posted by Chloe on .
I don't know how to solve this problem either. And yes, this is the question for this problem. I even double check it. Please help.
Chloroform (CHCL3) has a normal boiling point of
61 'C and an enthalpy of vaporization of 29.24 kJ/mol. What is the value of delta Gvap (in kJ) at 61'C for chloroform?
Chemistry - DrBob222, Tuesday, April 30, 2013 at 3:39pm
I wonder if this is the real problem? Isn't delta G = zero at the boiling point? It would have made more sense to ask for dS vap at the normal boiling point. And how could you calculate dG anyway with no dSvap given?
If that is the problem I would answer dG = 0 since you have an equilibrium at the normal boiling point and dG = 0 at equilibrium.
dG=? . how do I start this problem?
Do I use dG= -RT*ln*K?
dG=-(0.008314 kj/k*mol) (61+ 273.15) ln (29.24)??
I tried this and it is wrong.
There is no "starting it" if I'm right. dG = 0 since the solution at its normal boiling point is zero. There is no calculation. delta G = zero, period.
ok thank you!
Please follow up with my answer. I would like to know if I need to adjust my thinking.