Calculate the free energy change at 298 K for the reaction,
`Br_(2)(l)+ Cl_(2)(g) rarr 2BrCl(g)`. For the reaction `DeltaH^(@)=29.3` KJ & the entropies of `Br_(2)(l),Cl_(2)(g)` & `BrCl(g)` at the 298 K are 152.3,223.0,239.7 J `"mol^(-1)K^(-1)` respectively.

Options

A. `-1721.8` J
B. `-60321.8` J
C. `+60321.8` J
D. `+1721.8` J

Explanation

To calculate the free energy change (ΔG) at 298 K for the reaction:

\[ \text{Br}_2(l) + \text{Cl}_2(g) \rightarrow 2\text{BrCl}(g) \]

we will follow these steps:

### Step 1: Convert ΔH from kJ to J
Given:
\[ \Delta H^\circ = 29.3 \, \text{kJ} \]

To convert this to joules:
\[ \Delta H^\circ = 29.3 \, \text{kJ} \times 1000 \, \text{J/kJ} = 29300 \, \text{J} \]

### Step 2: Calculate the entropy change (ΔS)
The entropies of the substances at 298 K are given as:
- \( S(\text{Br}_2(l)) = 152.3 \, \text{J/mol·K} \)
- \( S(\text{Cl}_2(g)) = 223.0 \, \text{J/mol·K} \)
- \( S(\text{BrCl}(g)) = 239.7 \, \text{J/mol·K} \)

The entropy change (ΔS) for the reaction can be calculated using the formula:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

Calculating ΔS:
\[ S_{\text{products}} = 2 \times S(\text{BrCl}) = 2 \times 239.7 \, \text{J/mol·K} = 479.4 \, \text{J/mol·K} \]

\[ S_{\text{reactants}} = S(\text{Br}_2) + S(\text{Cl}_2) = 152.3 \, \text{J/mol·K} + 223.0 \, \text{J/mol·K} = 375.3 \, \text{J/mol·K} \]

Now, substituting the values:
\[ \Delta S = 479.4 \, \text{J/mol·K} - 375.3 \, \text{J/mol·K} = 104.1 \, \text{J/mol·K} \]

### Step 3: Calculate ΔG using the Gibbs free energy equation
The Gibbs free energy change (ΔG) is calculated using the formula:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S \]

Substituting the values:
- \( \Delta H^\circ = 29300 \, \text{J} \)
- \( T = 298 \, \text{K} \)
- \( \Delta S = 104.1 \, \text{J/mol·K} \)

Calculating:
\[ \Delta G^\circ = 29300 \, \text{J} - (298 \, \text{K} \times 104.1 \, \text{J/mol·K}) \]

Calculating \( T \Delta S \):
\[ T \Delta S = 298 \times 104.1 = 31025.8 \, \text{J} \]

Now substituting back:
\[ \Delta G^\circ = 29300 \, \text{J} - 31025.8 \, \text{J} \]
\[ \Delta G^\circ = -1725.8 \, \text{J} \]

### Final Result
Thus, the free energy change at 298 K for the reaction is:
\[ \Delta G^\circ \approx -1721.8 \, \text{J} \]