilibrium composition should favor products.7 Free Energy 59) For the evaporation of water during perspiration on a hot.7 kJ and ΔS° = -150. ΔH. the equilibrium composition should favor reactants.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. B) ΔG° = -332 kJ.7 Free Energy 57) Which statement is true about the formation of CaCO3(s) from CaO(s) and CO2(g) at 1. D) ΔG° = -166 kJ.7 Free Energy 58) The signs of ΔG. the equilibrium composition should favor reactants. Answer: B Topic: Section 16. Answer: C Topic: Section 16.00 atm? CaO(s) + CO2(g) → CaCO3(s) ΔH° = -178. the equilibrium composition should favor products. dry day. D) The reaction is not spontaneous at any temperature. A) -4.7 kJ B) -72.8 kJ D) -532. 2 NH3(g) → N2(g) + 3 H2(g) A) 0.60) For the reaction below ∆G° = + 33. C) the separate reactants in their standard states are completely converted to separate products in their standard states.7 J/K. D) When a reaction reaches equilibrium.464 K B) 166 K C) 298 K D) 464 K Answer: D Topic: Section 16. Estimate the temperature at which this reaction becomes spontaneous.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. the faster the reaction.7 Free Energy 61) Calculate the standard free energy change at 25°C for the reaction 2 NO(g) + O2(g) → 2 NO2(g).0 kJ. Answer: C Topic: Section 16. C) The standard state for solutes is the pure solute at 1 atm. B) The larger ΔG°. D) the spontaneous reaction occurs. and ∆S° = + 198.8 Standard Free-Energy Changes for Reactions 15 .6 kJ Answer: B Topic: Section 16. ΔY° for a reaction refers to the change in Y for the process in which A) the mixed reactants at 1 atm go to equilibrium at 1 atm. ΔG° = 0. |ΔG| decreases.8 Standard Free-Energy Changes for Reactions 62) For any thermodynamic function Y. B) the separate reactants at 1 atm go to equilibrium at 1 atm. ∆H° = + 92.2 kJ. Answer: A Topic: Section 16.6 kJ C) -157. 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. 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.9 Standard Free Energies of Formation 16 .8 Standard Free-Energy Changes for Reactions 65) Calculate the standard free energy for the reaction given. 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. Answer: C Topic: 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: A Topic: Section 16.8 kJ C) -1404.8 kJ Answer: C Topic: 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: B Topic: Section 16. 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. 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l) A) -465.8 kJ D) -2069.2 kJ B) -797. 9 Standard Free Energies of Formation 71) In general. the process is spontaneous. Calculate ΔG° for the reaction. B) ΔG° = 28 kJ.10 Free-Energy Changes and the Composition of the Reaction Mixture 17 . C) process of forming the compound from the elements is exothermic. Answer: C Topic: Section 16. C2H2(g) C) C2H2(g) D) CH3OH(l) Answer: C Topic: Section 16.68) A positive value of ΔG°f for a solid compound at 25°C means the A) compound cannot exist at 25°C and 1 atm. D) process of forming the compound from the stable elements at 25°C and 1 atm is nonspontaneous. the process is spontaneous. B) compound must be a liquid or a gas at 25°C and 1 atm. and determine if the reaction is spontaneous at 25°C if the pressure of SiCl4(g) is 1 atm. Answer: D Topic: Section 16. C) ΔG° = -564 kJ.