Answer to Question #94358 in General Chemistry for India Washington

Question

The vapor pressure and enthalpy of vaporization of an unknown substance are measured to be 222.6 torr and 36.4 kJ/mol, respectively at room temperature (298.15 K). Estimate the normal boiling point of the substance at an atmospheric pressure of 760 torr.

Answer 1

ln(P2/P1) = ∆Hvap/R (1/T1 - 1/T2)

P1 = 222.6 torr

P2 = 760 torr

T1 = 298.15

T2 = ?

∆Hvap = 36.4 kJ/mol = 36,400 J/mol

R = 8.314 J/K-mol

ln(760/222.6) = 36,400/8.314 (1/298.15 - 1/T2)

1.228 = 4378(0.00335 - 1/T2)

1.228 = 14.7 - 4378/T2

-13.47 = -4378/T2

T2 = 325K