Consider the following reaction in aqueous solution
`5Br^(-) (aq) + BrO_(3)^(-) (aq) + 6H^(+)(aq) to 3Br_(2)(aq) + 3H_(2)O(l)`
If the rate of appearance of `Br_(2)` at a particular time during the reaction is `0.025 M"sec"^(-1)`, what is the rate of disappearance (in `"Msec"^(-1)`) of `Br^(-)` at that time?

Options

A. `0.025 "Msec"^(-1)`
B. `0.042 "Msec"^(-1)`
C. `0.075 "Msec"^(-1)`
D. `0.125 "Msec"^(-1)`

Answer

To solve the problem, we need to determine the rate of disappearance of the bromide ion (Br^(-)) based on the rate of appearance of bromine (Br2) in the given reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Stoichiometry**:
The given reaction is:
\[ 5Br^(-) (aq) + BrO_3^(-) (aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l) \]
From the balanced equation, we can see the stoichiometric coefficients:
- For Br^(-): 5
- For Br2: 3

2. **Understand Rate of Reaction**:
The rate of appearance of Br2 is given as:
\[ \text{Rate of appearance of } Br_2 = \frac{d[Br_2]}{dt} = 0.025 \, M \, s^{-1} \]
The rate of disappearance of Br^(-) can be expressed in terms of the rate of appearance of Br2 using the stoichiometric coefficients.

3. **Relate the Rates Using Stoichiometry**:
The relationship between the rates can be expressed as:
\[ -\frac{1}{5} \frac{d[Br^-]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} \]
This equation indicates that for every 3 moles of Br2 produced, 5 moles of Br^(-) are consumed.

4. **Solve for the Rate of Disappearance of Br^(-)**:
Rearranging the equation gives:
\[ \frac{d[Br^-]}{dt} = -\frac{5}{3} \frac{d[Br_2]}{dt} \]
Substituting the given rate of appearance of Br2:
\[ \frac{d[Br^-]}{dt} = -\frac{5}{3} \times 0.025 \]

5. **Calculate the Rate**:
\[ \frac{d[Br^-]}{dt} = -\frac{5 \times 0.025}{3} = -\frac{0.125}{3} = -0.04167 \, M \, s^{-1} \]
Rounding this to three significant figures gives:
\[ \frac{d[Br^-]}{dt} \approx -0.042 \, M \, s^{-1} \]

### Final Answer:
The rate of disappearance of Br^(-) is approximately **0.042 M s^(-1)**.