(D) The rate of the reaction is determined by the slow step $(II)$:$r = k_2 [N_2O_2] [H_2]$Since $N_2O_2$ is an intermediate,we express its concentrat

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(D) The rate of the reaction is determined by the slow step $(II)$:$r = k_2 [N_2O_2] [H_2]$Since $N_2O_2$ is an intermediate,we express its concentration using the equilibrium constant $(K_{eq})$ from the fast step $(I)$:$K_{eq} = \frac{[N_2O_2]}{[NO]^2} \implies [N_2O_2] = K_{eq} [NO]^2$Substituting this into the rate expression:$r = k_2 (K_{eq} [NO]^2) [H_2]$Let $k = k_2 K_{eq}$,then the rate law is:$r = k [NO]^2 [H_2]$

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

For the reaction $2H_2 + 2NO \to N_2 + 2H_2O$,the following mechanism has been proposed:
$I$. $2NO \rightleftharpoons N_2O_2$ (fast)
$II$. $N_2O_2 + H_2 \xrightarrow{k_2} N_2O + H_2O$ (slow)
$III$. $N_2O + H_2 \to N_2 + H_2O$ (fast)
What will be the rate law of this reaction?

A
$k[NO][H_2]^2$
B
$k[NO][H_2]$
C
$k[NO]^0[H_2]^3$
D
$k[NO]^2[H_2]$

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Class 12 Questions

Class 12

Chemistry Questions

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

What is the molecularity of a reaction? Explain its types with examples.

Medium

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$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?

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The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:

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The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-

Medium

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The following data are obtained for a reaction,$X + Y \rightarrow$ Products.

$Expt.$ $[X]_0 / mol \ L^{-1}$ $[Y]_0 / mol \ L^{-1}$ $Rate / mol \ L^{-1} s^{-1}$

$1$ $0.25$ $0.25$ $1.0 \times 10^{-6}$

$2$ $0.50$ $0.25$ $4.0 \times 10^{-6}$

$3$ $0.25$ $0.50$ $8.0 \times 10^{-6}$

The overall order of the reaction is:

MediumKVPY 2012

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$Expt.$	$[X]_0 / mol \ L^{-1}$	$[Y]_0 / mol \ L^{-1}$	$Rate / mol \ L^{-1} s^{-1}$
$1$	$0.25$	$0.25$	$1.0 \times 10^{-6}$
$2$	$0.50$	$0.25$	$4.0 \times 10^{-6}$
$3$	$0.25$	$0.50$	$8.0 \times 10^{-6}$

For the reaction $2H_2 + 2NO \to N_2 + 2H_2O$,the following mechanism has been proposed:$I$. $2NO \rightleftharpoons N_2O_2$ (fast)$II$. $N_2O_2 + H_2 \xrightarrow{k_2} N_2O + H_2O$ (slow)$III$. $N_2O + H_2 \to N_2 + H_2O$ (fast)What will be the rate law of this reaction?

Similar Questions

What is the molecularity of a reaction? Explain its types with examples.

$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?

The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:

The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-

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For the reaction $2H_2 + 2NO \to N_2 + 2H_2O$,the following mechanism has been proposed:
$I$. $2NO \rightleftharpoons N_2O_2$ (fast)
$II$. $N_2O_2 + H_2 \xrightarrow{k_2} N_2O + H_2O$ (slow)
$III$. $N_2O + H_2 \to N_2 + H_2O$ (fast)
What will be the rate law of this reaction?