Reactions involving gold have been of particular intrests to alchemists. Consider the following reactions ,
`Au(OH)_(3)+4HCl toHAuCl_(4)+3H_(2)O`
`" "DeltaH=-28kcal`
`Au(OH)_(3)+4HBr toHAuBr_(4)+3H_(2)O`
`" "DeltaH=-36.8kcal`
In an experiment there was an absorption of 0.44 kcal when one mole of `HAuBr_(4)` was mixed with 4 moles of HCl. Then the fraction `HAuBr_(4)` converted into `HAuCl_(4)` : (percentage conversion)

Options

A. 0.05
B. 0.06
C. 0.07
D. 0.08

Answer

To solve the problem, we need to analyze the reactions and calculate the fraction of `HAuBr4` that is converted into `HAuCl4` based on the given enthalpy changes and the energy absorbed during the reaction.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes:**
- Reaction 1:
\[ Au(OH)_3 + 4HCl \rightarrow HAuCl_4 + 3H_2O \quad \Delta H_1 = -28 \text{ kcal} \]
- Reaction 2:
\[ Au(OH)_3 + 4HBr \rightarrow HAuBr_4 + 3H_2O \quad \Delta H_2 = -36.8 \text{ kcal} \]

2. **Write the Reaction for the Displacement of Bromine by Chlorine:**
- The reaction we are interested in is:
\[ HAuBr_4 + 4HCl \rightarrow HAuCl_4 + 4HBr \]
- We need to find the enthalpy change (\(\Delta H\)) for this reaction.

3. **Calculate the Enthalpy Change for the Displacement Reaction:**
- The enthalpy change for the displacement reaction can be calculated as:
\[ \Delta H = \Delta H_1 - \Delta H_2 \]
- Substituting the values:
\[ \Delta H = (-28 \text{ kcal}) - (-36.8 \text{ kcal}) = -28 + 36.8 = 8.8 \text{ kcal} \]

4. **Determine the Energy Absorbed in the Experiment:**
- The problem states that there was an absorption of 0.44 kcal when 1 mole of `HAuBr4` was mixed with 4 moles of `HCl`.

5. **Calculate the Fraction of `HAuBr4` Converted to `HAuCl4`:**
- The fraction of `HAuBr4` converted can be calculated using the formula:
\[ \text{Fraction converted} = \frac{\text{Energy absorbed}}{\text{Total enthalpy change}} = \frac{0.44 \text{ kcal}}{8.8 \text{ kcal}} \]
- Simplifying this gives:
\[ \text{Fraction converted} = \frac{0.44}{8.8} = 0.05 \]

6. **Convert the Fraction to Percentage:**
- To express this as a percentage:
\[ \text{Percentage conversion} = 0.05 \times 100 = 5\% \]

### Final Answer:
The fraction of `HAuBr4` converted into `HAuCl4` is **0.05** (or **5%**).