Edexcel International A Level Chemistry: Entropy (YCH11) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers Topic 12A: Entropy of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 12.1 to 12.11. Topic 12 appears in the International A2 (IA2) unit named Unit 4: Rates, Equilibria and Further Organic Chemistry. Learn the content first from the entropy study guide and the entropy revision notes. Mark off each outcome you have practised on the Edexcel Chemistry checklist; every other topic is on the course hub.

Use the data given. S⦵ values are in J mol⁻¹ K⁻¹ and enthalpy changes in kJ mol⁻¹.

Questions

1. Ammonium nitrate dissolves in water at room temperature even though the process is endothermic. State what this shows about enthalpy change and feasibility, and name the other factor that must be considered. [2]

2. Entropy is described as a measure of disorder.

(a) Describe two ways in which a system can become more disordered at the particle level. [2] (b) State the entropy of a perfect crystal at 0 K and explain your answer. [2]

3. Place these three substances in order of increasing standard entropy at 298 K, and explain the order: Hg(l), C(diamond), Ne(g). [3]

4. A scented vapour is released in one corner of a still, closed room. Explain, in terms of entropy, why the scent soon reaches every part of the room. [2]

5. For each change below, predict whether ΔSsystem is positive or negative and justify your prediction.

(a) C₆H₁₂(l) → C₆H₁₂(s) [1] (b) 2CO(g) + O₂(g) → 2CO₂(g) [1] (c) NaNO₃(s) → Na⁺(aq) + NO₃⁻(aq) [1] (d) Ag₂CO₃(s) → Ag₂O(s) + CO₂(g) [1]

6. For an exothermic reaction, ΔH = −30 kJ mol⁻¹.

(a) Calculate ΔSsurroundings at 200 K and at 600 K. [2] (b) Explain what your answers show about the effect of temperature on the feasibility of this reaction if its ΔSsystem is negative. [2]

7. Silver(I) oxide decomposes on heating:

2Ag₂O(s) → 4Ag(s) + O₂(g) ΔH = +62.2 kJ mol⁻¹

S⦵ values: Ag₂O(s) 121.3; Ag(s) 42.6; O₂(g) 205.0.

(a) Calculate ΔSsystem. [2] (b) Calculate the temperature above which the decomposition becomes feasible. [2]

8. Steam passed over hot coke (carbon) forms a fuel gas mixture:

C(s) + H₂O(g) → CO(g) + H₂(g) ΔH = +131 kJ mol⁻¹

S⦵ values: C(s) 5.7; H₂O(g) 188.7; CO(g) 197.6; H₂(g) 130.6.

(a) Calculate ΔSsystem. [2] (b) Calculate ΔSsurroundings and ΔStotal at 298 K. Does the reaction go at room temperature? [3] (c) Find the lowest temperature at which ΔStotal becomes positive. State one assumption you make. [2]

9. Ethanol can be made by the direct hydration of ethene:

C₂H₄(g) + H₂O(g) → C₂H₅OH(g) ΔH = −45.7 kJ mol⁻¹

S⦵ values: C₂H₄(g) 219.3; H₂O(g) 188.7; C₂H₅OH(g) 281.6.

(a) Calculate ΔSsystem and explain its sign. [2] (b) Calculate ΔStotal at 298 K. [2] (c) Above what temperature does ΔStotal for this reaction become negative under standard conditions? [2] (d) The industrial process runs well above this temperature. Suggest why a high temperature is still used. [1]

10. A student mixes solid barium hydroxide octahydrate with solid ammonium chloride in a beaker standing on a wet wooden block.

Ba(OH)₂·8H₂O(s) + 2NH₄Cl(s) → BaCl₂(s) + 2NH₃(g) + 10H₂O(l) ΔH = +164 kJ mol⁻¹

S⦵ values: Ba(OH)₂·8H₂O(s) 427.0; NH₄Cl(s) 94.6; BaCl₂(s) 123.7; NH₃(g) 192.5; H₂O(l) 69.9.

