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S5 Chemistry Practice Paper: Reaction Rates, Chemical Equilibrium and Acid-Base Titration (12 questions, 100 marks, 75 minutes)
S5 Chemistry: Reaction Rates, Equilibrium and Acid-Base Titration
S5 chemistry paper: rates and rate equations, reaction orders, collision theory, equilibrium and Kc, Le Chatelier, pH, Ka, buffers and titration. Full mark scheme.
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Paper details Level: S5 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Rates of Reaction (30 marks)
Section A
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. For the reaction A -> products at 25 C, the concentration of A falls from 0.80 mol dm-3 to 0.50 mol dm-3 in 60 s. (a) Calculate the average rate of consumption of A over this interval; (b) if the rate stays constant, what is [A] after a further 40 s? (6 marks) | 1. (a) Average rate = d[A]/dt = (0.80 – 0.50)/60 = 0.30/60 = 5.0e-03 mol dm-3 s-1 (3 marks); (b) [A] = 0.50 – 5.0e-03 x 40 = 0.30 mol dm-3 (3 marks). Marking note: the rate must be positive and carry units; reversing the subtraction loses 1 mark. |
| 2 | 2. A decomposition reaction is first order with a half-life of 25.0 min. (a) Calculate the rate constant k in min-1; (b) how long does it take for the concentration to fall to one eighth of its initial value? (7 marks) | 2. (a) For a first-order reaction t(1/2) = ln2/k, so k = 0.693/25.0 = 0.0277 = 0.028 min-1 (4 marks); (b) one eighth of the initial value requires 3 half-lives, t = 3 x 25.0 = 75.0 min (3 marks). Marking note: stating “3 half-lives” without a time earns 1 mark only. |
| 3 | 3. For the reaction aA + bB -> products the following initial rates were measured (rate in mol dm-3 s-1): run (1) [A] = 0.10, [B] = 0.10, rate 2.0e-3; run (2) [A] = 0.10, [B] = 0.20, rate 8.0e-3; run (3) [A] = 0.20, [B] = 0.10, rate 4.0e-3. (a) Deduce the order with respect to A and to B; (b) write the rate equation; (c) calculate the rate constant k with its units. (8 marks) | 3. (a) Comparing runs (1) and (2): [A] unchanged, [B] doubled and the rate rises from 2.0e-3 to 8.0e-3 (factor 4), so the order in B is 2; comparing (1) and (3): [B] unchanged, [A] doubled and the rate doubles, so the order in A is 1 (3 marks); (b) rate = k[A][B]^2 (2 marks); (c) k = 2.0e-3/(0.10 x 0.10^2) = 2.0, units dm6 mol-2 s-1 (3 marks). Marking note: the orders must be deduced from the rate ratios. |
| 4 | 4. The rate of a reaction roughly doubles for every 10 K rise in temperature. The temperature is raised from 300 K to 330 K. (a) By what factor does the rate increase at 330 K compared with 300 K? (b) If k = 3.0e-4 s-1 at 300 K, find k at 330 K. (c) Explain, in terms of collision theory, why a higher temperature increases the rate. (9 marks) | 4. (a) Three 10 K steps, so the rate at 330 K is 2^3 = 8 times that at 300 K (3 marks); (b) k(330 K) = 3.0e-4 x 8 = 2.4e-03 s-1 (3 marks); (c) a higher temperature raises the mean kinetic energy of the molecules, so a much larger fraction has energy at or above the activation energy; the collision frequency also rises, giving far more effective collisions (3 marks). Marking note: (c) must mention both activation energy and effective collisions. |
Section B: Chemical Equilibrium (30 marks)
Section B
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. In a closed 2.0 dm3 vessel, 0.50 mol of N2O4 decomposes to equilibrium: N2O4(g) <=> 2NO2(g). At equilibrium 0.20 mol of N2O4 remains. (a) Find the amount of NO2 at equilibrium; (b) find Kc, including its unit. (6 marks) | 5. (a) N2O4 decomposed = 0.50 – 0.20 = 0.30 mol; by the 1:2 stoichiometry n(NO2) = 2 x 0.30 = 0.60 mol (3 marks); (b) [N2O4] = 0.20/2.0 = 0.10 and [NO2] = 0.60/2.0 = 0.30 mol dm-3, so Kc = 0.30^2/0.10 = 0.90 mol dm-3 (3 marks). Marking note: omitting the division by 2.0 dm3 makes (b) invalid; a wrong unit loses 1 mark. |
