MHTCET 15 May 2024 Shift 2 Paper

Concept:For this reaction, the rate law connects the observed rate to the concentrations of A and B raised to their respective orders, enabling the calculation of the rate constant k.Explanation:The reaction is first order in A and second order in B, so the rate law is written as:$$\text{Rate} = k[A]^{1}[B]^{2}$$Substitute the given rate and concentrations into this equation:$$3.6 \times 10^{-2} = k(0.2)^{1}(0.1)^{2}$$Since $(0.1)^{2} = 0.01$, the equation simplifies to:$$3.6 \times 10^{-2} = k \times 0.2 \times 0.01$$Multiplying $0.2$ and $0.01$ gives $0.002$, so:$$3.6 \times 10^{-2} = k \times 0.002$$Divide both sides by $0.002$ to isolate k:$$k = \frac{3.6 \times 10^{-2}}{0.002} = 18$$The overall order of the reaction is $1 + 2 = 3$, so the unit of k is derived as $\mathrm{mol}^{-2} \mathrm{dm}^{6} \mathrm{sec}^{-1}$.Answer:The rate constant is $k = 18 \mathrm{~mol}^{-2} \mathrm{dm}^{6} \mathrm{sec}^{-1}$, which corresponds to option C.