Question

Find the unit of the rate constant of a reaction represented with a rate equation, $\text{rate}=k[A]^{1/2}[B]^{3/2}$

• A. \( \text{mol}^{-1} \text{ L s}^{-1} \)
• B. \( \text{s}^{-1} \)
• C. \( \text{mol L}^{-1} \text{ s}^{-1} \)
• D. \( \text{mol}^{-2} \text{ L}^2 \text{ s}^{-1} \)
• E. \( \text{mol}^{-3} \text{ L}^3 \text{ s}^{-1} \)

Hint:

A quick way to remember: The sum of the exponents of 'mol' and 'L' in the unit of \(k\) always equals 0, and the exponent of 'L' is always \((n-1)\).

Answer 1

The Correct Option is A

Concept: The general unit for the rate constant \(k\) for a reaction of order \(n\) is given by: \[ \text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1} \] The overall order of the reaction \(n\) is the sum of the powers of the concentration terms in the rate law.

Step 1: Calculate the overall order of the reaction \(n\).
From the rate equation \( \text{rate} = k[A]^{1/2}[B]^{3/2} \): \[ n = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2 \] The reaction is of the second order.

Step 2: Substitute the value of \(n\) into the general unit formula.
For \(n = 2\): \[ \text{Unit of } k = (\text{mol L}^{-1})^{1-2} \text{ s}^{-1} \] \[ \text{Unit of } k = (\text{mol L}^{-1})^{-1} \text{ s}^{-1} \]

Step 3: Simplify the expression.
\[ \text{Unit of } k = \text{mol}^{-1} \text{ L s}^{-1} \]