Q67 / 240 · KCET 2026 · Chemistry
For a 1st order change $R \rightarrow P$, the concentration of Reactant R changes from $0.1$ M to $0.025$ M in $40$ minutes. The rate of reaction when the concentration of R is $0.01$ M is ____
- A$1.73 \times 10^{-5}\text{ M min}^{-1}$
- Correct answerB$3.47 \times 10^{-4}\text{ M min}^{-1}$
- C$3.47 \times 10^{-5}\text{ M min}^{-1}$
- D$1.73 \times 10^{-4}\text{ M min}^{-1}$
Solution
The concentration of the reactant changes from $0.1\text{ M}$ to $0.025\text{ M}$, which is $\dfrac{1}{4}$ of the initial concentration.
This corresponds to $2$ half-lives.
$2 t_{1/2} = 40\text{ min} \Rightarrow t_{1/2} = 20\text{ min}$
The rate constant $k$ for a first-order reaction is:
$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 0.03465\text{ min}^{-1}$
The rate of reaction when $[R] = 0.01\text{ M}$ is:
$\text{Rate} = k[R]$
$\text{Rate} = 0.03465 \times 0.01 = 3.465 \times 10^{-4}\text{ M min}^{-1}$
$\text{Rate} \approx 3.47 \times 10^{-4}\text{ M min}^{-1}$
Answer: $3.47 \times 10^{-4}\text{ M min}^{-1}$
This corresponds to $2$ half-lives.
$2 t_{1/2} = 40\text{ min} \Rightarrow t_{1/2} = 20\text{ min}$
The rate constant $k$ for a first-order reaction is:
$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 0.03465\text{ min}^{-1}$
The rate of reaction when $[R] = 0.01\text{ M}$ is:
$\text{Rate} = k[R]$
$\text{Rate} = 0.03465 \times 0.01 = 3.465 \times 10^{-4}\text{ M min}^{-1}$
$\text{Rate} \approx 3.47 \times 10^{-4}\text{ M min}^{-1}$
Answer: $3.47 \times 10^{-4}\text{ M min}^{-1}$
Hence, the correct answer is option: 'B'.