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  • For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction.       <div style="float:left; width:

For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction.

 

 

 

 
 
 
 
 

Answers (1)

For a 1st order reaction

        t = \frac{2.303}{k} \log \frac{[R]_o}{[R]_o - [ R]}

        t_{99\%} = \frac{2.303}{k} \log \frac{[R]_o}{[R]_o - 0.99[ R]}

        t_{99\%} = \frac{2.303}{k} \log 100

       t_{99\%} = \frac{2.303}{k} \times2\log 10\quad -(1)

        t_{90\%} = \frac{2.303}{k} \log \frac{[R]_o}{[R]_o - 0.90[ R]}

        t_{90\%} = \frac{2.303}{k} \log 10\quad -(2)

Dividing (1) by (2)

    \frac{t_{99\%}}{t_{90\%}} = 2

    {t_{99\%}} = 2{t_{90\%}}

Posted by

Sumit Saini

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