If 75 % of a first order reaction was complet | vedclass

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $75 \%$ completion,$[A]_t = [A]_0

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$60$

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(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $75 \%$ completion,$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$,so $k = \frac{2.303}{90} \log \frac{1}{0.25} = \frac{2.303}{90} \log 4 = \frac{2.303 	imes 2 \log 2}{90} = \frac{2.303 	imes 0.6}{90} \ min^{-1}$.For $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$,so $t = \frac{2.303}{k} \log \frac{1}{0.4} = \frac{2.303}{k} \log 2.5$.Substituting $k$: $t = \frac{2.303 	imes 90}{2.303 	imes 0.6} 	imes 0.4 = \frac{90 	imes 0.4}{0.6} = \frac{36}{0.6} = 60 \ minutes$.

If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........
(Take: $\log 2=0.30 ; \log 2.5=0.40)$

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If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........(Take: $\log 2=0.30 ; \log 2.5=0.40)$