Question

Show that the time required for 99.9% completion in a first order reaction is 10 times of half-life (t_1/2) of the reaction [log 2 = 0.3010, log 10 = 1].

Answer 1

First-order reaction ${\displaystyle k=\frac{2.303}{t}\log \frac{a}{a-x}}$

where a = original amount

a − x = amount left betund

t = time period

k = rate constant

We also know that

${\displaystyle k=\frac{0.693}{t_{1/2}}}$

t_1/2 = half-life period.

Now, x = 99.9% and x = 50%

We can say ${\displaystyle t=\frac{2.303}{k}\log \frac{a}{a-x}}$

or ${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{2.303/k\log /100-99.9}{2.303/k\log /100-50}}$

or ${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{\log \frac{100}{0.1}}{\log \frac{100}{50}}}$

${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{-\log 1000}{\log 2}}$

${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{-\log 1\times {10}^{-3}}{\log 2}}$

${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{3}{0.3010}}$

${\displaystyle \frac{t_{99.9\%}}{t_{50\%}}=\frac{3}{0.3}}$

thus, t_99.9% = 10 × t_50%