Question

A first order reaction completes 20% in 10 minutes. Calculate the time taken for 75% completion.

Hint:

For first-order kinetics, times for given \(%\) completion depend only on the fraction remaining, not on initial concentration. Use \(t=\frac{1}{k}\ln\!\left(\frac{1}{\text{fraction remaining}}\right)\).

Answer 1

For a first-order reaction: \[ k=\frac{1}{t}\ln\!\left(\frac{[A]_0}{[A]}\right) \] After 10 min, 20% is complete \(\Rightarrow\) 80% remains: \([A]/[A]_0=0.80\). Hence \[ k=\frac{1}{10}\ln\!\left(\frac{1}{0.80}\right) = \frac{1}{10}\ln(1.25)\approx 0.0223~\text{min}^{-1}. \] For 75% completion, 25% remains: \([A]/[A]_0=0.25\). Time required: \[ t=\frac{1}{k}\ln\!\left(\frac{1}{0.25}\right)=\frac{1}{k}\ln(4) =\frac{1.3863}{0.0223}\approx 6.21\times10^1~\text{min} \] \[ \boxed{t \approx 62~\text{minutes}} \]