A chemist claims that the following reaction is feasible at 298 K
`SF_(6)(g) + 8 HI(g) rarr H_(2)S( g) + 6 HF (g) + 4I_(2)(g)`
Verify his claim . Given that `DeltaG_(f)^(@)` for `SF_(6)(g) , HI(g) , H_(2)S(g)` and `HF (g)` are - 991.61, 1.30, -33.01 and -270.73kJ `mol^(-1)` respectively.

Answer

To verify the claim that the reaction

\[ \text{SF}_6(g) + 8 \text{HI}(g) \rightarrow \text{H}_2\text{S}(g) + 6 \text{HF}(g) + 4 \text{I}_2(g) \]

is feasible at 298 K, we need to calculate the change in Gibbs free energy (\( \Delta G \)) for the reaction. The reaction is considered feasible if \( \Delta G < 0 \).

### Step 1: Write the expression for \( \Delta G \)

The change in Gibbs free energy for the reaction can be calculated using the formula:

\[ \Delta G = \Delta G_f^{\text{products}} - \Delta G_f^{\text{reactants}} \]

### Step 2: Identify the Gibbs free energy of formation values

From the given data:
- \( \Delta G_f^{\text{SF}_6(g)} = -991.61 \, \text{kJ/mol} \)
- \( \Delta G_f^{\text{HI}(g)} = 1.30 \, \text{kJ/mol} \)
- \( \Delta G_f^{\text{H}_2\text{S}(g)} = -33.01 \, \text{kJ/mol} \)
- \( \Delta G_f^{\text{HF}(g)} = -270.73 \, \text{kJ/mol} \)
- \( \Delta G_f^{\text{I}_2(g)} = 0 \, \text{kJ/mol} \) (since it is in its standard state)

### Step 3: Calculate \( \Delta G_f \) for the products

The total Gibbs free energy of formation for the products is:

\[ \Delta G_f^{\text{products}} = \Delta G_f^{\text{H}_2\text{S}} + 6 \times \Delta G_f^{\text{HF}} + 4 \times \Delta G_f^{\text{I}_2} \]

Substituting the values:

\[ \Delta G_f^{\text{products}} = -33.01 + 6 \times (-270.73) + 4 \times 0 \]
\[ = -33.01 - 1624.38 + 0 \]
\[ = -1657.39 \, \text{kJ} \]

### Step 4: Calculate \( \Delta G_f \) for the reactants

The total Gibbs free energy of formation for the reactants is:

\[ \Delta G_f^{\text{reactants}} = \Delta G_f^{\text{SF}_6} + 8 \times \Delta G_f^{\text{HI}} \]

Substituting the values:

\[ \Delta G_f^{\text{reactants}} = -991.61 + 8 \times 1.30 \]
\[ = -991.61 + 10.40 \]
\[ = -981.21 \, \text{kJ} \]

### Step 5: Calculate \( \Delta G \)

Now, substituting the values into the \( \Delta G \) equation:

\[ \Delta G = \Delta G_f^{\text{products}} - \Delta G_f^{\text{reactants}} \]
\[ = -1657.39 - (-981.21) \]
\[ = -1657.39 + 981.21 \]
\[ = -676.18 \, \text{kJ} \]

### Step 6: Conclusion

Since \( \Delta G = -676.18 \, \text{kJ} < 0 \), the reaction is feasible at 298 K.

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