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

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(A) Let the rate law be $Rate = k [A]^x [B_2]^y$. From experiment $(1)$ and $(2)$,$[A]_0$ is constant $(0.50 \ M)$ and $[B_2]_0$ is doubled ($0.50 \ M

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$Rate = k [B_2]$

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(A) Let the rate law be $Rate = k [A]^x [B_2]^y$. From experiment $(1)$ and $(2)$,$[A]_0$ is constant $(0.50 \ M)$ and $[B_2]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate increases from $1.6 	imes 10^{-4}$ to $3.2 	imes 10^{-4}$ (doubled). Therefore,$2^y = 2$,which implies $y = 1$. From experiment $(2)$ and $(3)$,$[B_2]_0$ is constant $(1.00 \ M)$ and $[A]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate remains $3.2 	imes 10^{-4}$ (no change). Therefore,$2^x = 1$,which implies $x = 0$. Substituting the values of $x$ and $y$ into the rate law,we get $Rate = k [A]^0 [B_2]^1 = k [B_2]$.

$Exp.$	$[A]_0$	$[B_2]_0$	$Rate \ (mol \ L^{-1} \ s^{-1})$
$(1)$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$(2)$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$(3)$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

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