Q22 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
A first order reaction take $30$ minutes for $75\%$ decomposition, calculate its rate constant ?
- A$0.0238\ \text{minute}^{-1}$
- Correct answerB$0.0463\ \text{minute}^{-1}$
- C$0.0715\ \text{minute}^{-1}$
- D$0.0957\ \text{minute}^{-1}$
Solution
For a first order reaction, the integrated rate law is given by:
$k = \dfrac{2.303}{t} \log \left( \dfrac{a}{a-x} \right)$
Given that the reaction is $75\%$ complete in $30$ minutes:
$t = 30\ \text{minutes}$
$x = 0.75a$
$a - x = a - 0.75a = 0.25a$
Substituting these values into the rate equation:
$k = \dfrac{2.303}{30} \log \left( \dfrac{a}{0.25a} \right)$
$k = \dfrac{2.303}{30} \log (4)$
$k = \dfrac{2.303}{30} \times 0.6020$
$k = \dfrac{1.386}{30} = 0.0462\ \text{minute}^{-1}$
Alternatively, $75\%$ decomposition corresponds to two half-lives:
$t_{75\%} = 2 \times t_{1/2} \Rightarrow 30 = 2 \times t_{1/2} \Rightarrow t_{1/2} = 15\ \text{minutes}$
$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{15} = 0.0462\ \text{minute}^{-1}$
The closest given option is $0.0463\ \text{minute}^{-1}$.
Answer: $0.0463\ \text{minute}^{-1}$
$k = \dfrac{2.303}{t} \log \left( \dfrac{a}{a-x} \right)$
Given that the reaction is $75\%$ complete in $30$ minutes:
$t = 30\ \text{minutes}$
$x = 0.75a$
$a - x = a - 0.75a = 0.25a$
Substituting these values into the rate equation:
$k = \dfrac{2.303}{30} \log \left( \dfrac{a}{0.25a} \right)$
$k = \dfrac{2.303}{30} \log (4)$
$k = \dfrac{2.303}{30} \times 0.6020$
$k = \dfrac{1.386}{30} = 0.0462\ \text{minute}^{-1}$
Alternatively, $75\%$ decomposition corresponds to two half-lives:
$t_{75\%} = 2 \times t_{1/2} \Rightarrow 30 = 2 \times t_{1/2} \Rightarrow t_{1/2} = 15\ \text{minutes}$
$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{15} = 0.0462\ \text{minute}^{-1}$
The closest given option is $0.0463\ \text{minute}^{-1}$.
Answer: $0.0463\ \text{minute}^{-1}$
Hence, the correct answer is option: 'B'.