For the reaction 2A + B rightarrow text{Produ | vedclass

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(B) The rate of the reaction is determined by the slow step,which is the rate-determining step $(RDS)$.Rate $= k[A_2][B]$.Since $A_2$ is an intermedia

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$3$

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(B) The rate of the reaction is determined by the slow step,which is the rate-determining step $(RDS)$.Rate $= k[A_2][B]$.Since $A_2$ is an intermediate,we express its concentration using the equilibrium constant $K_{eq}$ from the fast step:$K_{eq} = \frac{[A_2]}{[A]^2} \implies [A_2] = K_{eq}[A]^2$.Substituting this into the rate equation:Rate $= k \cdot K_{eq}[A]^2[B] = k'[A]^2[B]^1$.The order of the reaction is the sum of the powers of the concentration terms in the rate law:Order $= 2 + 1 = 3$.

For the reaction $2A + B \rightarrow \text{Product}$,the following mechanism is given. What is the order of the reaction?
$2A \rightleftharpoons A_2$ (fast)
$A_2 + B \rightarrow P$ (slow)

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Which one of the following is wrongly matched?

The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
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Then the correct statement is:

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For the reaction $2A + B \rightarrow \text{Product}$,the following mechanism is given. What is the order of the reaction? $2A \rightleftharpoons A_2$ (fast)$A_2 + B \rightarrow P$ (slow)