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

Explanation

To determine the relationship between the rate of appearance of bromine (Br₂) and the rate of disappearance of bromide ions (Br⁻) in the given reaction:

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

we will use the stoichiometry of the reaction.

### Step-by-Step Solution:

1. **Identify the Stoichiometric Coefficients:**
- In the balanced equation, the stoichiometric coefficients are:
- For BrO₃⁻: 1
- For Br⁻: 5
- For Br₂: 3
- For H₂O: 3

2. **Write the Rate Expressions:**
- The rate of a reaction can be expressed in terms of the change in concentration of reactants and products. For a general reaction:
\[ aA + bB \rightarrow cC + dD \]
The rate can be expressed as:
\[ -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \]

3. **Apply to the Given Reaction:**
- For the given reaction, we can write:
\[ -\frac{1}{1} \frac{d[\text{BrO}_3^-]}{dt} = -\frac{1}{5} \frac{d[\text{Br}^-]}{dt} = \frac{1}{3} \frac{d[\text{Br}_2]}{dt} \]

4. **Relate the Rates:**
- From the expression above, we can relate the rate of disappearance of bromide ions to the rate of appearance of bromine:
\[ \frac{d[\text{Br}_2]}{dt} = \frac{3}{5} \left(-\frac{d[\text{Br}^-]}{dt}\right) \]

5. **Final Relationship:**
- Thus, we can conclude that the rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ions (Br⁻) as follows:
\[ \text{Rate of appearance of Br}_2 = \frac{3}{5} \times \text{Rate of disappearance of Br}^- \]