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

We know, for first-order reaction,

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

Since the reaction is 75% complete in 32 min we can write,

\(k = \frac{2.303}{32}log\frac{100}{100 − 75}\)

⇒ \(k = \frac{2.303}{32}log\frac{100}{25}\)

⇒ \(k = \frac{2.303}{32}log(4)\)

⇒ \(k = \frac{2.303}{32} × 0.602\)

∴ \(k = 0.0433 min^{-1}\)

Now when the reaction is 50% complete, the time required will be

\(t_{50} = \frac{2.303}{k}log\frac{a}{a − x}\)

⇒\(t_{50} = \frac{2.303}{0.0433}log\frac{100}{100 − 50}\)

⇒\(t_{50} = \frac{2.303}{0.0433}log\frac{100}{50}\)

⇒\(t_{50} = \frac{2.303}{0.0433}log 2\)

⇒\(t_{50} = \frac{2.303}{0.0433} × 0.301\)

⇒\(t_{50} = 16.01\)

∴ \(t_{50} ≈ 16 min\)