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.

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 \(c\).