Question

The time required for $60\%$ completion of a first order reaction is $50$ min. The time required for $93.6\%$ completion of the same reaction will be

• A. 100 min
• B. 83.8 min
• C. 50 min
• D. 150 min

Answer 1

Answer: (D)

150 min

Answer 2

For a first order reaction the, the rate constant is given by $k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$ Given, at $50 \,min , 60 \%$ of the reaction is completed $\therefore k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$ $=\frac{2.303}{50} \log \frac{100}{40}$ $=\frac{2.303}{50} \times 0.397$ So, when $93.6 \%$ of the reaction is completed, $\Rightarrow \frac{2.303}{50} \times 0.397$ $=\frac{2.303}{t} \log \frac{100}{6.4}$ $\Rightarrow \frac{2.303}{50} \times 0.397$ $=\frac{2.303}{t} \times 1.19$ $\Rightarrow t \approx 150\, min$