Question

4. Prove that, "The time required for 99% completion of a first order reaction is double than the
time required for 90% completion."
​

Answer 1

To prove:

The time required for 99% completion of a first order reaction is double than the time required for 90% completion.

Proof:

For 99% completion :

$\therefore \: \rm{t_{99\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{1 - \frac{99}{100} } ) }$

$= > \: \rm{t_{99\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{1 - 0.99} ) }$

$= > \: \rm{t_{99\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{0.01} ) }$

$= > \: \rm{t_{99\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{ {10}^{ - 2} } ) }$

$= > \: \rm{t_{99\%} = \dfrac{1}{k} \: \: ln( {10}^{2} ) }$

$= > \: \rm{t_{99\%} =2 \times \bigg \{ \dfrac{1}{k} \: \: ln( 10 ) \bigg \}}$

For 90% completion:

$\therefore \: \rm{t_{90\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{1 - \frac{90}{100} } ) }$

$= > \: \rm{t_{90\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{1 - 0.9} ) }$

$= > \: \rm{t_{90\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{0.1} ) }$

$= > \: \rm{t_{90\%} = \dfrac{1}{k} \: \: ln( \dfrac{1}{ {10}^{ - 1} } ) }$

$= > \: \rm{t_{90\%} = \dfrac{1}{k} \: \: ln(10 ) }$

So, we can say that :

$\boxed{ \bf{ \huge{t_{99\%} = 2 \times t_{90\%}}}}$