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

Concept: For a reaction: \[ aA + bB \rightarrow \text{Products} \] Rate law is: \[ \text{Rate} = k[A]^m[B]^n \] where:
• \(m\) = order with respect to A
• \(n\) = order with respect to B Orders are determined experimentally by comparing rate changes with concentration changes.

Step 1: Determine order with respect to A Compare experiment 1 and 2: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{4 \times 10^{-3}}{2 \times 10^{-3}} = 2 \] Concentration change: \[ \frac{[A]_2}{[A]_1} = \frac{0.2}{0.1} = 2 \] [B] remains constant. Thus: \[ 2 = 2^m \] \[ m = 1 \] So, order with respect to A is: \[ 1 \]

Step 2: Determine order with respect to B Compare experiment 2 and 3: \[ \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{1.6 \times 10^{-2}}{4 \times 10^{-3}} = 4 \] Concentration change: \[ \frac{[B]_3}{[B]_2} = \frac{0.2}{0.1} = 2 \] [A] remains constant. Thus: \[ 4 = 2^n \] \[ n = 2 \] So, order with respect to B is: \[ 2 \]

Step 3: Write final order pair \[ (m,n) = (1,2) \] Final Answer: \[ \boxed{(1,2)} \]