The experimental data for the reaction 2A + B | vedclass

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Question

(C) The rate law is given by $R = K[A]^{\alpha}[B_2]^{\beta}$.From the experimental data:$1.6 	imes 10^{-4} = K[0.50]^{\alpha}[0.50]^{\beta}$ ......

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$r = K[B_2]$

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(C) The rate law is given by $R = K[A]^{\alpha}[B_2]^{\beta}$.From the experimental data:$1.6 	imes 10^{-4} = K[0.50]^{\alpha}[0.50]^{\beta}$ ...... $(i)$$3.2 	imes 10^{-4} = K[0.50]^{\alpha}[1.00]^{\beta}$ ...... $(ii)$$3.2 	imes 10^{-4} = K[1.00]^{\alpha}[1.00]^{\beta}$ ...... $(iii)$Divide $(iii)$ by $(ii)$:$\frac{3.2 	imes 10^{-4}}{3.2 	imes 10^{-4}} = \frac{K[1.00]^{\alpha}[1.00]^{\beta}}{K[0.50]^{\alpha}[1.00]^{\beta}}$$1 = (2)^{\alpha} \implies \alpha = 0$.Divide $(ii)$ by $(i)$:$\frac{3.2 	imes 10^{-4}}{1.6 	imes 10^{-4}} = \frac{K[0.50]^{\alpha}[1.00]^{\beta}}{K[0.50]^{\alpha}[0.50]^{\beta}}$$2 = (2)^{\beta} \implies \beta = 1$.Thus,the rate law is $R = K[A]^0[B_2]^1$,which simplifies to $R = K[B_2]$.

Exp.	$[A] \ (mol \ L^{-1})$	$[B_2] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ S^{-1})$
$1$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$2$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$3$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

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