MHTCET 13 Apr 2026 Shift 1 Paper

Concept:The unit of the rate constant depends on the overall order of the reaction.Explanation:For a reaction of order $n$, the rate law is written as:$$\text{Rate} = k[\text{Reactant}]^n$$Here, $k$ is the rate constant and $[\text{Reactant}]$ is the molar concentration.The unit of rate for any reaction is always:$$\mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{s}^{-1}$$Rearranging the rate law gives the unit of $k$:$$k = \frac{\text{Rate}}{[\text{Reactant}]^n}$$Substituting the units, the unit of $k$ becomes:$$(\mathrm{mol} \ \mathrm{dm}^{-3})^{1-n} \ \mathrm{s}^{-1}$$For a zero-order reaction, $n = 0$.So the unit of $k$ becomes:$$\mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{s}^{-1}$$This exactly matches the unit given in the question.Therefore, the reaction whose rate constant has unit $\mathrm{mol} \ \mathrm{dm}^{-3} \ \mathrm{s}^{-1}$ is a zero-order reaction.Answer:Option A: Zero-order reaction.