For the reaction A to B, the rate increases b | vedclass

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Question

Rate for this reaction can be written as : Rate $= k [ A ]^{ n }$; where $n$ is the order of the reaction.

Let the initial concentration of $A$ be

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Rate for this reaction can be written as : Rate $= k [ A ]^{ n }$; where $n$ is the order of the reaction.

Let the initial concentration of $A$ be $P$.

$\Longrightarrow 	ext { Rate }= kP ^{ n }= r _1 \quad \ldots . 	ext { (1) }$

Now, it is said that when we increase the concentration of $A$ to $1.5 P$, rate of reaction becomes $2.25$ times the initial rate

$\Rightarrow 	ext { Rate }= k (1.5 P )^{ n }=2.25 r _1 \quad \ldots 	ext { (2) }$

On dividing $(2)$ by $(1)$, we get

$(1.5)^{ n }=2.25$

$\Rightarrow n =2$

Hence, 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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