(A) The initial rate is $r_1 = K [A]^2[B]$.When the concentration of $A$ is doubled $([A]_2 = 2[A])$ and the concentration of $B$ is halved $([B]_2 =

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(A) The initial rate is $r_1 = K [A]^2[B]$.When the concentration of $A$ is doubled $([A]_2 = 2[A])$ and the concentration of $B$ is halved $([B]_2 = \frac{[B]}{2})$,the new rate $r_2$ is:$r_2 = K (2[A])^2 \times (\frac{[B]}{2})$$r_2 = K (4[A]^2) \times (\frac{[B]}{2})$$r_2 = 2 \times K [A]^2[B] = 2 \times r_1$Therefore,the reaction rate becomes double.

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

Medium

If a reaction has the experimental rate expression $\text{rate} = K [A]^2[B]$,what happens to the reaction rate if the concentration of $A$ is doubled and the concentration of $B$ is halved?

A
Rate becomes double
B
Rate becomes eight times
C
Rate becomes tripled
D
No change in rate

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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If a reaction has the experimental rate expression $\text{rate} = K [A]^2[B]$,what happens to the reaction rate if the concentration of $A$ is doubled and the concentration of $B$ is halved?

Similar Questions

The given plots represent the variation of the concentration of a reactant $R$ with time for two different reactions $(i)$ and $(ii).$ The respective orders of the reactions are

For a reaction,$I^{-} + OCl^{-} \to IO^{-} + Cl^{-}$ in an aqueous medium,the rate of reaction is given by $\frac{d[IO^{-}]}{dt} = k[I^{-}][OCl^{-}]$. The overall order of reaction is

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