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(D) Let the order of reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.The rate law is given by: $	ext{Rate} = k[A]^x[B]^y$Using the

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Rate $= k[A][B]^2$

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(D) Let the order of reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.The rate law is given by: $	ext{Rate} = k[A]^x[B]^y$Using the data from the table:$I. \ 6.0 	imes 10^{-3} = k(0.1)^x(0.1)^y$$II. \ 7.2 	imes 10^{-2} = k(0.3)^x(0.2)^y$$III. \ 2.88 	imes 10^{-1} = k(0.3)^x(0.4)^y$$IV. \ 2.40 	imes 10^{-2} = k(0.4)^x(0.1)^y$Dividing Eq. $(I)$ by Eq. $(IV)$:$\frac{6.0 	imes 10^{-3}}{2.40 	imes 10^{-2}} = \left(\frac{0.1}{0.4}ight)^x \left(\frac{0.1}{0.1}ight)^y$$0.25 = (0.25)^x \implies x = 1$Dividing Eq. $(II)$ by Eq. $(III)$:$\frac{7.2 	imes 10^{-2}}{2.88 	imes 10^{-1}} = \left(\frac{0.3}{0.3}ight)^x \left(\frac{0.2}{0.4}ight)^y$$0.25 = (0.5)^y \implies (0.5)^2 = (0.5)^y \implies y = 2$Therefore,the rate law is: $	ext{Rate} = k[A]^1[B]^2 = k[A][B]^2$.

$Run$	$[A] / mol \ L^{-1}$	$[B] / mol \ L^{-1}$	Initial rate of formation of $D / mol \ L^{-1} \ min^{-1}$
$I.$	$0.1$	$0.1$	$6.0 \times 10^{-3}$
$II.$	$0.3$	$0.2$	$7.2 \times 10^{-2}$
$III.$	$0.3$	$0.4$	$2.88 \times 10^{-1}$
$IV.$	$0.4$	$0.1$	$2.40 \times 10^{-2}$

$P_{A_0} \text{ (atm)}$	$0.1$	$0.025$
$t_{1/2} \text{ (sec)}$	$100$	$50$

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