MHTCET 20 Apr 2026 Shift 2 Paper

Concept:For a first-order reaction, the rate depends linearly on the concentration of the single reactant. The rate law is written as Rate $= k[A]$, where $k$ is the rate constant and $[A]$ is the molar concentration of reactant $A$.Explanation:The first-order rate law is given by:$$\text{Rate} = k[A]$$Rearranging this equation to solve for the rate constant $k$:$$k = \frac{\text{Rate}}{[A]}$$Substitute the given values into the equation:$$k = \frac{5.4 \times 10^{-6} \ \mathrm{mol \ dm^{-3} \ s^{-1}}}{0.3 \ \mathrm{M}}$$Perform the division of the numerical coefficients:$$\frac{5.4}{0.3} = 18$$Therefore, the rate constant becomes:$$k = 18 \times 10^{-6} \ \mathrm{s^{-1}}$$Express this result in standard scientific notation:$$k = 1.80 \times 10^{-5} \ \mathrm{s^{-1}}$$This value matches option D from the given choices.Answer:$$k = 1.80 \times 10^{-5} \ \mathrm{s^{-1}}$$Option D is the correct answer.