(a) State what happens to the water between the beaker and the block, and why. [1] (b) Calculate ΔSsystem. [2] (c) Calculate ΔSsurroundings at 298 K. [2] (d) Calculate ΔStotal at 298 K and state what it shows. [2] (e) Explain how this endothermic reaction can occur at room temperature even though ΔSsurroundings is negative. [1]

11. An ionic salt Z dissolves in water: ΔsolH = +40.0 kJ mol⁻¹. S⦵ values: Z(s) 140.0; cation(aq) 102.0; anion(aq) 158.0.

(a) Explain why dissolving an ionic solid involves two opposing entropy effects. [2] (b) Calculate ΔStotal for dissolving Z at 298 K, and state whether Z dissolves to form a standard solution at this temperature. [3] (c) Calculate the temperature above which this dissolving becomes feasible. [1]

12. Ethanol boils at 351 K. Its enthalpy change of vaporisation at this temperature is +38.6 kJ mol⁻¹.

(a) State the value of ΔStotal for boiling at 351 K and explain why. [2] (b) Calculate ΔSsystem for vaporisation. [1] (c) Paper is thermodynamically unstable in air, yet a book can last for centuries. Use the ideas of thermodynamic and kinetic stability to explain this. [2]

Answers

1. Enthalpy change alone does not decide whether a change occurs [1]. Entropy must also be considered: the total entropy change must be positive [1]. [2] Examiner insight: link entropy to the total entropy change being positive; the bare word “entropy” may not be enough for the second mark.

2. (a) The molecules or particles spread out into more possible arrangements [1]. Energy quanta are shared out between the molecules in more possible ways [1]. (b) Zero [1]. There is only one possible arrangement: every particle is fixed in place and there is no thermal energy to disperse [1]. Examiner insight: the specification links disorder to both particles and energy quanta, so an answer that only mentions particles moving about is likely to earn only one of the two marks in (a).

3. C(diamond) < Hg(l) < Ne(g) [1]. Diamond is a solid whose atoms only vibrate in fixed positions in a rigid lattice [1]. A liquid’s particles can move past each other, and a gas’s particles move freely through the whole volume, so the gas has the most possible arrangements [1]. [3] Examiner insight: an order with no reasoning tied to the state of each substance is unlikely to earn full credit; refer to particle freedom for each state.

4. Random motion of the molecules spreads them out [1]. There are far more arrangements with the molecules spread through the room than concentrated in one corner, so spreading increases the entropy, which is the natural direction of change [1]. [2] Examiner insight: saying the molecules “want to” spread earns no credit; rest the explanation on random motion and the number of possible arrangements.

5. (a) Negative: liquid becomes solid, so the particles lose freedom of movement [1]. (b) Negative: 3 mol of gas become 2 mol of gas [1]. (c) Positive (usually): ions spread out from the lattice into solution [1]. (d) Positive: a gas is produced from a solid [1]. Examiner insight: give the sign and the reason together for each part; a correct sign with no reason is unlikely to be credited.

6. (a) At 200 K: ΔSsurroundings = −(−30 000)/200 = +150 J mol⁻¹ K⁻¹ [1]. At 600 K: −(−30 000)/600 = +50 J mol⁻¹ K⁻¹ [1]. (b) As temperature rises, ΔSsurroundings gets smaller, so it contributes less to ΔStotal [1]. With a negative ΔSsystem, the reaction is feasible only below a certain temperature [1]. Examiner insight: in (a), forgetting to convert kJ to J gives 0.15 and 0.05; with the unit J mol⁻¹ K⁻¹ these are wrong by a factor of 1000.

7. (a) ΔSsystem = [4(42.6) + 205.0] − 2(121.3) = 375.4 − 242.6 [1] = +132.8 J mol⁻¹ K⁻¹ [1]. (b) At the changeover, ΔStotal = 0, so T = ΔH/ΔSsystem = 62 200/132.8 [1] = 468 K [1]. Feasible above 468 K. Examiner insight: writing 62.2/132.8 = 0.468 K shows a unit error; convert ΔH to J mol⁻¹ before dividing.

