(D) The rate law for the reaction is given by: $Rate = k[A]^2[B]^0 = k[A]^2$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concent

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(D) The rate law for the reaction is given by: $Rate = k[A]^2[B]^0 = k[A]^2$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = 0.5[B])$,the new rate $(Rate')$ is: $Rate' = k(2[A])^2(0.5[B])^0 = k(4[A]^2)(1) = 4k[A]^2$. Comparing the new rate with the initial rate,we get $Rate' = 4 \times Rate$. Therefore,the rate becomes $4$ times the original rate.

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

Easy

$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?

GSEB 2024 All GSEB

A
$1/2$ times
B
$2$ times
C
$8$ times
D
$4$ times

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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

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$A$ reaction $A + B \rightarrow C$ is second order in $A$ and zeroth order in $B$. How is the rate affected when the concentration of $A$ is doubled and $B$ is halved?

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Which one of the following is wrongly matched?

For any reaction,the rate constant $K = 2.3 \times 10^{-5} \ mol^{-3/2} \ L^{3/2} \ S^{-1}$. The order of the reaction is . . . . . . .

In the reaction $X \to Y$,if the concentration of reactant $X$ is increased by $1.5$ times,the rate of reaction increases by $1.837$ times. Determine the order of the reaction with respect to $X$.

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