In a first-order reaction, the concentration of the reactant decreases from 0.8 Μ to 0.4 Μ in 15 minutes. The time is taken for the concentration to change from 0.1 Μ to 0.025 Μ is:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 30 minutes.
Order is one here as concentration changes from 0.8 Μ to 0.4 Μ (50%) in 15 minutes, this half-life is 15 minutes = \(T_{50}\)
A change from 0.1 Μ to 0.025 Μ is 75% and for a first-order reaction
\(T_{50} = 15 min\)
\(k = \frac{0.693}{T_{50}}\)
\(k = 0.0462\)
\(\alpha = 0.1M\)
\((\alpha − x) = 0.025 M\)
For first-order reaction
\(k = \frac{2.303}{t}log\frac{\alpha}{\alpha − x}\)
\(⇒ 0.0462 = \frac{2.303}{t}log\frac{0.1}{0.025}\)
\(⇒ 0.0462 = \frac{2.303}{t}\times 0.602\)
\(⇒ t = \frac{2.303 × 0.602}{0.0462}\)
\(⇒ t \approx 30 min\)