Question

For the first order reaction $A \rightarrow B$, the half life is $30 min$ The time taken for $75 \%$ completion of the reaction is min (Nearest integer) Given : $\log _g 2=03010$ $ \log 3=04771 $ $ \log 5=06989$

Answer 1

Correct Answer: 60

1. For a first-order reaction, the time required for completion is given by:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}. \]
2. For 75\% completion, \([A] = \frac{1}{4}[A]_0\). Substituting:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{\frac{1}{4}[A]_0} = \frac{2.303}{k} \log 4. \]
3. Use the relation between half-life and rate constant:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30}. \]
4. Substituting \(k\) and \(\log 4 = 0.602\):
\[ t = \frac{2.303 \times 0.602}{0.693 / 30} = 60 \, \text{min}. \]
Thus, the time taken is 60 min. For first-order reactions, the time for completion depends logarithmically on the fraction of the reaction completed.