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(C) The rate law for the reaction is given by: $	ext{Rate} = k[A]^n$,where $n$ is the order of the reaction.Let the initial concentration of $A$ be $

Vedclass · Accepted Answer

$2$

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(C) The rate law for the reaction is given by: $	ext{Rate} = k[A]^n$,where $n$ is the order of the reaction.Let the initial concentration of $A$ be $[A]_1 = P$ and the initial rate be $r_1 = kP^n$.When the concentration is increased by a factor of $1.5$,the new concentration is $[A]_2 = 1.5P$.The new rate becomes $r_2 = 2.25 r_1 = k(1.5P)^n$.Dividing the expression for $r_2$ by $r_1$ gives: $\frac{r_2}{r_1} = \frac{k(1.5P)^n}{kP^n} = (1.5)^n$.Since $\frac{r_2}{r_1} = 2.25$,we have $(1.5)^n = 2.25$.Since $(1.5)^2 = 2.25$,it follows that $n = 2$.Therefore,the order of the reaction is $2$.

For the reaction $A \to B$,the rate increases by a factor of $2.25$ when the concentration of $A$ is increased by $1.5$. What is the order of the reaction?

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