| 6 | 6. At 700 K, Kc = 54 for H2(g) + I2(g) <=> 2HI(g). A 1.0 dm3 vessel is charged with 0.10 mol H2, 0.10 mol I2 and 0.20 mol HI. (a) Calculate the reaction quotient Qc; (b) state the direction of the net reaction and justify your answer. (7 marks) | 6. (a) Qc = [HI]^2/([H2][I2]) = 0.20^2/(0.10 x 0.10) = 4.0 (4 marks); (b) Qc = 4.0 is less than Kc = 54, so the system is not at equilibrium and the net reaction proceeds forwards (towards HI) until Qc reaches 54 (3 marks). Marking note: answering “to the right” without comparing Qc with Kc earns 1 mark. |
| 7 | 7. In a 1.0 dm3 vessel, 1.0 mol H2 and 1.0 mol I2 are allowed to reach equilibrium, at which point 1.50 mol of HI is present. (a) Find the equilibrium concentrations of H2 and I2; (b) calculate Kc. (8 marks) | 7. (a) Forming 1.50 mol HI consumes 0.75 mol H2 and 0.75 mol I2, so [H2] = [I2] = 1.0 – 0.75 = 0.25 mol dm-3 (4 marks); (b) Kc = 1.50^2/(0.25 x 0.25) = 36.0 (4 marks). Marking note: an ICE table earns 2 method marks even if the arithmetic is wrong. |
| 8 | 8. In the Haber process N2(g) + 3H2(g) <=> 2NH3(g), dH = -92 kJ mol-1. The conversion of nitrogen is 78% at 400 C and 62% at 500 C. (a) Calculate the percentage decrease in conversion; (b) use Le Chatelier principle to explain the effect of temperature on the yield; (c) explain why industry nevertheless operates at about 450 C. (9 marks) | 8. (a) Decrease = (78 – 62)/78 x 100% = 20.5% (3 marks); (b) the forward reaction is exothermic (dH < 0), so raising the temperature shifts the equilibrium in the endothermic (reverse) direction and the NH3 yield falls (3 marks); (c) 450 C is a compromise: a lower temperature favours the yield but makes the reaction too slow and the iron catalyst insufficiently active, so a moderate temperature is used together with high pressure (3 marks). Marking note: (c) must discuss the trade-off between rate or catalyst activity and yield. |
Section C: Acids, Bases and Titration (40 marks)
Section C
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. Calculate the pH at 25 C of 0.020 mol dm-3 HCl(aq), assuming complete ionisation, and state the assumptions you make. (8 marks) | 9. HCl is a strong acid and ionises completely, so [H+] = [HCl] = 0.020 mol dm-3 (2 marks); pH = -log(0.020) = 1.70 (4 marks); assumptions: the contribution from the self-ionisation of water is negligible and activity coefficients are taken as 1 (2 marks). Marking note: 1.70 or 1.7 are both accepted. |
| 10 | 10. A 0.10 mol dm-3 solution of the weak acid HA at 25 C has Ka = 1.8e-5. Calculate (a) [H+]; (b) the pH; (c) the degree of ionisation as a percentage. (9 marks) | 10. (a) Ka = [H+]^2/[HA], so [H+] = sqrt(1.8e-5 x 0.10) = 1.34e-03 mol dm-3 (3 marks); (b) pH = -log(1.34e-03) = 2.87 (3 marks); (c) degree of ionisation = 1.34e-03/0.10 x 100% = 1.34% (3 marks). Marking note: the assumption x << 0.10 must be stated; solving the quadratic and obtaining the same value loses nothing. |
| 11 | 11. A buffer is made by mixing equal volumes of 0.10 mol dm-3 HA (pKa = 4.76) and 0.050 mol dm-3 NaA. (a) Calculate the pH of the buffer; (b) explain why the pH changes very little when a small amount of NaOH is added. (11 marks) | 11. (a) On mixing equal volumes the ratio is unchanged, [A-]/[HA] = 0.050/0.10 = 0.50, so pH = pKa + log(0.50) = 4.76 – 0.30 = 4.46 (6 marks); (b) added OH- reacts with HA to form A-, so [HA] falls slightly and [A-] rises slightly; the ratio hardly changes and hence the pH is almost unchanged (5 marks). Marking note: (b) requires the neutralisation equation for full marks. |
| 12 | 12. 20.0 cm3 of hydrochloric acid of unknown concentration is titrated with 0.100 mol dm-3 NaOH; 25.0 cm3 of the alkali is used to reach the end point. (a) Calculate the amount of NaOH neutralised; (b) calculate the concentration of the acid; (c) what volume of 0.050 mol dm-3 NaOH would be needed instead? (12 marks) | 12. (a) n(NaOH) = 0.100 x 25.0/1000 = 2.50e-03 mol (4 marks); (b) HCl + NaOH -> NaCl + H2O in a 1:1 ratio, so n(HCl) = 2.50e-03 mol and [HCl] = 2.50e-03/0.0200 = 0.125 mol dm-3 (4 marks); (c) halving the concentration doubles the volume: V = 2.50e-03/0.050 x 1000 = 50.0 cm3 (4 marks). Marking note: ignoring the 1:1 ratio loses 2 marks; leaving the volume in dm3 loses 1 mark. |
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