Question

* Show that the time required for 99.9% completion of a 1st order reactant in three time the time required for 90% completion​

Answer 1

the half life of a first order reaction is given by the formula

$\boxed{t = 2.303 \: log( \frac{a}{a - x} ) }$

Half life for 99.9% completion :

Substituting X as 99.9 / 100 we get ,

$\implies{t(99.9) = 2.303 \: log( \frac{a}{a - \frac{99.9a}{100} } ) }$

$\implies{t(99.9) = 2.303 \: log( \frac{100a}{0.1a} ) }$

$\implies{t(99.9) = 2.303 \: log( {10}^{3} ) }$

$\implies{t(99.9) = 2.303 \: \times 3 \times log( 10 ) }$

$\implies{t(99.9) = 2.303 \: \times 3....(1)}$

Half life for 90% completion :

Substituting X as 90/ 100 we get ,

$\implies{t(90) = 2.303 \: log( \frac{a}{a - \frac{90a}{100} } ) }$

$\implies{t(90) = 2.303 \: log( \frac{100a}{10a} ) }$

$\implies{t(90) = 2.303 \: log(10)}$

$\implies{t(90) = 2.303...(2)}$

Now dividing (1) and (2) we get ,

$\implies{ \frac{t(99.9)}{t(90)} = \frac{3 \times 2.303}{2.303} }$

$\implies{ \frac{t(99.9)}{t(90)} =3}$

hence proved ,

$\boxed {t(99.9) = 3 \times t(90)}$