MHTCET 13 Apr 2026 Shift 1 Paper

Concept:The rate law expresses how the reaction rate depends on the concentrations of the reactants. We can substitute the given concentration and rate values into the rate law to obtain the rate constant $k$.Explanation:The given rate law is:$$\text{rate} = k[A][B]^{2}$$We are given:$$\text{rate} = 0.25\ \mathrm{M\,s^{-1}},\quad [A] = 1\ \mathrm{M},\quad [B] = 0.2\ \mathrm{M}$$Substitute these values into the rate law:$$0.25 = k(1)(0.2)^{2}$$Calculate $(0.2)^2$:$$(0.2)^2 = 0.04$$So the equation becomes:$$0.25 = k \times 0.04$$Solve for $k$:$$k = \frac{0.25}{0.04} = 6.25$$The overall order of the reaction is $1 + 2 = 3$.For a third-order reaction, the unit of the rate constant is $\mathrm{M^{-2}\,s^{-1}}$.Therefore, the rate constant is:$$k = 6.25\ \mathrm{M^{-2}\,s^{-1}}$$Answer:The rate constant of the reaction at $25^\circ\mathrm{C}$ is $6.25\ \mathrm{M^{-2}\,s^{-1}}$, which corresponds to option A.