For the reaction A to text{Products},the rate | vedclass

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(C) The rate law for the reaction is given by $Rate = k[A]^n$,where $n$ is the order of the reaction.Let the initial rate be $R_1 = k[A]^n$.When the c

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$2$

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(C) The rate law for the reaction is given by $Rate = k[A]^n$,where $n$ is the order of the reaction.Let the initial rate be $R_1 = k[A]^n$.When the concentration is halved,the new concentration is $[A]' = \frac{[A]}{2}$.The new rate is $R_2 = k(\frac{[A]}{2})^n$.Given that $R_2 = \frac{R_1}{4}$,we have $\frac{R_1}{4} = k(\frac{[A]}{2})^n$.Substituting $R_1 = k[A]^n$,we get $\frac{k[A]^n}{4} = k(\frac{[A]}{2})^n$.$\frac{1}{4} = (\frac{1}{2})^n$.$(\frac{1}{2})^2 = (\frac{1}{2})^n$.Therefore,$n = 2$.

For the reaction $A \to \text{Products}$,the rate of reaction becomes one-fourth when the concentration of $A$ is halved. What is the order of the reaction?

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