Match List-I with List-II List-I Order of reaction List-II Expression…

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

List-I Order of reaction	List-II Expressions/Unit
(A) Unit of zero order	(II) $mol\, L^{-1} s^{-1}$
(B) Unit of First order	(III) $s^{-1}$
(C) Integrated Rate Law for zero order	(I) $kt= [R]_0-[R]_t$
(D) Integrated Rate Law for First order	(IV) $[R]= [R]_0e^{-kt}$

(A) Unit of Zero Order — (II) $\text{mol L}^{-1} \text{s}^{-1}$

In a zero-order reaction, the rate is independent of concentration ($Rate = k[R]^0 = k$). Therefore, the units of the rate constant $k$ are the same as the units of the rate itself, which is concentration over time: $\text{mol L}^{-1} \text{s}^{-1}$.

(B) Unit of First Order — (III) $\text{s}^{-1}$

In a first-order reaction, the rate is proportional to the concentration ($Rate = k[R]^1$). Rearranging for $k$ gives $k = Rate / [R]$. Using units: $(\text{mol L}^{-1} \text{s}^{-1}) / (\text{mol L}^{-1}) = \mathbf{s^{-1}}$.

(C) Integrated Rate Law for Zero Order — (I) $kt = [R]_0 - [R]_t$

The differential rate law for a zero-order reaction is $-d[R]/dt = k$. Integrating this from $t=0$ to $t$ gives $[R]_t = -kt + [R]_0$. Rearranging this to show the change in concentration over time gives $kt = [R]_0 - [R]_t$.

(D) Integrated Rate Law for First Order — (IV) $[R] = [R]_0 e^{-kt}$

The differential rate law for a first-order reaction is $-d[R]/dt = k[R]$. Integrating this yields $\ln[R] = -kt + \ln[R]_0$. Converting this from logarithmic form to exponential form gives the expression $[R] = [R]_0 e^{-kt}$, which describes the exponential decay of the reactant.