Question

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℃?

Answer 1

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)

* Now using the formula In the late [t, ] * Where, P = Vapour pressure at boling point = latm T2 = Boling point = 351.55K P = Napowr pressure at 25°C T, = 25°C = (273+25)K= 298k AH = enthalpy: 38.6 kus - 38600ml R = Gas constant = 8.314 Putting the all value in formula molk in calm ) (98.600 mo s (8.3142, 12932- 351.56 2.303 log (adem) – 4642.77 (5.11x1044) 2-303 10g (4 atm ) = 2.3732 log(+atm ) - 11030479 Taking antilog on both sides Ialom - 167.030419 latm = 10.7270

P_{​​​​​​1} = 1atm/10.7270

P_{​​​​​​1} = 0.093atm

Hence, vapour pressure of ethanol at 25°C = 0.093atm