Q27 / 120 · VITEEE 2015 · Chemistry
The concentrations of the reactant $A$ in the reacntion $A \rightarrow B$ at different times are given below
$\begin{array}{ll}\text { Concentration }(\boldsymbol{M}) & \text { Time (Minutes) } \\ 0.069 & 0 \\ 0.052 & 17 \\ 0.035 & 34 \\ 0.018 & 51\end{array}$
The rate constant of the reaction according to the correct order of reaction is
- Correct answerA$0.001 \mathrm{M} / \mathrm{min}$
- B$0.001 \mathrm{~min}^{-1}$
- C$0.001 \mathrm{~min} / \mathrm{M}$
- D$0.001 \mathrm{M}^{-1} \mathrm{~min}^{-1}$
Solution
\begin{array}{c|c|c}
\hline Interval & Conc.change & Rate \\
\hline 0-17 \mathrm{~min} & 0.069-0.052=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline 17-34 \mathrm{~min} & 0.052-0.035=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline 34-51 \mathrm{~min} & 0.035-0.018=0.017 \mathrm{M} & \frac{0.017}{17}=0.001 \\
\hline
\end{array}
$\because$ Rate remains constant. So, it is independent of concentration, the reaction is of zero order.
According to rate law
Rate $=\mathrm{K}(\text { conc. })^{0}=0.001 \mathrm{M} / \mathrm{min}$
Hence, the correct answer is option: 'A'.