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). C) ΔG goes to zero. as a reaction goes to equilibrium A) ΔG decreases. the process is nonspontaneous. B) ΔG°f decreases. Answer: C Topic: 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). D) ΔG° decreases. CH3OH(l) B) C8H18(l). D) ΔG° = -564 kJ. A) ΔG° = 28 kJ. the process is nonspontaneous. Answer: C Topic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture 73) For a reaction at constant temperature. but ΔG° remains constant.10 Free-Energy Changes and the Composition of the Reaction Mixture 74) At high temperatures boron carbide vaporizes according to the equation B4C(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 pB C) ΔG = ΔG° + 4 R T ln pB D) ΔG = ΔG° .4 R T ln pB Answer: C Topic: Section 16. but ΔG° remains constant. NO(g) + O3(g) ⇌ NO2(g) + O2(g). D) ΔG decreases. C) ΔG increases. Calculate ΔG under the following conditions: A) -159 kJ B) -167 kJ C) -198 kJ D) -236 kJ Answer: A Topic: Section 16. B) ΔG and ΔG° decrease. ΔG° = -198 kJ for the reaction.10 Free-Energy Changes and the Composition of the Reaction Mixture 18 . as Q increases A) ΔG and ΔG° increase.72) At 25°C. C) ΔG = -220 kJ. R T = 21.10 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. B) if Q = 1.ΔG°f [MgF2 (s)]} + RT ln ([Mg2+] [F-]2/ [MgF2]) B) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F. C) at STP.75) At 2600 K.400 mol of C(s).6 kJ.(aq)] .10 Free-Energy Changes and the Composition of the Reaction Mixture 78) If Q increases A) ΔG increases and the reaction becomes more spontaneous. D) ΔG decreases and the reaction becomes less spontaneous. nonspontaneous. C) ΔG decreases and the reaction becomes more spontaneous. B) ΔG = -270 kJ.ΔG°f [MgF2 (s)]} + RT ln ([Mg2+] [F-])2) C) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F.00 mol B4C(s). 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. At this temperature. spontaneous.ΔG°f [MgF2 (s)]} + RT ln Ksp Answer: B Topic: Section 16. Answer: C Topic: Section 16.(aq)]} + RT ln ([Mg2+] [F-]2) D) ΔG = {ΔG°f [Mg2+ (aq)] + 2 ΔG°f [F. and B(g) at a partial pressure of 1. spontaneous.(aq)] . D) ΔG = -220 kJ. 0.0 × 10-5 atm. Answer: B Topic: Section 16.10 Free-Energy Changes and the Composition of the Reaction Mixture 19 . nonspontaneous.(aq)] . D) at the start of the reaction. B) ΔG increases and the reaction becomes less spontaneous.10 Free-Energy Changes and the Composition of the Reaction Mixture 76) ΔG = ΔG° for a reaction A) if Q = K. Answer: B Topic: Section 16. A) ΔG = -270 kJ. boron carbide vaporizes according to B4C(s) ⇌ 4 B(g) + C(s) At 2500 K. C) S is maximized. the equilibrium pressure of B(g) is 0. Calculate for this process. A) 832 kJ B) 799 kJ C) 281 kJ D) 247 kJ Answer: A Topic: Section 16.79) What is the relationship between ΔG. and Kp for a reaction involving gases? A) ΔG = Qp/Kp B) ΔG = Kp/Qp C) ΔG = RTln(Qp/Kp) D) ΔG = RTln(Kp/Qp) Answer: C Topic: Section 16.0342 mm Hg over a mixture of 0. B) ΔG° = 0. A) Q = 1. Answer: D Topic: Section 16. D) G is minimized.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 × 10-5 Answer: C Topic: Section 16.11 Free Energy and Chemical Equilibrium 80) When equilibrium is reached at constant temperature and pressure. Qp.300 mol B4C(s) and 0.11 Free Energy and Chemical Equilibrium 20 .500 mol C(s).11 Free Energy and Chemical Equilibrium 82) At high temperatures. NH4HS(s) ⇌ H2S(g) + NH3(g)ΔH° = 83. C) K is between 0 and 1. A) -7.6 × 10-4 and Kp should increase as the temperature rises.67 kJ C) +3.2 × 103 and Kp should increase as the temperature rises.4 × 103 B) 1. B) K = 0. C) Kp = 1. Calculate ΔG° at 90°C for the reaction 2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g).11 Free Energy and Chemical Equilibrium 21 . D) K > 1.11 Free Energy and Chemical Equilibrium 85) If ΔG° is negative for a reaction.2 × 103 and Kp should decrease as the temperature rises.1 × 10-3 D) 7. Answer: C Topic: