In the reaction
`BrO^(-3)(aq) + 5Br^(-) (aq) + 6H^(+) rarr 3Br_(2)(1) + 3H_(2)O(1)`
The rate of appearance of bromine `(Br_(2))` is related to rate of disapperance of bromide ions as folllwoing :

Options

A. `(d[Br_(2)])/(d t) = (3)/(5)(d[Br^(-)])/(d t)`
B. `(d[Br_(2)])/(d t) = -(3)/(5)(d[Br^(-)])/(d t)`
C. `(d[Br_(2)])/(d t) = -(5)/(3)(d[Br^(-)])/(d t)`
D. `(d[Br_(2)])/(d t) = (5)/(3)(d[Br^(-)])/(d t)`

Explanation

To solve the problem, we need to relate the rate of appearance of bromine (Br₂) to the rate of disappearance of bromide ions (Br⁻) based on the stoichiometry of the given reaction:

**Given Reaction:**
\[ \text{BrO}_3^{-} (aq) + 5 \text{Br}^- (aq) + 6 \text{H}^+ \rightarrow 3 \text{Br}_2 (l) + 3 \text{H}_2O (l) \]

### Step-by-Step Solution:

1. **Identify the Stoichiometric Coefficients:**
- From the balanced equation, we see:
- 5 moles of Br⁻ are consumed for every 3 moles of Br₂ produced.

2. **Write the Rate Expressions:**
- The rate of disappearance of bromide ions (Br⁻) can be expressed as:
\[ -\frac{1}{5} \frac{d[\text{Br}^-]}{dt} \]
- The rate of appearance of bromine (Br₂) can be expressed as:
\[ \frac{1}{3} \frac{d[\text{Br}_2]}{dt} \]

3. **Set the Rates Equal:**
- According to the stoichiometry of the reaction, we can relate these two rates:
\[ -\frac{1}{5} \frac{d[\text{Br}^-]}{dt} = \frac{1}{3} \frac{d[\text{Br}_2]}{dt} \]

4. **Rearranging the Equation:**
- To find the relationship between the rates, we can rearrange the equation:
\[ \frac{d[\text{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\text{Br}^-]}{dt} \]

5. **Conclusion:**
- This means that the rate of appearance of bromine (Br₂) is proportional to the rate of disappearance of bromide ions (Br⁻) with a factor of \(-\frac{3}{5}\).

### Final Answer:
The rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ions (Br⁻) as follows:
\[ \frac{d[\text{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\text{Br}^-]}{dt} \]