(g) Irreversible Series Reactions

Consider the following elementary series reaction. \[A \stackrel{k_1}{\rightarrow} R \stackrel{k_2}{\rightarrow} S\] with \(C_{R0}=C_{S0}=0\), and \(C_{A0}=C_A+C_R+C_S\). \[\begin{align*} -r_A &= -\frac{dC_A}{dt} = k_1C_A \\ r_R &= \frac{dC_R}{dt} = k_1C_A - k_2C_R \\ r_S &= \frac{dC_S}{dt} = k_2C_R \end{align*}\] Integrated form of the above equations are: \[\begin{align*} C_A &= C_{A0}\exp(-k_1t) \nonumber \\ C_R &= C_{A0}k_1\left[\frac{\exp(-k_1t)}{k_2-k_1} + \frac{\exp(-k_2t)}{k_1-k_2} \right] \nonumber \\ C_S &= C_{A0}\left[1+\frac{k_2}{k_1-k_2}\exp(-k_1t) + \frac{k_1}{k_2-k_1}\exp(-k_2t) \right] \nonumber \\ \text{For $k_1\gg k_2$, we get} \\ C_S &\approx C_{A0}[1-\exp(-k_2t)] \tag*{(13)} \\ \text{For $k_1\ll k_2$, we get} \\ C_S &\approx C_{A0}[1-\exp(-k_1t)] \tag*{(14)}\end{align*}\] From Eqns.(13) and (14), we could observe that the rate is controlled by the slower rate constant. Hence, it is the slowest of the steps in a series reaction that governs the overall rate of reaction. Fig.(11) shows the trend of the reaction. 


\[\left.{ \begin{array}{rl} t_{\text{max}} =& \dfrac{\ln(k_2/k_1)}{k_2-k_1} \\ \ \\ \frac{C_{R,\text{max}}}{C_{A0}} =& \left(\dfrac{k_1}{k_2}\right)^{k_2/(k_2-k_1)} \end{array}} \right\} \text{for $k_1\ne k_2$}\] and, \[\left.{ \begin{array}{rl} t_{\text{max}} =& \dfrac{1}{k} \\ \ \\ \frac{C_{R,\text{max}}}{C_{A0}} =& \dfrac{1}{e} \end{array}} \right\} \text{for $k_1 = k_2=k$}\]