Question

Time required for 99.9% completion of a first order reaction is _____ time the time required for completion of 90% reaction.(nearest integer).

Answer 1

The rate constant $K$ for a first-order reaction is given by: $K = \frac{1}{t} \ln\left(\frac{100}{100 - \text{completion percentage}}\right)$
For 99.9% completion:
$t_{99.9\%} = \frac{\ln(1000)}{K}$
For 90% completion:
$t_{90\%} = \frac{\ln(10)}{K}$
Ratio of times:
$\frac{t_{99.9\%}}{t_{90\%}} = \frac{\ln(1000)}{\ln(10)} = \frac{3 \ln(10)}{\ln(10)} = 3$

Answer 2

The rate constant $K$ for a first-order reaction is given by: $K = \frac{1}{t} \ln\left(\frac{100}{100 - \text{completion percentage}}\right)$
For 99.9% completion:
$t_{99.9\%} = \frac{\ln(1000)}{K}$
For 90% completion:
$t_{90\%} = \frac{\ln(10)}{K}$
Ratio of times:
$\frac{t_{99.9\%}}{t_{90\%}} = \frac{\ln(1000)}{\ln(10)} = \frac{3 \ln(10)}{\ln(10)} = 3$