In the following reaction, how is the rate of appearance of the underlined Product related to the rate of disappearance of the underlined reactant ?
`BrO_(3)^(ɵ)(aq) + 5underline(Br)^(ɵ)(aq) + 6H^(o+) (aq) rarr 3underline(Br_(2))(l) + 3H_(2)O(l)`

Options

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

Explanation

To solve the problem, we need to relate the rate of appearance of the product \( \text{Br}_2 \) to the rate of disappearance of the reactant \( \text{Br}^- \) in the given reaction:

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

### Step 1: Write the Rate Expressions

The rate of a reaction can be expressed in terms of the change in concentration of reactants and products over time. For the reactants and products involved in this reaction, we can write:

\[ \text{Rate of disappearance of } \text{Br}^- = -\frac{1}{5} \frac{d[\text{Br}^-]}{dt} \]

\[ \text{Rate of appearance of } \text{Br}_2 = \frac{1}{3} \frac{d[\text{Br}_2]}{dt} \]

### Step 2: Relate the Rates Using Stoichiometry

From the balanced chemical equation, we see that 5 moles of \( \text{Br}^- \) produce 3 moles of \( \text{Br}_2 \). Therefore, we can relate the rates of disappearance of \( \text{Br}^- \) and appearance of \( \text{Br}_2 \) using their stoichiometric coefficients:

\[ -\frac{1}{5} \frac{d[\text{Br}^-]}{dt} = \frac{1}{3} \frac{d[\text{Br}_2]}{dt} \]

### Step 3: Cross Multiply to Find the Relationship

To find a direct relationship between the rates, we can cross-multiply:

\[ 3 \left(-\frac{d[\text{Br}^-]}{dt}\right) = 5 \frac{d[\text{Br}_2]}{dt} \]

This simplifies to:

\[ \frac{d[\text{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\text{Br}^-]}{dt} \]

### Conclusion

Thus, the rate of appearance of \( \text{Br}_2 \) is related to the rate of disappearance of \( \text{Br}^- \) by the equation:

\[ \frac{d[\text{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\text{Br}^-]}{dt} \]