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(D) The rate law is given by $R = k[X]^x[Y]^y$.From experiment $1$ and $2$,keeping $[Y]$ constant:$\frac{R_2}{R_1} = \frac{k[0.50]^x[0.25]^y}{k[0.25]^

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$5$

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(D) The rate law is given by $R = k[X]^x[Y]^y$.From experiment $1$ and $2$,keeping $[Y]$ constant:$\frac{R_2}{R_1} = \frac{k[0.50]^x[0.25]^y}{k[0.25]^x[0.25]^y} = \frac{4.0 	imes 10^{-6}}{1.0 	imes 10^{-6}}$$2^x = 4$ $\Rightarrow 2^x = 2^2$ $\Rightarrow x = 2$.From experiment $1$ and $3$,keeping $[X]$ constant:$\frac{R_3}{R_1} = \frac{k[0.25]^x[0.50]^y}{k[0.25]^x[0.25]^y} = \frac{8.0 	imes 10^{-6}}{1.0 	imes 10^{-6}}$$2^y = 8$ $\Rightarrow 2^y = 2^3$ $\Rightarrow y = 3$.The overall order of the reaction is $x + y = 2 + 3 = 5$.

$Expt.$	$[X]_0 / mol \ L^{-1}$	$[Y]_0 / mol \ L^{-1}$	$Rate / mol \ L^{-1} s^{-1}$
$1$	$0.25$	$0.25$	$1.0 \times 10^{-6}$
$2$	$0.50$	$0.25$	$4.0 \times 10^{-6}$
$3$	$0.25$	$0.50$	$8.0 \times 10^{-6}$

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