Any suggestions on how to approach this problem? I can't rearrange the equation to get what I want?

Question

The vapor pressure of dichloromethane, , at 0.0 C is 134 mmHg. The normal boiling point of dichloromethane is 40.0 C. Calculate the molar heat of vaporization.
ln(P2)=ln(P1)+(heat of vaporization/R[gas constant]) (1/T1-1/T2) **R=8.3145 j/(mol*K)

Answer 1

this is a variation of the Clausius-Clapeyron equation

you need to use algebra to solve for the heat of vaporization:

heat of vaporization = [(ln P2 - ln P1)(R)] / [(1/T1) - (1/T2)]

P2 is standard pressure = 760 mm Hg
P1 is 134 mm Hg
T1 is 0 celsius but must be converted to K (273 K)
T2 is 40 celsius or 313 K
R is the universal gas constant = 8.3145 J/mole K

your answer should have units of J/mole which can be converted to kJ/mole

according to my calculator, the answer is 30800 J/mole or 30.8 kJ/mole

according to the Handbook of Chemistry and Physics which gives a value around 28.0, this is a reasonable answer

Answer 2

There should be another form of the same equation in the text, but even if you can't find it, you should be able to calculate the heat of vaporization from what you have already. If you just substitute everything in to the equation, keeping the temperatures in K and keeping the right T with the right P. Also you need to remember that at the boiling point, the vapor pressure is atmospheric pressure which I think you'll need to express in kPa.

Hope this helps...

Answer 3

heat =[R/{(1/T1)-(1/T2)}]* ln(P2/P1)