(B) The general rate law is given by: $\text{Rate} = K[A]^x[B]^y$. When the concentration of $A$ is doubled,the rate increases by $4$ times $(2^2 = 4)

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(B) The general rate law is given by: $\text{Rate} = K[A]^x[B]^y$. When the concentration of $A$ is doubled,the rate increases by $4$ times $(2^2 = 4)$,which implies the order with respect to $A$ is $x = 2$. When the concentration of $B$ is doubled,the rate remains unchanged,which implies the order with respect to $B$ is $y = 0$. Substituting these values into the rate law: $\text{Rate} = K[A]^2[B]^0 = K[A]^2$.

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

Medium

For the reaction $A + B \rightarrow$ products,doubling the concentration of $A$ increases the reaction rate by four times,but doubling the concentration of $B$ has no effect on the reaction rate. What is the rate law?

A
Rate $= K[A][B]$
B
Rate $= K[A]^2$
C
Rate $= K[A]^2[B]$
D
Rate $= K[A]^2[B]^2$

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Class 12

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Chapter

Chemical Kinetics

Subtopic

Rate law , Rate constant , Order of Reaction and Molecularity

$A$ chemical reaction proceeds through the following steps:
Step-$I$: $2A \rightleftharpoons X$ (fast)
Step-$II$: $X + B \rightarrow Y$ (slow)
Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)
The rate law for the overall reaction is:

Medium

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Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.
Exp. $[A]$ $[B]$ Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$ $0.1$ $0.1$ $2 \times 10^{-3}$
$2$ $0.4$ $0.1$ $0.8 \times 10^{-2}$
$3$ $0.1$ $0.2$ $1.6 \times 10^{-2}$
(in $.5$)

Exp.	$[A]$	$[B]$	Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$0.8 \times 10^{-2}$
$3$	$0.1$	$0.2$	$1.6 \times 10^{-2}$

Medium

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For a reaction scheme $A$ $\xrightarrow{k_1} B$ $\xrightarrow{k_2} C$,if the rate of formation of $B$ is set to be zero,then the concentration of $B$ is given by:

DifficultJEE MAIN 2019

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Select the incorrect option :

Medium

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For the reaction,$CH_3Br_{(aq)} + OH_{(aq)}^{-} \rightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$,the rate law is $\text{rate} = k[CH_3Br][OH^{-}]$. What is the change in the rate of reaction if the concentration of both reactants is doubled?

MediumMHT CET 2023

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For the reaction $A + B \rightarrow$ products,doubling the concentration of $A$ increases the reaction rate by four times,but doubling the concentration of $B$ has no effect on the reaction rate. What is the rate law?

Similar Questions

$A$ chemical reaction proceeds through the following steps:Step-$I$: $2A \rightleftharpoons X$ (fast)Step-$II$: $X + B \rightarrow Y$ (slow)Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)The rate law for the overall reaction is:

Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.Exp.$[A]$$[B]$Rate of reaction $(mol \ L^{-1} \ s^{-1})$$1$$0.1$$0.1$$2 \times 10^{-3}$$2$$0.4$$0.1$$0.8 \times 10^{-2}$$3$$0.1$$0.2$$1.6 \times 10^{-2}$ (in $.5$)

For a reaction scheme $A$ $\xrightarrow{k_1} B$ $\xrightarrow{k_2} C$,if the rate of formation of $B$ is set to be zero,then the concentration of $B$ is given by:

Select the incorrect option :

For the reaction,$CH_3Br_{(aq)} + OH_{(aq)}^{-} \rightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$,the rate law is $\text{rate} = k[CH_3Br][OH^{-}]$. What is the change in the rate of reaction if the concentration of both reactants is doubled?

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$A$ chemical reaction proceeds through the following steps:
Step-$I$: $2A \rightleftharpoons X$ (fast)
Step-$II$: $X + B \rightarrow Y$ (slow)
Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)
The rate law for the overall reaction is:

Determine the order of reaction on the basis of following data for the reaction $A + B \to C$.
Exp. $[A]$ $[B]$ Rate of reaction $(mol \ L^{-1} \ s^{-1})$
$1$ $0.1$ $0.1$ $2 \times 10^{-3}$
$2$ $0.4$ $0.1$ $0.8 \times 10^{-2}$
$3$ $0.1$ $0.2$ $1.6 \times 10^{-2}$
(in $.5$)