Unit 9 · Thermodynamics and ElectrochemistryPractice · 热力学与电化学
Lessons covered:U9.1 Introduction to EntropyU9.2 Absolute Entropy and Entropy ChangeU9.3 Gibbs Free Energy and Thermodynamic FavorabilityU9.4 Thermodynamic and Kinetic ControlU9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)U9.6 Coupled ReactionsU9.7 Galvanic (Voltaic) and Electrolytic CellsU9.8 Cell Potential and Free EnergyU9.9 Cell Potential Under Nonstandard ConditionsU9.10 Electrolysis and Faraday's Law
- 1Multiple choice
In which of these processes does the entropy of the system decrease (ΔS < 0)?
ExplanationThree gas molecules become two, so there are fewer gas molecules and fewer possible arrangements: ΔS < 0. A solid dissolving, a solid releasing gases, and a liquid evaporating all spread the particles out, so ΔS > 0.
Review: U9.1 Introduction to Entropy
- 2Numeric
Calculate ΔS° for N₂(g) + 3H₂(g) → 2NH₃(g). S° (J mol⁻¹ K⁻¹): N₂(g) 191.6, H₂(g) 130.7, NH₃(g) 192.8.
J/KExplanationΔS° = ΣnS°(products) − ΣnS°(reactants) = 2 × 192.8 − (191.6 + 3 × 130.7) = 385.6 − 583.7 = −198.1 J/K. Four gas molecules become two, so ΔS° is negative, as predicted. Elements have nonzero S° and can’t be left out.
- 3Numeric
For C(s) + CO₂(g) → 2CO(g), ΔH° = +172.47 kJ and ΔS° = +175.86 J/K. Use ΔG° = ΔH° − TΔS° to find ΔG° in kJ at 25 °C (298 K).
kJExplanationConvert ΔS° to kJ/K first: 0.17586 kJ/K. ΔG° = 172.47 − 298 × 0.17586 = 172.47 − 52.41 = +120.1 kJ. ΔG° > 0, so the reaction is not favorable at 25 °C (free energies of formation give +120.06 kJ, the same). Forgetting the unit conversion gives a huge negative number.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 4Numeric
For C(s) + CO₂(g) → 2CO(g), ΔH° = +172.47 kJ and ΔS° = +175.86 J/K. Assuming these don’t change with temperature, above what temperature, in kelvins, does the reaction become thermodynamically favorable?
KExplanationΔH° > 0 and ΔS° > 0, so at high temperature TΔS° outweighs ΔH° and ΔG° turns negative. Setting ΔG° = 0: T = ΔH° ÷ ΔS° = 172470 J ÷ 175.86 J/K = 981 K (about 708 °C). That is why red-hot coke in a blast furnace turns CO₂ into CO.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 5Multiple choice
Burning natural gas, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), has ΔG° of about −818 kJ, yet when a gas stove is turned on without lighting it, methane mixes with air and doesn’t burn. Which is the best explanation?
ExplanationA very negative ΔG° means the reaction is thermodynamically favorable with an enormous K, but it says nothing about speed. A spark supplies energy so some molecules get over the activation barrier, and the heat they release keeps more molecules reacting. ΔG° and K depend on temperature, not on whether there is a spark.
- 6Numeric
At 25 °C (298 K), ΔG° = −29.45 kJ/mol for CO(g) + 2H₂(g) → CH₃OH(l). What is the equilibrium constant K at this temperature? R = 8.314 J/(mol·K).
× 10Explanationln K = −ΔG° ÷ RT = 29450 ÷ (8.314 × 298) = 11.89, so K = e11.89 = 1.5 × 105. ΔG° < 0 gives K > 1: products are favored. Convert ΔG° to J/mol.
- 7Multiple choice
To make titanium, TiO₂ is first converted to TiCl₄. TiO₂(s) + 2Cl₂(g) → TiCl₄(l) + O₂(g) has ΔG° = +151.6 kJ, which is unfavorable, so industry adds coke at the same time: 2C(s) + O₂(g) → 2CO(g), ΔG° = −274.3 kJ. Why does adding coke let TiO₂ be converted to TiCl₄?
ExplanationThe O₂ made in the first reaction is used up at once by the coke (the shared intermediate). Adding the equations: TiO₂(s) + 2Cl₂(g) + 2C(s) → TiCl₄(l) + 2CO(g), ΔG° = +151.6 + (−274.3) = −122.7 kJ, so the overall reaction is favorable. ΔG° for the first reaction by itself hasn’t changed, and coke is a reactant, not a catalyst.
Review: U9.6 Coupled Reactions
- 8Multiple choice
A zinc strip in 1 M Zn(NO₃)₂ and a silver strip in 1 M AgNO₃ are joined by a KNO₃ salt bridge and a wire to make a galvanic cell. E°: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V; Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V. Which statement is correct?
ExplanationZinc, with the lower E°, is oxidized more easily, so it is the anode: Zn → Zn²⁺ + 2e⁻. Ag⁺ is reduced at the silver electrode, the cathode. Electrons flow through the wire from anode to cathode. On the silver side Ag⁺ is used up, so K⁺ cations from the salt bridge move in to replace the positive charge, while NO₃⁻ moves toward the zinc side.
- 9Numeric
For Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), E°: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V; Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V. Calculate ΔG° in kJ/mol. F = 96485 C/mol.
kJ/molExplanationE°cell = E°(cathode) − E°(anode) = 0.80 − (−0.76) = 1.56 V. Two electrons are transferred (one Zn gives up 2; two Ag⁺ take 1 each). ΔG° = −nFE° = −2 × 96485 × 1.56 = −301033 J/mol ≈ −301 kJ/mol. Doubling the silver half-reaction doesn’t double its E°.
- 10Multiple choice
A galvanic cell made of zinc in a Zn²⁺ solution and silver in an Ag⁺ solution runs Zn + 2Ag⁺ → Zn²⁺ + 2Ag, with E° = 1.56 V. If [Zn²⁺] = 1.0 M and [Ag⁺] = 0.010 M, how does the cell potential E compare with E°?
ExplanationQ = [Zn²⁺] ÷ [Ag⁺]² = 1.0 ÷ 0.010² = 1.0 × 10⁴ > 1: there is less reactant (Ag⁺), so the driving force is smaller and E < E°. By the Nernst equation, E = 1.56 − (0.0592/2) × log(1.0 × 10⁴) = 1.44 V, still positive, so the cell still works.
- 11Numeric
A steel part is nickel-plated by electrolyzing NiSO₄ solution with a current of 2.50 A for 1.00 h. What is the greatest mass of nickel that can deposit on the cathode? M(Ni) = 58.69 g/mol, F = 96485 C/mol.
gExplanationq = It = 2.50 × 3600 = 9000 C; n(e⁻) = 9000 ÷ 96485 = 0.0933 mol. Ni²⁺ + 2e⁻ → Ni, so n(Ni) = 0.0933 ÷ 2 = 0.0466 mol and m = 0.0466 × 58.69 = 2.74 g. Time must be in seconds; forgetting to divide by 2 doubles the answer.
- 12Multiple choice
For a redox reaction, which set of relationships always goes together?
ExplanationΔG° = −nFE°, so a positive E° means a negative ΔG°. ΔG° = −RT ln K, so a negative ΔG° means ln K > 0 and K > 1. All three describe the same thing: under standard conditions the forward reaction is favorable.
Review: U9.5 Free Energy and Equilibrium (ΔG° = −RT ln K)U9.8 Cell Potential and Free Energy
Your missed questions were saved automatically to your mistake notebook · 0
Your best score counts only full runs through the set, not retries of missed questions.