Calculate the standard free energy change for the reaction :
`H_(2)(g) + I_(2)(g) to 2HI(g) DeltaH^(@) = 51.9 kJ "mol"^(-1)` Given : `S^(@)(H_(2)) = 130.6 J K^(-1) "mol"^(-1)`
`S^(@)(I_(2)) = 116.7 J K^(-1) "mol"^(-1)` and `S^(@)(HI) = 206.3 J K^(-1) "mol"^(-1)`.

Explanation

To calculate the standard free energy change (ΔG°) for the reaction:

\[ H_2(g) + I_2(g) \rightarrow 2HI(g) \]

we will use the Gibbs free energy equation:

\[ \Delta G° = \Delta H° - T \Delta S° \]

### Step 1: Calculate ΔS°
We need to find the change in entropy (ΔS°) for the reaction. The formula for ΔS° is:

\[ \Delta S° = S°_{\text{products}} - S°_{\text{reactants}} \]

For our reaction:

- Products: 2 moles of HI
- Reactants: 1 mole of H₂ and 1 mole of I₂

Using the given standard entropy values:

- \( S°(H_2) = 130.6 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S°(I_2) = 116.7 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S°(HI) = 206.3 \, \text{J K}^{-1} \text{mol}^{-1} \)

Now, calculate ΔS°:

\[ \Delta S° = [2 \times S°(HI)] - [S°(H_2) + S°(I_2)] \]

Substituting the values:

\[ \Delta S° = [2 \times 206.3] - [130.6 + 116.7] \]

\[ \Delta S° = [412.6] - [247.3] = 165.3 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 2: Convert ΔH° to J/mol
The given ΔH° is:

\[ \Delta H° = 51.9 \, \text{kJ mol}^{-1} = 51.9 \times 10^3 \, \text{J mol}^{-1} = 51900 \, \text{J mol}^{-1} \]

### Step 3: Use the Gibbs Free Energy Equation
Now we can calculate ΔG° using the temperature T = 298 K (room temperature):

\[ \Delta G° = \Delta H° - T \Delta S° \]

Substituting the values:

\[ \Delta G° = 51900 \, \text{J mol}^{-1} - (298 \, \text{K} \times 165.3 \, \text{J K}^{-1} \text{mol}^{-1}) \]

Calculating \( T \Delta S° \):

\[ T \Delta S° = 298 \times 165.3 = 49283.4 \, \text{J mol}^{-1} \]

Now substituting this back into the ΔG° equation:

\[ \Delta G° = 51900 - 49283.4 = 2616.6 \, \text{J mol}^{-1} \]

### Step 4: Convert ΔG° to kJ/mol
Finally, convert ΔG° to kJ/mol:

\[ \Delta G° = \frac{2616.6}{1000} = 2.6166 \, \text{kJ mol}^{-1} \]

### Final Answer
Thus, the standard free energy change for the reaction is:

\[ \Delta G° \approx 2.62 \, \text{kJ mol}^{-1} \]

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