In acidic medium, the rate of reaction between `BrO_(3)^(ɵ)` and `Br^(ɵ)` is given by the expression
`(-d[BrO_(3)^(ɵ)])/(dt) =k[BrO_(3)^(ɵ)][Br^(ɵ)][H^(o+)]^(2)`

Options

A. The rate constant of overall reaction is `4s^(-1)`.
B. The rate of reaction is independent of the concentration of acid.
C. The change in `pH` of the solution will no affect the rate.
D. Doubling the concentration of `H^(o+)` ions will increase the reaction rate by `4` times.

Explanation

To solve the problem, we need to analyze the rate expression given for the reaction between \( \text{BrO}_3^- \) and \( \text{Br}^- \) in acidic medium:

\[ -\frac{d[\text{BrO}_3^-]}{dt} = k[\text{BrO}_3^-][\text{Br}^-][\text{H}^+]^2 \]

### Step 1: Identify the order of the reaction with respect to each reactant
The rate expression shows how the rate of reaction depends on the concentration of each reactant:

- For \( \text{BrO}_3^- \): The exponent is 1, so the order with respect to \( \text{BrO}_3^- \) is 1.
- For \( \text{Br}^- \): The exponent is also 1, so the order with respect to \( \text{Br}^- \) is 1.
- For \( \text{H}^+ \): The exponent is 2, so the order with respect to \( \text{H}^+ \) is 2.

### Step 2: Calculate the overall order of the reaction
The overall order of the reaction is the sum of the individual orders:

\[ \text{Overall order} = 1 (\text{for } \text{BrO}_3^-) + 1 (\text{for } \text{Br}^-) + 2 (\text{for } \text{H}^+) = 4 \]

### Step 3: Determine the units of the rate constant \( k \)
The units of the rate constant \( k \) depend on the overall order of the reaction. The general formula for the units of \( k \) is:

\[ \text{Units of } k = \text{(concentration)}^{1 - n} \cdot \text{(time)}^{-1} \]

Where \( n \) is the overall order of the reaction. Since we found \( n = 4 \):

\[ \text{Units of } k = \text{mol L}^{-1(1-4)} \cdot \text{s}^{-1} = \text{mol}^{-3} \cdot \text{L}^3 \cdot \text{s}^{-1} \]

### Step 4: Analyze the options given in the question
Now we can analyze the options based on our findings:

- **Option A**: Incorrect, as the unit of \( k \) is \( \text{mol}^{-3} \cdot \text{L}^3 \cdot \text{s}^{-1} \), not \( 4 \, \text{s}^{-1} \).
- **Option B**: Incorrect, as the rate does depend on the concentration of \( \text{H}^+ \) (to the power of 2).
- **Option C**: Incorrect, since changing the pH affects the concentration of \( \text{H}^+ \) and thus the rate.
- **Option D**: Correct, as doubling the concentration of \( \text{H}^+ \) will increase the rate by \( 2^2 = 4 \) times.

### Final Conclusion
The correct answer is **Option D**.

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