Consecutive Reactions
The reactions in which the reactant forms an intermediate and the intermediate forms the product in one or many subsequent reactions are called consecutive or sequential reactions. Every stage has its own reactant and rate constant. In such reactions, the product is not formed directly from the starting reactant.
Let us consider a simple reaction profile:
Where, A = reactant, B = intermediate, and C = product.
Initially, only the reactant 'A' is present. As the reaction starts, the concentration of reactant 'A' decreases and produces an intermediate 'B' through the $k_1$ rate constant. When 'B' is formed, it sequentially produces the product 'C' through the $k_2$ rate constant. After the complete conversion, only 'C' remains while the concentrations of A and B approach zero.
The overall rate of reaction depends upon the relative magnitude of these two ($k_1$ and $k_2$) rate constants.
During the course of the reaction, the concentrations of 'A', 'B', and 'C' vary as a function of time. The concentration of 'A' decreases exponentially, while the concentration of intermediate 'B' increases initially, reaches a maximum value, and then declines. The concentration of final product 'C' increases continuously, asymptotically reaching the value of the initial concentration of 'A'.
Suppose that the initial concentration of reactant 'A' is $[A]_0$, while the concentrations of 'A', 'B', and 'C' after time 't' are $[A]$, $[B]$, and $[C]$ respectively. Therefore:
$[A]_0 = [A] + [B] + [C]$ ----- (Eq. X)1. Rate law in terms of [A]
The rate of disappearance of reactant 'A' in the given system is expressed by the first-order differential relation:
$-d[A]/dt = k_1 [A]$ ----- (Eq. 1)$-d[A]/[A] = k_1 \cdot dt$ ----- (Eq. 2)
Integrating both sides of the differential equation yields:
$-\ln[A] = k_1 t + \text{constant}$ ----- (Eq. 3)At $t = 0$, $[A] = [A]_0$, which evaluates the constant integration parameter as $-\ln[A]_0$. Substituting this back:
$-\ln[A] = k_1 t - \ln[A]_0$ ----- (Eq. 4)$\ln[A]_0 - \ln[A] = k_1 t$ ----- (Eq. 5)
$\ln([A]/[A]_0) = -k_1 t$ ----- (Eq. 7)
$[A]/[A]_0 = e^{-k_1 t}$ ----- (Eq. 9)
Thus, the integrated rate equation for the initial reactant is:
$[A] = [A]_0 \cdot e^{-k_1 t}$ ----- (Eq. 10)2. Rate law in terms of [B]
The net rate of formation of the intermediate transient species 'B' is equal to its rate of production from A minus its rate of consumption to form C:
$d[B]/dt = k_1 [A] - k_2 [B]$Substituting the integrated value of $[A]$ from Eq. 10 establishes a first-order linear differential equation:
$d[B]/dt + k_2 [B] = k_1 [A]_0 \cdot e^{-k_1 t}$Integrating this equation using an integrating factor method gives:
$[B] = [A]_0 \left( \frac{k_1}{k_2 - k_1} \right) \left( e^{-k_1 t} - e^{-k_2 t} \right)$ ----- (Eq. 11)3. Rate law in terms of [C]
The rate of product tracking can be found by substituting the expressions for $[A]$ and $[B]$ back into the mass balance constraint (Eq. X):
$[C] = [A]_0 - [A] - [B]$$[C] = [A]_0 - [A]_0 \cdot e^{-k_1 t} - [A]_0 \left( \frac{k_1}{k_2 - k_1} \right) \left( e^{-k_1 t} - e^{-k_2 t} \right)$ ----- (Eq. 12)
$[C] = [A]_0 \left\{ 1 - e^{-k_1 t} - \frac{k_1}{k_2 - k_1} e^{-k_1 t} + \frac{k_1}{k_2 - k_1} e^{-k_2 t} \right\}$ ----- (Eq. 14)
$[C] = [A]_0 \left\{ 1 - \frac{k_2 e^{-k_1 t} - k_1 e^{-k_1 t} + k_1 e^{-k_1 t} - k_1 e^{-k_2 t}}{k_2 - k_1} \right\}$ ----- (Eq. 16)
Simplifying the numerator items yields the final integrated product expression:
$[C] = [A]_0 \left[ 1 - \frac{1}{k_2 - k_1} \left( k_2 e^{-k_1 t} - k_1 e^{-k_2 t} \right) \right]$ ----- (Eq. 18)Maximum Concentration of Intermediate B
At the highest peak concentration of intermediate B, the change in its rate becomes zero ($d[B]/dt = 0$). Differentiating Eq. 11 with respect to time:
$d[B]/dt = [A]_0 \left( \frac{k_1}{k_2 - k_1} \right) \left( -k_1 e^{-k_1 t} + k_2 e^{-k_2 t} \right) = 0$ ----- (Eq. 19)$k_1 e^{-k_1 t_{\max}} = k_2 e^{-k_2 t_{\max}}$
$\frac{k_1}{k_2} = e^{(k_1 - k_2)t_{\max}}$ ----- (Eq. 21)
Taking the natural logarithm isolates the time required to reach peak intermediate concentration ($t_{\max}$):
$t_{\max} = \frac{1}{k_1 - k_2} \ln\left(\frac{k_1}{k_2}\right)$ ----- (Eq. 23)Substituting $t_{\max}$ back into the integrated concentration framework of Eq. 11 provides the peak intermediate value:
$[B]_{\max} = [A]_0 \left( \frac{k_2}{k_1} \right)^{\frac{k_2}{k_1 - k_2}}$ ----- (Eq. 24)Rate Law Limiting Cases
Case 1: When $k_2 \gg k_1$
In this condition, the intermediate B converts to product C almost instantly upon formation. Because $k_1$ is small relative to $k_2$, Eq. 18 simplifies to $[C] \approx [A]_0(1 - e^{-k_1 t})$.
The concentration of the transient intermediate remains extremely low and practically constant throughout the reaction lifecycle, validating the use of the Steady-State Approximation (SSA).
Case 2: When $k_1 \gg k_2$
In this alternative condition, reactant A rapidly transforms entirely into intermediate B before any significant product C can assemble. Here, the rate-limiting step shifts, and the product profile reduces to tracking $[C] \approx [A]_0(1 - e^{-k_2 t})$.
Common Laboratory Example: The step-by-step basic saponification mechanism of a symmetrical diester in an alkaline environment.