Isomerisation of gaseous cyclobutene to butad | vedclass

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Question

(A) For a first order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given $K = 3.3 	imes 10^{-4} \ s^{-1

Vedclass · Accepted Answer

$116.67$

Vedclass · Answer

(A) For a first order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given $K = 3.3 	imes 10^{-4} \ s^{-1}$ and the reaction is $90 \%$ complete,so $[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$. Substituting the values: $3.3 	imes 10^{-4} = \frac{2.303}{t} \log \frac{[A]_0}{0.10[A]_0}$ $3.3 	imes 10^{-4} = \frac{2.303}{t} \log(10)$ Since $\log(10) = 1$: $t = \frac{2.303}{3.3 	imes 10^{-4}} \ s$ $t \approx 0.6978 	imes 10^4 \ s = 6978 \ s$ To convert time into minutes: $t = \frac{6978}{60} \ min \approx 116.3 \ min$ Rounding to the nearest provided option,the answer is $116.67 \ min$.

Isomerisation of gaseous cyclobutene to butadiene is a first order reaction. At $T \ K$,the rate constant of the reaction is $3.3 \times 10^{-4} \ s^{-1}$. What is the time required (in $min$) to complete $90 \%$ of this reaction at the same temperature? $(\log 2 = 0.3)$

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