In the reaction A rightarrow B,if the concent | vedclass

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(C) The rate law for the reaction is given by: $	ext{Rate} = k[A]^x$ ...$(1)$When the concentration of $A$ is doubled,the rate increases by $1.59$ ti

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$\frac{2}{3}$

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(C) The rate law for the reaction is given by: $	ext{Rate} = k[A]^x$ ...$(1)$When the concentration of $A$ is doubled,the rate increases by $1.59$ times:$1.59 	imes 	ext{Rate} = k[2A]^x$ ...$(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{1.59 	imes 	ext{Rate}}{	ext{Rate}} = \frac{k[2A]^x}{k[A]^x}$$1.59 = (2)^x$Taking $\log$ on both sides:$\log(1.59) = x \log(2)$$0.2014 = x 	imes 0.3010$$x = \frac{0.2014}{0.3010} \approx 0.669$Since $0.669 \approx \frac{2}{3}$,the order of the reaction is $\frac{2}{3}$.

In the reaction $A \rightarrow B$,if the concentration of $A$ is doubled,then the reaction rate increases by $1.59$ times. What will be the order of the reaction?

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