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(A) The rate of a reaction is determined by the slowest step in the mechanism. The rate law for the slow step is $R = k_3[A][B]$. Since $B$ is an inte

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$\propto [A]^2$

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(A) The rate of a reaction is determined by the slowest step in the mechanism. The rate law for the slow step is $R = k_3[A][B]$. Since $B$ is an intermediate formed in the fast equilibrium step $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$,we have the equilibrium constant $K_{eq} = \frac{[B]}{[A]} = \frac{k_1}{k_2}$. This gives $[B] = \frac{k_1}{k_2}[A]$. Substituting this into the rate expression: $R = k_3[A](\frac{k_1}{k_2}[A]) = \frac{k_3 k_1}{k_2}[A]^2$. Thus,the rate is proportional to $[A]^2$.

$[A_2]$	$[B]$	$-d[A_2]/dt$
$0.1$	$0.2$	$1 \times 10^{-2} \, M \, s^{-1}$
$0.2$	$0.2$	$2 \times 10^{-2} \, M \, s^{-1}$
$0.2$	$0.4$	$8 \times 10^{-2} \, M \, s^{-1}$

For a hypothetical reaction,$A \rightarrow C$. The mechanism is: $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$ (Fast),$A + B \xrightarrow{k_3} C$ (Slow). The rate law for this reaction is:

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