P12.1. Acrolein + water exhibits an atmospheric (1 bar) azeotrope at 97.4 wt% acrolein and 52.4°C.
a. Determine the values of Aij for the van Laar equation that match this bubble-point pressure at the same liquid and vapor compositions and temperature. (ANS. 2.97, 2.21)
(You may use the shortcut vapor pressure equation for acrolein: Tc = 506 K; Pc = 51.6 bar; and ω = 0.330; MW = 56.)
b. Tabulate P at 326.55 K and x = {0.1,0.3,0.5} via the van Laar equation using the A12 and A21 determined above. (ANS. P = {1.11, 1.15, 1.03}
P12.2. The system α-epichlorohydrin(1) + n-propanol(2) exhibits an azeotrope at 760 mmHg and 96°C containing 16 mol% epichlorohydrin. Use the van Laar theory to estimate the composition of the vapor in equilibrium with a 90 mol% epichlorohydrin liquid solution at 96°C. (α-epichlorohydrin has the formula C3H5ClO, and IUPAC name 1-chloro-2,3-epoxypropane. Its vapor pressure can be approximated by: log10Psat = 8.0270–2007/T, where Psat is in mmHg and T is in Kelvin. You can use the shortcut vapor pressure equation for n-propanol.) (ANS. 0.72, 0.63 bar)
P12.3. The following free energy model has been suggested for a particularly unusual binary liquid-liquid mixture. Derive the expression for the activity coefficient of component 1,
where, and .
(ANS. )
P12.4. The Scatchard-Hildebrand model can be extended to multicomponent mixtures in the following manner. Setting aij = (aiiajj)½ and aii/Vi = Viδ2 Eqn. 12.7 can be rewritten as
Recognizing that the quadratic term is separable and simplifying the square-root of the square:
where <δ> ≡ ΣxiViδi/V ≡ ΣΦiδi and V ≡ ΣxiVi.
This result can be made even simpler by adding and subtracting V<δ>2 and rearranging to obtain:
where we have substituted the definition of <δ> in the second term and the definition of V in the third.
Collecting all terms in a common summation, we obtain:
This is a remarkably simple result. Derive an equally simple expression for the activity coefficient of a component in a multicomponent mixture. (Hint: It is easier to start with Eqn. 12.65.) (ANS. Eqn. 12.54)
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