(A) In a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.The slow s

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(A) In a multi-step reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.The slow step is: $2B \xrightarrow{k} B_2$.The rate law for this step is: $Rate = k[B]^2$.Comparing this to the general rate law expression $Rate = k[A]^x[B]^y$:Order with respect to $A$ $(x)$ = $0$.Order with respect to $B$ $(y)$ = $2$.Overall order of the reaction = $x + y = 0 + 2 = 2$.Therefore,the correct option is $A$.

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

Medium

The mechanism of the reaction $A + 2B \to D$ is given by:
$2B \xrightarrow{k} B_2$ [Slow]
$B_2 + A \to D$ [Fast]
The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:

A
$Rate = k[B]^2, 0, 2, 2$
B
$Rate = k[A]^1[B]^2, 1, 2, 3$
C
$Rate = k[A]^2, 0, 2, 2$
D
$Rate = k[A]^2[B]^1, 1, 2, 3$

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Class 12

Chemistry Questions

Chapter

Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

The rate of a gaseous reaction is given by the expression $K [A] [B]$. If the volume of the reaction vessel is suddenly reduced to $1/4$ of the initial volume,the reaction rate relating to the original rate will be

Medium

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For a reaction,$AB_5 \to AB + 4B$,the rate can be expressed in the following ways:
$-\frac{d[AB_5]}{dt} = K[AB_5]$ ; $\frac{d[B]}{dt} = K_1[AB_5]$
What is the correct relation between $K$ and $K_1$?

Medium

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The initial reaction $2AB + B \to A_2B_3$ is carried out in vessels of volume $1 \ dm^3$ and $2 \ dm^3$ respectively,using equal moles of reactants. The ratio of the reaction rates $(r_1/r_2)$ is:

Medium

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If $a$ is the initial concentration and $n$ is the order of the reaction and the half-life period is $T$,then:

Medium

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During the kinetic study of the reaction,$2A + B \rightarrow C + D,$ the following results were obtained:

$Run$ $[A] / mol \ L^{-1}$ $[B] / mol \ L^{-1}$ Initial rate of formation of $D / mol \ L^{-1} \ min^{-1}$

$I.$ $0.1$ $0.1$ $6.0 \times 10^{-3}$

$II.$ $0.3$ $0.2$ $7.2 \times 10^{-2}$

$III.$ $0.3$ $0.4$ $2.88 \times 10^{-1}$

$IV.$ $0.4$ $0.1$ $2.40 \times 10^{-2}$

Based on the above data,which one of the following is correct?

AIPMT 2010

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$Run$	$[A] / mol \ L^{-1}$	$[B] / mol \ L^{-1}$	Initial rate of formation of $D / mol \ L^{-1} \ min^{-1}$
$I.$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II.$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III.$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV.$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

The mechanism of the reaction $A + 2B \to D$ is given by:$2B \xrightarrow{k} B_2$ [Slow]$B_2 + A \to D$ [Fast]The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:

Similar Questions

The rate of a gaseous reaction is given by the expression $K [A] [B]$. If the volume of the reaction vessel is suddenly reduced to $1/4$ of the initial volume,the reaction rate relating to the original rate will be

For a reaction,$AB_5 \to AB + 4B$,the rate can be expressed in the following ways:$-\frac{d[AB_5]}{dt} = K[AB_5]$ ; $\frac{d[B]}{dt} = K_1[AB_5]$What is the correct relation between $K$ and $K_1$?

The initial reaction $2AB + B \to A_2B_3$ is carried out in vessels of volume $1 \ dm^3$ and $2 \ dm^3$ respectively,using equal moles of reactants. The ratio of the reaction rates $(r_1/r_2)$ is:

If $a$ is the initial concentration and $n$ is the order of the reaction and the half-life period is $T$,then:

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The mechanism of the reaction $A + 2B \to D$ is given by:
$2B \xrightarrow{k} B_2$ [Slow]
$B_2 + A \to D$ [Fast]
The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:

For a reaction,$AB_5 \to AB + 4B$,the rate can be expressed in the following ways:
$-\frac{d[AB_5]}{dt} = K[AB_5]$ ; $\frac{d[B]}{dt} = K_1[AB_5]$
What is the correct relation between $K$ and $K_1$?