The unit of rate constant of a reaction with the rate law, Rate = $k[A]^{3/2}[B]^{-1}$ is

$mol^{1/2}\, L^{-1/2}\, s^{-1}$

The correct answer is Option (4) → $mol^{1/2}\, L^{-1/2}\, s^{-1}$

We need to calculate the unit of rate constant k step by step.

Given rate law:

$\text{Rate} = k [A]^{3/2} [B]^{-1}$

• Rate unit: $\text{mol L}^{-1} \text{s}^{-1}$
• Concentration unit: [A] and [B] → $\text{mol L}^{-1}$

Step 1: Express units

$\text{mol L}^{-1} \text{s}^{-1} = k \times (\text{mol L}^{-1})^{3/2} \times (\text{mol L}^{-1})^{-1}$

$\text{mol L}^{-1} \text{s}^{-1} = k \times (\text{mol}^{3/2} \text{L}^{-3/2}) \times (\text{mol}^{-1} \text{L}^{1})$

$\text{mol L}^{-1} \text{s}^{-1} = k \times (\text{mol}^{3/2 - 1} \text{L}^{-3/2 + 1})$

$\text{mol L}^{-1} \text{s}^{-1} = k \times (\text{mol}^{1/2} \text{L}^{-1/2})$

Step 2: Solve for k

$k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^{1/2} \text{L}^{-1/2}} = \text{mol}^{1 - 1/2} \text{L}^{-1 + 1/2} \text{s}^{-1}$

$k = \text{mol}^{1/2} \text{L}^{-1/2} \text{s}^{-1}$