Question

$75 \%$ of a first order reaction was completed in $32\, min$, when was $50 \%$ of the reaction completed?

• A. 24 min
• B. 16 min
• C. 8 min
• D. 48 min

Answer 1

Answer: (B)

16 min

Answer 2

First order reaction, $k=\frac{2.303}{t} \log \frac{a}{a-x}$ Here, $\therefore a=100$ $\therefore x=75$ $\because a-x=(100-75)$ $k=\frac{2.303}{32} \log \frac{100}{25}$ $=\frac{2.303}{32} \log 4$ $=\frac{2.303 \times \log 2^{2}}{32}$ $=\frac{2.303 \times 2 \log 2}{32}$ $=\frac{2.303 \times 2 \times 0.3010}{32}$ $=\frac{2 \times 0.63}{32}$ $\therefore t_{1 / 2}=\frac{0.693}{k}$ $\therefore t_{1 / 2}=\frac{0.693}{2 \times 0.693 / 32}4$ $=16\, \min$