8. (a) ΔSsystem = (197.6 + 130.6) − (5.7 + 188.7) = 328.2 − 194.4 [1] = +133.8 J mol⁻¹ K⁻¹ [1]. (b) ΔSsurroundings = −131 000/298 = −439.6 J mol⁻¹ K⁻¹ [1]. ΔStotal = 133.8 + (−439.6) = −305.8 J mol⁻¹ K⁻¹ [1]. Since ΔStotal is negative, the reaction does not go at room temperature [1]. (c) T = 131 000/133.8 = 979 K [1]. Assumption: ΔH and ΔSsystem do not change significantly with temperature [1]. Examiner insight: base the conclusion in (b) on the sign of ΔStotal; saying “not feasible because it is endothermic” gives the wrong reason even when the value is correct.

9. (a) ΔSsystem = 281.6 − (219.3 + 188.7) = −126.4 J mol⁻¹ K⁻¹ [1]. Negative because 2 mol of gas become 1 mol of gas [1]. (b) ΔSsurroundings = +45 700/298 = +153.4 J mol⁻¹ K⁻¹; ΔStotal = 153.4 − 126.4 [1] = +27.0 J mol⁻¹ K⁻¹ [1]. (c) T = 45 700/126.4 [1] = 362 K (above this, ΔStotal is negative) [1]. (d) A higher temperature gives a faster rate [1]. Examiner insight: in (d), an answer about entropy or yield misses the point; the question is about rate, which feasibility calculations do not predict.

10. (a) It freezes: the endothermic reaction takes in heat from the surroundings [1]. (b) ΔSsystem = [123.7 + 2(192.5) + 10(69.9)] − [427.0 + 2(94.6)] = 1207.7 − 616.2 [1] = +591.5 J mol⁻¹ K⁻¹ [1]. (c) ΔSsurroundings = −164 000/298 [1] = −550.3 J mol⁻¹ K⁻¹ [1]. (d) ΔStotal = 591.5 + (−550.3) = +41.2 J mol⁻¹ K⁻¹ [1]. It is positive, so the reaction is feasible at 298 K [1]. (e) ΔSsystem is large and positive (solids give a gas and a liquid) and outweighs the negative ΔSsurroundings [1]. Examiner insight: in (b), check that each coefficient is applied; missing the 10 in front of H₂O(l) is a common slip that changes the sign of ΔStotal.

11. (a) Spreading the ions from the lattice through the water increases entropy [1]. Water molecules become ordered around the ions, which decreases entropy [1]. (b) ΔSsystem = (102.0 + 158.0) − 140.0 = +120.0; ΔSsurroundings = −40 000/298 = −134.2 J mol⁻¹ K⁻¹ [1]. ΔStotal = 120.0 − 134.2 = −14.2 J mol⁻¹ K⁻¹ [1]. Negative, so Z does not dissolve to form a standard solution at 298 K [1]. (c) T = 40 000/120.0 = 333 K [1]. Examiner insight: in (a), two separate effects are needed for two marks; “the ions spread out” alone is unlikely to earn both.

12. (a) ΔStotal = 0 [1], because boiling liquid and its vapour are at equilibrium at 351 K, with no overall tendency to change either way [1]. (b) ΔSsystem = 38 600/351 = +110.0 J mol⁻¹ K⁻¹ [1]. (c) Paper is thermodynamically unstable because ΔStotal for its reaction with oxygen is positive, so the reaction is feasible [1]. It is kinetically stable because the activation energy is high, so the rate at room temperature is negligible [1]. Examiner insight: in (c), kinetic stability must be explained with activation energy or rate; repeating an energy or entropy argument does not explain why the reaction is slow.

Where marks are usually lost

• Mixing kJ mol⁻¹ and J mol⁻¹ K⁻¹ in one calculation, leaving an answer out by a factor of 1000.
• Losing the minus sign in ΔSsurroundings = −ΔH/T.
• Leaving out coefficients when summing standard entropies.
• Basing a feasibility conclusion on ΔH or ΔSsystem instead of ΔStotal.
• Quoting a changeover temperature with no “above” or “below”.
• Giving a sign for ΔSsystem with no reason.
• Explaining kinetic stability with entropy instead of activation energy.

Next steps

• Recap with the entropy revision notes.
• Reread the worked examples in the entropy study guide.
• See every Edexcel International A Level Chemistry resource on the course hub.
• Mark your progress on the Edexcel Chemistry checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 12: Entropy and Energetics, part 12A: Entropy, outcomes 12.1 to 12.11.