Question

The following data is for a reaction between reactants A and B: \[ \begin{array}{|c|c|c|} \hline \text{Rate (mol L}^{-1}\text{s}^{-1}) & [A] & [B] \hline 2 \times 10^{-3} & 0.1\,M & 0.1\,M \hline 4 \times 10^{-3} & 0.2\,M & 0.1\,M \hline 1.6 \times 10^{-2} & 0.2\,M & 0.2\,M \hline \end{array} \] The order of the reaction with respect to A and B, respectively are:

• A. \(1,0\)
• B. \(0,1\)
• C. \(1,2\)
• D. \(2,1\)

Hint:

While finding reaction order, compare only those experiments where one reactant concentration changes and the other remains constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The order of a reaction represents the power to which the concentration of a reactant is raised in the rate law.
Let the rate law be: $Rate = k [A]^x [B]^y$.
To find $x$ (order w.r.t A) and $y$ (order w.r.t B), we compare experiments where one concentration is varied while the other is kept constant.
Step 2: Key Formula or Approach:
We will use the ratio of the rate equations from the provided data:
(1) $2 \times 10^{-3} = k (0.1)^x (0.1)^y$
(2) $4 \times 10^{-3} = k (0.2)^x (0.1)^y$
(3) $1.6 \times 10^{-2} = 16 \times 10^{-3} = k (0.2)^x (0.2)^y$
Step 3: Detailed Explanation:
To find order w.r.t A (x):
Compare Exp (1) and Exp (2) where $[B]$ is constant at $0.1 M$.
\[ \frac{Rate_2}{Rate_1} = \frac{k(0.2)^x (0.1)^y}{k(0.1)^x (0.1)^y} \]
\[ \frac{4 \times 10^{-3}}{2 \times 10^{-3}} = \left( \frac{0.2}{0.1} \right)^x \]
\[ 2 = 2^x \implies x = 1 \]
The reaction is first order with respect to A.

To find order w.r.t B (y):
Compare Exp (2) and Exp (3) where $[A]$ is constant at $0.2 M$.
\[ \frac{Rate_3}{Rate_2} = \frac{k(0.2)^1 (0.2)^y}{k(0.2)^1 (0.1)^y} \]
\[ \frac{16 \times 10^{-3}}{4 \times 10^{-3}} = \left( \frac{0.2}{0.1} \right)^y \]
\[ 4 = 2^y \implies 2^2 = 2^y \implies y = 2 \]
The reaction is second order with respect to B.

The overall order of the reaction is $1 + 2 = 3$, and the rate law is $Rate = k[A]^1[B]^2$.
Step 4: Final Answer:
The orders with respect to A and B are 1 and 2, respectively.