Q27 / 150 · MHT CET 2026 (13 April Shift 1) · Chemistry
In first order reaction, 20 % concentration remains after 10 min. What would be the rate constant of reaction if $\log_{10}(5) = 0.6989$?
- A$1.609$ min$^{-1}$
- B$6.989$ min$^{-1}$
- C$16.09$ min$^{-1}$
- Correct answerD$0.1609$ min$^{-1}$
Solution
For a first order reaction, the rate constant $k$ is given by the formula:
$k = \dfrac{2.303}{t} \log_{10} \left( \dfrac{[A]_0}{[A]_t} \right)$
Given that $20\%$ concentration remains after $10$ min, we have:
$t = 10$ min
$[A]_t = \dfrac{20}{100} [A]_0 = 0.2 [A]_0$
Substituting these values into the formula:
$k = \dfrac{2.303}{10} \log_{10} \left( \dfrac{[A]_0}{0.2 [A]_0} \right)$
$k = \dfrac{2.303}{10} \log_{10}(5)$
Given $\log_{10}(5) = 0.6989$:
$k = \dfrac{2.303}{10} \times 0.6989$
$k = 0.2303 \times 0.6989$
$k = 0.1609$ min$^{-1}$
$k = \dfrac{2.303}{t} \log_{10} \left( \dfrac{[A]_0}{[A]_t} \right)$
Given that $20\%$ concentration remains after $10$ min, we have:
$t = 10$ min
$[A]_t = \dfrac{20}{100} [A]_0 = 0.2 [A]_0$
Substituting these values into the formula:
$k = \dfrac{2.303}{10} \log_{10} \left( \dfrac{[A]_0}{0.2 [A]_0} \right)$
$k = \dfrac{2.303}{10} \log_{10}(5)$
Given $\log_{10}(5) = 0.6989$:
$k = \dfrac{2.303}{10} \times 0.6989$
$k = 0.2303 \times 0.6989$
$k = 0.1609$ min$^{-1}$
Hence, the correct answer is option: 'D'.