DrBob222,
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.
To solve this problem, we need to use the equation ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the enthalpy of vaporization, T is the temperature in Kelvin, and ΔS is the change in entropy.
Since we are given the enthalpy of vaporization (ΔH) as 29.24 kJ/mol, and we want to find ΔG at the boiling point of chloroform which is 61°C, we need to convert the temperature to Kelvin. To do this, we use the equation K = °C + 273.15.
Therefore, 61°C + 273.15 = 334.15K.
Now we can calculate ΔG using the equation ΔG = ΔH - TΔS. However, the problem states that no ΔS is given, so we cannot directly calculate ΔG using this equation because we don't have all the required values.
It seems like there might be some missing information or a mistake in the question. If you have any additional information or clarity on the question, please provide it so that I can assist you better.