Rate equationsAQA A-Level Chemistry: Topic test

1. 1
   In acidic solution, bromate(V) ions oxidise bromide ions: BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l). The experimentally determined rate equation is rate = k[BrO₃⁻][Br⁻][H⁺]².

   (a)

   What is the overall order of the reaction?

   [1 mark]

   • A3
   • B4
   • C5
   • D12

   (b)

   What are the units of the rate constant, k?

   [1 mark]

   • Amol dm⁻³ s⁻¹
   • Bmol⁻¹ dm³ s⁻¹
   • Cmol⁻² dm⁶ s⁻¹
   • Dmol⁻³ dm⁹ s⁻¹

   (c)

   The concentration of H⁺ is doubled and the concentration of Br⁻ is halved. The concentration of BrO₃⁻ is unchanged. Calculate the factor by which the rate changes.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Peroxodisulfate ions oxidise iodide ions: S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq). A student adds a fixed small amount of sodium thiosulfate and some starch to each reaction mixture at constant temperature and measures the time, t, for the blue-black colour to appear. Experiment 1: [S₂O₈²⁻] = 0.040 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, t = 40 s. Experiment 2: [S₂O₈²⁻] = 0.080 mol dm⁻³, [I⁻] = 0.040 mol dm⁻³, t = 20 s. Experiment 3: [S₂O₈²⁻] = 0.040 mol dm⁻³, [I⁻] = 0.020 mol dm⁻³, t = 80 s.

   (a)

   What is the order of reaction with respect to S₂O₈²⁻?

   [1 mark]

   • AFirst order
   • BZero order
   • CSecond order
   • DThird order

   (b)

   Why does the time, t, give a measure of the initial rate?

   [1 mark]

   • AThe thiosulfate acts as a catalyst, so the time is proportional to the rate
   • BThe colour appears when all of the peroxodisulfate has been used up
   • CThe same amount of iodine is produced in every experiment before the colour appears, so rate is proportional to 1/t
   • DThe starch reacts with the peroxodisulfate and slows the reaction by a fixed factor

   (c)

   Deduce the order of reaction with respect to I⁻ and write the rate equation for the reaction.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The decomposition of dinitrogen pentoxide in an inert solvent, 2N₂O₅ → 4NO₂ + O₂, is first order with respect to N₂O₅, so rate = k[N₂O₅]. At 298 K the rate constant k = 3.46 × 10⁻⁵ s⁻¹ and the activation energy Ea = 103 kJ mol⁻¹. The rate constant varies with temperature according to k = Ae^(−Ea/RT), where R = 8.31 J K⁻¹ mol⁻¹.

   (a)

   Calculate the value of the Arrhenius constant, A, at 298 K. Give your answer to two significant figures and include units.

   [3 marks]

   (b)

   Use your value of A to calculate the rate constant at 318 K. Hence calculate the initial rate of decomposition at 318 K when [N₂O₅] = 0.0150 mol dm⁻³.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   In acidic solution, hydrogen peroxide oxidises iodide ions: H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l). Initial rates were measured at constant temperature. Experiment 1: [H₂O₂] = 0.020 mol dm⁻³, [I⁻] = 0.010 mol dm⁻³, [H⁺] = 0.10 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: [H₂O₂] = 0.040 mol dm⁻³, [I⁻] = 0.010 mol dm⁻³, [H⁺] = 0.10 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [H₂O₂] = 0.020 mol dm⁻³, [I⁻] = 0.020 mol dm⁻³, [H⁺] = 0.10 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 4: [H₂O₂] = 0.020 mol dm⁻³, [I⁻] = 0.010 mol dm⁻³, [H⁺] = 0.20 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹.

   (a)

   Use the data to deduce the order of reaction with respect to each of H₂O₂, I⁻ and H⁺, write the rate equation, and calculate the value of the rate constant, k, including its units.

   [6 marks]

   (b)

   The rate equation is rate = k[H₂O₂][I⁻]. A three-step mechanism is proposed in which step 1 is the rate-determining step.
   Step 1: H₂O₂ + I⁻ → IO⁻ + H₂O
   Step 2: IO⁻ + H⁺ → HOI
   Step 3: HOI + H⁺ + I⁻ → I₂ + H₂O
   Explain how the mechanism is consistent with the rate equation and with the overall equation. Explain why a mechanism whose rate-determining step involved H⁺ would not be consistent with the experimental results.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   Nitrogen monoxide reacts with chlorine: 2NO(g) + Cl₂(g) → 2NOCl(g). The rate equation is rate = k[NO]²[Cl₂]. At 300 K the rate constant k = 2.6 × 10³ mol⁻² dm⁶ s⁻¹.

