For the reaction A + B to text{products},what | vedclass

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(B) The rate law is given by: $	ext{Rate} = K[A]^{x}[B]^{y}$From experiment $(1)$: $5 	imes 10^{-4} = K[2.5 	imes 10^{-4}]^{x}[3 	imes 10^{-5}]^{y

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$2$ with respect to $A$ and $1$ with respect to $B$

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(B) The rate law is given by: $	ext{Rate} = K[A]^{x}[B]^{y}$From experiment $(1)$: $5 	imes 10^{-4} = K[2.5 	imes 10^{-4}]^{x}[3 	imes 10^{-5}]^{y} \quad \dots (i)$From experiment $(2)$: $4 	imes 10^{-3} = K[5 	imes 10^{-4}]^{x}[6 	imes 10^{-5}]^{y} \quad \dots (ii)$From experiment $(3)$: $1.6 	imes 10^{-2} = K[1 	imes 10^{-3}]^{x}[6 	imes 10^{-5}]^{y} \quad \dots (iii)$Dividing $(iii)$ by $(ii)$:$\frac{1.6 	imes 10^{-2}}{4 	imes 10^{-3}} = \left(\frac{1 	imes 10^{-3}}{5 	imes 10^{-4}}ight)^{x} \implies 4 = 2^{x} \implies x = 2$Dividing $(ii)$ by $(i)$:$\frac{4 	imes 10^{-3}}{5 	imes 10^{-4}} = \left(\frac{5 	imes 10^{-4}}{2.5 	imes 10^{-4}}ight)^{2} \cdot \left(\frac{6 	imes 10^{-5}}{3 	imes 10^{-5}}ight)^{y}$$8 = 2^{2} \cdot 2^{y} \implies 8 = 4 \cdot 2^{y} \implies 2 = 2^{y} \implies y = 1$Thus,the order with respect to $A$ is $2$ and with respect to $B$ is $1$.

$Exp.$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$2.5 \times 10^{-4}$	$3 \times 10^{-5}$	$5 \times 10^{-4}$
$2$	$5 \times 10^{-4}$	$6 \times 10^{-5}$	$4 \times 10^{-3}$
$3$	$1 \times 10^{-3}$	$6 \times 10^{-5}$	$1.6 \times 10^{-2}$

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