A substance has a ΔHv = 20.0kJ/mol. It has a vapor pressure of 0.800 atm at -2 degrees celsius. What's the normal boiling point?

Question

(Use the Clausius-Clapeyron equation in its two point form) R= 0.08206L·atm/mol·K = 8.314 J/mol·K

Answer 1

Clausius-Clapeyron equation: ln (P1/P2) = -∆Hvap/R (1/T1 - 1/T2)

P1 = 0.800 atm

T1 = -2ºC + 273 = 271K

P2 = 1.0 atm (normal conditions)

T2 = ?

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (note change to kJ to be consistent with units of ∆Hvap)

∆Hvap = 20.0 kJ/mol

ln (0.800/1.00) = -20.0/0.008314 (1/271 - 1/T2)

-0.223 = -2406(0.00369 - 1/T2)

-0.223 = -8.88 + 2406/T2

8.657 = 2406/T2

T2 = 278K = normal boiling point at 1.00 atm pressure