For the reaction `H^(+) + BrO_(3)^(-) + 3 Br^(-) to 5 Br_(2) + H_(2) O` which of the following relation correctly represents the consumption & formation of reactants and products:

Options

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

Answer

To solve the problem, we need to analyze the given chemical reaction and derive the correct relationship between the consumption and formation of reactants and products. The reaction is:

\[ H^+ + BrO_3^- + 3 Br^- \rightarrow 5 Br_2 + H_2O \]

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- Reactants: \( H^+, BrO_3^-, 3 Br^- \)
- Products: \( 5 Br_2, H_2O \)

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

3. **Apply to the Given Reaction**:
For our specific reaction:
- For \( Br^- \): The stoichiometric coefficient is 3, so:
\[ -\frac{1}{3} \frac{d[Br^-]}{dt} \]
- For \( Br_2 \): The stoichiometric coefficient is 5, so:
\[ \frac{1}{5} \frac{d[Br_2]}{dt} \]

4. **Set Up the Rate Equation**:
From the above expressions, we can equate the rates:
\[ -\frac{1}{3} \frac{d[Br^-]}{dt} = \frac{1}{5} \frac{d[Br_2]}{dt} \]

5. **Rearranging the Equation**:
To express \( \frac{d[Br^-]}{dt} \) in terms of \( \frac{d[Br_2]}{dt} \):
\[ \frac{d[Br^-]}{dt} = -\frac{3}{5} \frac{d[Br_2]}{dt} \]

6. **Final Expression**:
The final relationship we derived is:
\[ \frac{d[Br^-]}{dt} = -\frac{3}{5} \frac{d[Br_2]}{dt} \]

### Conclusion:
The correct option that represents the consumption and formation of reactants and products is:
**Option B: \( \frac{d[Br^-]}{dt} = -\frac{3}{5} \frac{d[Br_2]}{dt} \)**.