Match List-I with List-II List-I (Order of reaction) List-II (Unit of…

The correct answer is Option (3) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)

List-I (Order of reaction)	List-II (Unit of rate constant, k)
(A) nth order reaction	(I) $(mol\, L^{-1})^{1-n}\, s^{-1}$
(B) zero order reaction	(II) $mol\, L^{-1}\, s^{-1}$
(C) first order reaction	(IV) $s^{-1}$
(D) second order reaction	(III) $mol^{-1}\, L\, s^{-1}$

Complete Statement:

Units of rate constant ($k$) change with the order of reaction because rate always has the unit $\text{mol L}^{-1} \text{s}^{-1}$.

We start from the rate law:

$\text{Rate} = k [\text{Concentration}]^{n}$

Rate unit is always $\text{mol L}^{-1} \text{s}^{-1}$.

(A) $n^{th}$ Order Reaction

$\text{Rate} = k [C]^{n}$

So,

$k = \frac{\text{Rate}}{[C]^{n}}$

• Unit of rate = $\text{mol L}^{-1} \text{s}^{-1}$
• Unit of concentration = $\text{mol L}^{-1}$

Therefore, the unit of $k$ becomes:

$(\text{mol L}^{-1})^{1-n} \text{s}^{-1}$

So, $n^{th}$ order $\rightarrow (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$.

(B) Zero Order Reaction ($n = 0$)

$\text{Rate} = k [C]^{0}$

But $[C]^{0} = 1$

So $\text{Rate} = k$

Unit of $k$ = Unit of rate

$= \text{mol L}^{-1} \text{s}^{-1}$

(C) First Order Reaction ($n = 1$)

$\text{Rate} = k [C]$

$k = \frac{\text{Rate}}{[C]}$

• Unit of rate = $\text{mol L}^{-1} \text{s}^{-1}$
• Unit of concentration = $\text{mol L}^{-1}$

They cancel out, leaving only: $\text{s}^{-1}$

(D) Second Order Reaction ($n = 2$)

$\text{Rate} = k [C]^{2}$

$k = \frac{\text{Rate}}{[C]^{2}}$

$\text{Unit} = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^{2} \text{L}^{-2}} = \mathbf{L \text{ mol}^{-1} \text{s}^{-1}} \text{}$

Final Matching Summary:

• $n^{\text{th}}$ order $\rightarrow (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$
• Zero order $\rightarrow \text{mol L}^{-1} \text{s}^{-1}$
• First order $\rightarrow \text{s}^{-1}$
• Second order $\rightarrow \text{L mol}^{-1} \text{s}^{-1}$