For the non-stoichiometric reaction 2A + B to | vedclass

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(B) Let the order of reaction with respect to $A$ and $B$ be $\alpha$ and $\beta$ respectively.The rate law is given by: $\frac{d[C]}{dt} = k[A]^{\alp

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$\frac{d[C]}{dt} = k[A]$

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(B) Let the order of reaction with respect to $A$ and $B$ be $\alpha$ and $\beta$ respectively.The rate law is given by: $\frac{d[C]}{dt} = k[A]^{\alpha}[B]^{\beta}$From the given data:$1.2 	imes 10^{-3} = k[0.1]^{\alpha}[0.1]^{\beta} \quad (i)$$1.2 	imes 10^{-3} = k[0.1]^{\alpha}[0.2]^{\beta} \quad (ii)$$2.4 	imes 10^{-3} = k[0.2]^{\alpha}[0.1]^{\beta} \quad (iii)$Dividing $(ii)$ by $(i)$:$\frac{1.2 	imes 10^{-3}}{1.2 	imes 10^{-3}} = \frac{k[0.1]^{\alpha}[0.2]^{\beta}}{k[0.1]^{\alpha}[0.1]^{\beta}}$$1 = 2^{\beta} \implies \beta = 0$Dividing $(iii)$ by $(i)$:$\frac{2.4 	imes 10^{-3}}{1.2 	imes 10^{-3}} = \frac{k[0.2]^{\alpha}[0.1]^{\beta}}{k[0.1]^{\alpha}[0.1]^{\beta}}$$2 = 2^{\alpha} \implies \alpha = 1$Thus,the rate law is: $\frac{d[C]}{dt} = k[A]^1[B]^0 = k[A]$.

Initial Conc. $(A)$	Initial Conc. $(B)$	Initial rate of formation of $C \ (mol \ L^{-1} \ s^{-1})$
$0.1 \ M$	$0.1 \ M$	$1.2 \times 10^{-3}$
$0.1 \ M$	$0.2 \ M$	$1.2 \times 10^{-3}$
$0.2 \ M$	$0.1 \ M$	$2.4 \times 10^{-3}$

$[A_2]_0$	$[B_2]_0$	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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