For the vaporization of benzene, ΔHvap = 30.7 kJ/mol and ΔSvap = 87.0 J/(K·mol). Does benzene boil at 75 °C and 1 atm pressure?

• A. No, because the enthalpy change is too high at this temperature.
• B. Yes, because the entropy change is positive at this temperature.
• C. Yes, because the Gibbs free energy change is negative at this temperature.
• D. No, because the Gibbs free energy change is positive at this temperature.

Answer: (C)

Verified step by step guidance

1. First, understand that the boiling point of a substance is the temperature at which its vapor pressure equals the external pressure. At this point, the Gibbs free energy change (ΔG) for the phase transition is zero.
2. Use the Gibbs free energy equation: ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
3. Convert the given temperature from Celsius to Kelvin by adding 273.15. So, T = 75 + 273.15 K.
4. Substitute the given values into the Gibbs free energy equation: ΔG = 30.7 kJ/mol - (T in Kelvin) * 87.0 J/(K·mol). Note that you need to convert ΔHvap from kJ/mol to J/mol by multiplying by 1000.
5. Determine if ΔG is negative, zero, or positive. If ΔG is zero, benzene boils at this temperature and pressure. If ΔG is negative, the process is spontaneous, indicating that benzene can boil at this temperature and pressure. If ΔG is positive, benzene does not boil at this temperature and pressure.