Let the rate law be $\text{Rate} = k[\text{CH}_3\text{COCH}_3]^x[\text{Br}_2]^y[\text{H}^+]^z$.

Comparing experiment 1 and 2: When $[\text{Br}_2]$ is doubled ($0.05 \text{ M}$ to $0.10 \text{ M}$) while keeping other concentrations constant, the rate remains unchanged ($5.7 \times 10^{-5}\text{ M s}^{-1}$). Thus, the order with respect to $\text{Br}_2$ is 0 ($y = 0$).

Comparing experiment 2 and 3: When $[\text{H}^+]$ is doubled ($0.05 \text{ M}$ to $0.10 \text{ M}$) keeping other concentrations constant, the rate is approximately doubled (from $5.7 \times 10^{-5}$ to $1.2 \times 10^{-4} \approx 1.14 \times 10^{-4}$). Thus, the order with respect to $\text{H}^+$ is 1 ($z = 1$).

Comparing experiment 1 and 4: $[\text{CH}_3\text{COCH}_3]$ is increased by $4/3$ times ($0.30$ to $0.40$), $[\text{Br}_2]$ is unchanged, and $[\text{H}^+]$ is increased by $4$ times ($0.05$ to $0.20$). The rate becomes $\frac{3.1 \times 10^{-4}}{5.7 \times 10^{-5}} \approx 5.44$ times the initial rate. Based on the previously determined orders, the expected new rate would be proportional to the changes: $\text{Rate}_4 = \text{Rate}_1 \times \left(\frac{0.40}{0.30}\right)^x \times \left(\frac{0.20}{0.05}\right)^1$ $3.1 \times 10^{-4} = 5.7 \times 10^{-5} \times \left(\frac{4}{3}\right)^x \times 4$ $\left(\frac{4}{3}\right)^x = \frac{3.1 \times 10^{-4}}{5.7 \times 10^{-5} \times 4} = \frac{3.1}{22.8} \approx 1.36$ Since $\frac{4}{3} \approx 1.33$, we get $x = 1$.

Therefore, the overall rate equation is $\text{Rate} = k[\text{CH}_3\text{COCH}_3]^1[\text{Br}_2]^0[\text{H}^+]^1 = k[\text{CH}_3\text{COCH}_3][\text{H}^+]$.