(A) The rate law for the reaction is $r = k[A][B]^2$. Let the initial concentrations be $[A] = x$ and $[B] = y$. Initial rate $r_1 = k(x)(y)^2 = kxy^2

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(A) The rate law for the reaction is $r = k[A][B]^2$. Let the initial concentrations be $[A] = x$ and $[B] = y$. Initial rate $r_1 = k(x)(y)^2 = kxy^2$. New concentrations are $[A]' = 4x$ and $[B]' = y/2$. New rate $r_2 = k(4x)(y/2)^2 = k(4x)(y^2/4) = kxy^2$. Since $r_2 = r_1$,the rate remains constant.

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

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For the reaction $A + 2B \rightleftharpoons 2C + D$,if the concentration of $A$ is increased four times and the concentration of $B$ is reduced to half,the rate of the reaction will be:

A
remain constant.
B
become half.
C
become double.
D
become $1/4$ of the original.

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For the reaction $A + 2B \rightleftharpoons 2C + D$,if the concentration of $A$ is increased four times and the concentration of $B$ is reduced to half,the rate of the reaction will be:

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