Acentric Factor with Mixtures

[Jen0715]

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!!!

 

[Latexman]

The equation of state (EOS) you select specifies the mixing rule to use for acentric factor.

Good luck,
Latexman

 

[Jen0715]

Well I am eventually using the SRK EOS to find Z-factor. But, I am using Kay's Rule to find the Pr and Tr.

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

 

[Latexman]

I just looked at my 3rd Edition of Sherwood, Reid, and Prausnitz and I don't think you need a mixture [ω] for the original Soave modification of the Redkich-Kwong EOS. You do need pure component [ω]'s to calculate F_i and F_j, which are then used to calculate F_m.

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

 

[TrentFGuidry]

If you are going to use the Soave Redlich Kwong or the Peng Robinson equation of state, you are probably best off using the usual mixing rules for those equations rather than using pseudo critical values.

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