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A compound A in solution reacts by two paths at once, as many organic compounds do when one…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A compound A in solution reacts by two paths at once, as many organic compounds do when one starting material can give two products. Both paths are first order in A, and B and C are stable: they do not react further or turn into each other.

\displaystyle \text{A} \xrightarrow{k_1} \text{B} \qquad\qquad \text{A} \xrightarrow{k_2} \text{C}

At $300\ \text{K}$ , a solution starts with $[\text{A}]_0 = 0.500\ \text{mol L}^{-1}$ and no B or C. The figure shows how the three concentrations change with time. The curves are drawn to scale for this run. The ratio $[\text{B}]/[\text{C}]$ stays at $3.00$ all through the run, and A has a half-life of $40.0\ \text{min}$ . The activation energies of the two paths are $E_1 = 80.0\ \text{kJ mol}^{-1}$ and $E_2 = 100.0\ \text{kJ mol}^{-1}$ , and their pre-exponential factors do not depend on temperature. A chemist who follows only the loss of A sees one rate constant, $k = k_1 + k_2$ . The activation energy found from it is defined by $E_{\text{app}} = -R\,\dfrac{d\ln k}{d(1/T)}$ . Take $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$ and $\ln 2 = 0.693$ .

A graph of concentration, in moles per litre, against time, in minutes, from 0 to 160 minutes. The curve for A starts at 0.500 and falls towards zero. The curves for B and C start at zero and rise towards steady values, the curve for B always three times as high as the curve for C.

What is $k_1$ at $300\ \text{K}$ ?
What is $k_2$ at $300\ \text{K}$ ?
What is $[\text{B}]$ after $60.0\ \text{min}$ ?
What is $E_{\text{app}}$ at $300\ \text{K}$ ?
The same run is repeated at $320\ \text{K}$ . What is the ratio $[\text{B}]/[\text{C}]$ now?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The half-life fixes the sum $k_1 + k_2$ and the product ratio fixes their ratio. Then follow each path with its own activation energy.

(a) and (b). $k = k_1 + k_2 = \dfrac{0.693}{40.0}\ \text{min}^{-1} = 0.01733\ \text{min}^{-1}$ . Since $\dfrac{d[\text{B}]}{dt} = k_1[\text{A}]$ and $\dfrac{d[\text{C}]}{dt} = k_2[\text{A}]$ , the ratio $\dfrac{[\text{B}]}{[\text{C}]} = \dfrac{k_1}{k_2} = 3.00$ . So $k_1 = \tfrac{3}{4}k = 0.01300\ \text{min}^{-1}$ and $k_2 = \tfrac{1}{4}k = 0.00433\ \text{min}^{-1}$ .

(c) B after $60.0\ \text{min}$ . A falls as $[\text{A}] = [\text{A}]_0e^{-kt}$ , with $kt = 1.040$ and $e^{-kt} = 0.3536$ . B takes the fraction $0.75$ of the A that has reacted:

\displaystyle [\text{B}] = 0.75 \times 0.500 \times 0.6464\ \text{mol L}^{-1} = 0.242\ \text{mol L}^{-1}

(d) The apparent activation energy. $\dfrac{d\ln k}{d(1/T)} = \dfrac{1}{k}\left(\dfrac{dk_1}{d(1/T)} + \dfrac{dk_2}{d(1/T)}\right) = -\dfrac{k_1E_1 + k_2E_2}{Rk}$ . So $E_{\text{app}}$ is the average of $E_1$ and $E_2$ weighted by the share of each path:

\displaystyle E_{\text{app}} = 0.75 \times 80.0 + 0.25 \times 100.0 = 85.0\ \text{kJ mol}^{-1}

(e) At $320\ \text{K}$ . The pre-exponential factors cancel in the ratio of the ratios:

\displaystyle \frac{(k_1/k_2)_{320}}{(k_1/k_2)_{300}} = \exp\left[\frac{E_2 - E_1}{R}\left(\frac{1}{320} - \frac{1}{300}\right)\right] = \exp\left[\frac{20000}{8.314}\left(\frac{1}{320} - \frac{1}{300}\right)\right] = e^{-0.5012} = 0.6058

So $\dfrac{[\text{B}]}{[\text{C}]} = 3.00 \times 0.6058 = 1.82$ . Heating favours the path with the higher activation energy, so the share of C grows.

Check. Integrating the three rate equations in small steps gives the same $[\text{B}]$ , and differentiating $\ln(k_1 + k_2)$ numerically gives the same $E_{\text{app}}$ .

Watch out. $E_{\text{app}}$ is not the plain average $90.0\ \text{kJ mol}^{-1}$ . The faster path counts for more, and the weighting itself changes with temperature.

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