The units of rate constant of four reactions are given below. Arrange…

The correct answer is Option (4) → (B), (C), (A), (D)

The general unit for a rate constant (k) for an $n^{\text{th}}$ order reaction is:

Units of $k=(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$ or more commonly $\text{L}^{n-1}\,\text{mol}^{1-n}\,\text{s}^{-1}$

Let's find the order (n) for each unit:

(A) $s^{-1}$

Compare to ($\text{mol L}^{-1})^{1-n}\text{s}^{-1}$. For units of $\text{mol L}^{-1}$ to disappear, $1-n=0$.

$1-n=0\Rightarrow n=1$

Order = 1 (First Order)

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

Compare to ($\text{mol L}^{-1})^{1-n}\text{s}^{-1}$. For $\text{mol L}^{-1}$ to have a power of 1, $1-n=1$.

$1-n=1\Rightarrow n=0$

Order = 0 (Zero Order)

(C) $\text{mol}^{1/2}\text{L}^{-1/2}\text{s}^{-1}$

This is equivalent to $(\text{mol L}^{-1})^{1/2}\text{s}^{-1}.$

$1-n=\frac{1}{2}\Rightarrow n=1-\frac{1}{2}=\frac{1}{2}$

Order = $\frac{1}{2}$ (Half Order)$ = 0.5

(D) $\text{L}^{1/2}\text{mol}^{-1/2}\text{s}^{-1}$

This is equivalent to $(\text{mol L}^{-1})^{-1/2}\text{s}^{-1}.$

$1-n=-\frac{1}{2}\Rightarrow n=1+\frac{1}{2}=\frac{3}{2}$

Order = $\frac{3}{2}$ (One and a half Order)$ = 1.5

Increasing Order of Reaction: 0 < 0.5 < 1 < 1.5

So, (B) < (C) < (A) < (D)