(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g

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(B) Step $(i)$: $NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$ (Fast equilibrium)Step $(ii)$: $NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$ (Slow,rate-determining step)The rate law is determined by the slow step: $\text{Rate} = k[NOBr_2][NO]$.Since $NOBr_2$ is an intermediate,we express its concentration using the equilibrium constant $K_C$ from step $(i)$: $K_C = \frac{[NOBr_2]}{[NO][Br_2]}$.Therefore,$[NOBr_2] = K_C[NO][Br_2]$.Substituting this into the rate law: $\text{Rate} = k \cdot K_C[NO][Br_2][NO] = k'[NO]^2[Br_2]$.The order of the reaction with respect to $NO_{(g)}$ is $2$.

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

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The following mechanism has been proposed for the reaction of $NO$ with $Br_2$ to form $NOBr$:
$NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$
$NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$
If the second step is the rate-determining step,the order of the reaction with respect to $NO_{(g)}$ is:

AIEEE 2006 All AIEEE

A
$3$
B
$2$
C
$1$
D
$0$

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Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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The rate law for the reaction $RCl + NaOH_{(aq)} \to ROH + NaCl$ is given by $\text{Rate} = K_1[RCl]$. The rate of the reaction will be

DifficultIIT 1988

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Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.
Exp. $[A]$ $[B]$ Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$ $0.1$ $0.1$ $2 \times 10^{-3}$
$2$ $0.4$ $0.1$ $0.8 \times 10^{-2}$
$3$ $0.1$ $0.2$ $1.6 \times 10^{-2}$
(in $.5$)

Exp.	$[A]$	$[B]$	Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$0.8 \times 10^{-2}$
$3$	$0.1$	$0.2$	$1.6 \times 10^{-2}$

Medium

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For the reaction $3 \, A_{(g)} \xrightarrow{K} B_{(g)} + C_{(g)}$,$K$ is $10^{-14} \, L/mol \cdot min$. If $[A] = 0.5 \, M$,then the value of $-\frac{d[A]}{dt}$ (in $M/sec$) is:

Difficult

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The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

EasyGUJCET 2023

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For the following reaction: $NO_{2(g)} + CO_{(g)} \to NO_{(g)} + CO_{2(g)}$,the rate law is: $\text{Rate} = k [NO_2]^2$. If $0.1 \ mol$ of gaseous carbon monoxide is added at constant temperature to the reaction mixture,which of the following statements is true?

MediumAIIMS 2016

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The rate law for the reaction $RCl + NaOH_{(aq)} \to ROH + NaCl$ is given by $\text{Rate} = K_1[RCl]$. The rate of the reaction will be

Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.Exp.$[A]$$[B]$Rate of reaction $(mol \ L^{-1} \ s^{-1})$$1$$0.1$$0.1$$2 \times 10^{-3}$$2$$0.4$$0.1$$0.8 \times 10^{-2}$$3$$0.1$$0.2$$1.6 \times 10^{-2}$ (in $.5$)

For the reaction $3 \, A_{(g)} \xrightarrow{K} B_{(g)} + C_{(g)}$,$K$ is $10^{-14} \, L/mol \cdot min$. If $[A] = 0.5 \, M$,then the value of $-\frac{d[A]}{dt}$ (in $M/sec$) is:

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur? Experiment $[S]$ $[Nu]$ Rate $1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

For the following reaction: $NO_{2(g)} + CO_{(g)} \to NO_{(g)} + CO_{2(g)}$,the rate law is: $\text{Rate} = k [NO_2]^2$. If $0.1 \ mol$ of gaseous carbon monoxide is added at constant temperature to the reaction mixture,which of the following statements is true?

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Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.
Exp. $[A]$ $[B]$ Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$ $0.1$ $0.1$ $2 \times 10^{-3}$
$2$ $0.4$ $0.1$ $0.8 \times 10^{-2}$
$3$ $0.1$ $0.2$ $1.6 \times 10^{-2}$
(in $.5$)

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$