MHTCET 16 Apr 2026 Shift 2 Paper

Concept:For a first-order reaction, the integrated rate law relates $k$, time, and the concentration ratio $[A]_0/[A]_t$.Explanation:For a first-order reaction, the rate constant $k$ is calculated using the formula:$$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$$Here, $[A]_0 = 1.0\ \mathrm{N}$, $[A]_t = 0.25\ \mathrm{N}$, and $t = 276\ \mathrm{s}$.Substitute these values into the equation:$$k = \frac{2.303}{276} \log \frac{1.0}{0.25}$$Simplify the concentration ratio:$$\frac{1.0}{0.25} = 4$$Thus, the equation becomes:$$k = \frac{2.303}{276} \log 4$$Given $\log 4 = 0.6021$, calculate the numerator:$$2.303 \times 0.6021 = 1.386$$Now divide by the time:$$k = \frac{1.386}{276} = 0.0050\ \mathrm{s}^{-1}$$Since the concentration units cancel, any unit such as $\mathrm{N}$ or $\mathrm{M}$ gives the same result.Answer:The rate constant is $0.0050\ \mathrm{s}^{-1}$, which matches Option B.