Section 16. B) Kp = 8.67 kJ D) +7.2 × 102 C) 8. Answer: D Topic: Section 16. C) K is between 0 and 1.11 Free Energy and Chemical Equilibrium 86) If ΔG° is positive for a reaction. A) K < 0.11 Free Energy and Chemical Equilibrium 84) What is K if ΔG° = -18. At equilibrium the total pressure of the gases produced is 0.85 kJ B) -3.545 atm.5 kJ at 25°C.6 × 10-4 and Kp should decrease as the temperature rises.0 kJ for a reaction at 25°? A) 1. D) K > 1.3 × 10-4 Answer: A Topic: Section 16. B) K = 0.85 kJ Answer: D Topic: Section 16.47 kJ and ΔG° = 17. D) Kp = 1. A) K < 0. Answer: A Topic: Section 16.83) Solid NaHCO3 is heated to 90°C. A) Kp = 8.11 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. ΔG = C) ΔH = 0. ΔS = +. what are the signs (+. ΔG = Answer: B Topic: Key Concept Problems 22 . and ΔG for this process? A) ΔH = +. D) the reverse reaction is spontaneous and the system is near equilibrium. -.11 Free Energy and Chemical Equilibrium 89) In figure (1) below argon atoms. or 0) of ΔH. ΔG = + D) ΔH = -. ΔG = + B) ΔH = 0. are in separate compartments. and neon atoms. Answer: D Topic: Section 16. A) the forward reaction is spontaneous and the system is far from equilibrium. ΔS.88) If ΔG is small and positive. represented by shaded spheres. Assuming that argon and neon behave as ideal gases. ΔS = -. ΔS = +. Figure (2) shows the equilibrium state of the system after the stopcock separating the two compartments is opened. 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. ΔS = -. represented by unshaded spheres. what are the signs (+. ΔS. represented by shaded spheres. and chlorine molecules. becomes crystalline iodine solid I2(s): I2(g): → I2(s). ΔG = + B) ΔH = 0. -. ΔS = -. Figure (2) shows the equilibrium state of the system after the stopcock separating the two compartments is opened. ΔS = +. ΔS = -. are in separate compartments. ΔG = + D) ΔH = -. ΔG = Answer: B Topic: Key Concept Problems 91) The figure represents the spontaneous deposition of iodine in which iodine vapor. ΔG = + B) ΔH = +. ΔS = -. represented by unshaded spheres. ΔS. ΔG = + D) ΔH = -. or 0) of ΔH. ΔG = C) ΔH = -.90) In figure (1) below oxygen molecules. ΔG = C) ΔH = 0. ΔS = +. ΔS = +. ΔS = +. and ΔG for this process? A) ΔH = +. I2(g). ΔG = Answer: D Topic: Key Concept Problems 23 . What are the signs (+ or -) of ΔH. and ΔG for this process? A) ΔH = +. Assuming the oxygen and the chlorine behave as ideal gases. ΔS = -. ΔG = Answer: B Topic: Key Concept Problems 93) An ideal gas is expanded at constant temperature. ΔG = C) ΔH = 0. and ΔG for this system? A) ΔH = +. ΔS = +. N2(g): N2(l) → N2(g). or 0) of ΔH. ΔS. N2(l). ΔG = + B) ΔH = 0. ΔS = +. ΔG = C) ΔH = -. ΔS = -. ΔS = +. -. ΔS = -. What are the signs (+ or -) of ΔH. ΔG = + B) ΔH = +. ΔG = Answer: B Topic: Key Concept Problems 24 . becomes gaseous nitrogen.92) The figure represents the spontaneous evaporation of nitrogen in which liquid nitrogen. ΔG = + D) ΔH = -. What are the signs (+. ΔS = -. ΔS = +. ΔS = -. and ΔG for this process? A) ΔH = +. ΔG = + D) ΔH = -. ΔS. and ΔG for this process? A) ΔH = +. ΔG = + B) ΔH = +.ΔG = Answer: D Topic: Key Concept Problems 25 . which is nonspontaneous at 25°C. ΔS = -. 96) What are the signs (+. ΔS = +. ΔG = + D) ΔH = -. ΔS. What are the signs (+ or -) of ΔH. ΔS = -. ΔS. ΔS = +. or -) of ΔH. ΔG = Answer: A Topic: Key Concept Problems 95) The figure above represents the reaction O2(g) → 2O(g). ΔS = +. ΔG = C) ΔH = -. ΔS = -. Answer: C Topic: Key Concept Problems The figure below represents the spontaneous reaction of H2 (shaded spheres) with O2 (unshaded spheres) to produce gaseous H2O. ΔG = C) ΔH = -. and ΔG for this process? A) ΔH = +. ΔS = +. D) spontaneous at all temperatures. 