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(B) The rate law for the reaction is given by $Rate = k[X]^n$,where $n$ is the order of the reaction.Let the initial rate be $R_1 = k[X]^n$ and the in

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$1.5$

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(B) The rate law for the reaction is given by $Rate = k[X]^n$,where $n$ is the order of the reaction.Let the initial rate be $R_1 = k[X]^n$ and the initial concentration be $[X]$.When the concentration is increased by $1.5$ times,the new concentration is $[X]' = 1.5[X]$ and the new rate is $R_2 = 1.837 R_1$.Substituting these into the rate law: $1.837 R_1 = k(1.5[X])^n$.Dividing the two equations: $1.837 = (1.5)^n$.Taking the logarithm on both sides: $\log(1.837) = n \log(1.5)$.$0.2641 = n 	imes 0.1761$.$n = \frac{0.2641}{0.1761} \approx 1.5$.Therefore,the order of the reaction with respect to $X$ is $1.5$.

$[A_0]$	$0.1 \ M$	$0.025 \ M$
$t_{1/2} \ (s)$	$100 \ s$	$50 \ s$

In the reaction $X \to Y$,if the concentration of reactant $X$ is increased by $1.5$ times,the rate of reaction increases by $1.837$ times. Determine the order of the reaction with respect to $X$.

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