DEVIATIONS FROM RAOULT’S LAW

E. DEVIATIONS FROM RAOULT’S LAW

Consider two species k and j that form a binary mixture. The attraction force between similar molecules of species k is denoted as F kk and between dissimilar molecules as F kj . The following scenarios ensue: (1) F kj =F kk so that the ideal solution model and Raoult’s Law ap- ply, e.g., toluene–benzene mixtures and mixtures of adjacent homologous series; (2) F kj >F kk implying a nonideal solution in which contraction occurs upon mixing, e.g., acetone–water mixtures and other examples of hydrogen bonding; and (3) F kj <F kk , which corresponds to a nonideal solution in which the volume expands upon mixing, e.g., ethanol–hexane and other polar–non polar liquids. In case of the second scenario, since the intermolecular attraction forces are stronger between k–j pairs than between k–k molecular pairs, the vapor pressure of species k can be lower than that predicted using Raoult’s Law, which is referred to as a nega- tive deviation from the Law. In case (3) the attraction forces are lower, and a larger amount of vapor may be produced as compared with the Raoult’s Law prediction, i.e., both the second and third scenarios suggest that we must involve activity coefficients, γ k(l) . It will now be

shown that p sat

(43) where γ

k = γ k(l) X k,l p k .

id

k(l) = ˆf k(l) (T,P)/ ˆf k(l) (T,P) = ˆf k(l) (T,P)/(X k(l) f k(l) (T,P)).

1. Evaluation of the Activity Coefficient

We have previously employed the ideal solution model to predict the vapor pressure of a component k in an ideal solution. If the measured component vapor pressure differs from that prediction, then it is apparent that the ideal solution model is not valid. We can determine the activity coefficient (that represents the degree of non-ideality from the measured vapor pressure data) as follows.

Figure 10: Henry’s constant H(T) for the solubility of gases in water. (From S. S. Zum- dahl, Chemistry, DC Heath and Company, Lexington, Mass, 1986. With permission.)

(44) At phase equilibrium,

γ k = ˆf k / ˆf k,id = ˆf k /X k f k (T,P).

ˆf k(g) = γ k(g) X k f k(g) (T,P) = ˆf k(l) and

γ k(g) X k f k(g) (T,P)= ˆf k(l) = γ k (l) X k,l f k(l) (T,P).

(46) Since the vapor is assume to be an ideal gas mixture γ k(g) =1, and f k(g)( T,P) = P. Therefore,

p k = γ k(l) X k,l f k(l)( T,P), where

k,l (T,P) = f k(l) (T,P ) POY ≈f k(l) (T,P )=P , i.e.,

(48) Thus, γ k(l) is a measure of the deviation from Raoult’s law. With respect to the measured vapor

p k = γ k(l) X k,l sat f k(l)( T,P) = γ k(l) X k,l P = γ k(l) p k,Raoult .

pressure at a specified value of X k , namely, γ k(l) =p k /(X k,l p sat k ), or γ k,l =p k /p k,Raoult .

Recall that the γ k ’s for any phase are related to ( g E /( R T)) (see Eqs. (114) to (117), Chapter 8). If,

(50) then we obtain the Van Laar Equations:

X 1(l) X 2(l) R T/ g E = B´ + C´ (X 1(l) -X 2(l) ),

ln γ

1 = A(T) X

2(l) /((A(T)/B(T)) X 1(l) +X 2(l) ) , ln γ 2 = B(T) X 1(l) /((X 1(l) + (B(T)/A(T)) X 2

2(l) ) , where

A (T) = 1/(B´- C´), and (52) B(T) = 1/(B´ + C´).

(53) The constants A and B are generally weak functions of temperature, and for the bi-

nary mixture, one can solve for A and B as

A = ln γ (1+ (X

ln γ )) 1(l) 2 2(l) 2(l) 1(l) 1(l) , (54)

ln γ )/(X

(55) At the azeotropic condition, at which X k(l) =X k , Eq. (47 ) assumes the form

B= ln γ 2(l) (1+ (X 1(l) ln γ 1(l) )/(X 2(l) ln γ 2(l) )) 2 .

(56) Thus, γ

p k =X k(l) P= γ k(l) X k(l) p sat k .

k(l)( X k,azeotropic )= P/ p k (T). (57) The Wilson equation for activity coefficient has the form

sat

2 ln 2 1(l) = AX 2,l /((A/B) X 1(l) +X 2(l) ) , ln γ 2 (58)

= BX 2 1(l) /(X 1(l) +(B/A)X 2(l) ) 2 ,

ln γ 2(l) = – ln(X 2(l) +X 1(l) A 21 )+ X 1(l)( A 12 /(X 1(l) +X 2(l) A 12 )

21 /(X –A 2(l) +X 1(l) A 21 )).

As X 1(l) → 0, ln γ 1(l) = – ln A 12 +1–A 21 ,and ln γ 2(l) = – ln A 21 + 1– A 12 , where

A ij = (v j /v i ) exp (– a ij /R T), and v j denotes the molal volume of species j, and a ij is a known constant that is independent of

composition and temperature, where a ij ≈a ji .