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 high-temperature decomposition of ethane gas (C₂H₆) into ethylene (C₂H₄) and hydrogen gas (H₂), we will analyze the thermodynamic properties involved, particularly focusing on the changes in Gibbs free energy (ΔG) and entropy (ΔS) for the reaction.

### Step-by-Step Solution:

1. **Write the Reaction**:
The decomposition of ethane can be represented as:
\[ C_2H_6(g) \rightarrow C_2H_4(g) + H_2(g) \]

2. **Identify the Change in Moles**:
- Reactants: 1 mole of C₂H₆
- Products: 1 mole of C₂H₄ + 1 mole of H₂ = 2 moles
- Change in moles (Δn) = moles of products - moles of reactants = 2 - 1 = 1

3. **Determine the Change in Entropy (ΔS)**:
Since the number of moles of products is greater than that of reactants, the entropy change (ΔS) is expected to be 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:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, ΔG = 0, and we can rearrange this to:
\[ \Delta H = T \Delta S \]

5. **Analyze the Conditions**:
- If ΔS is positive, for ΔG to be positive (meaning the reaction is non-spontaneous), it implies that ΔH must be greater than TΔS.
- Therefore, we conclude that:
\[ \Delta H > T \Delta S \]

6. **Determine the Equilibrium Constant (K)**:
The Gibbs free energy change at standard conditions can also be related to the equilibrium constant (K) of the reaction:
\[ \Delta G^\circ = -RT \ln K \]
If ΔG is positive, then K must be less than 1, indicating that the products are less favored at equilibrium.

7. **Conclusion**:
Based on the analysis, we can summarize the findings:
- ΔS > 0 (entropy increases)
- ΔH > ΔG (enthalpy is greater than Gibbs free energy)
- K < 1 (the reaction favors reactants at equilibrium)

### Final Answer:
The correct statements regarding the thermodynamic properties of the reaction are:
- ΔG° < ΔH°
- ΔS > 0
- K < 1