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. ΔS = -. ΔG = + D) ΔH = -. C) spontaneous at high temperatures and nonspontaneous at low temperatures.94) The figure above represents the nonspontaneous reaction O2(g) → 2O(g). ΔG = + B) ΔH = +. Kp? A) Qp < Kp B) Qp = Kp = 1 C) Qp = Kp ≠ 1 D) Qp > Kp Answer: A Topic: Key Concept Problems 99) For initial state 2 what is the relationship between the reaction quotient. 98) For initial state 1 what is the relationship between the reaction quotient. C) spontaneous at high temperatures and nonspontaneous at low temperatures.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. The following pictures represent two possible initial states and the equilibrium state of the system. and the equilibrium constant. and the equilibrium constant. Qp. Answer: B Topic: Key Concept Problems Consider the reaction 2A(g) ⇌ A2(g). Qp. D) spontaneous at all temperatures. Kp? A) Qp < Kp B) Qp = Kp = 1 C) Qp = Kp ≠ 1 D) Qp > Kp Answer: D Topic: Key Concept Problems 26 . ΔS = -. ΔG = + D) ΔH = -. ΔG = C) ΔH = -. ΔS = +. ΔS = -. ΔG = Answer: D Topic: Key Concept Problems 101) What are the signs (+ or -) of ΔH. ΔG = Answer: B Topic: 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: D Topic: 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: A Topic: Key Concept Problems 27 . ΔG = + D) ΔH = -.100) What are the signs (+ or -) of ΔH. ΔG = + B) ΔH = +. ΔG = + B) ΔH = +. ΔS = +. ΔS = -. and ΔG when the system spontaneously goes from initial state 2 to the equilibrium state? A) ΔH = +. ΔG = C) ΔH = -. ΔS = +. ΔS. ΔS = +. ΔS. ΔS = -. and ΔG when the system spontaneously goes from initial state 1 to the equilibrium state? A) ΔH = +. and spontaneous at f. D) spontaneous at d. C) spontaneous at d. B) nonspontaneous at d. Answer: C Topic: Key Concept Problems 106) According to the diagram above. B) ΔG° is positive and the equilibrium composition is rich in reactants. Answer: B Topic: Key Concept Problems 28 . at equilibrium at e.104) Which of the above reaction mixtures is ΔG of reaction = ΔG ° ? A) (1) B) (2) C) (3) D) (4) Answer: C Topic: Key Concept Problems 105) According to the diagram above. and f. and nonspontaneous at f. at equilibrium at e. the forward reaction is A) nonspontaneous at d and e. D) ΔG° is negative and the equilibrium composition is rich is reactants. C) ΔG° is negative and the equilibrium composition is rich in products. A) ΔG° is positive and the equilibrium composition is rich in products. and spontaneous at f. e. Answer: A Topic: Key Concept Problems 108) The following pictures represent three equilibrium mixtures for the interconversion of A.c. ΔG °(3) = + C) ΔG °(1) = 0. ΔG °(2) = 0. B. respectively. and C molecules (unshaded spheres) into X. ΔG °(2) = 0. ΔG °(3) = 0 B) ΔG °(1) = -. D) ΔG° is negative and is equal to a . ΔG °(2) = +. Option: algorithmic 29 .c. What is the sign of ΔG ° for each of the three reactions? A) ΔG °(1) = -. and Z molecules (shaded spheres).107) According to this diagram. ΔG °(2) = -. 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: B Topic: Section 16. A) ΔG° is positive and is equal to a . ΔG °(3) = + D) ΔG °(1) = +. Y.b.3 Entropy and Probability Algo. ΔG °(3) = Answer: B Topic: Key Concept Problems 16. B) ΔG° is positive and is equal to b . ΔS = R ln (Vf/Vi).c. C) ΔG° is negative and is equal to b .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. 