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Reaction Kinetics: Question 8

Syllabus 26.1, 26.2

Structured A2 9 marks

A reaction between two aqueous compounds, E and F, produces G and H: E(aq)+2F(aq)→G(aq)+H(aq)\text{E(aq)} + 2\text{F(aq)} \rightarrow \text{G(aq)} + \text{H(aq)} The initial rate of this reaction was measured for three combinations of starting concentrations, at the same constant temperature.

Experiment [E] / mol dm⁻³ [F] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.20 0.20 3.0×10−43.0\times10^{-4}
2 0.40 0.20 6.0×10−46.0\times10^{-4}
3 0.20 0.40 3.0×10−43.0\times10^{-4}

(a) Deduce the order of reaction with respect to E and with respect to F, explaining your reasoning. [3]

(b) Write the rate equation for this reaction, and state the overall order of reaction. [1]

Two mechanisms have been proposed for this reaction, each consisting of two steps:

Mechanism 1: Step 1: E→I\text{E} \rightarrow \text{I} (slow) Step 2: I+2F→G+H\text{I} + 2\text{F} \rightarrow \text{G} + \text{H} (fast)

Mechanism 2: Step 1: E+F→I\text{E} + \text{F} \rightarrow \text{I} (slow) Step 2: I+F→G+H\text{I} + \text{F} \rightarrow \text{G} + \text{H} (fast)

(c) Deduce, with reference to the rate equation from part (b), which of these two mechanisms is consistent with the experimental data. Explain your reasoning, including why the other mechanism is not consistent. [3]

(d) Identify the role of species I in the mechanism you chose in part (c), and explain why F, although a reactant in the overall equation, does not appear in the rate equation. [2]

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Worked solution

Part (a): Deducing the order with respect to E and F

Order with respect to E (Experiments 1 and 2, where [F][\text{F}] is unchanged at 0.20 mol dm−30.20\text{ mol dm}^{-3}): [E]2[E]1=0.400.20=2rate2rate1=6.0×10−43.0×10−4=2\frac{[\text{E}]_2}{[\text{E}]_1} = \frac{0.40}{0.20} = 2 \qquad \frac{\text{rate}_2}{\text{rate}_1} = \frac{6.0\times10^{-4}}{3.0\times10^{-4}} = 2 Doubling [E][\text{E}] doubles the rate, so the reaction is first order with respect to E.

Order with respect to F (Experiments 1 and 3, where [E][\text{E}] is unchanged at 0.20 mol dm−30.20\text{ mol dm}^{-3}): [F]3[F]1=0.400.20=2rate3rate1=3.0×10−43.0×10−4=1\frac{[\text{F}]_3}{[\text{F}]_1} = \frac{0.40}{0.20} = 2 \qquad \frac{\text{rate}_3}{\text{rate}_1} = \frac{3.0\times10^{-4}}{3.0\times10^{-4}} = 1 Doubling [F][\text{F}] leaves the rate unchanged, so the reaction is zero order with respect to F.

Part (b): Rate equation and overall order

Since F is zero order, it does not appear in the rate equation: rate=k[E]\text{rate} = k[\text{E}] Overall order =1+0=1= 1 + 0 = 1 (first order overall).

Part (c): Matching the rate equation to a mechanism

The rate equation for a rate-determining (slow) step is written directly from the species reacting in that step alone, not from the overall balanced equation.

  • Mechanism 1’s slow step is E→I\text{E} \rightarrow \text{I}, involving only one molecule of E. This gives a predicted rate equation of rate=k[E]\text{rate} = k[\text{E}], which is first order in E and zero order in F (F does not appear at all, since it only reacts in the fast second step), this exactly matches the experimental rate equation from part (b).
  • Mechanism 2’s slow step is E+F→I\text{E} + \text{F} \rightarrow \text{I}, involving one molecule of E and one of F. This would give a predicted rate equation of rate=k[E][F]\text{rate} = k[\text{E}][\text{F}], first order in both E and F. This contradicts the experimental result that F is zero order, so Mechanism 2 is not consistent with the data.

Mechanism 1 is therefore the mechanism consistent with the experimental rate equation.

Part (d): The role of I, and why F is absent from the rate equation

In Mechanism 1, species I is a reaction intermediate: it is formed in Step 1 (the slow, rate-determining step) and is then completely used up in Step 2 (the fast step) to form the products G and H. Because it is produced and consumed within the mechanism, I does not appear in the overall balanced equation.

F does not appear in the rate equation because F only takes part in the fast second step, which occurs after the rate-determining step. The overall rate of a multi-step reaction is controlled entirely by the concentrations of the species involved in the slow, rate-determining step; since F is absent from that step, changing [F][\text{F}] has no effect on the observed rate, even though F is consumed overall.

Final answers

  • (a) First order with respect to E; zero order with respect to F.
  • (b) rate=k[E]\text{rate} = k[\text{E}]; overall order =1= 1.
  • (c) Mechanism 1 is consistent (its slow step, E→I\text{E}\rightarrow\text{I}, gives rate =k[E]=k[\text{E}]); Mechanism 2 would require F to be first order, which contradicts the data.
  • (d) I is a reaction intermediate (formed then consumed); F is absent from the rate equation because it reacts only in the fast step, after the rate-determining step.