Chemical reaction kinetics: Second-order kinetics
Kinetics of the reaction A + B → C
The chemical reaction is of the type \[\text{A}+\text{B }{\mathop{\longrightarrow}\limits_{}^{k}} \text{ C}\] with reaction rate constant \(k\). When we assume that this is an elementary reaction, then we have the following system of differential equations for the concentrations of A and B: \[ \left\{\;\begin{aligned} \frac{\dd[\text{A}]}{\dd t}\;&= -k\, [\text{A}]\, [\text{B}] \\ \\ \frac{\dd[\text{B}]}{\dd t}\;&=-k\, [\text{A}]\, [\text{B}]\end{aligned} \right.\]
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Suppose that \(x(t)\) is the amount of reactant A per unit volume that has reacted already at time \(t\). Then \(x(t)\) is also the amount of reactant B per unit of volume that has reacted already at time \(t\). If we denote the initial concentrations of reactant A and B as \(a\) and \(b\), respectively, then we have \([\text{A}]=a-x\) and \([\text{B}]=b-x\), and we can summarize the system of differential equations to the following single differential equation for \(x(t)\) \[\begin{aligned}\frac{\dd x}{\dd t} &=-\frac{\dd(a-x)}{\dd t}\\ \\ &=-\frac{\dd[\text{A}]}{\dd t}\\ \\ &= k\, [\text{A}]\, [\text{B}]\\ \\&=k\,(a-x)(b-x)\end{aligned}\]
We distinguish two cases.
a = b In this case we have \[\frac{\dd x}{\dd t}=k\,(a-x)^2\] and \[[A]=\frac{a}{1+k\,a\, t}\]
a ≠ b Also in this, the time course of the concentrations of reactants A and B can be expressed in exact formulas, but this is beyond the learning goals (although we learn all the required techniques). We only give a simulation to play with.
