A Integrating Factor Method for Consecutive Reactions

In analyzing consecutive reactions, we encounter the following differential equation for the intermediate species B:

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

This is a first-order linear differential equation of the form:

\[\frac{\mathrm{d}y}{\mathrm{d}t} + P(t)y = Q(t)\]

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

The integrating factor method involves multiplying both sides by \(\mathrm{e}^{\int P(t)\mathrm{d}t}\). Here:

\[\int P(t)\mathrm{d}t = \int k_2\mathrm{d}t = k_2t\]

So our integrating factor is \(\mathrm{e}^{k_2t}\). Multiplying both sides of our original equation:

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

The left side is now the derivative of \([\mathrm{B}]\mathrm{e}^{k_2t}\):

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

Integrating both sides:

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

Solving for [B]:

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

Using the initial condition [B]₀ = 0 at t = 0:

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

Therefore:

\[C = -\frac{k_1[\mathrm{A}]_0}{k_2-k_1}\]

And our final solution is:

\[[\mathrm{B}] = \frac{k_1[\mathrm{A}]_0}{k_2-k_1}(\mathrm{e}^{-k_1t} - \mathrm{e}^{-k_2t})\]

This method of solution is generally applicable to first-order linear differential equations that arise in chemical kinetics.