6.1: The Law of Mass Action
The law of mass action describes the rate at which chemicals interact in reactions. It is assumed that different chemical molecules come into contact by collision before reacting, and that the collision rate is directly proportional to the number of molecules of each reacting species. Suppose that two chemicals \(A\) and \(B\) react to form a product chemical \(C\), written as
\[A+B \stackrel{k}{\rightarrow} C, \nonumber \]
with \(k\) the rate constant of the reaction. For simplicity, we will use the same symbol \(C\), say, to refer to both the chemical \(C\) and its concentration. The law of mass action says that \(d C / d t\) is proportional to the product of the concentrations \(A\) and \(B\), with proportionality constant \(k\). That is,
\[\dfrac{d C}{d t}=k A B \label{6.1.2} \]
Similarly, the law of mass action enables us to write equations for the time-derivatives of the reactant concentrations \(A\) and \(B\) :
\[\dfrac{d A}{d t}=-k A B, \quad \dfrac{d B}{d t}=-k A B \nonumber \]
Notice that when using the law of mass action to find the rate-of-change of a concentration, the chemical that the arrow points towards is increasing in concentration (positive sign), the chemical that the arrow points away from is decreasing in concentration (negative sign). The product of concentrations on the right-hand-side is always that of the reactants from which the arrow points away, multiplied by the rate constant that is on top of the arrow.
Equation \ref{6.1.2} can be solved analytically using conservation laws. Each reactant, original and converted to product, is conserved since one molecule of each reactant gets converted into one molecule of product. Therefore,
\[\begin{aligned} \dfrac{d}{d t}(A+C) &=0 \quad & & A+C=A_{0} \\[4pt] \dfrac{d}{d t}(B+C) &=0 \quad & & \Longrightarrow & B+C &=B_{0} \end{aligned} \nonumber \]
where \(A_{0}\) and \(B_{0}\) are the initial concentrations of the reactants, and no product is present initially. Using the conservation laws, Equation \ref{6.1.2} becomes
\[\dfrac{d C}{d t}=k\left(A_{0}-C\right)\left(B_{0}-C\right), \text { with } C(0)=0 \nonumber \]
which may be integrated by separating variables. After some algebra, the solution is determined to be
\[C(t)=A_{0} B_{0} \dfrac{e^{\left(B_{0}-A_{0}\right) k t}-1}{B_{0} e^{\left(B_{0}-A_{0}\right) k t}-A_{0}} \nonumber \]
which is a complicated expression with the simple limits
\[\lim _{t \rightarrow \infty} C(t)= \begin{cases}A_{0} & \text { if } A_{0}<B_{0} \\[4pt] B_{0} & \text { if } B_{0}<A_{0}\end{cases} \nonumber \]
The reaction stops after one of the reactants is depleted; and the final concentration of the product is equal to the initial concentration of the depleted reactant.
If we also include the reverse reaction,
\[A+B \stackrel{k_{+}}{\stackrel{k}{k_{-}}} C_{1} \nonumber \]
then the time-derivative of the product is given by
\[\dfrac{d C}{d t}=k_{+} A B-k_{-} C \nonumber \]
Notice that \(k_{+}\)and \(k_{-}\)have different units. At equilibrium, \(\dot{C}=0\), and using the conservation laws \(A+C=A_{0}, B+C=B_{0}\), we obtain
\[\left(A_{0}-C\right)\left(B_{0}-C\right)-\dfrac{k_{-}}{k_{+}} C=0 \nonumber \]
from which we define the equilibrium constant \(K_{e q}\) by
\[K_{e q}=k_{-} / k_{+} \nonumber \]
which has units of concentration. Therefore, at equilibrium, the concentration of the product is given by the solution of the quadratic equation
\[C^{2}-\left(A_{0}+B_{0}+K_{e q}\right) C+A_{0} B_{0}=0 \nonumber \]
with the extra condition that \(0<C<\min \left(A_{0}, B_{0}\right) .\) For instance, if \(A_{0}=B_{0} \equiv R_{0}\), then at equilibrium,
\[C=R_{0}-\dfrac{1}{2} K_{e q}\left(\sqrt{1+4 R_{0} / K_{e q}}-1\right) \nonumber \]
If \(K_{e q} \ll R_{0}\), then \(A\) and \(B\) have a high affinity, and the reaction proceeds mainly to \(C\), with \(C \rightarrow R_{0}\).
Below are two interesting reactions. In reaction (ii), \(A\) is assumed to be held at a constant concentration.
\[A+X \stackrel{k_{+}}{\stackrel{\longrightarrow}{k_{-}}} 2 X \nonumber \]
(ii)
\[A+X \stackrel{k_{1}}{\rightarrow} 2 X, \quad X+Y \stackrel{k_{2}}{\rightarrow} 2 Y, \quad Y \stackrel{k_{3}}{\rightarrow} B \nonumber \]
Can you write down the equations for \(\dot{X}\) in reaction (i), and \(\dot{X}\) and \(\dot{Y}\) in reaction (ii)? When normalized properly, the equations from reaction (ii) reduce to the LotkaVolterra predator-prey equations introduced in \(\S 1.4\). The chemical concentrations \(X\) and \(Y\), therefore, oscillate in time like predators and their prey.