Initial concentrations of both the reactants of a purely second order reaction (where ) are equal. I — Chemical Kinetics Chemistry Question
Question
Initial concentrations of both the reactants of a purely second order reaction $A+B \rightarrow \text{Products}$ (where $[A]_0 = [B]_0 = a$) are equal. It is observed that 60% of the reaction gets completed in $3000 \text{ seconds}$. How much time will be logically taken for 20% completion of the identical reaction?
💡 Solution & Explanation
The integrated rate law for a second order reaction with equal initial concentrations is $k = \frac{1}{t} \left[\frac{1}{a-x} - \frac{1}{a}\right] = \frac{x}{at(a-x)}$. For 60% completion, $x = 0.6a$ at $t = 3000 \text{ s}$. Thus, $k = \frac{0.6a}{a(3000)(0.4a)} = \frac{0.6}{1200 a} = \frac{1}{2000 a}$. For 20% completion, $x = 0.2a$. $k = \frac{0.2a}{a(t)(0.8a)} = \frac{0.2}{0.8 a t} = \frac{1}{4 a t}$. Equating the two $k$ expressions: $\frac{1}{2000 a} = \frac{1}{4 a t} \Rightarrow t = 500 \text{ s}$.