Consider the following reaction,the rate expr | vedclass

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(A) Given the reaction $A + B ightarrow C$ and the rate law $	ext{rate} = k[A]^{1/2}[B]^{1/2}$.Since the initial concentrations $[A]_0 = [B]_0 = 1

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) Given the reaction $A + B ightarrow C$ and the rate law $	ext{rate} = k[A]^{1/2}[B]^{1/2}$.Since the initial concentrations $[A]_0 = [B]_0 = 1 \ M$,at any time $t$,$[A] = [B]$.Substituting this into the rate law: $	ext{rate} = k[A]^{1/2}[A]^{1/2} = k[A]$.This confirms the reaction follows first-order kinetics with respect to $A$.The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given $k = 4.6 	imes 10^{-2} \ s^{-1}$,$[A]_0 = 1 \ M$,and $[A]_t = 0.1 \ M$.Substituting the values: $4.6 	imes 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}$.$4.6 	imes 10^{-2} = \frac{2.303}{t} 	imes 1$.$t = \frac{2.303}{4.6 	imes 10^{-2}} \approx 50.06 \ s$.Rounding to the nearest integer,$t = 50 \ s$.

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