Ethanol has an enthalpy of evaporation of 38.6kJ/mol, and a normal boiling point of 78.4oC. What is the vapor pressure of ethanol at 15oC?

Question

Hint: the normal boiling point is when the vapor pressure reaches 1 atm. Use,
. Pay attention to use right units

Answer 1

equation: ln(P2/P1)=(delta H/R)((1/T1)-(1/T1)
1st:T1=78.4+273=354.4K
T2= 15+273=286K
R=8.3154J/mol
delta H=38.6*1000=38600J/mol
2nd: ln(P2/1atm)=((8600J/mol/8.3145J/mol)((1/354.4)-(1/286))= 3.0211
3rd: natural e^3.0211= 20.513
4th: (P2/1atm)=(1/20.513)
5th: P2=(1atm/20.513)=answer: 4.8750x10^-2

Answer 2

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