(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

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

(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
$r = K_2 \frac{K_1}{K_{-1}} [H_2] [I_2]^2$

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Class 12

Chemistry Questions

Chapter

Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

Write the unit of the rate constant for the following reactions:
$1.$ $\frac{1}{2}$ order
$2.$ $\frac{3}{2}$ order

Easy

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For the decomposition of a compound $AB$ at $600 \ K$,the following data were obtained:
$[AB] \ (mol \ dm^{-3})$ Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$
$0.20$ $2.75 \times 10^{-8}$
$0.40$ $11.0 \times 10^{-8}$
$0.60$ $24.75 \times 10^{-8}$

The order for the decomposition of $AB$ is:

DifficultKCET 2009

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The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is:

Difficult

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Which among the following is a false statement?

Easy

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Consider a reaction $aG + bH \rightarrow$ Products. When the concentration of both reactants $G$ and $H$ is doubled,the rate increases by $8$ times. However,when the concentration of $G$ is doubled keeping the concentration of $H$ fixed,the rate is doubled. The overall order of the reaction is:

MediumIIT 2007

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$[AB] \ (mol \ dm^{-3})$	Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$
$0.20$	$2.75 \times 10^{-8}$
$0.40$	$11.0 \times 10^{-8}$
$0.60$	$24.75 \times 10^{-8}$

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

Write the unit of the rate constant for the following reactions:$1.$ $\frac{1}{2}$ order$2.$ $\frac{3}{2}$ order

For the decomposition of a compound $AB$ at $600 \ K$,the following data were obtained: $[AB] \ (mol \ dm^{-3})$Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$$0.20$$2.75 \times 10^{-8}$$0.40$$11.0 \times 10^{-8}$$0.60$$24.75 \times 10^{-8}$ The order for the decomposition of $AB$ is:

The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$Then the correct statement is:

Which among the following is a false statement?

Consider a reaction $aG + bH \rightarrow$ Products. When the concentration of both reactants $G$ and $H$ is doubled,the rate increases by $8$ times. However,when the concentration of $G$ is doubled keeping the concentration of $H$ fixed,the rate is doubled. The overall order of the reaction is:

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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.

Write the unit of the rate constant for the following reactions:
$1.$ $\frac{1}{2}$ order
$2.$ $\frac{3}{2}$ order

The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is: