Q28 / 150 · MHT CET 2026 (11 April Shift 1) · Chemistry
A first order reaction complete $60$% in $20$ minutes. How long will the reaction take to complete $84$%?
- A$54$ min
- B$68$ min
- Correct answerC$40$ min
- D$76$ min
Solution
For a first order reaction, the rate constant is given by $k = \dfrac{2.303}{t} \log\left(\dfrac{a}{a-x}\right)$
For $60\%$ completion in $20$ minutes:
$k = \dfrac{2.303}{20} \log\left(\dfrac{100}{100-60}\right) = \dfrac{2.303}{20} \log(2.5)$
For $84\%$ completion in time $t$:
$k = \dfrac{2.303}{t} \log\left(\dfrac{100}{100-84}\right) = \dfrac{2.303}{t} \log(6.25)$
Equating the two expressions for $k$:
$\dfrac{1}{20} \log(2.5) = \dfrac{1}{t} \log(2.5^2)$
$\dfrac{1}{20} \log(2.5) = \dfrac{2}{t} \log(2.5)$
$t = 40$ min
Answer: $40$ min
For $60\%$ completion in $20$ minutes:
$k = \dfrac{2.303}{20} \log\left(\dfrac{100}{100-60}\right) = \dfrac{2.303}{20} \log(2.5)$
For $84\%$ completion in time $t$:
$k = \dfrac{2.303}{t} \log\left(\dfrac{100}{100-84}\right) = \dfrac{2.303}{t} \log(6.25)$
Equating the two expressions for $k$:
$\dfrac{1}{20} \log(2.5) = \dfrac{1}{t} \log(2.5^2)$
$\dfrac{1}{20} \log(2.5) = \dfrac{2}{t} \log(2.5)$
$t = 40$ min
Answer: $40$ min
Hence, the correct answer is option: 'C'.