B Integrated Rate Law for Consecutive Reactions

In Lecture 4, we wrote the differential rate equation for the intermediate B in the consecutive reaction \(\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}\). The derivation below is not examinable, but it shows where the integrated rate law comes from and why the concentration profile of B has its characteristic shape. Substituting the first-order solution \([\mathrm{A}]_t = [\mathrm{A}]_0\mathrm{e}^{-k_1t}\) gives:

\[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = k_1[\mathrm{A}]_0\mathrm{e}^{-k_1t} - k_2[\mathrm{B}]\]

This cannot be solved by straightforward separation of variables, because both terms on the right-hand side involve time-dependent quantities. Instead, we use the integrating factor method.

B.1 Identifying the Standard Form

Rearranging to collect the \([\mathrm{B}]\) terms on the left gives a first-order linear ODE of the standard form \(\mathrm{d}y/\mathrm{d}t + P(t)\,y = Q(t)\):

\[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} + k_2[\mathrm{B}] = k_1[\mathrm{A}]_0\mathrm{e}^{-k_1t}\]

where \(P(t) = k_2\) (a constant) and \(Q(t) = k_1[\mathrm{A}]_0\mathrm{e}^{-k_1t}\).

B.2 Applying the Integrating Factor

The integrating factor is \(\mu(t) = \mathrm{e}^{\int P(t)\,\mathrm{d}t} = \mathrm{e}^{k_2t}\). Multiplying both sides by \(\mathrm{e}^{k_2t}\):

\[\mathrm{e}^{k_2t}\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} + k_2[\mathrm{B}]\mathrm{e}^{k_2t} = k_1[\mathrm{A}]_0\mathrm{e}^{(k_2-k_1)t}\]

The key step is recognising that the left-hand side is the derivative of a product:

\[\frac{\mathrm{d}}{\mathrm{d}t}\!\left([\mathrm{B}]\mathrm{e}^{k_2t}\right) = k_1[\mathrm{A}]_0\mathrm{e}^{(k_2-k_1)t}\]

Integrating both sides with respect to \(t\):

\[[\mathrm{B}]\mathrm{e}^{k_2t} = \frac{k_1[\mathrm{A}]_0}{k_2-k_1}\mathrm{e}^{(k_2-k_1)t} + \alpha\]

where \(\alpha\) is the constant of integration. Dividing through by \(\mathrm{e}^{k_2t}\):

\[[\mathrm{B}]_t = \frac{k_1[\mathrm{A}]_0}{k_2-k_1}\mathrm{e}^{-k_1t} + \alpha\mathrm{e}^{-k_2t}\]

B.3 Applying the Initial Condition

At \(t = 0\), \([\mathrm{B}]_0 = 0\) (we start with pure A), so:

\[0 = \frac{k_1[\mathrm{A}]_0}{k_2-k_1} + \alpha\]

giving \(\alpha = -k_1[\mathrm{A}]_0/(k_2-k_1)\). Substituting back:

\[\begin{equation} [\mathrm{B}]_t = \frac{k_1[\mathrm{A}]_0}{k_2-k_1}\left(\mathrm{e}^{-k_1t} - \mathrm{e}^{-k_2t}\right) \tag{B.1} \end{equation}\]

B.4 Interpreting the Result

The concentration of B is the difference of two decaying exponentials — one decaying at rate \(k_1\) (reflecting the supply of B from A) and the other at rate \(k_2\) (reflecting the removal of B to form C). At short times, the \(\mathrm{e}^{-k_1t}\) term dominates and \([\mathrm{B}]\) rises; at longer times, \([\mathrm{A}]\) is depleted and consumption of B overtakes its formation, so \([\mathrm{B}]\) falls back towards zero. This gives the characteristic rise-and-fall profile discussed in Lecture 4.

Note that Eqn. (B.1) requires \(k_1 \neq k_2\); if the two rate constants are equal, the factor \(1/(k_2 - k_1)\) diverges and the derivation above breaks down. The equal-rate-constant case requires a separate treatment (using L’Hôpital’s rule or re-solving the ODE directly), and gives \([\mathrm{B}]_t = k_1[\mathrm{A}]_0\,t\,\mathrm{e}^{-k_1t}\).