Final answer:
The change in Gibbs free energy (∆G) for the vaporization of benzene at 80 ℃ is calculated using the Gibbs free energy equation. By inserting the values ∆H = 30.7 kJ/mol, T = 353.15 K (after converting from ℃ to K), and ∆S = 87.0 J/K·mol (converted to kJ), the result is that ∆G ≈ -0.0 kJ/mol when rounded to one decimal place.
Step-by-step explanation:
The question is asking for the calculation of the change in the Gibbs free energy, denoted as ∆G, for the vaporization of benzene at a temperature of 80 ℃. To calculate ∆G for the vaporization of benzene, we can use the Gibbs free energy equation:
∆G = ∆H - T∆S
Where:
- ∆H is the enthalpy change of vaporization
- T is the temperature in Kelvin
- ∆S is the entropy change of vaporization
To convert the temperature from Celsius to Kelvin, we add 273.15:
T(K) = 80 + 273.15 = 353.15 K
Now we substitute the given values into the Gibbs free energy equation:
∆G = 30.7 kJ/mol - (353.15 K × 87.0 J/K·mol)
Note: We need to convert ∆S from J to kJ to match the units of ∆H.
∆G = 30.7 kJ/mol - (353.15 K × 0.087 kJ/K·mol)
∆G = 30.7 kJ/mol - 30.724 kJ/mol
∆G ≈ -0.0 kJ/mol
Therefore, the change in Gibbs free energy (∆G) for the vaporization of benzene at 80 ℃ is approximately -0.0 kJ/mol when rounded to one decimal place.