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)),),(0.30,0.05,0.05,1.5 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)]`
B. rate `= k[CH_(3)COCH_(3)] [Br_(2)] [H^(+)]^(2)`
C. rate `= k[CH_(3)COCH_(3)] [Br_(2)][H^(+)]`
D. rate `= k[CH_(3)COCH_(3)] [H_(+)]`

Explanation

To solve the problem of determining the order of the reaction involving the bromination of acetone, we will analyze the given kinetic data step by step.

### Step 1: Write the Rate Law Expression
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [\text{CH}_3COCH_3]^a [\text{Br}_2]^b [\text{H}^+]^c \]
where \( k \) is the rate constant, and \( a \), \( b \), and \( c \) are the orders of the reaction with respect to each reactant.

### Step 2: Set Up the Equations from the Given Data
Using the provided initial rates and concentrations, we can set up equations based on the rate law.

1. For the first set of data:
\[ 1.5 \times 10^{-5} = k (0.30)^a (0.05)^b (0.05)^c \] (Equation 1)

2. For the second set of data:
\[ 5.7 \times 10^{-5} = k (0.30)^a (0.10)^b (0.05)^c \] (Equation 2)

3. For the third set of data:
\[ 1.2 \times 10^{-4} = k (0.30)^a (0.10)^b (0.10)^c \] (Equation 3)

4. For the fourth set of data:
\[ 3.1 \times 10^{-4} = k (0.40)^a (0.05)^b (0.20)^c \] (Equation 4)

### Step 3: Compare Equations to Determine the Order
To find the values of \( a \), \( b \), and \( c \), we will compare the equations.

#### Finding \( b \):
- Compare Equation 1 and Equation 2:
\[ \frac{5.7 \times 10^{-5}}{1.5 \times 10^{-5}} = \frac{k (0.30)^a (0.10)^b (0.05)^c}{k (0.30)^a (0.05)^b (0.05)^c} \]
Simplifying gives:
\[ \frac{5.7}{1.5} = \frac{(0.10)^b}{(0.05)^b} \implies \frac{5.7}{1.5} = 2^b \]
Solving for \( b \):
\[ 3.8 \approx 2^b \implies b \approx 0 \]

#### Finding \( c \):
- Compare Equation 2 and Equation 3:
\[ \frac{1.2 \times 10^{-4}}{5.7 \times 10^{-5}} = \frac{k (0.30)^a (0.10)^b (0.10)^c}{k (0.30)^a (0.10)^b (0.05)^c} \]
Simplifying gives:
\[ \frac{1.2}{5.7} = \frac{(0.10)^c}{(0.05)^c} \implies \frac{1.2}{5.7} = 2^c \]
Solving for \( c \):
\[ 0.21 \approx 2^c \implies c \approx 1 \]

#### Finding \( a \):
- Compare Equation 3 and Equation 4:
\[ \frac{3.1 \times 10^{-4}}{1.2 \times 10^{-4}} = \frac{k (0.40)^a (0.05)^b (0.20)^c}{k (0.30)^a (0.10)^b (0.10)^c} \]
Simplifying gives:
\[ \frac{3.1}{1.2} = \frac{(0.40)^a (0.20)^c}{(0.30)^a (0.10)^c} \]
This leads to:
\[ \frac{3.1}{1.2} \approx \frac{(0.40)^a (0.20)^1}{(0.30)^a (0.10)^1} \]
Solving for \( a \):
\[ 2.58 \approx \left(\frac{0.40}{0.30}\right)^a \cdot 2 \implies a \approx 1 \]

### Step 4: Summarize the Orders
From our calculations:
- \( a = 1 \)
- \( b = 0 \)
- \( c = 1 \)

### Final Rate Law Expression
The overall rate law is:
\[ \text{Rate} = k [\text{CH}_3COCH_3]^1 [\text{Br}_2]^0 [\text{H}^+]^1 \]
This indicates that the reaction is first order with respect to acetone and hydrogen ions, and zero order with respect to bromine.