Kinetics of the reaction A + B → C

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

$[A]=\frac{a}{1+k\,a\, t}$

For math enthusuasts we give a proof.

We rewrite

$\frac{\dd x}{\dd t}=k\,(a-x)^2$ as

$\frac{x'(t)}{\bigl(a-x(t)\bigr)^2}=k$ and accocrding to the calculation rules for differentiating as

$\left(\frac{1}{a-x(t)}\right)'=k$ So

$\frac{1}{a-x(t)}=k\,t+C$ for some constant $C$ . Substitution of $t=0$ gives

$C=\frac{1}{a}$ Thus:

$\frac{1}{a-x(t)}=k\,t+\frac{1}{a} = \frac{k\,a\,t+1}{a}$ But then we have

$a-x(t)=\frac{a}{k\,a\,t+1}$ i.e.

$[A]=\frac{a}{1+k\,a\, t}$

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.

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$k$

$[A]$

$[B]$