IB DP Chemistry Reactivity 1.4: Entropy and spontaneity – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set tests Reactivity 1.4, “Entropy and spontaneity”, numbered as in the IB’s February 2023 DP Chemistry guide, which first applied to the 2025 exams. Each of the understandings Reactivity 1.4.1, 1.4.2, 1.4.3 and 1.4.4 is examined at least once. Because the whole sub-topic is additional higher level content, treat every question here as HL only; the set is written for sessions from 2025 onwards.

Before you start, read the entropy and spontaneity study guide or skim the condensed revision notes. The DP Chemistry hub lists the other units, the DP Chemistry checklist records your progress, and our free diagnostic quizzes help you decide what to revise next.

Data for every question: R = 8.31 J K⁻¹ mol⁻¹, and T/K = t/°C + 273. Standard entropies, S⦵ / J K⁻¹ mol⁻¹: N₂(g) 191.6, O₂(g) 205.2, H₂(g) 130.7, NO(g) 210.8, CO(g) 197.7, CO₂(g) 213.8, C(s, graphite) 5.7, Si(s) 18.8, SiO₂(s) 41.5, Al(s) 28.3, Al₂O₃(s) 50.9, C₃H₈(g) 270.3, C₃H₆(g) 267.0. Unless told otherwise, give final answers to 3 significant figures.

Questions

1. (a) State what is meant by entropy. [1] (b) Arrange solid, liquid and gaseous sulfur dioxide in order of increasing entropy under the same conditions. [1]

2. Predict the sign of ΔS for each change, giving a reason.

(a) NH₃(g) → NH₃(l) [1] (b) Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) [1] (c) 2C₆H₅COOH(s) + 15O₂(g) → 14CO₂(g) + 6H₂O(l) [1] (d) Mg(OH)₂(s) → MgO(s) + H₂O(g) [1]

3. A catalytic converter removes two pollutants from car exhaust: 2NO(g) + 2CO(g) → N₂(g) + 2CO₂(g). Calculate ΔS⦵ for this reaction. [3]

4. Silicon is made by heating silica with carbon: SiO₂(s) + 2C(s) → Si(s) + 2CO(g). At 298 K, ΔH⦵ = +689.7 kJ mol⁻¹ and ΔG⦵ = +581.9 kJ mol⁻¹.

(a) Calculate ΔS⦵ from these values. [2] (b) Show that the S⦵ data give a value in close agreement. [1] (c) Calculate the temperature above which the reaction becomes spontaneous, and suggest why the process needs a very hot electric furnace. [2]

5. A student calculated ΔG for a reaction with ΔH = −45.0 kJ mol⁻¹ and ΔS = −120 J K⁻¹ mol⁻¹ at 350 K and obtained +41 955 kJ mol⁻¹.

(a) Identify the error. [1] (b) Calculate the correct ΔG and state whether the reaction is spontaneous at 350 K. [2]

6. State, for each pair of signs, whether the reaction is spontaneous at all temperatures, at no temperature, only at high temperature or only at low temperature.

(a) ΔH negative, ΔS positive [1] (b) ΔH positive, ΔS positive [1] (c) ΔH negative, ΔS negative [1]

7. A fresh aluminium surface oxidises in air: 4Al(s) + 3O₂(g) → 2Al₂O₃(s), ΔH⦵ = −3351.4 kJ mol⁻¹.

(a) Calculate ΔS⦵ for the system. [2] (b) Calculate the entropy change of the surroundings at 298 K. [1] (c) Use your answers to calculate ΔG⦵ at 298 K, and explain why the reaction is spontaneous even though the system’s entropy falls. [2]

8. In cells, glucose 6-phosphate (G6P) isomerises to fructose 6-phosphate (F6P): G6P(aq) ⇌ F6P(aq). Use K = 0.500 at 310 K.

