The enthalpy of vaporization of Substance X is 21.0 kj/mol and its normal boiling point is 19. C Calculate the vapor pressure of X at -98. C Round your answer to 2 significant digits.

Answer 1

Clausius Clapeyron equation:

ln (P2/P1) = -∆Hvap / R (1/T2 - 1/T1)

P1 = 1 atm (normal pressure)

T1 = 19ºC + 273 = 292K

P2 = ?

T2 = -98ºC + 273 = 175K

Ea = 21.0 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

Substituting the values and solving for P2, we have ...

ln (P2/1) = -21.0/0.008314 (1/175 - 1/292)

ln P2 = -2526 (0.00571 - 0.00342)

ln P2 = -2525 * 000229

ln P2 = -5.78

P2 = 3.1x10^-3 atm = 0.0031 atm