In a standard second-order reaction mechanism , exactly 20% of the reactant is found to be consumed — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For second order, $k = \frac{1}{t} \left( \frac{1}{a-x} - \frac{1}{a} \right)$. When 20% is complete, $x = 0.2a$. $k \times 10 = \frac{1}{0.8a} - \frac{1}{a} = \frac{1.25}{a} - \frac{1}{a} = \frac{0.25}{a}$. Thus, $k \cdot a = \frac{0.25}{10} = 0.025 \text{ min}^{-1}$. The half life $t_{1/2}$ for second order is $t_{1/2} = \frac{1}{ka} = \frac{1}{0.025} = 40 \text{ minutes}$.