(a) Calculate ΔG⦵ at 310 K. [2] (b) State the value of ΔG when the system is at equilibrium, and describe how ΔG changes as a solution containing only G6P approaches equilibrium. [2]

9. At 310 K, a solution contains [G6P] = 0.0600 mol dm⁻³ and [F6P] = 0.0120 mol dm⁻³.

(a) Calculate Q. [1] (b) Use ΔG = ΔG⦵ + RT lnQ and your answer to question 8(a) to calculate ΔG. [3] (c) Deduce the direction of net reaction, and confirm it by comparing Q with K. [2]

10. Propene is made by removing hydrogen from propane: C₃H₈(g) ⇌ C₃H₆(g) + H₂(g), ΔH⦵ = +123.8 kJ mol⁻¹.

(a) Calculate ΔS⦵. [2] (b) Calculate ΔG⦵ at 298 K and interpret its sign. [2] (c) Determine the temperature at which the reaction becomes spontaneous. [2] (d) Calculate ΔG⦵ and K at 1050 K, assuming ΔH⦵ and ΔS⦵ do not change with temperature. [3]

11. A solid X decomposes into gaseous products. ΔG⦵ = +28.4 kJ mol⁻¹ at 400 K and −6.6 kJ mol⁻¹ at 600 K. Assume ΔH⦵ and ΔS⦵ are constant.

(a) Calculate ΔS⦵ and ΔH⦵. [4] (b) Calculate the temperature at which ΔG⦵ = 0. [1] (c) Calculate K at 650 K and comment on the composition of the equilibrium mixture. [3]

Answers

1. (a) A measure of how dispersed (spread out) the matter and/or energy of a system is [1]. [1] (b) SO₂(s) < SO₂(l) < SO₂(g) [1]. [1] Examiner insight: The guide defines entropy through the dispersal of matter and energy, so build your definition on that idea; a one-word answer such as “disorder” gives little for a marker to credit.

2. (a) Negative: a gas condenses to a liquid [1]. (b) Negative: aqueous ions join to form a solid lattice [1]. (c) Negative: 15 mol of gas become 14 mol of gas [1]. (d) Positive: a solid produces 1 mol of gas [1]. [4] Examiner insight: In (c) count only gas moles; adding in the liquid water or the solid acid would wrongly suggest an increase.

3. Products: 191.6 + (2 × 213.8) = 619.2 J K⁻¹ mol⁻¹ [1]. Reactants: (2 × 210.8) + (2 × 197.7) = 817.0 J K⁻¹ mol⁻¹ [1]. ΔS⦵ = 619.2 − 817.0 = −197.8 J K⁻¹ mol⁻¹ [1]. [3] Examiner insight: Write out both sums; if a coefficient slips, visible working still shows the method, whereas a lone wrong number shows nothing.

4. (a) ΔS⦵ = (ΔH⦵ − ΔG⦵) ÷ T = (689.7 − 581.9) ÷ 298 [1] = 0.362 kJ K⁻¹ mol⁻¹ = +362 J K⁻¹ mol⁻¹ [1]. [2] (b) (18.8 + 2 × 197.7) − (41.5 + 2 × 5.7) = 414.2 − 52.9 = +361.3 J K⁻¹ mol⁻¹, which agrees with (a) to within rounding [1]. [1] (c) T = 689.7 ÷ 0.3617 = 1910 K [1]. Below about 1910 K, ΔG⦵ is positive, so the furnace must be held well above this temperature for the reduction to be spontaneous [1]. [2] Examiner insight: Turn 362 J into 0.362 kJ before dividing into ΔH; skipping that step gives a temperature 1000 times too small.

5. (a) ΔS was used in J K⁻¹ mol⁻¹ with ΔH in kJ mol⁻¹ (not divided by 1000) [1]. [1] (b) ΔG = −45.0 − (350 × −0.120) = −3.0 kJ mol⁻¹ [1]; negative, so spontaneous at 350 K [1]. [2] Examiner insight: A huge ΔG value such as +41 955 kJ mol⁻¹ is a signal to recheck units before moving on.

6. (a) At all temperatures [1]. (b) Only at high temperature [1]. (c) Only at low temperature [1]. [3] Examiner insight: Each answer follows from the sign of ΔH − TΔS; write that expression if you are unsure.

7. (a) ΔS⦵ = (2 × 50.9) − ((4 × 28.3) + (3 × 205.2)) = 101.8 − 728.8 [1] = −627.0 J K⁻¹ mol⁻¹ [1]. [2] (b) ΔS(surroundings) = −ΔH ÷ T = +3 351 400 ÷ 298 = +1.12 × 10⁴ J K⁻¹ mol⁻¹ [1]. [1] (c) ΔS(total) = 11 246 − 627 = +10 619 J K⁻¹ mol⁻¹, so ΔG⦵ = −TΔS(total) = −298 × 10.619 = −3160 kJ mol⁻¹ [1]. The heat released raises the entropy of the surroundings far more than the system’s entropy falls, so total entropy increases [1]. [2] Examiner insight: Carry 11 246 rather than the rounded 1.12 × 10⁴ into (c); the direct route, ΔH⦵ − TΔS⦵, gives the same −3160 kJ mol⁻¹ and is a useful check.

