Q127 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Chemistry
For the first order reaction, 80 % of the reactant decomposes in 60 minutes. Calculate the half life of the reaction.
- A40 min
- Correct answerB26 min
- C35 min
- D20 min
Solution
For a first order reaction, the rate constant $k$ is given by
$k = \dfrac{2.303}{t} \log \left( \dfrac{a}{a-x} \right)$
Given $t = 60$ min and $80\%$ decomposition, $x = 0.8a$ and $a-x = 0.2a$.
$k = \dfrac{2.303}{60} \log \left( \dfrac{a}{0.2a} \right) = \dfrac{2.303}{60} \log 5$
The half-life $t_{1/2}$ is given by
$t_{1/2} = \dfrac{0.693}{k} = \dfrac{2.303 \times \log 2}{k}$
Substituting the value of $k$:
$t_{1/2} = \dfrac{2.303 \times 0.301}{\dfrac{2.303 \times 0.699}{60}}$
$t_{1/2} = \dfrac{0.301 \times 60}{0.699} \approx 25.83 \text{ min}$
Rounding to the nearest integer, $t_{1/2} \approx 26 \text{ min}$.
Answer: 26 min
$k = \dfrac{2.303}{t} \log \left( \dfrac{a}{a-x} \right)$
Given $t = 60$ min and $80\%$ decomposition, $x = 0.8a$ and $a-x = 0.2a$.
$k = \dfrac{2.303}{60} \log \left( \dfrac{a}{0.2a} \right) = \dfrac{2.303}{60} \log 5$
The half-life $t_{1/2}$ is given by
$t_{1/2} = \dfrac{0.693}{k} = \dfrac{2.303 \times \log 2}{k}$
Substituting the value of $k$:
$t_{1/2} = \dfrac{2.303 \times 0.301}{\dfrac{2.303 \times 0.699}{60}}$
$t_{1/2} = \dfrac{0.301 \times 60}{0.699} \approx 25.83 \text{ min}$
Rounding to the nearest integer, $t_{1/2} \approx 26 \text{ min}$.
Answer: 26 min
Hence, the correct answer is option: 'B'.