For the reaction X_{2(g)} + Y_{2(g)} rightarr | vedclass

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(C) The rate of reaction is defined as $Rate = -\frac{d[X_{2}]}{dt} = -\frac{d[Y_{2}]}{dt} = \frac{1}{2} \frac{d[XY]}{dt}$.Let the rate law be $Rate =

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$2.5 	imes 10^{-3}$

Vedclass · Answer

(C) The rate of reaction is defined as $Rate = -\frac{d[X_{2}]}{dt} = -\frac{d[Y_{2}]}{dt} = \frac{1}{2} \frac{d[XY]}{dt}$.Let the rate law be $Rate = k[X_{2}]^{a}[Y_{2}]^{b}$.From the data:$1$) $5 	imes 10^{-6} = k(0.1)^{a}(0.1)^{b}$$2$) $10^{-5} = k(0.2)^{a}(0.1)^{b}$Dividing $(2)$ by $(1)$: $2 = 2^{a} \implies a = 1$.$3$) $4 	imes 10^{-5} = k(0.2)^{a}(0.2)^{b}$Dividing $(3)$ by $(2)$: $4 = 2^{b} \implies b = 2$.Thus,$Rate = k[X_{2}][Y_{2}]^{2}$.The rate of appearance of $XY$ is $\frac{d[XY]}{dt} = 2 	imes Rate = 2k[X_{2}][Y_{2}]^{2}$.Using the first data point: $5 	imes 10^{-6} = 2k(0.1)(0.1)^{2} = 2k(10^{-3})$.$k = \frac{5 	imes 10^{-6}}{2 	imes 10^{-3}} = 2.5 	imes 10^{-3} \ M^{-2} \ sec^{-1}$.

$[X_{2}] \ (M)$	$[Y_{2}] \ (M)$	Rate of appearance of $XY \ (M \ sec^{-1})$
$0.1$	$0.1$	$5 \times 10^{-6}$
$0.2$	$0.1$	$10^{-5}$
$0.2$	$0.2$	$4 \times 10^{-5}$

$[A_2]_0$	$[B_2]_0$	Initial rate of formation of $AB$ $(mol \ L^{-1} s^{-1})$
$0.1 \ M$	$0.1 \ M$	$2.5 \times 10^{-4}$
$0.2 \ M$	$0.1 \ M$	$5.0 \times 10^{-4}$
$0.2 \ M$	$0.2 \ M$	$1.0 \times 10^{-3}$

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