(C) The rate law for the reaction $2A + B \rightarrow A_2B$ is given by $Rate = k[A]^x[B]^y$. Assuming the reaction is elementary,the rate law is $Rat

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(C) The rate law for the reaction $2A + B \rightarrow A_2B$ is given by $Rate = k[A]^x[B]^y$. Assuming the reaction is elementary,the rate law is $Rate = k[A]^2[B]^1$. Let the initial rate be $R_1 = k[A]^2[B]$. When the concentration of $A$ is doubled $(2[A])$ and $B$ is halved $([B]/2)$,the new rate $R_2$ is: $R_2 = k(2[A])^2([B]/2) = k(4[A]^2)([B]/2) = 2k[A]^2[B]$. Comparing $R_2$ with $R_1$,we get $R_2 = 2R_1$. Therefore,the rate of the reaction increases by $2$ times.

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

Easy

For the reaction $2A + B \rightarrow A_2B$,if the concentration of reactant $A$ is doubled and the concentration of reactant $B$ is halved,the rate of the reaction will:

A
Increase by $4$ times
B
Decrease by $2$ times
C
Increase by $2$ times
D
Remain the same

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

For the reaction $A + B \rightarrow \text{Product}$,the rate law is given by $\text{Rate} = K[A]^1[B]^2$. Which of the following statements is incorrect?

Easy

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The reaction $2 \ NO + Br_2 \rightarrow 2 \ NOBr$ takes place through the mechanism given below:
$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)
$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)
The overall order of the reaction is $.....$.

MediumJEE MAIN 2023

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$A$ reaction was found to be $2nd$ order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled,with everything else kept the same,the rate of reaction will

MediumAIEEE 2006

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Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is

Difficult

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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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For the reaction $2A + B \rightarrow A_2B$,if the concentration of reactant $A$ is doubled and the concentration of reactant $B$ is halved,the rate of the reaction will:

Similar Questions

For the reaction $A + B \rightarrow \text{Product}$,the rate law is given by $\text{Rate} = K[A]^1[B]^2$. Which of the following statements is incorrect?

The reaction $2 \ NO + Br_2 \rightarrow 2 \ NOBr$ takes place through the mechanism given below:$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)The overall order of the reaction is $.....$.

$A$ reaction was found to be $2nd$ order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled,with everything else kept the same,the rate of reaction will

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$takes place in two steps :$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$The order of the reaction is

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 $K_1$?

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The reaction $2 \ NO + Br_2 \rightarrow 2 \ NOBr$ takes place through the mechanism given below:
$NO + Br_2 \Leftrightarrow NOBr_2$ (fast)
$NOBr_2 + NO \rightarrow 2 \ NOBr$ (slow)
The overall order of the reaction is $.....$.

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is

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 $K_1$?