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Reaction Kinetics: Question 4

Syllabus 26.1

Structured A2 10 marks

In an investigation into the kinetics of the reaction between two aqueous reactants, A and B, A(aq)+2B(aq)→C(aq)+D(aq)\text{A(aq)} + 2\text{B(aq)} \rightarrow \text{C(aq)} + \text{D(aq)} the initial rate of reaction was measured for three different combinations of starting concentrations, at the same constant temperature. The results are shown below.

Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 2.0×10−32.0\times10^{-3}
2 0.20 0.10 4.0×10−34.0\times10^{-3}
3 0.20 0.20 16.0×10−316.0\times10^{-3}

(a) Using Experiments 1 and 2, deduce the order of reaction with respect to A. Using Experiments 2 and 3, deduce the order of reaction with respect to B. Show your reasoning in each case. [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]=0.30 mol dm−3[\text{A}] = 0.30\text{ mol dm}^{-3} and [B]=0.15 mol dm−3[\text{B}] = 0.15\text{ mol dm}^{-3}. [2]

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Worked solution

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

Order with respect to A (compare Experiments 1 and 2, where [B][\text{B}] is unchanged at 0.10 mol dm−30.10\text{ mol dm}^{-3}):

[A]2[A]1=0.200.10=2rate2rate1=4.0×10−32.0×10−3=2\frac{[\text{A}]_2}{[\text{A}]_1} = \frac{0.20}{0.10} = 2 \qquad \frac{\text{rate}_2}{\text{rate}_1} = \frac{4.0\times10^{-3}}{2.0\times10^{-3}} = 2

Doubling [A][\text{A}] doubles the rate, so rate ∝[A]1\propto [\text{A}]^1: the reaction is first order with respect to A.

Order with respect to B (compare Experiments 2 and 3, where [A][\text{A}] is unchanged at 0.20 mol dm−30.20\text{ mol dm}^{-3}):

[B]3[B]2=0.200.10=2rate3rate2=16.0×10−34.0×10−3=4\frac{[\text{B}]_3}{[\text{B}]_2} = \frac{0.20}{0.10} = 2 \qquad \frac{\text{rate}_3}{\text{rate}_2} = \frac{16.0\times10^{-3}}{4.0\times10^{-3}} = 4

Doubling [B][\text{B}] quadruples the rate (4=224 = 2^2), so rate ∝[B]2\propto [\text{B}]^2: the reaction is second order with respect to B.

Part (b): Rate equation and overall order

rate=k[A][B]2\text{rate} = k[\text{A}][\text{B}]^2

Overall order =1+2=3= 1 + 2 = 3 (third 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, [B]=0.10[\text{B}]=0.10, rate =2.0×10−3=2.0\times10^{-3}):

k=rate[A][B]2=2.0×10−30.10×(0.10)2=2.0×10−30.10×0.010=2.0×10−31.0×10−3=2.0k = \frac{\text{rate}}{[\text{A}][\text{B}]^2} = \frac{2.0\times10^{-3}}{0.10 \times (0.10)^2} = \frac{2.0\times10^{-3}}{0.10 \times 0.010} = \frac{2.0\times10^{-3}}{1.0\times10^{-3}} = 2.0

Check using Experiment 3’s data instead ([A]=0.20[\text{A}]=0.20, [B]=0.20[\text{B}]=0.20, rate =16.0×10−3=16.0\times10^{-3}):

k=16.0×10−30.20×(0.20)2=16.0×10−30.20×0.040=16.0×10−38.0×10−3=2.0k = \frac{16.0\times10^{-3}}{0.20 \times (0.20)^2} = \frac{16.0\times10^{-3}}{0.20 \times 0.040} = \frac{16.0\times10^{-3}}{8.0\times10^{-3}} = 2.0

Both experiments give the same value, confirming k=2.0k = 2.0.

Units: for a third-order reaction, k=rate[A][B]2k = \dfrac{\text{rate}}{[\text{A}][\text{B}]^2} has units

mol dm−3s−1(mol dm−3)×(mol dm−3)2=mol dm−3s−1mol3dm−9=mol1−3dm−3−(−9)s−1=mol−2dm6s−1\frac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})\times(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3}\text{s}^{-1}}{\text{mol}^3\text{dm}^{-9}} = \text{mol}^{1-3}\text{dm}^{-3-(-9)}\text{s}^{-1} = \text{mol}^{-2}\text{dm}^{6}\text{s}^{-1}

So k=2.0 mol−2dm6s−1k = 2.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}.

Part (d): Predicting the rate for new concentrations

Using the rate equation from part (b) with [A]=0.30 mol dm−3[\text{A}]=0.30\text{ mol dm}^{-3} and [B]=0.15 mol dm−3[\text{B}]=0.15\text{ mol dm}^{-3}:

rate=k[A][B]2=2.0×0.30×(0.15)2=2.0×0.30×0.0225\text{rate} = k[\text{A}][\text{B}]^2 = 2.0 \times 0.30 \times (0.15)^2 = 2.0 \times 0.30 \times 0.0225

Working left to right: 0.30×0.0225=0.006750.30 \times 0.0225 = 0.00675, then 2.0×0.00675=0.01352.0 \times 0.00675 = 0.0135.

Check by grouping differently: 2.0×0.30=0.602.0 \times 0.30 = 0.60, then 0.60×0.0225=0.01350.60 \times 0.0225 = 0.0135, the same result, confirming the calculation.

rate=1.35×10−2 mol dm−3s−1\text{rate} = 1.35\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}

Final answers

  • (a) First order with respect to A; second order with respect to B.
  • (b) rate=k[A][B]2\text{rate} = k[\text{A}][\text{B}]^2; overall order =3= 3.
  • (c) k=2.0 mol−2dm6s−1k = 2.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}.
  • (d) rate=1.35×10−2 mol dm−3s−1\text{rate} = 1.35\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}.