MHTCET 13 Apr 2026 Shift 1 Paper

Concept:For a first-order reaction, the rate depends linearly on the concentration of one reactant, and its rate constant is found using the integrated rate law: $k = \frac{2.303}{t} \log_{10} \frac{[R]_0}{[R]}$. Here, $[R]_0$ is the initial concentration, $[R]$ is the concentration after time $t$, and $\log_{10}$ is the common logarithm.Explanation:It is given that only $20\%$ of the initial concentration remains after $t = 10$ min.Since only the ratio $\frac{[R]_0}{[R]}$ is needed, percentage values can be used directly.Take $[R]_0 = 100$ and $[R] = 20$.Therefore, $\frac{[R]_0}{[R]} = \frac{100}{20} = 5$.Substitute $t = 10$ min and this ratio into the first-order rate equation:$$k = \frac{2.303}{10} \log_{10}(5)$$Given $\log_{10}(5) = 0.6989$, we get:$$k = \frac{2.303}{10} \times 0.6989 = \frac{1.609}{10} = 0.1609 \text{ min}^{-1}$$Answer:The rate constant of the reaction is $0.1609 \text{ min}^{-1}$.Hence, the correct option is D: $0.1609 \text{ min}^{-1}$.