Q32 / 110 · VITEEE 2024 · Chemistry
$A(g) \longrightarrow B(g)$ is a first order reaction. The initial concentration of $A$ is $0.2 \mathrm{~mol} \mathrm{~L}^{-1}$. After 10 minutes the concentration of $B$ is found to be $0.18 \mathrm{~mol} \mathrm{~L}^{-1}$. The rate constant (in $\min ^{-1}$ ) for the reaction is
- A$0.2303 \mathrm{~min}^{-1}$
- Correct answerB$2.303 \mathrm{~min}^{-1}$
- C$0.693 \mathrm{~min}^{-1}$
- D$0.01 \mathrm{~min}^{-1}$
Solution
$A(g) \longrightarrow B(g)$
Given, concentration of $A,[A]_0=0.2 \mathrm{~mol} \mathrm{~L}^{-1}$
Concentration of $B$ after 10 minute
$=0.18 \mathrm{~mol} \mathrm{~L}^{-1}$
$[B]$ formed $=[A]$ reacted
Hence, concentration of $A$ after 10 min
$=0.20-0.18=0.02 \mathrm{~mol} \mathrm{~L}^{-1}$
Rate constant is given by formula.
$\begin{aligned} k & =\frac{2.303}{t} \log \frac{[A]_0}{[A]}=\frac{2.303}{10} \log \frac{0.20}{0.02} \\ & =\frac{2.303}{10} \log 100 \\ k & =\frac{2.303}{10} \log 10^{10} \\ k & =2.303 \mathrm{~min}^{-1}\end{aligned}$
Hence, the correct answer is option: 'B'.