For the reaction
`5Br^(-)(aq)+BrO_(3)^(-)(aq)+6H^(+)(aq)rarr 3Br_(2)(aq)+3H_(2)O(l)`
the reate expression was found to be `-(d[BrO^(3-)])/(dt)=k[Br^(-)][H^(+)]^(2)[BrO_(3)^(-)]`
Which of the following statements is /are correct?
I. Doubling the intial concentration of all the reactants will increase the reaction rate by a factor of 8.
II. Unit of rate constant of the reaction in a buffer solution is `"min"^(-1)`
III. Doubling the concentration of all the reactants at the same time will increase the reaction rate by a factor of 16
IV. rate of conversion of `BrO_(3)^(-)` and rate of disappearance of `Br^(-)` are the same

Options

A. I and II
B. II and III
C. II and IV
D. III only

Answer

To solve the question regarding the reaction and the rate expression, we will analyze each statement step by step.

### Given Reaction:
\[ 5Br^(-)(aq) + BrO_3^(-)(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l) \]

### Rate Expression:
\[ -\frac{d[BrO_3^-]}{dt} = k[Br^-][H^+]^2[BrO_3^-] \]

### Analyzing the Statements:

**Statement I:** Doubling the initial concentration of all the reactants will increase the reaction rate by a factor of 8.

1. **Current Concentration:** Let the initial concentrations be:
- \([Br^-] = [Br^-]_0\)
- \([H^+] = [H^+]_0\)
- \([BrO_3^-] = [BrO_3^-]_0\)

2. **Doubling Concentrations:**
- New concentrations will be:
\[ [Br^-] = 2[Br^-]_0, \quad [H^+] = 2[H^+]_0, \quad [BrO_3^-] = 2[BrO_3^-]_0 \]

3. **New Rate Calculation:**
\[ \text{New Rate} = k(2[Br^-]_0)(2[H^+]_0)^2(2[BrO_3^-]_0) \]
\[ = k(2[Br^-]_0)(4[H^+]_0^2)(2[BrO_3^-]_0) = 16k[Br^-]_0[H^+]_0^2[BrO_3^-]_0 \]
- The original rate was:
\[ \text{Original Rate} = k[Br^-]_0[H^+]_0^2[BrO_3^-]_0 \]

4. **Factor of Increase:**
\[ \text{Factor} = \frac{\text{New Rate}}{\text{Original Rate}} = \frac{16k[Br^-]_0[H^+]_0^2[BrO_3^-]_0}{k[Br^-]_0[H^+]_0^2[BrO_3^-]_0} = 16 \]
- **Conclusion:** This statement is **incorrect** (it should be 16, not 8).

**Statement II:** Unit of rate constant of the reaction in a buffer solution is "min"^(-1).

1. **Order of Reaction Calculation:**
- From the rate expression, the order is:
\[ 1 \text{ (for } Br^-) + 2 \text{ (for } H^+) + 1 \text{ (for } BrO_3^-) = 4 \]

2. **Unit of Rate Constant (k):**
- The unit of rate is concentration/time, which is \(mol \, L^{-1} \, min^{-1}\).
- Thus, the unit of \(k\) is:
\[ k = \frac{(mol \, L^{-1})}{(mol \, L^{-1})^4} = L^3 \, mol^{-3} \, min^{-1} \]
- **Conclusion:** This statement is **incorrect** (the unit is not "min"^(-1)).

**Statement III:** Doubling the concentration of all the reactants at the same time will increase the reaction rate by a factor of 16.

- As calculated in Statement I, doubling all reactants leads to an increase in rate by a factor of 16.
- **Conclusion:** This statement is **correct**.

**Statement IV:** Rate of conversion of \(BrO_3^-\) and rate of disappearance of \(Br^-\) are the same.

1. **Rate Expressions:**
- Rate of disappearance of \(BrO_3^-\):
\[ -\frac{d[BrO_3^-]}{dt} = k[Br^-][H^+]^2[BrO_3^-] \]
- Rate of disappearance of \(Br^-\):
\[ -\frac{1}{5}\frac{d[Br^-]}{dt} = k[Br^-][H^+]^2[BrO_3^-] \]
- This means:
\[ \frac{d[Br^-]}{dt} = -5k[Br^-][H^+]^2[BrO_3^- \]
- The rates are not the same due to the stoichiometric coefficients.

2. **Conclusion:** This statement is **incorrect**.

### Final Answer:
- Only **Statement III** is correct.