The vapor pressure of Y is 4.560 atm at 440K and 13.680 atm at 515K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(l) = Y(g).
G440o = -5.55 kJ/mol
G515o = -11.20 kJ/mol
Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values
delta S = J/(mol-K)
delta H = kJ/mol
What is the normal boiling point of Y in degrees Celcius?
t = oC
What is the vapor pressure of Y at 545 K?
Po = atm
ANSWER:
For this reaction, the equilibrium constant is
At the same time, the free energy is defined as
where R = gas constant = 8.314 J/mol.K and T = temperature in K
Then, replacing
Now, we can calculate the standard free energy at each temperature:
To find the standar enthalpy of vaporization we use the Clausius-Clapeyron equation:
And the standard entrophy is
or
The nomral boiling point (at P = 1 atm) is calculated using the Clausius-Clapeyron equation:
Finally, the vapor pressure of Y at 545 K is:
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