Question

For benzene, the ∆H° of vaporization is 30.72 kJ/mol and the ∆S° of vaporization is 86.97 J/mol･K. At 1.00 atm and 231.0 K, what is the ∆G° of vaporization for benzene, in kJ/mol?

Answer 1

The ΔG° of vaporization for benzene at 1.00 atm and 231.0 K is -15.99 kJ/mol.

Step-by-step explanation:

To calculate the ΔG° of vaporization for benzene, we can use the equation ΔG° = ΔH° - T×ΔS°, where ΔH° is the enthalpy of vaporization, T is the temperature in Kelvin, and ΔS° is the entropy change of vaporization.

Given: ΔH° = 30.72 kJ/mol, ΔS° = 86.97 J/mol·K, T = 231.0 K

Convert ΔH° from kJ to J: 30.72 kJ/mol = 30.72 × 103 J/mol

Now substitute the values into the equation to find ΔG°: ΔG° = 30.72 × 103 J/mol - (231.0 K)(86.97 J/mol·K)

ΔG° = -15,986.32 J/mol

Finally, convert ΔG° from J/mol to kJ/mol: -15,986.32 J/mol = -15.98632 kJ/mol