Using `Delta_(f)H^(@)` and `S_(m)^(@)` calculate the standard Gibbs energy of formation, `Delta_(f)G^(@)` for each following :
(a)`CS_(2)(l)`
(b) `N_(2)H_(4)(l)`
`Delta_(f)H^(@)(CS_(2)) = 89.70 kJ "mol"^(-1), Delta_(r)S^(@)(CS_(2)) = 151.34 J K^(-1) "mol"^(-1)`
`Delta_(f)H^(@)(N_(2)H_(4)) = 50.63 kJ "mol"^(-1)`, `Delta_(r)S^(@)N_(2)H_(2) = 121.21 J K^(-1) "mol"^(-1)`.

Explanation

To calculate the standard Gibbs energy of formation (Δ_fG°) for the compounds CS₂ and N₂H₄, we will use the following relation:

\[ \Delta_fG° = \Delta_fH° - T \Delta_rS° \]

where:
- Δ_fG° = standard Gibbs energy of formation
- Δ_fH° = standard enthalpy of formation
- T = temperature in Kelvin
- Δ_rS° = standard entropy change

### Step-by-Step Solution:

#### Part (a): For CS₂ (l)

1. **Identify Given Values:**
- Δ_fH°(CS₂) = 89.70 kJ/mol
- Δ_rS°(CS₂) = 151.34 J/K·mol

2. **Convert Δ_fH° from kJ to J:**
\[ \Delta_fH°(CS₂) = 89.70 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 89700 \, \text{J/mol} \]

3. **Use Room Temperature (T = 25°C):**
\[ T = 25 + 273 = 298 \, \text{K} \]

4. **Calculate TΔ_rS°:**
\[ T \Delta_rS°(CS₂) = 298 \, \text{K} \times 151.34 \, \text{J/K·mol} = 45000.32 \, \text{J/mol} \]

5. **Calculate Δ_fG° for CS₂:**
\[ \Delta_fG°(CS₂) = \Delta_fH°(CS₂) - T \Delta_rS°(CS₂) \]
\[ \Delta_fG°(CS₂) = 89700 \, \text{J/mol} - 45000.32 \, \text{J/mol} = 44699.68 \, \text{J/mol} \]

6. **Convert Δ_fG° back to kJ:**
\[ \Delta_fG°(CS₂) = \frac{44699.68 \, \text{J/mol}}{1000} = 44.70 \, \text{kJ/mol} \]

#### Part (b): For N₂H₄ (l)

1. **Identify Given Values:**
- Δ_fH°(N₂H₄) = 50.63 kJ/mol
- Δ_rS°(N₂H₄) = 121.21 J/K·mol

2. **Convert Δ_fH° from kJ to J:**
\[ \Delta_fH°(N₂H₄) = 50.63 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 50630 \, \text{J/mol} \]

3. **Use Room Temperature (T = 25°C):**
\[ T = 25 + 273 = 298 \, \text{K} \]

4. **Calculate TΔ_rS°:**
\[ T \Delta_rS°(N₂H₄) = 298 \, \text{K} \times 121.21 \, \text{J/K·mol} = 36000.58 \, \text{J/mol} \]

5. **Calculate Δ_fG° for N₂H₄:**
\[ \Delta_fG°(N₂H₄) = \Delta_fH°(N₂H₄) - T \Delta_rS°(N₂H₄) \]
\[ \Delta_fG°(N₂H₄) = 50630 \, \text{J/mol} - 36000.58 \, \text{J/mol} = 14629.42 \, \text{J/mol} \]

6. **Convert Δ_fG° back to kJ:**
\[ \Delta_fG°(N₂H₄) = \frac{14629.42 \, \text{J/mol}}{1000} = 14.63 \, \text{kJ/mol} \]

### Final Results:
- Δ_fG°(CS₂) = 44.70 kJ/mol
- Δ_fG°(N₂H₄) = 14.63 kJ/mol