U9.3 · Gibbs Free Energy and Thermodynamic FavorabilityPractice · 吉布斯自由能与热力学有利性
Lessons covered:U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 1Numeric
Calculate ΔG° at 25 °C for CH₄(g) + H₂O(g) → CO(g) + 3H₂(g). ΔG°f (kJ/mol): CH₄(g) −50.5, H₂O(g) −228.59, CO(g) −137.15.
kJExplanationΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants) = (−137.15 + 3 × 0) − (−50.5 + (−228.59)) = +141.94 kJ. H₂ is an element with ΔG°f = 0. ΔG° > 0, so it is unfavorable at 25 °C; industry runs it hot.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 2Numeric
For Br₂(l) → Br₂(g), ΔH° = +30.91 kJ and ΔS° = +93.27 J/K. Assuming these don’t change with temperature, estimate the boiling point of bromine in kelvins.
KExplanationAt the boiling point, liquid and vapor (1 bar) are at equilibrium, so ΔG° = 0: T = ΔH° ÷ ΔS° = 30910 J ÷ 93.27 J/K = 331 K, about 58 °C, very close to the measured 58.8 °C. Below this ΔG° > 0; above it ΔG° < 0.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 3Multiple choice
A reaction has ΔH° < 0 and ΔS° < 0. Under what conditions is it thermodynamically favorable?
ExplanationΔG° = ΔH° − TΔS°. The negative ΔH° helps; −TΔS° is positive and hurts, more so as T rises. At low temperatures ΔH° dominates and ΔG° < 0; once T > ΔH°/ΔS°, ΔG° turns positive.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 4Numeric
A blast-furnace reaction, Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g), has ΔH° = +492.64 kJ and ΔS° = +543.08 J/K. Assuming these don’t change with temperature, find ΔG° in kJ at 1200 K.
kJExplanationΔG° = ΔH° − TΔS° = 492.64 − 1200 × 0.54308 = 492.64 − 651.70 = −159 kJ. At 25 °C, ΔG° = +330.8 kJ, unfavorable; making lots of CO gas gives a large ΔS°, so at high temperature −TΔS° wins.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 5Multiple choice
A reaction has ΔG° = −45 kJ/mol at 25 °C. Which statement is correct?
ExplanationΔG° < 0 means the forward direction is favored under standard conditions and K > 1 (here about 10⁸). It says nothing about rate, and it isn’t “complete conversion”: a little reactant remains at equilibrium. ΔG° changes with temperature, so without ΔH° and ΔS° you can’t claim it is favorable at every temperature.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 6Multiple choice
The graph shows ΔG° for against temperature (assume ΔH° and ΔS° don’t change with temperature). Above what temperature is the reaction thermodynamically favorable under standard conditions?
ExplanationThe reaction is favorable when ΔG° < 0. The line crosses zero near 325 K, and ΔG° is negative above that. For the line ΔG° = ΔH° − TΔS°, the slope is −ΔS° (it slopes down, so ΔS° = +175.9 J/K, as more gas molecules form), and extrapolating to 0 K gives ΔH° = +57.2 kJ. The crossing is at T = ΔH°/ΔS° = 57.2 ÷ 0.1759 ≈ 325 K. At 298 K, ΔG° is still positive (about +4.8 kJ/mol).
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
- 7Multiple choice
The graph shows ΔG° against temperature for four reactions (lines P, Q, R, and S). Which reaction is thermodynamically favorable at every temperature?
ExplanationΔG° = ΔH° − TΔS°, so a line’s intercept at T = 0 is ΔH° and its slope is −ΔS°. Q has a negative intercept (ΔH° < 0) and slopes down (ΔS° > 0), so it stays below zero at every temperature. P (ΔH° > 0, ΔS° > 0) is favorable only at high temperature, S (ΔH° < 0, ΔS° < 0) only at low temperature, and R (ΔH° > 0, ΔS° < 0) at no temperature.
Review: U9.3 Gibbs Free Energy and Thermodynamic Favorability
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