Question

Use the following data for water at 25 degrees C to answer this question. DeltaHfm(H2O)(l) = -285.830 kj/mol, DeltaHfm(H2O)(g) = -241.818 kJ/mol, DeltaGfm(H2O)(l) = -237.129kJ/mol, DeltaGfm(H2O)(g) = -228.572 kJ/mol

A) calculate the vapor pressure of water at 25 degrees celcius (hint: calculate the equilibrium constant for the "reaction" H2O(l) to H2O(g))
B) calculate the normal boiling point of water (pH2O)= 1 atm
C) why does the answer to part b differ from 100 degrees C, the correct normal boiling point of water?

Answer 1

Clausius-Clapeyron equation

2.303 log(P2/P1) = (ΔH_vap/R)((1/T1) - (1/T2))

• ΔH_vap: The enthalpy of vaporization of the liquid.
• R: The real gas content, or 8.314 J/(K × Mol).
• T1: The temperature at which the vapor pressure is known (or the starting temperature.)=25+273=298K
• T2: The temperature at which the vapor pressure is to be found (or the final temperature)=100+273=373
• P1 and P2: The vapor pressures at the temperatures T1 and T2, respectively. P2 =760 torr at the boiling point

• log(760/P1) = (41076/2.303*8.314)((1/298) - (1/373))

• =41076(373-298)/(2.303*8.314*298*373)

  log760-logP1=1.4475

  logP1=2.8808-1.4475=1.4333

  P1=27.12 torr

  for question number (b)

  log(P2/P1) = (ΔH_vap/R)((1/T1) - (1/T2))

• log(760/27.12) = (41076/2.303*8.314)((1/298) - (1/T2))

  1.44=41076(T2-298)/(2.303*8.314*298T2)

  0.2T2=T2-298

  0.8T2=298

  T2=372.5K

  (C) internal hydrogen bonding in H2O molecules may be reason for the deviation.