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(C) Let the rate law be $	ext{Rate} = k [CH_3COCH_3]^x [Br_2]^y [H^+]^z$.Comparing experiments $(1)$ and $(2)$: $[CH_3COCH_3]$ and $[H^+]$ are consta

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Rate $= k [CH_3COCH_3][H^+]$

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(C) Let the rate law be $	ext{Rate} = k [CH_3COCH_3]^x [Br_2]^y [H^+]^z$.Comparing experiments $(1)$ and $(2)$: $[CH_3COCH_3]$ and $[H^+]$ are constant,while $[Br_2]$ doubles. The rate remains $5.7 	imes 10^{-5} M s^{-1}$. Thus,$y = 0$.Comparing experiments $(2)$ and $(3)$: $[CH_3COCH_3]$ is constant,$[Br_2]$ is constant,and $[H^+]$ doubles. The rate increases from $5.7 	imes 10^{-5}$ to $1.14 	imes 10^{-4}$ (a factor of $2$). Thus,$z = 1$.Comparing experiments $(1)$ and $(4)$: $[Br_2]$ is constant. $[CH_3COCH_3]$ increases by $4/3$ and $[H^+]$ increases by $4$. The rate increases from $5.7 	imes 10^{-5}$ to $3.04 	imes 10^{-4}$ (a factor of $\approx 5.33$). Since $5.33 = (4/3) 	imes 4$,we find $x = 1$.Therefore,the rate law is $	ext{Rate} = k [CH_3COCH_3][H^+]$.

$[CH_3COCH_3]$	$[Br_2]$	$[H^+]$
$0.30$	$0.05$	$0.05$
$0.30$	$0.10$	$0.05$
$0.30$	$0.10$	$0.10$
$0.40$	$0.05$	$0.20$

Experiment	$[Cl_2] \ (M)$	$[NO] \ (M)$	Rate $(mol \ L^{-1} \sec^{-1})$
$I$	$0.05$	$0.05$	$1 \times 10^{-3}$
$II$	$0.15$	$0.05$	$3 \times 10^{-3}$
$III$	$0.05$	$0.15$	$9 \times 10^{-3}$

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