The heat of vaporization of a compound at 300. K is 27.7 kJ/mol and its vapor pr
ID: 790091 • Letter: T
Question
The heat of vaporization of a compound at 300. K is 27.7 kJ/mol and its vapor pressure is 540.5 torr. Determine the following thermodynamic properties of the vaporization at 300K:
The heat of vaporization of a compound at 300. K is 27.7 kJ/mol and its vapor pressure is 540.5 torr. Determine the following thermodynamic properties of the vaporization at 300K: delta G degree = delta S degree = J/mol-K Assume that delta H degree and delta S degree are temperature independent and estimate the normal boiling point of the compound to the nearest degree Celcius.t = degree C
Explanation / Answer
delta S = (Heat of Vaporization)/Temperature = 27.7*1000/300 = 92.33 J/mol-K
The Clausius Clayperon Equation
Ln(P_2/P_1) = -(Delta_H/R) (1/T_2 - 1/T_1)
P_2 = 540.5 torr
540.5 torr = 760 torr
T_2 = 300K
Delta_H = 27.7*1000 J
R= 8.314 J/mol K
ln(540.5/760) = -(27700/8.314)((1/300)-(1/T_1))
solving T_1 = 309.498 K = 36.498 degree Celsius is the boiling point.
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