The problems are on the entropy Gibbs energy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Entropy measures how many ways particles and energy can be arranged; it rises from solid to liquid to gas. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on entropy Gibbs energy, from which has more entropy? to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the conversion of units, the substitution and the result with its unit in bold. Where a problem asks you to explain, the key reasoning is given in a sentence so you can compare it with your own wording.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R1.4
Problem 1 Which has more entropy?
1. Liquid water: particles can move more freely.2. CO₂(g): gases have far more freedom than solids.
3. NaCl(aq): separated ions spread through the solution.
4. Two moles of gas: more particles give more ways to arrange the energy.
Problem 2 Predicting the sign of ΔS
1. Positive: a gas is formed from liquid.2. Negative: 4 mol of gas become 2 mol.
3. Positive: a gas is formed from solid.
4. Negative: a gas condenses to liquid.
Problem 3 Entropy of alkanes
1. Entropy increases as the molecules get larger.2. 229.6 − 186.3 = +43.3 J K⁻¹ mol⁻¹; 270.3 − 229.6 = +40.7 J K⁻¹ mol⁻¹.
3. Bigger molecules have more atoms and more ways to vibrate, rotate and bend, so energy can be arranged in more ways.
Problem 4 Gibbs energy of a hand warmer
1. ΔS = −120 ÷ 1000 = −0.120 kJ K⁻¹ mol⁻¹.2. ΔG = −85.0 − 298 × (−0.120) = −85.0 + 35.8 = −49.2 kJ/mol.
3. Spontaneous, because ΔG is negative.
Problem 5 Meaning of the sign of ΔG
1. Spontaneous.2. Non-spontaneous (forward direction).
3. At equilibrium.
4. The reverse reaction is spontaneous, because its ΔG has the opposite sign.
Problem 6 Entropy change for water formation
1. Products: 2 × 69.9 = 139.8. Reactants: 2 × 130.7 + 205.0 = 466.4.2. ΔS° = 139.8 − 466.4 = −326.6 J K⁻¹ mol⁻¹.
3. Three moles of gas become a liquid, so the system becomes much more ordered.
Problem 7 Four sign combinations
1. At all temperatures (both terms make ΔG negative; ΔG is negative at low T).2. Low T only.
3. High T only.
4. Never, since ΔH is positive and −TΔS is also positive.
Problem 8 Burning methane
1. Products: 213.8 + 2 × 69.9 = 353.6. Reactants: 186.3 + 2 × 205.0 = 596.3. ΔS° = −242.7 J K⁻¹ mol⁻¹.2. ΔG = −890 − 298 × (−0.2427) = −890 + 72.3 = −818 kJ/mol.
3. Spontaneous, but it needs a spark because the activation energy is high.
Problem 9 Melting point of ice
1. ΔG = 0 at equilibrium between solid and liquid.2. T = ΔH ÷ ΔS = 6010 ÷ 22.0 = 273 K (0 °C).
3. ΔG = 6.01 − 290 × 0.0220 = 6.01 − 6.38 = −0.37 kJ/mol, so spontaneous.
Problem 10 Boiling point of ethanol
1. T = 38 600 ÷ 110 = 351 K.2. 351 − 273 = 78 °C.
3. Above 351 K the −TΔS term outweighs ΔH, so ΔG becomes negative and the liquid boils.
Problem 11 Dissolving sodium chloride
1. ΔS° = (59.0 + 56.5) − 72.1 = +43.4 J K⁻¹ mol⁻¹.2. ΔG = 3.9 − 298 × 0.0434 = 3.9 − 12.9 = −9.0 kJ/mol.
3. The positive entropy change as ions spread through the solution makes −TΔS larger than ΔH.
Problem 12 From ΔG to the equilibrium constant
1. ΔG° = −8.31 × 298 × ln(4000) = −2476 × 8.294 = −20 540 J = −20.5 kJ/mol.2. ln K = −15 000 ÷ 2476 = −6.057, so K = 2.3 × 10⁻³.
3. The first reaction (K is greater than 1 and ΔG° is negative).
Problem 13 Reading a ΔG–T graph
1. At T = 0, ΔG = ΔH = +60 kJ/mol.2. Gradient = −ΔS. From +60 at 0 K to −60 at 600 K, gradient = −120 ÷ 600 = −0.200, so ΔS = +0.200 kJ K⁻¹ mol⁻¹ (200 J K⁻¹ mol⁻¹).
3. ΔG = 0 at 300 K; the reaction is spontaneous above 300 K.
Problem 14 Entropy of the surroundings
1. ΔS(surr) = 1 367 000 ÷ 298 = +4587 J K⁻¹ mol⁻¹.2. ΔS(total) = 4587 − 138.4 = +4449 J K⁻¹ mol⁻¹.
3. −298 × 4.449 = −1325.8 kJ/mol; directly, ΔG = −1367 − 298 × (−0.1384) = −1367 + 41.2 = −1326 kJ/mol. They agree.
Problem 15 Decomposition of hydrogen peroxide
1. ΔS° = (2 × 69.9 + 205.0) − 2 × 109.6 = 344.8 − 219.2 = +125.6 J K⁻¹ mol⁻¹; ΔG = −196 − 298 × 0.1256 = −233 kJ/mol.2. ΔG shows only that the reaction is feasible. The activation energy is high without a catalyst (such as MnO₂ or an enzyme), so the rate is negligible.
Problem 16 Baking soda decomposition
1. ΔS° = (135.0 + 188.8 + 213.8) − 2 × 101.7 = 537.6 − 203.4 = +334.2 J K⁻¹ mol⁻¹.2. ΔG = 135.6 − 298 × 0.3342 = 135.6 − 99.6 = +36.0 kJ/mol: not spontaneous at room temperature.
3. T = 135 600 ÷ 334.2 = 406 K (133 °C).
Problem 17 The contact process
1. ΔS° = 2 × 256.8 − (2 × 248.2 + 205.0) = 513.6 − 701.4 = −187.8 J K⁻¹ mol⁻¹.2. At 298 K: −198 + 298 × 0.1878 = −142 kJ/mol. At 1200 K: −198 + 1200 × 0.1878 = +27.4 kJ/mol.
3. T = 198 000 ÷ 187.8 = 1054 K.
4. At low T equilibrium favours SO₃ but the rate is too slow; around 700 K with a catalyst is a compromise between yield and rate.
Problem 18 Dinitrogen tetroxide equilibrium
1. ΔG = 57.2 − 298 × 0.1758 = 57.2 − 52.4 = +4.8 kJ/mol.2. ln K = −4812 ÷ (8.31 × 298) = −1.943; K = 0.143.
3. K = 1 when ΔG = 0: T = 57 200 ÷ 175.8 = 325 K. On heating, the equilibrium shifts to NO₂ and the gas turns darker brown.
Problem 19 Three reactions on one graph
1. Line A starts at −120 and falls; line B starts at +50 and rises; line C starts at +90 and falls steeply, as labelled in the figure.2. A is spontaneous at all temperatures (ΔH negative, ΔS positive); B at none (ΔH positive, ΔS negative).
3. T = 90 000 ÷ 200 = 450 K; ΔG(298 K) = 90 − 298 × 0.200 = +30.4 kJ/mol.
Problem 20 Spot the errors
1. ΔH is in kJ but ΔS is in J: the units were not matched, so the TΔS term is 1000 times too large.2. ΔG = −50.0 − 298 × (−0.100) = −50.0 + 29.8 = −20.2 kJ/mol, spontaneous.
3. T = ΔH ÷ ΔS = −50 000 ÷ −100 = 500 K; above 500 K, ΔG is positive.
