A Integrated Rate Law for Two Different Reactants

In Lecture 2, we stated without proof the integrated rate law for a reaction between two different species, \(\mathrm{A} + \mathrm{B} \rightarrow \text{products}\), with rate law \(\nu = k[\mathrm{A}][\mathrm{B}]\). Here we derive that result. The derivation is not examinable, but it illustrates how mass balance can be used to reduce a two-variable problem to a single-variable integral.

A.1 Setting Up the Problem

Both concentrations change with time, so the differential rate law

\[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k[\mathrm{A}][\mathrm{B}]\]

contains two unknowns. To integrate, we need to express \([\mathrm{B}]\) in terms of \([\mathrm{A}]\). For 1:1 stoichiometry, mass balance (Lecture 1) tells us that A and B are consumed in equal amounts, so at any time \(t\):

\[[\mathrm{A}]_0 - [\mathrm{A}]_t = [\mathrm{B}]_0 - [\mathrm{B}]_t\]

Rearranging gives \([\mathrm{B}]_t = [\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}]_t\). Writing \(\Delta = [\mathrm{B}]_0 - [\mathrm{A}]_0\) for the difference in initial concentrations, the differential equation becomes:

\[\frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]([\mathrm{A}] + \Delta)} = -k\,\mathrm{d}t\]

A.2 Solving by Partial Fractions

The left-hand side can be decomposed using partial fractions:

\[\frac{1}{[\mathrm{A}]([\mathrm{A}] + \Delta)} = \frac{1}{\Delta}\!\left(\frac{1}{[\mathrm{A}]} - \frac{1}{[\mathrm{A}] + \Delta}\right)\]

Integrating both sides between \(t = 0\) and \(t\):

\[\begin{align*} \frac{1}{\Delta}\left[\ln[\mathrm{A}] - \ln([\mathrm{A}] + \Delta)\right]_{[\mathrm{A}]_0}^{[\mathrm{A}]_t} &= -kt \\[6pt] \frac{1}{\Delta}\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0} - \ln\frac{[\mathrm{B}]_t}{[\mathrm{B}]_0}\right) &= -kt \end{align*}\]

where in the second line we have used \([\mathrm{A}]_t + \Delta = [\mathrm{B}]_t\) and \([\mathrm{A}]_0 + \Delta = [\mathrm{B}]_0\). Rearranging gives the final result:

\[\begin{equation} kt = \frac{1}{[\mathrm{B}]_0 - [\mathrm{A}]_0}\ln\!\left(\frac{[\mathrm{B}]_t / [\mathrm{B}]_0}{[\mathrm{A}]_t / [\mathrm{A}]_0}\right) \tag{A.1} \end{equation}\]

A.3 The Equal-Concentration Limit

This result requires \([\mathrm{A}]_0 \neq [\mathrm{B}]_0\) (i.e. \(\Delta \neq 0\)); otherwise the partial fraction decomposition breaks down (we cannot divide by zero). If the initial concentrations are equal, then by mass balance \([\mathrm{A}]_t = [\mathrm{B}]_t\) at all times, so the rate law becomes \(\mathrm{d}[\mathrm{A}]/\mathrm{d}t = -k[\mathrm{A}]^2\). This is exactly the single-reactant second-order case already solved in Lecture 2 (Eqn. (2.4)).