Q21 / 125 · BITSAT 2011 · Chemistry
During the kinetic study of the reaction, $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$, following results were obtained:
$$
\begin{array}{|c|c|c|c|}
\hline \text { Run } & {[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l}
\text { Initial rate of } \\
\text { formation of } \\
\mathrm{D} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}
\end{array} \\
\hline \text { I } & 0.1 & 0.1 & 6.0 \times 10^{-1} \\
\hline \text { II } & 0.3 & 0.2 & 7.2 \times 10^{-1} \\
\hline \text { III } & 0.3 & 0.4 & 28.8 \times 10^{-1} \\
\hline \text { IV } & 0.4 & 0.1 & 24.0 \times 10^{-1} \\
\hline
\end{array}
$$
Based on the above data which one of the following is correct?
- Arate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]$
- Brate $=k[\mathrm{~A}][\mathrm{B}]$
- Crate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}$
- Correct answerDrate $=k[\mathrm{~A}][\mathrm{B}]^{2}$
Solution
In case of (II) and (III) Keeping concentration of [A] constant, when the concentration of $[\mathrm{B}]$ is doubled, the rate quadruples. Hence it is second order with respect to $\mathrm{B}$. In case of $\mathrm{I} \&$ IV Keeping the concentration of [B] constant, when the concentration of $[\mathrm{A}]$ is increased four times, rate also increases four times. Hence, the order with respect to $\mathrm{A}$ is one. Hence Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$
Hence, the correct answer is option: 'D'.