The molar heat of vaporization of benzene is 31kj/mol. Calculate the entropy change for the liquid vapor equilibrium. At 1 atm pressure, benzene boils at 80.1 degrees Celsius.
To calculate the entropy change for the liquid-vapor equilibrium of benzene, we need to use the equation:
ΔS = ΔHvap / T
Where:
ΔS is the entropy change
ΔHvap is the molar heat of vaporization
T is the temperature in Kelvin
First, we need to convert the boiling temperature of benzene from Celsius to Kelvin:
T = 80.1 + 273.15 = 353.25 K
Now, we can calculate the entropy change:
ΔS = 31 kJ/mol / 353.25 K
ΔS ≈ 0.0878 kJ/(mol·K)
Therefore, the entropy change for the liquid-vapor equilibrium of benzene is approximately 0.0878 kJ/(mol·K).