The time taken for 10 % completion of a first | vedclass

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Question

(A) For a first order reaction,the rate equation is $t = \frac{2.303}{k} \log \frac{100}{100 - x}$.For $10 \%$ completion $(x = 10)$: $20 = \frac{2.30

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) For a first order reaction,the rate equation is $t = \frac{2.303}{k} \log \frac{100}{100 - x}$.For $10 \%$ completion $(x = 10)$: $20 = \frac{2.303}{k} \log \frac{100}{90} = \frac{2.303}{k} \log (10/9)$ $(i)$.For $19 \%$ completion $(x = 19)$: $t = \frac{2.303}{k} \log \frac{100}{100 - 19} = \frac{2.303}{k} \log \frac{100}{81} = \frac{2.303}{k} \log (10/9)^2 = 2 	imes \frac{2.303}{k} \log (10/9)$ (ii).Dividing equation (ii) by equation $(i)$:$\frac{t}{20} = \frac{2 	imes \frac{2.303}{k} \log (10/9)}{\frac{2.303}{k} \log (10/9)} = 2$.Therefore,$t = 20 	imes 2 = 40 \ min$.

The time taken for $10 \%$ completion of a first order reaction is $20 \ min$. Then,for $19 \%$ completion,the reaction will take (in $min$)

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