Question

For a first order reaction, the starting material reduces to \( \dfrac{1}{4} \) of its initial value after 20 min.
The rate constant of the reaction will be: (Given \( \ln 4 = 1.38 \))

• A. 0.069 min$^{-1}$
• B. 1.3844 min$^{-1}$
• C. 4.3465 min$^{-1}$
• D. 2.1674 min$^{-1}$

Hint:

In first-order kinetics, use the formula \( k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]} \right) \) to calculate the rate constant when initial and final concentrations are known.

Answer 1

The Correct Option is A

For a first-order reaction, the integrated rate law is:
\[ k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]}\right) \]
Given:
Initial concentration = $[A]_0$
Final concentration = $\dfrac{1}{4}[A]_0$
Time, $t = 20$ min
\[ k = \frac{1}{20} \ln\left(\frac{[A]_0}{[A]_0/4}\right) = \frac{1}{20} \ln(4) \]
\[ k = \frac{1}{20} \times 1.38 = 0.069 \ \text{min}^{-1} \]
Thus, the rate constant of the reaction is 0.069 min$^{-1}$.