Acentric Factor with Mixtures
Acentric Factor with Mixtures
(OP)
Hi,
I'm trying to figure out how to calculate the acentric factor for a solution with 50 mole% benzene and 50 mole% toluene...
Thank you!!!
I'm trying to figure out how to calculate the acentric factor for a solution with 50 mole% benzene and 50 mole% toluene...
Thank you!!!





RE: Acentric Factor with Mixtures
Good luck,
Latexman
RE: Acentric Factor with Mixtures
I was thinking I found the mixture acentric factor much like Kay's Rule, but wasn't sure if that was right..
I was thinking:
acentric factor (w) = 0.5(w,benzene) + 0.5(w,toluene)
What do you think?
Thank you so much!
Jenna
RE: Acentric Factor with Mixtures
However, you may be using a different modification that I don't know about. Most EOS's use a vapor fraction average, except for some EOS's that are specifically for liquid mixtures, those use a volume fraction average.
Good luck,
Latexman
RE: Acentric Factor with Mixtures
Here is an outline of what I am talking about for the SRK.
For the pure components:
a = (0.42748 (R Tc) * (R Tc) / Pc) * [1 + (.480 +1.574w -0.176ww)*(1-sqrt(T/Tc)]^2
b=0.08664*(R Tc)/Pc
a and b would be calculated at a given temperature for both benzene and toluene using their critical temperature, critical pressure, and accentric factors.
After you have a and b for both pure benzene and toluene, mixing rules can be applied to get the mixture a and b.
The usual mixing rules are linear in b and quadratic in a, although sometimes mixing rules using quadratic bs are used, especially by Prausnitz.
For the linear b / quadratic a mixing rules:
a = sum i sum j of xi * xj * aij
where aij= (1-cij)*sqrt(ai)*sqrt(aj). cij is usally close to zero.
For your 50/50 mole % benzene and toluene case, this would give, assuming cij is 0:
a = x1*x1* a1 + 2*x1*x2 * sqrt(a1) * sqrt(a2) + x2*x2*a2
a=.25 * a1 + .5 * sqrt(a1) * sqrt(a2) + .25 * a2.
b=sum i of xi*bi
For your 50/50 mole % benzene and toluene case, this would give:
b=x1*b1 + x2*b2
b=0.5*b1+0.5*b2
Trent F Guidry