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Reaction Kinetics: Question 7

Syllabus 26.1

Structured A2 10 marks

An investigation into the kinetics of a reaction between three aqueous reactants, A, B and C, A(aq)+B(aq)+C(aq)→D(aq)\text{A(aq)} + \text{B(aq)} + \text{C(aq)} \rightarrow \text{D(aq)} measured the initial rate of reaction for four different combinations of starting concentrations, at the same constant temperature. The results are shown below.

Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ [C] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 0.10 4.0×10−34.0\times10^{-3}
2 0.20 0.10 0.10 8.0×10−38.0\times10^{-3}
3 0.10 0.20 0.10 4.0×10−34.0\times10^{-3}
4 0.10 0.10 0.20 8.0×10−38.0\times10^{-3}

(a) Using an appropriate pair of experiments in each case, deduce the order of reaction with respect to A, with respect to B, and with respect to C. Show your reasoning. [3]

(b) Write the rate equation for this reaction, and state the overall order of reaction. [2]

(c) Calculate the rate constant, kk, for this reaction at this temperature, including its units. [3]

(d) Calculate the initial rate of reaction, in mol dm⁻³ s⁻¹, for an experiment in which [A]=0.25 mol dm−3[\text{A}] = 0.25\text{ mol dm}^{-3}, [B]=0.50 mol dm−3[\text{B}] = 0.50\text{ mol dm}^{-3} and [C]=0.30 mol dm−3[\text{C}] = 0.30\text{ mol dm}^{-3}. [2]

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Worked solution

Part (a): Deducing the order with respect to A, B and C

Each comparison must hold two of the three concentrations constant while only one changes.

Order with respect to A (Experiments 1 and 2, where [B][\text{B}] and [C][\text{C}] are both unchanged at 0.10 mol dm−30.10\text{ mol dm}^{-3}): [A]2[A]1=0.200.10=2rate2rate1=8.0×10−34.0×10−3=2\frac{[\text{A}]_2}{[\text{A}]_1} = \frac{0.20}{0.10} = 2 \qquad \frac{\text{rate}_2}{\text{rate}_1} = \frac{8.0\times10^{-3}}{4.0\times10^{-3}} = 2 Doubling [A][\text{A}] doubles the rate, so the reaction is first order with respect to A.

Order with respect to B (Experiments 1 and 3, where [A][\text{A}] and [C][\text{C}] are both unchanged at 0.10 mol dm−30.10\text{ mol dm}^{-3}): [B]3[B]1=0.200.10=2rate3rate1=4.0×10−34.0×10−3=1\frac{[\text{B}]_3}{[\text{B}]_1} = \frac{0.20}{0.10} = 2 \qquad \frac{\text{rate}_3}{\text{rate}_1} = \frac{4.0\times10^{-3}}{4.0\times10^{-3}} = 1 Doubling [B][\text{B}] leaves the rate unchanged, so the reaction is zero order with respect to B.

Order with respect to C (Experiments 1 and 4, where [A][\text{A}] and [B][\text{B}] are both unchanged at 0.10 mol dm−30.10\text{ mol dm}^{-3}): [C]4[C]1=0.200.10=2rate4rate1=8.0×10−34.0×10−3=2\frac{[\text{C}]_4}{[\text{C}]_1} = \frac{0.20}{0.10} = 2 \qquad \frac{\text{rate}_4}{\text{rate}_1} = \frac{8.0\times10^{-3}}{4.0\times10^{-3}} = 2 Doubling [C][\text{C}] doubles the rate, so the reaction is first order with respect to C.

Part (b): Rate equation and overall order

Since B is zero order, [B]0=1[\text{B}]^0 = 1 and B drops out of the rate equation entirely: rate=k[A][C]\text{rate} = k[\text{A}][\text{C}] Overall order =1+0+1=2= 1 + 0 + 1 = 2 (second order overall).

Part (c): Calculating the rate constant, k

Rearranging for kk and substituting Experiment 1’s data ([A]=0.10[\text{A}]=0.10, [C]=0.10[\text{C}]=0.10, rate =4.0×10−3=4.0\times10^{-3}): k=rate[A][C]=4.0×10−30.10×0.10=4.0×10−31.0×10−2=0.40k = \frac{\text{rate}}{[\text{A}][\text{C}]} = \frac{4.0\times10^{-3}}{0.10 \times 0.10} = \frac{4.0\times10^{-3}}{1.0\times10^{-2}} = 0.40

Check using Experiment 4’s data instead ([A]=0.10[\text{A}]=0.10, [C]=0.20[\text{C}]=0.20, rate =8.0×10−3=8.0\times10^{-3}): k=8.0×10−30.10×0.20=8.0×10−32.0×10−2=0.40k = \frac{8.0\times10^{-3}}{0.10 \times 0.20} = \frac{8.0\times10^{-3}}{2.0\times10^{-2}} = 0.40 Both give the same value, confirming k=0.40k = 0.40.

Units: for a second-order reaction, k=rate[A][C]k = \dfrac{\text{rate}}{[\text{A}][\text{C}]} has units mol dm−3s−1(mol dm−3)×(mol dm−3)=mol dm−3s−1mol2dm−6=mol1−2dm−3−(−6)s−1=mol−1dm3s−1\frac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})\times(\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3}\text{s}^{-1}}{\text{mol}^2\text{dm}^{-6}} = \text{mol}^{1-2}\text{dm}^{-3-(-6)}\text{s}^{-1} = \text{mol}^{-1}\text{dm}^{3}\text{s}^{-1}

So k=0.40 mol−1dm3s−1k = 0.40\text{ mol}^{-1}\text{dm}^{3}\text{s}^{-1}.

Part (d): Predicting the rate for new concentrations

Since B does not appear in the rate equation, its value (0.50 mol dm−30.50\text{ mol dm}^{-3}) is not used in the calculation: rate=k[A][C]=0.40×0.25×0.30\text{rate} = k[\text{A}][\text{C}] = 0.40 \times 0.25 \times 0.30

Working left to right: 0.25×0.30=0.0750.25 \times 0.30 = 0.075, then 0.40×0.075=0.0300.40 \times 0.075 = 0.030.

Check by grouping differently: 0.40×0.25=0.100.40 \times 0.25 = 0.10, then 0.10×0.30=0.0300.10 \times 0.30 = 0.030, the same result, confirming the calculation.

rate=3.0×10−2 mol dm−3s−1\text{rate} = 3.0\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}

Final answers

  • (a) First order with respect to A; zero order with respect to B; first order with respect to C.
  • (b) rate=k[A][C]\text{rate} = k[\text{A}][\text{C}]; overall order =2= 2.
  • (c) k=0.40 mol−1dm3s−1k = 0.40\text{ mol}^{-1}\text{dm}^{3}\text{s}^{-1}.
  • (d) rate=3.0×10−2 mol dm−3s−1\text{rate} = 3.0\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}.