(A) The rate law for the reaction is given by: $Rate = k[NO]^2[O_2]$.When the volume of the vessel is reduced to half,the concentration of each gaseou

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Increase $8$ times its initial value.

(A) The rate law for the reaction is given by: $Rate = k[NO]^2[O_2]$.When the volume of the vessel is reduced to half,the concentration of each gaseous reactant doubles because $Concentration = \text{moles} / \text{Volume}$.Let the initial concentrations be $[NO]_i$ and $[O_2]_i$. The new concentrations are $[NO]_f = 2[NO]_i$ and $[O_2]_f = 2[O_2]_i$.The new rate $Rate_f = k(2[NO]_i)^2(2[O_2]_i) = k \times 4[NO]_i^2 \times 2[O_2]_i = 8 \times k[NO]_i^2[O_2]_i$.Therefore,the new rate is $8$ times the initial rate.

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

Medium

For the reaction $2NO_{(g)} + O_{2_{(g)}} \rightarrow 2NO_{2_{(g)}}$,the volume of the reaction vessel is suddenly reduced to half of its original volume. If the reaction is first order with respect to $O_2$ and second order with respect to $NO$,the rate of the reaction will:

A
Increase $8$ times its initial value.
B
Increase $4$ times its initial value.
C
Become $1/4$ of its initial value.
D
Become $1/8$ of its initial value.

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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

$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$

Initial $[A_2]$ Initial $[B_2]$ $R.O.R. (r) \ M s^{-1}$

$0.2$ $0.2$ $0.04$

$0.1$ $0.4$ $0.04$

$0.2$ $0.4$ $0.08$

Order of reaction with respect to $A_2$ and $B_2$ are respectively:

Initial $[A_2]$	Initial $[B_2]$	$R.O.R. (r) \ M s^{-1}$
$0.2$	$0.2$	$0.04$
$0.1$	$0.4$	$0.04$
$0.2$	$0.4$	$0.08$

Medium

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Write the unit of the rate constant for the following reactions:
$1.$ $5/2$ order
$2.$ $n$ order

Easy

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Which one of the following statements regarding the order of a reaction is not correct?

Medium

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The reaction rate between two substances $A$ and $B$ is expressed as: $\text{rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,the ratio of the new rate to the initial rate will be:

MediumAIEEE 2012

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The rate constant for the reaction $2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-5} \, sec^{-1}$. If the rate is $2.40 \times 10^{-5} \, M \, sec^{-1}$,then the concentration of $N_2O_5$ (in $M$) is?

Medium

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For the reaction $2NO_{(g)} + O_{2_{(g)}} \rightarrow 2NO_{2_{(g)}}$,the volume of the reaction vessel is suddenly reduced to half of its original volume. If the reaction is first order with respect to $O_2$ and second order with respect to $NO$,the rate of the reaction will:

Similar Questions

$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$ Initial $[A_2]$ Initial $[B_2]$ $R.O.R. (r) \ M s^{-1}$ $0.2$ $0.2$ $0.04$ $0.1$ $0.4$ $0.04$ $0.2$ $0.4$ $0.08$ Order of reaction with respect to $A_2$ and $B_2$ are respectively:

Write the unit of the rate constant for the following reactions:$1.$ $5/2$ order$2.$ $n$ order

Which one of the following statements regarding the order of a reaction is not correct?

The reaction rate between two substances $A$ and $B$ is expressed as: $\text{rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,the ratio of the new rate to the initial rate will be:

The rate constant for the reaction $2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-5} \, sec^{-1}$. If the rate is $2.40 \times 10^{-5} \, M \, sec^{-1}$,then the concentration of $N_2O_5$ (in $M$) is?

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$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$

Initial $[A_2]$ Initial $[B_2]$ $R.O.R. (r) \ M s^{-1}$

$0.2$ $0.2$ $0.04$

$0.1$ $0.4$ $0.04$

$0.2$ $0.4$ $0.08$

Order of reaction with respect to $A_2$ and $B_2$ are respectively:

Write the unit of the rate constant for the following reactions:
$1.$ $5/2$ order
$2.$ $n$ order