Entropy and Gibbs Free Energy: Quiz
Seven questions on entropy changes, ΔG = ΔH − TΔS, spontaneity and the link between ΔG° and K, with explanations.
Entropy and Gibbs Free Energy: Quiz
7 questions
3 mol of gas become 2 mol, so the entropy decreases. The others increase disorder or the moles of gas.
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Answer: 2SO₂(g) + O₂(g) → 2SO₃(g)
3 mol of gas become 2 mol, so the entropy decreases. The others increase disorder or the moles of gas.
T = ΔH ÷ ΔS = 50.0 kJ ÷ 0.200 kJ/K = 250. K. Above this, TΔS outweighs ΔH.
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Answer: 250. K
T = ΔH ÷ ΔS = 50.0 kJ ÷ 0.200 kJ/K = 250. K. Above this, TΔS outweighs ΔH.
ΔG = ΔH − TΔS = (negative) − (positive) is negative for every T.
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Answer: at all temperatures
ΔG = ΔH − TΔS = (negative) − (positive) is negative for every T.
ΔS = −0.200 kJ/K; ΔG = −100. kJ − (300. K)(−0.200 kJ/K) = −100. kJ + 60.0 kJ = −40.0 kJ. Forgetting to convert J to kJ gives the large wrong answer.
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Answer: −40.0 kJ
ΔS = −0.200 kJ/K; ΔG = −100. kJ − (300. K)(−0.200 kJ/K) = −100. kJ + 60.0 kJ = −40.0 kJ. Forgetting to convert J to kJ gives the large wrong answer.
ΔG° = −RT ln K: positive ΔG° means ln K is negative, so K is less than 1 (reactants favoured).
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Answer: less than 1
ΔG° = −RT ln K: positive ΔG° means ln K is negative, so K is less than 1 (reactants favoured).
For CaCO₃ → CaO + CO₂, ΔH = +179.2 kJ and ΔS = +160.2 J/K, so ΔG becomes negative above about 1119 K.
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Answer: ΔH and ΔS are both positive, so ΔG is only negative at high T
For CaCO₃ → CaO + CO₂, ΔH = +179.2 kJ and ΔS = +160.2 J/K, so ΔG becomes negative above about 1119 K.
That is the second law. Spontaneous reactions can be slow and can be endothermic (e.g. ice melting at 25 °C).
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Answer: it increases the total entropy of the universe
That is the second law. Spontaneous reactions can be slow and can be endothermic (e.g. ice melting at 25 °C).