Successive reactions: A → B → C

Chemical reaction kinetics: Kinetics of multi-step reactions

Successive reactions: A → B → C

We consider two successive chemical reactions of type $\text{A }{\mathop{\longrightarrow}\limits_{}^{k_1}} \text{ B }{\mathop{\longrightarrow}\limits_{}^{k_2}} \text{ C}$ with reaction rate constants $k_1$ and $k_2$ . When we assume that both $\text{A }{\mathop{\longrightarrow}\limits_{}^{k_1}} \text{ B}$ as $\text{B }{\mathop{\longrightarrow}\limits_{}^{k_2}} \text{ C}$ are elementary reactions, then according to the law of mass action we can write down the following system of differential equations for the concentrations of A, B and C: $\left\{\begin{aligned} \frac{\dd[\text{A}]}{\dd t} &= -k_1\, [\text{A}] \\ \\ \frac{\dd[\text{B}]}{\dd t} &=k_1\, [\text{A}]-k_2\, [\text{B}]\\ \\ \frac{\dd[\text{C}]}{\dd t} &= k_2\, [\text{B}]\end{aligned} \right.$

The concentrations of each substance can be calculated exacly, given the initial concentrations $[\text{A}]_0$ , $[\text{B}]_0$ , $[\text{C}]_0$ . We examine the case $[\text{B}]_0=0, [\text{C}]_0=0$ : $\left\{\begin{aligned} {[\text{A}]} &= [\text{A}]_0e^{-k_1t} \\ \\ [\text{B}] &= \left\{\begin{array}{lr} [\text{A}]_0k_1\,t\,e^{-k_1t} & \text{if }k_1=k_2 \\ \\ \dfrac{ [\text{A}]_0\,k_1}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right) & \text{if }k_1\neq k_2 \end{array} \right. \\ \\ [\text{C}] &= [\text{A}]_0-[\text{A}]-[\text{B}]\end{aligned} \right.$

For the math enthusiast we give a proof.

We use $x(t)$ and $y(t)$ for the concentrations $[\text{A}]$ and $[\text{B}]$ , respectively, and we assume $[\text{A}]_0=a, [\text{B}]_0=0, [\text{C}]_0=0$ . The differential equation $\frac{\dd x}{\dd t}=-k_1\, x$ describes exponential decay and the solution equals $x=a\,e^{-k_1t}$ Substitution in the differential equation $\dfrac{\dd y}{\dd t}=k_1\,x-k_2\,y$ gives $\frac{\dd y}{\dd t}=a\,k_1\,e^{-k_1t}-k_2\,y$ Thus: $\frac{\dd y}{\dd t}+k_2\,y=a\,k_1\,e^{-k_1t}$ We multiply the left- and right-hand side with $e^{k_2t}$ and get $e^{k_2t}\frac{\dd y}{\dd t}+k_2\,e^{k_2t}\,y=a\,k_1\,e^{(k_2-k_1)t}$ Using the calculation rules for differentiation we get: $\frac{\dd}{\dd t}\left(e^{k_2t}\,y\right)=a\,k_1\,e^{(k_2-k_1)t}$ So: $e^{k_2t}\,y=a\,k_1\!\int e^{(k_2-k_1)t}\,dt$ For computing the integral we distinguish two cases.

If $k_1=k_2$ , then we really have $e^{k_2t}\,y=a\,k_1\!\int 1\,dt$ and we get $e^{k_2t}\,y=a\,k_1\,t+C$ for some constant $C$ . From $y(0)=0$ follows $C=0$ . So in this case, the solution can be written as $y=a\,k_1\,t\,e^{-k_1t}$

If $k_1\neq k_2$ , then we have an exponential function as integrand. The integral can be computed and the result ist: $e^{k_2t}\,y=\frac{a\,k_1}{k_2-k_1}e^{(k_2-k_1)t}+C$ From $y(0)=0$ follows $0=\frac{a\,k_1}{k_2-k_1}+C$ Thus: $C= -\frac{a\,k_1}{k_2-k_1}$ So: $e^{k_2t}\,y=\frac{a\,k_1}{k_2-k_1}\left(e^{(k_2-k_1)t}-1\right)$ Rewriting leads to: $y=\frac{a\,k_1}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)$

We provide a simulation to explore the successive reaction. For example, verify that if $k_1\ll k_2$ the concentration of substance B remains low because B is rapidly converted into C, and that the kinetics differs little from that of $\text{A}{\mathop{\longrightarrow}\limits_{}^{k_1}}\text{C}$ . In contrast, $k_2\ll k_1$ implies that a high concentration of substance B is achieved because the second reaction $\text{B }{\mathop{\longrightarrow}\limits_{}^{k_2}}\text{ C}$ is slow, and that substance B only starts to disappears when there is almost no substance A left for conversion.

0,0

0.50

$k_1$

0.25

$k_2$

$[C]$

$[A]$

$[B]$

A concrete example of two successive reactions is the thermal dissociation of dimethylether $\text{CH}_3\text{OCH}_3\longrightarrow \text{CH}_4 + \text{CH}_2\text{O}\longrightarrow \text{CH}_4 + \text{CO} +\text{H}_2$

However, this kinetic model is also applicable in quantitative pharmacokinetics. For example, in case of an oral administration of a drug we commonly deal with a so-called two-compartment model. This consists of the gastro-intestinal tract from which the pharmacon (the active ingredient of a drug) is absorbed, with the absorption rate constant $k_a$ , in what is called the central compartment. From this central compartment the drug will be eliminated with elimination rate constant $k_e$ . The picture below visualizes the compartmental model.

Let $A_\text{GI tract}$ and $A$ be the amount of the pharmacon in the gastro-intestinal tract and in the central compartment, repsectively. Then the following hold: $\left\{\begin{aligned} \frac{dA_\text{GI tract}}{\dd t} &= -k_aA_\text{GI tract} \\ \\ \frac{dA}{\dd t} &=k_aA_\text{GI tract}-k_eA\end{aligned} \right.$ For the amount of the drug in the central compartment we have then the biexponential formula $A=c\cdot (e^{-k_et}-e^{-k_at})$ for some positive constant $c$ .