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. II: ΔS = positive Answer: C Topic: Section 16.0 moles of O2(g) increases from 1. Option: algorithmic 4) Under which of the following conditions would one mole of He have the highest entropy.2 atm. Option: algorithmic 3) Predict the sign of ΔS for each of the following systems. II. S? A) 17°C and 15 L B) 127°C and 15 L C) 17°C and 25 L D) 127°C and 25 L Answer: D Topic: Section 16. Option: algorithmic 30 .50 R ln (Vf/Vi) B) ΔS = -4.0 atm to 1.3 Entropy and Probability Algo. II: ΔS = negative D) I: ΔS = positive. S°.0 moles of O2(g) increases from 44 L to 52 L. which occur at constant temperature I.3 Entropy and Probability Algo.4 Entropy and Temperature Algo. II: ΔS = negative B) I: ΔS = negative.50 V at constant temperature? A) ΔS = 4.50 Answer: C Topic: Section 16. A) I: ΔS = negative.50 D) ΔS = -R ln 4. The volume of 2. Option: algorithmic 5) Which one of the following would be expected to have the lowest standard molar entropy.50 R ln (Vf/Vi) C) ΔS = R ln 4.5 Standard Molar Entropies and Standard Entropies of Reaction Algo. II: ΔS = positive C) I: ΔS = positive. at 25°C? A) C10H22(s) B) C10H22(l) C) C14H30(s) D) C14 OH(l) Answer: A Topic: Section 16. The pressure of 2. Answer: 1670 Topic: Section 16. Answer: + 209.2 Enthalpy.0 mole sample of gas at STP has a ________ entropy than 1.3 Entropy and Probability 4) The entropy of water at 25° is ________ than the entropy of water at 35°C.1 Spontaneous Processes 2) The sign (+ or –) of ∆H is ________ and the sign (+ or –) of ∆S is ________ for the evaporation of water. At 25°C and 1 atm pressure the decomposition of water into hydrogen and oxygen is classified as ________. The standard free energy of formation of acetylene is ________ kJ/mol. S°.4 Entropy and Temperature 5) Standard molar entropies. Answer: +.7 Free Energy 8) Acetylene. Entropy. and ∆S° = + 15.7 Free Energy 7) A reaction has ∆G° = + 21. Answer: nonspontaneous.5 Standard Molar Entropies and Standard Entropies of Reaction 6) A reaction for which ∆H° = + 98.7 kJ/mol. and a standard entropy change for its formation from its elements.8 Standard Free-Energy Changes for Reactions 31 . Answer: higher Topic: Section 16.16.0 J/mol∙K can become spontaneous at a temperature of ________ K.0 kJ/mol. + Topic: Section 16. ∆H° = 226.9 Answer: – 242. and Spontaneous Processes: A Brief Review 3) A 1.8 J/K∙mol. Answer: less than Topic: Section 16.5 kJ/mol. C2H2. spontaneous Topic: Section 16.2 Topic: Section 16. in J/K∙mol. for this reaction is ________ J.8 kJ and ∆S° = + 141. has a standard enthalpy of formation.6 69. ∆H° = + 25. and the melting of ice is classified as ________.3 Short Answer Questions 1) Chemical and physical changes can be classified as spontaneous or nonspontaneous.0 mole of gas at 273 K and 835 mm Hg.2 205. Answer: nonspontaneous. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) 186.8 Topic: Section 16.0 213. ∆S°. ∆S° = 58.5 J/K is ________ (spontaneous or nonspontaneous) at low temperatures and ________ (spontaneous or nonspontaneous) at high temperatures. spontaneous Topic: Section 16. The standard entropy of reaction. are given below each reactant and product in the reaction shown below. At 25°C the equilibrium constant for this reaction .4 – 237.9 Standard Free Energies of Formation 10) The standard free energy for a reaction is ∆G° = – 33.8 0 – 394. The standard free energy of reaction. in kJ/mol.09 × 105 Topic: Section 16. ∆G°.0 kJ.9) Standard free energies of formation. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) – 50. ∆G°. for this reaction is ________ kJ. Answer: 6.0 Topic: Section 16.2 Answer: – 818.11 Free Energy and Chemical Equilibrium 32 . are given below each reactant and product in the reaction shown below. Kp = ________.