The enthalpy and entropy of vaporization of ethanol are 38.6 kJ/mol and 109.8 J/mol-K, respectively. What is the vapor pressure of ethanol at 25℃?
First of all we calculate Boling point of ethanol.
We know that,
∆S = ∆H/T
T = ∆H/∆S
Where,
∆S = entropy change = 109.8J/mol-K
∆H = change in enthalpy = 38.6kJ/mol = 38600J/mol
T = Boling point
Putting the all value in formula.
T = (38600J/mol)/(109.8J/mol-K)
T = 351.55K
step (2)
We know that , at Boling point vapour pressure of liquid = 1atm
Step (3)
P1 = 1atm/10.7270
P1 = 0.093atm
Hence, vapour pressure of ethanol at 25°C = 0.093atm
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