(B) The rate of the reaction is determined by the slow step: $r = K_2 [I]^2 [H_2]$. Since $I$ is an intermediate,we use the equilibrium step to expre

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$r = K_2 \frac{K_1}{K_{-1}} [H_2] [I_2]$

(B) The rate of the reaction is determined by the slow step: $r = K_2 [I]^2 [H_2]$. Since $I$ is an intermediate,we use the equilibrium step to express its concentration. For the fast equilibrium step: $K_{eq} = \frac{K_1}{K_{-1}} = \frac{[I]^2}{[I_2]}$. Therefore,$[I]^2 = \frac{K_1}{K_{-1}} [I_2]$. Substituting this into the rate expression: $r = K_2 (\frac{K_1}{K_{-1}} [I_2]) [H_2]$. Thus,$r = K_2 \frac{K_1}{K_{-1}} [H_2] [I_2]$.

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

Medium

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below:
$I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step)
${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step)
Determine the rate law for the reaction.

A
$r = K_2 K_1 [H_2] [I_2]$
B
$r = K_2 \frac{K_1}{K_{-1}} [H_2] [I_2]$
C
$r = K_2 \sqrt{\frac{K_1}{K_{-1}}} [H_2] [I_2]$
D
$r = K_2 \frac{K_1}{K_{-1}} [H_2] [I_2]^2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

Rate law , Rate constant , Order of Reaction and Molecularity

The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:
$(i)$ $NO_2 \xrightarrow{slow} NO + O$
$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$
$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$
Select the correct statement.

Medium

View Solution

For the reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$,the experimental data suggest,$\text{rate} = K[H_2][Br_2]^{1/2}$. The molecularity and order of the reaction are respectively:

Easy

View Solution

Write about elementary and complex reactions.

Medium

View Solution

The rate constants for first-order and second-order reactions have units of ..... respectively.

Easy

View Solution

For the reaction $A + B \to C + D$,if the concentration of $A$ is doubled without affecting the concentration of $B$,the rate doubles. If the concentration of $B$ is increased $9$ times without affecting the concentration of $A$,the rate becomes $3$ times. Determine the order of the reaction.

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below: $I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step) ${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step) Determine the rate law for the reaction.

Similar Questions

The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:$(i)$ $NO_2 \xrightarrow{slow} NO + O$$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$Select the correct statement.

For the reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$,the experimental data suggest,$\text{rate} = K[H_2][Br_2]^{1/2}$. The molecularity and order of the reaction are respectively:

Write about elementary and complex reactions.

The rate constants for first-order and second-order reactions have units of ..... respectively.

For the reaction $A + B \to C + D$,if the concentration of $A$ is doubled without affecting the concentration of $B$,the rate doubles. If the concentration of $B$ is increased $9$ times without affecting the concentration of $A$,the rate becomes $3$ times. Determine the order of the reaction.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below:
$I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step)
${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step)
Determine the rate law for the reaction.

The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:
$(i)$ $NO_2 \xrightarrow{slow} NO + O$
$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$
$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$
Select the correct statement.