Question

For the reaction, \( A + B \rightarrow \) products, the following data is obtained: A + B \( \rightarrow \) products, the following data is given: $$ \begin{array}{|c|c|c|c|} \hline \text{Exp. No.} & [A] \, (\text{mol L}^{-1}) & [B] \, (\text{mol L}^{-1}) & \text{Initial rate} \, (\text{mol L}^{-1} s^{-1}) \\ \hline 1 & 0.2 & 0.4 & 0.8 \\ 2 & 0.4 & 0.2 & 1.6 \\ 3 & 0.3 & 0.2 & 0.4 \\ \hline \end{array} $$ The rate law of the reaction is: 

• A. \( k[A][B] \)
• B. \( k[A] \)
• C. \( k[B]^2 \)
• D. \( k[A]^1[B]^2 \)

Hint:

To determine the rate law, compare the rate data from different experiments. The rate law can be determined by analyzing how the change in the concentration of reactants affects the rate of the reaction.

Answer 1

The Correct Option is D

We are given the initial rate data and we need to determine the rate law. The rate law is in the form: \[ r = k[A]^m[B]^n. \] - Comparing Experiments 1 and 2: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A_2]^m[B_2]^n}{k[A_1]^m[B_1]^n} = \frac{1.6}{0.8} = 2. \] This gives: \[ \frac{[A_2]^m[B_2]^n}{[A_1]^m[B_1]^n} = 2 \quad \Rightarrow \quad \frac{(0.4)^m(0.2)^n}{(0.2)^m(0.4)^n} = 2. \] Solving this gives \( m = 1 \) and \( n = 2 \). Thus, the rate law is \( r = k[A]^1[B]^2 \).