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Question

(A) From the rate law expression: $Rate = k [A]^x [B]^y$Using the data from experiments $I, II,$ and $III$:$6.00 	imes 10^{-3} = k (0.1)^x (0.1)^y \d

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$0.3, 0.4$

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(A) From the rate law expression: $Rate = k [A]^x [B]^y$Using the data from experiments $I, II,$ and $III$:$6.00 	imes 10^{-3} = k (0.1)^x (0.1)^y \dots(1)$$2.40 	imes 10^{-2} = k (0.1)^x (0.2)^y \dots(2)$$1.20 	imes 10^{-2} = k (0.2)^x (0.1)^y \dots(3)$Dividing $(3)$ by $(1)$: $2^x = 2 \implies x = 1$Dividing $(2)$ by $(1)$: $2^y = 4 \implies y = 2$Thus,the rate law is $Rate = k [A]^1 [B]^2$.For experiment $IV$:$7.20 	imes 10^{-2} = k (X)^1 (0.2)^2$Using $k$ from experiment $I$: $k = \frac{6.00 	imes 10^{-3}}{(0.1)(0.1)^2} = 6 \ L^2 \ mol^{-2} \ min^{-1}$$7.20 	imes 10^{-2} = 6 	imes X 	imes 0.04 \implies X = \frac{7.20 	imes 10^{-2}}{0.24} = 0.3 \ M$For experiment $V$:$2.88 	imes 10^{-1} = 6 	imes (0.3) 	imes Y^2$$Y^2 = \frac{2.88 	imes 10^{-1}}{1.8} = 0.16 \implies Y = 0.4 \ M$Therefore,$X = 0.3$ and $Y = 0.4$.

Experiment	$[A] / mol \ L^{-1}$	$[B] / mol \ L^{-1}$	Initial rate / $mol \ L^{-1} \ min^{-1}$
$I$	$0.1$	$0.1$	$6.00 \times 10^{-3}$
$II$	$0.1$	$0.2$	$2.40 \times 10^{-2}$
$III$	$0.2$	$0.1$	$1.20 \times 10^{-2}$
$IV$	$X$	$0.2$	$7.20 \times 10^{-2}$
$V$	$0.3$	$Y$	$2.88 \times 10^{-1}$

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