In a reaction A rightarrow B,if the concentra | vedclass

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(B) For a reaction $A ightarrow B$,the rate law is given by: $	ext{Rate} = k[A]^n$,where $n$ is the order of reaction. Given that when the concentr

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$1/2$

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(B) For a reaction $A ightarrow B$,the rate law is given by: $	ext{Rate} = k[A]^n$,where $n$ is the order of reaction. Given that when the concentration $[A]$ is increased by $9$ times,the rate increases by $3$ times. So,$3 	imes 	ext{Rate} = k(9[A])^n$. Dividing this by the original rate equation: $\frac{3 	imes 	ext{Rate}}{	ext{Rate}} = \frac{k(9[A])^n}{k[A]^n}$. $3 = 9^n$. Since $9 = 3^2$,we have $3 = (3^2)^n = 3^{2n}$. Comparing the exponents: $1 = 2n$,which gives $n = 1/2$. Therefore,the order of reaction is $1/2$.

In a reaction $A \rightarrow B$,if the concentration of reactant is increased by $9$ times,then the rate of reaction increases $3$ times. What is the order of reaction?

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