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 problem step by step, let's analyze the given reaction and the rate expression.

### Step 1: Understand the Rate Expression
The rate expression given is:
\[ -\frac{d[BrO_3^-]}{dt} = k[Br^-][H^+]^2[BrO_3^-] \]
Here, the orders of the reaction with respect to each reactant are:
- \(Br^-\): 1
- \(H^+\): 2
- \(BrO_3^-\): 1

### Step 2: Calculate the Effect of Doubling Concentrations
When we double the initial concentrations of all reactants, we can express the new concentrations as:
- \( [Br^-] = 2[Br^-] \)
- \( [H^+] = 2[H^+] \)
- \( [BrO_3^-] = 2[BrO_3^-] \)

Substituting these into the rate expression:
\[ -\frac{d[BrO_3^-]}{dt} = k(2[Br^-])(2[H^+])^2(2[BrO_3^-]) \]
Calculating the new rate:
\[ = k(2[Br^-])(4[H^+]^2)(2[BrO_3^-]) = k \cdot 16[Br^-][H^+]^2[BrO_3^-] \]
This shows that the new rate is 16 times the original rate.

### Step 3: Analyze Each Statement
1. **Statement I**: Doubling the initial concentration of all the reactants will increase the reaction rate by a factor of 8.
- **Conclusion**: Incorrect. The factor is 16, not 8.

2. **Statement II**: Unit of rate constant of the reaction in a buffer solution is \( \text{min}^{-1} \).
- The rate law has units of concentration per time, which is \( \text{mol L}^{-1} \text{s}^{-1} \). The unit of \( k \) can be derived from the rate law:
\[ k = \frac{\text{Rate}}{[Br^-][H^+]^2[BrO_3^-]} \implies k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})(\text{mol L}^{-1})^2} = \text{L}^2 \text{mol}^{-2} \text{s}^{-1} \]
In terms of minutes, this will be \( \text{min}^{-1} \) when considering the time unit conversion.
- **Conclusion**: Correct.

3. **Statement III**: Doubling the concentration of all the reactants at the same time will increase the reaction rate by a factor of 16.
- **Conclusion**: Correct, as calculated in Step 2.

4. **Statement IV**: Rate of conversion of \( BrO_3^- \) and rate of disappearance of \( Br^- \) are the same.
- The rates of disappearance are related by their stoichiometric coefficients. The rate of disappearance of \( BrO_3^- \) is given by:
\[ -\frac{1}{1} \frac{d[BrO_3^-]}{dt} \quad \text{and for } Br^- \text{ it is } -\frac{1}{5} \frac{d[Br^-]}{dt} \]
They are not equal due to different coefficients.
- **Conclusion**: Incorrect.

### Final Summary of Statements
- **I**: Incorrect
- **II**: Correct
- **III**: Correct
- **IV**: Incorrect