For the first-order isomerization reaction A | vedclass

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Question

(A) Step $1$: Calculate the concentration after $1 \ hour$ $(60 \ min)$.For a first-order reaction: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.$\l

Vedclass · Accepted Answer

$3.4354 	imes 10^{-3} \ M \ min^{-1}$

Vedclass · Answer

(A) Step $1$: Calculate the concentration after $1 \ hour$ $(60 \ min)$.For a first-order reaction: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.$\log \frac{[A]_0}{[A]_t} = \frac{kt}{2.303} = \frac{4.5 	imes 10^{-3} \ min^{-1} 	imes 60 \ min}{2.303} = 0.11726$.$\frac{[A]_0}{[A]_t} = 	ext{antilog}(0.11726) = 1.310$.$[A]_t = \frac{1 \ M}{1.310} = 0.7633 \ M$.Step $2$: Calculate the rate after $60 \ min$.Rate $= k[A]_t = 4.5 	imes 10^{-3} \ min^{-1} 	imes 0.7633 \ M = 3.435 	imes 10^{-3} \ M \ min^{-1}$.

For the first-order isomerization reaction $A \rightarrow B$,the rate constant is $4.5 \times 10^{-3} \ min^{-1}$. If the initial concentration of $A$ is $1 \ M$,find the rate of the reaction after $1 \ hour$.

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