Question

Show that in a first order reaction time required for completion of $99.9\%is10$ times of half-life $\left(t_{1/2}\right)$ of the reaction.

Answer 1

Given,

Let$[A{]}_0=100$

At time $t=t_{99.9\%};\left[A\right]=(100-99.9)=0.1$

Formula
For first order reaction,

$k=\frac{2.303}{t}\log \left(\frac{[A{]}_0}{\left[A\right]}\right)$

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

For $t_{99.9\%}$;

$t_{99.9\%}=\frac{2.303}{k}\log \left(\frac{100}{0.1}\right)$

$\Rightarrow t_{99.9\%}=\frac{2.303}{k}\log 1000$

$t_{99.9\%}=\frac{2.303}{k}\times 3$

$\Rightarrow t_{99.9\%}=\frac{6.9}{k}$

$t_{99.9\%}=10\times \frac{0.69}{k}$

Now, we know that for first order
reactions, half life is given as

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

So,

$t_{99.9\%}=10t_{1/2}$

Hence, in a first order reaction time
required for complection of $99.9\%$ is $10$
times of half-life $t_{1/2}$ of the reaction.