Industrial acetylene gas (ethyne: `C_(2)H_(2)`) is made by the high temperature decomposition of ethane gas: `C_(2)H_(6)`, at `300^(@)` C, according to the following equation: `C_(2)H_(6)(g) to C_(2)H_(5)(g) + 2H_(2)(g)`

Options

A. `Delta_(r)G^(@) gt Delta_(r)H^(@)` and `Delta_(r)S^(@) gt 0`
B. `Delta_(r)G^(@) lt Delta_(r)H^(@)` and `Delta_(r)S^(@)` lt 0
C. `Delta_(r)G^(@) lt Delta_(r)H^(@)` and `Delta_(r)S^(@) gt 0`
D. `Delta_(r)G^(@) gt Delta_(r)S^(@) lt 0`

Explanation

To solve the problem regarding the decomposition of ethane gas (C₂H₆) into ethyl radical (C₂H₅) and hydrogen gas (H₂) at high temperatures, we will analyze the thermodynamic parameters involved, specifically the change in Gibbs free energy (ΔG°), change in enthalpy (ΔH°), and change in entropy (ΔS°).

### Step-by-Step Solution:

1. **Write the Reaction**:
The decomposition of ethane can be represented as:
\[ C_2H_6(g) \rightarrow C_2H_5(g) + 2H_2(g) \]

2. **Identify the Change in Moles**:
- On the reactant side, we have 1 mole of C₂H₆.
- On the product side, we have 1 mole of C₂H₅ and 2 moles of H₂, totaling 3 moles.
- Therefore, the change in moles of gas is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 3 - 1 = 2 \]

3. **Determine the Change in Entropy (ΔS°)**:
Since the number of moles increases from 1 to 3, the change in entropy (ΔS°) is positive:
\[ \Delta S° > 0 \]

4. **Use the Gibbs Free Energy Equation**:
The relationship between Gibbs free energy (ΔG°), enthalpy (ΔH°), and entropy (ΔS°) is given by:
\[ \Delta G° = \Delta H° - T\Delta S° \]

5. **Analyze the Conditions**:
- Since the reaction occurs at high temperatures (300°C), we can expect that the term \(T\Delta S°\) will be significant.
- Given that ΔS° is positive, if ΔG° is positive, it implies that ΔH° must be greater than \(T\Delta S°\) to keep ΔG° positive.

6. **Conclusion**:
Based on the analysis, we conclude that:
- ΔG° is positive (indicating non-spontaneity under standard conditions).
- ΔH° is greater than \(T\Delta S°\).
- ΔS° is positive.

Thus, the correct condition for the reaction is that ΔH° > ΔG° and ΔS° > 0.

### Final Answer:
The option that matches these conditions is option number 3, where ΔH° is greater than ΔG° and ΔS° is greater than 0.