75% of a first-order reaction was completed in 32 min. When was 50% of the reaction completed?

16 min

The correct answer is option 2. 16 min.

For First order reaction,

\(k = \frac{2.303}{t}log \frac{a}{a − x}\)

When the reaction is 75% complete then,

\(x = \frac{75}{100}a = 0.75 a\)

\(∴ k = \frac{2.303}{32}log \frac{a}{a −0.75 a}\)

\(⇒ k = \frac{2.303}{32}log \frac{1}{0.25}\)

\(⇒ k = \frac{2.303}{32} ×0.602\)

\(⇒k = 0.04332\)

Also we know ,

\(t_{1/2} = \frac{0.693}{k}\)

\(t_{1/2} = \frac{0.693}{0.04332}\)

\(⇒ t_{1/2} = 15.99\)

\(∴ t_{1/2} ≈ 16 min\)