A first order reaction is 75% completed in 10 | vedclass

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Question

(B) 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 = 100 - 7

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(B) 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,$[A]_t = 100 - 75 = 25$,so $t_{75\%} = \frac{2.303}{K} \log \frac{100}{25} = \frac{2.303}{K} \log 4 = 100 \ min$  ...... $(i)$For $87.5\%$ completion,$[A]_t = 100 - 87.5 = 12.5$,so $t_{87.5\%} = \frac{2.303}{K} \log \frac{100}{12.5} = \frac{2.303}{K} \log 8$  ...... $(ii)$Dividing $(ii)$ by $(i)$:$\frac{t_{87.5\%}}{100} = \frac{\log 8}{\log 4} = \frac{\log 2^3}{\log 2^2} = \frac{3 \log 2}{2 \log 2} = 1.5$$t_{87.5\%} = 1.5 	imes 100 = 150 \ min$.

$A$ first order reaction is $75\%$ completed in $100 \ min$. How long will it take for its $87.5\%$ completion? (in $\min$)

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