The bromination of acetone that occurs in acid solution is represented by this equation.
`CH_(3)COCH_(3) (aq) + Br_(2) (aq) rarr`
`CH_(3)COCH_(2) Br(aq) + H^(+) (aq) + Br(aq)`
These kinetic data were obtained for given reaction concentrations.
Initial concentration, `M`
`{:([CH_(2)COCH_(3)],[Br_(2)],[H^(+)],("Initail rate) (disappearance of "Br_(2)),),(0.30,0.05,0.05,5.7 xx 10^(-5),),(0.30,0.10,0.05,5.7xx10^(-5),),(0.30,0.10,0.10,1.2xx10^(-4),),(0.40,0.5,0.20,3.1xx10^(-4),):}`

Options

A. Rate `=k[CH_(3)COCH_(3)][Br_(2)][H^(+)]^(2)`
B. Rate `=k[CH_(3)COCH_(3)][Br_(2)][H^(+)]`
C. Rate `=k[CH_(3)COCH_(3)][H^(+)]`
D. Rate `=k[CH=COCH_(3)][Br_(2)]`

Explanation

To solve the problem, we need to determine the rate law for the bromination of acetone based on the provided kinetic data. The rate law can be expressed as:

\[ \text{Rate} = k [\text{CH}_3\text{COCH}_3]^x [\text{Br}_2]^y [\text{H}^+]^z \]

### Step 1: Write the rate equations based on the provided data

From the given data, we have the following initial rates:

1. \( [\text{CH}_3\text{COCH}_3] = 0.30 \, M, [\text{Br}_2] = 0.05 \, M, [\text{H}^+] = 0.05 \, M \) → Rate = \( 5.7 \times 10^{-5} \)
2. \( [\text{CH}_3\text{COCH}_3] = 0.30 \, M, [\text{Br}_2] = 0.10 \, M, [\text{H}^+] = 0.05 \, M \) → Rate = \( 5.7 \times 10^{-5} \)
3. \( [\text{CH}_3\text{COCH}_3] = 0.30 \, M, [\text{Br}_2] = 0.10 \, M, [\text{H}^+] = 0.10 \, M \) → Rate = \( 1.2 \times 10^{-4} \)
4. \( [\text{CH}_3\text{COCH}_3] = 0.40 \, M, [\text{Br}_2] = 0.50 \, M, [\text{H}^+] = 0.20 \, M \) → Rate = \( 3.1 \times 10^{-4} \)

### Step 2: Set up equations for the rates

Using the rate law, we can set up equations for each of the experiments:

1. \( 5.7 \times 10^{-5} = k (0.30)^x (0.05)^y (0.05)^z \) (Equation 1)
2. \( 5.7 \times 10^{-5} = k (0.30)^x (0.10)^y (0.05)^z \) (Equation 2)
3. \( 1.2 \times 10^{-4} = k (0.30)^x (0.10)^y (0.10)^z \) (Equation 3)
4. \( 3.1 \times 10^{-4} = k (0.40)^x (0.50)^y (0.20)^z \) (Equation 4)

### Step 3: Compare equations to find \(y\)

From Equations 1 and 2, we can eliminate \(k\) and \(z\):

\[ \frac{5.7 \times 10^{-5}}{5.7 \times 10^{-5}} = \frac{(0.30)^x (0.05)^y (0.05)^z}{(0.30)^x (0.10)^y (0.05)^z} \]

This simplifies to:

\[ 1 = \frac{(0.05)^y}{(0.10)^y} \]

This implies:

\[ (0.10)^y = (0.05)^y \implies 2^y = 1 \implies y = 0 \]

### Step 4: Compare equations to find \(z\)

Next, we can use Equations 2 and 3 to find \(z\):

\[ \frac{1.2 \times 10^{-4}}{5.7 \times 10^{-5}} = \frac{(0.30)^x (0.10)^y (0.10)^z}{(0.30)^x (0.10)^y (0.05)^z} \]

This simplifies to:

\[ \frac{1.2}{5.7} = \frac{(0.10)^z}{(0.05)^z} \]

Calculating \( \frac{1.2}{5.7} \approx 0.2105 \):

\[ 0.2105 = 2^z \implies z \approx 0.5 \]

### Step 5: Find \(x\)

Now we can use Equation 4 to find \(x\):

\[ 3.1 \times 10^{-4} = k (0.40)^x (0.50)^0 (0.20)^{0.5} \]

Using the value of \(z\) we found and substituting \(y = 0\):

\[ 3.1 \times 10^{-4} = k (0.40)^x (0.20)^{0.5} \]

### Conclusion: Rate Law

After analyzing all the equations, we can conclude that the rate law for the reaction is:

\[ \text{Rate} = k [\text{CH}_3\text{COCH}_3]^x [\text{Br}_2]^0 [\text{H}^+]^{0.5} \]

Where \(x\) can be determined from further analysis or additional data.