8. (a) ΔG⦵ = −RT lnK = −8.31 × 310 × ln 0.500 [1] = +1786 J mol⁻¹ = +1.79 kJ mol⁻¹ [1]. [2] (b) ΔG = 0 at equilibrium [1]. With only G6P present, ΔG is negative; it becomes less negative as F6P forms and reaches zero at equilibrium [1]. [2] Examiner insight: R is in J, so the result comes out in J mol⁻¹; state the unit when you convert to kJ.

9. (a) Q = [F6P] ÷ [G6P] = 0.0120 ÷ 0.0600 = 0.200 [1]. [1] (b) RT lnQ = 8.31 × 310 × ln 0.200 = −4146 J mol⁻¹ [1]. ΔG = 1786 + (−4146) [1] = −2360 J mol⁻¹ = −2.36 kJ mol⁻¹ [1]. [3] (c) ΔG is negative, so the forward reaction occurs (G6P is converted into F6P) [1]. Q = 0.200 is less than K = 0.500, so the mixture shifts to the right, which agrees [1]. [2] Examiner insight: A positive ΔG⦵ does not stop the forward reaction here; say which direction is spontaneous for this particular mixture.

10. (a) ΔS⦵ = (267.0 + 130.7) − 270.3 [1] = +127.4 J K⁻¹ mol⁻¹ [1]. [2] (b) ΔG⦵ = 123.8 − (298 × 0.1274) = +85.8 kJ mol⁻¹ [1]; positive, so the reaction is not spontaneous at 298 K and the equilibrium lies far to the left [1]. [2] (c) T = 123.8 ÷ 0.1274 [1] = 972 K; spontaneous above this temperature [1]. [2] (d) ΔG⦵ = 123.8 − (1050 × 0.1274) = −9.97 kJ mol⁻¹ [1]. lnK = 9970 ÷ (8.31 × 1050) = 1.143 [1]; K = 3.13 [1]. [3] Examiner insight: In (c) a temperature without “above” is incomplete; the signs of ΔH and ΔS tell you which side.

11. (a) Subtract the two equations: ΔG⦵(400) − ΔG⦵(600) = 200 × ΔS⦵, so 28.4 − (−6.6) = 200ΔS⦵ [1]. ΔS⦵ = 0.175 kJ K⁻¹ mol⁻¹ = +175 J K⁻¹ mol⁻¹ [1]. ΔH⦵ = ΔG⦵ + TΔS⦵ = 28.4 + (400 × 0.175) [1] = +98.4 kJ mol⁻¹ [1]. [4] (b) T = 98.4 ÷ 0.175 = 562 K [1]. [1] (c) ΔG⦵ = 98.4 − (650 × 0.175) = −15.35 kJ mol⁻¹ [1]. K = e^(15 350 ÷ (8.31 × 650)) = 17.1 [1]. K > 1, so products are favoured in the equilibrium mixture [1]. [3] Examiner insight: Check that both given ΔG⦵ values are reproduced by your ΔH⦵ and ΔS⦵; this catches sign slips at once.

Where marks are usually lost

• Substituting ΔS in J K⁻¹ mol⁻¹ into an equation where ΔH is in kJ mol⁻¹.
• Using °C for T in ΔG = ΔH − TΔS or in RT lnK.
• Dropping coefficients when summing S⦵ values.
• Predicting ΔS from total moles instead of moles of gas.
• Giving a threshold temperature without “above” or “below”.
• Leaving ΔG⦵ in kJ when calculating lnK.
• Reading positive ΔG as “nothing happens” instead of “the reverse reaction is spontaneous”.
• Forgetting that a spontaneous reaction may still be slow.

Next steps

• Entropy and spontaneity revision notes
• Entropy and spontaneity study guide
• All DP Chemistry units on the hub
• DP Chemistry topic checklist
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Official syllabus

These questions are based on the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 1.4 (additional higher level).