[orenstrawberrybanana]

have problem solving this question- i need some help

the normal boiling point of benzene is 80.1oC and its triple point lies at 6.85oC and 0.052atm. use the clausius-clapeyron equation to estimate the molar enthalpy of vaporization. given that the molar enthalpy of sublimation of benzene is 44.05kj mol-1, calculate the molar enthalpy of fusion

[Hunt]

you just need to plug in the values. 

Take, Ln P_v = -delta_vap H / RT + C ( constant of integration)

plug in the coordinates of the trip pt and normal boil pt to solve 2 eq's of 2 uknowns . u find delta_vap H. Then ofcourse delta H sublimation = fusion + vaporization. You find the changei n enthalpy of fusion.