   (a)

   The concentration of NO is tripled while the concentration of Cl₂ is unchanged. By what factor does the rate increase?

   [1 mark]

   • A3
   • B6
   • C9
   • D27

   (b)

   The concentrations of both NO and Cl₂ are doubled. By what factor does the rate increase?

   [1 mark]

   • A8
   • B4
   • C6
   • D16

   (c)

   A mixture has [NO] = 0.0100 mol dm⁻³ and [Cl₂] = 0.0050 mol dm⁻³ at 300 K. Calculate the initial rate of reaction. Include units.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   Bromine reacts with methanoic acid in aqueous solution: Br₂(aq) + HCOOH(aq) → 2Br⁻(aq) + 2H⁺(aq) + CO₂(g). The solution is orange because of the bromine. A student keeps the temperature constant, uses a large excess of methanoic acid, and follows the reaction by measuring the colour intensity with a colorimeter at regular intervals so that a concentration–time graph for bromine can be plotted.

   (a)

   How is the initial rate of the reaction found from the concentration–time graph for bromine?

   [1 mark]

   • AFrom the area under the curve between time zero and the end of the reaction
   • BFrom the time taken for the concentration of bromine to halve
   • CBy dividing the initial concentration by the total time for the reaction
   • DFrom the gradient of a tangent drawn to the curve at time zero

   (b)

   In one experiment the concentration of bromine falls from 0.0080 mol dm⁻³ to 0.0040 mol dm⁻³ in 130 s, and then from 0.0040 mol dm⁻³ to 0.0020 mol dm⁻³ in a further 130 s. What is the order of reaction with respect to bromine?

   [1 mark]

   • AZero order
   • BFirst order
   • CSecond order
   • DThe order cannot be deduced because only one experiment was done

   (c)

   Describe how the student could use the concentration–time graph to confirm the order of reaction with respect to bromine without using half-lives.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   A polymer manufacturer uses an initiator that decomposes in a first order reaction. The rate constant was found to be 1.20 × 10⁻³ s⁻¹ at 310 K and 4.05 × 10⁻³ s⁻¹ at 325 K. The Arrhenius equation can be written ln k = −Ea/RT + ln A, where R = 8.31 J K⁻¹ mol⁻¹.

   (a)

   Calculate the activation energy, Ea, in kJ mol⁻¹.

   [3 marks]

   (b)

   A catalyst lowers the activation energy by 15.0 kJ mol⁻¹ and does not change A. Calculate the factor by which the rate constant increases at 325 K, and explain in terms of activation energy why the catalyst has this effect.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   The gas-phase reaction 2ICl(g) + H₂(g) → I₂(g) + 2HCl(g) was investigated at constant temperature by measuring initial rates. Experiment 1: [ICl] = 0.0150 mol dm⁻³, [H₂] = 0.0050 mol dm⁻³, initial rate = 3.6 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 2: [ICl] = 0.0300 mol dm⁻³, [H₂] = 0.0050 mol dm⁻³, initial rate = 7.2 × 10⁻⁷ mol dm⁻³ s⁻¹. Experiment 3: [ICl] = 0.0150 mol dm⁻³, [H₂] = 0.0150 mol dm⁻³, initial rate = 1.08 × 10⁻⁶ mol dm⁻³ s⁻¹.

   (a)

   Deduce the order of reaction with respect to each reactant, write the rate equation and calculate the rate constant, k, with its units. Hence calculate the initial rate when [ICl] = 0.0100 mol dm⁻³ and [H₂] = 0.0200 mol dm⁻³ at the same temperature.

   [6 marks]

   (b)

   A two-step mechanism is proposed.
   Step 1: ICl + H₂ → HI + HCl
   Step 2: HI + ICl → I₂ + HCl
   Use the rate equation to deduce which step is rate-determining and show that the mechanism is consistent with the overall equation. Identify the intermediate. Explain, with reference to activation energy, why increasing the temperature increases the rate constant, and state what a plot of ln k against 1/T would show about Ea.

   [6 marks]

   Total for question 8: 12 marks

End of questions