Question

The reaction represented by A → B follows first order kinetics. At a given temperature, 20\% of the reaction is completed in 223 s. The time taken to complete 50\% of the reaction at the same temperature is ............ s (rounded off to the nearest integer).

Hint:

In first order kinetics, half-life depends only on rate constant: $t_{1/2} = \ln 2 / k$. Always use given data to determine $k$ first.

Answer 1

Step 1: First order kinetics expression.
\[ \ln \frac{[A]_0}{[A]} = k t \] Fraction reacted = $x$. Then, \[ [A] = [A]_0 (1 - x) \] Step 2: For 20\% reaction.
At $t = 223$ s, $x = 0.2$, so $[A]/[A]_0 = 0.8$. \[ k = \frac{1}{t} \ln \frac{1}{0.8} \] \[ k = \frac{1}{223} \ln (1.25) \] \[ k = \frac{1}{223} \times 0.223 = 0.001 \, s^{-1} \] Step 3: Time for 50\% reaction.
For $x = 0.5$, $[A]/[A]_0 = 0.5$. \[ t = \frac{1}{k} \ln \frac{1}{0.5} \] \[ t = \frac{1}{0.001} \ln (2) \] \[ t = 1000 \times 0.693 = 693 \, s \] Recheck: using $k = 0.001002$ gives $t = 691$ s, but rounding to integer ≈ 693 s. Final Answer: \[ \boxed{693 \, s} \]