Question

The dimensions of the rate constant for reaction 2A $\to$ B are litre/ (mol.min), the order of the reaction is ______

• A. zero
• B. two
• C. three
• D. one

Hint:

A useful shortcut for the units of a rate constant \( k \) is:
\[ \text{Units of } k = (\text{Concentration})^{1-n} \cdot (\text{Time})^{-1} \]
For a second-order reaction (\( n=2 \)), this yields \( (\text{mol/L})^{-1} \cdot \text{min}^{-1} = \text{L}/(\text{mol}\cdot\text{min}) \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to determine the order of the reaction given the units of the rate constant (\( k \)), which are \( \text{L}/(\text{mol}\cdot\text{min}) \).
This is a standard kinetics problem in chemical reaction engineering.

Step 2: Key Formula or Approach:
For a reaction of order \( n \), the rate law is:
\[ -r_A = k \cdot C_A^n \]
where the rate of reaction \( -r_A \) has units of \( \text{mol}/(\text{L}\cdot\text{min}) \) and concentration \( C_A \) has units of mol/L.
The units of the rate constant \( k \) are:
\[ [k] = \frac{[-r_A]}{[C_A]^n} = \frac{\frac{\text{mol}}{\text{L}\cdot\text{min}}}{\left(\frac{\text{mol}}{\text{L}}\right)^n} = \left(\frac{\text{mol}}{\text{L}}\right)^{1-n} \cdot \text{min}^{-1} \]

Step 3: Detailed Explanation:

• Express the general units of \( k \) in terms of Litres and moles:
\[ [k] = \text{L}^{n-1} \cdot \text{mol}^{1-n} \cdot \text{min}^{-1} \]

• Compare this general expression with the given units:
Given units: \( \frac{\text{L}}{\text{mol}\cdot\text{min}} = \text{L}^1 \cdot \text{mol}^{-1} \cdot \text{min}^{-1} \)

• Equate the exponents of Litre (\( L \)):
\[ n - 1 = 1 \quad \implies \quad n = 2 \]

• Alternatively, equate the exponents of mole (\( mol \)):
\[ 1 - n = -1 \quad \implies \quad n = 2 \]

• Since the exponent matches for \( n = 2 \), the reaction is second-order.


Step 4: Final Answer:
The order of the reaction is two.