Nonideal Mixtures and Solutions

1. Nonideal Mixtures and Solutions

Rewriting Eq. (26),

(27) We define the term

Σln ˆα ν k k ≤ –Σν k g k (T,P)/( R T).

(28) which is constant at specified values of T and P and Eq. (27) assumes the form

K(T,P) = Π ˆα ν k k = exp (– Σν k g k (T,P)/( R T))

Π ˆα ν k k ≤ K(T,P), or (28a) K(T,P) ≥ Π ˆα ν k k (28b)

The physical meaning of this relation is as follows. Consider an ideal gas mixture of 5 kmole of CO, 3 kmole of O 2 and 4 kmole of CO 2 in a PCW assembly in which a reaction proceeds at fixed (T,P). Here, ˆa k = X k and K(T,P) remains constant while X k changes. Equation (28) must be satisfied as the reaction proceeds and the equality holds good at chemical equilibrium.

Unlike the superheated steam tables in which properties are tabulated as functions of (T,P), K(T,P) is tabulated typically only at P 0 , the standard pressure, since simple relations are available (particularly for ideal gases) to relate K(T,P) to K(T,P 0 ). Such a relation is provided below. Recall from Chapter 8 that for a species k at a state characterized by specified (T,P)

g k (T,P) = g k (T,P o )+( R T ln (f (T,P)/f k k (T,P o )),

(29) where f . is the fugacity of species k. If that species is an ideal gas, f k (T,P) = P and f k (T,P 0 )=P 0 .

The second term on the RHS represents the deviation from this behavior at P o . Selecting P o =1 bar,

(30) using this relation in Eq. (28),

k (T,P) = g k (T) + ( R T ln (f (T,P)/f k k (T, 1bar)), so that

K(T,P) = exp(– Σν k o g k (T)/( R T)) exp(– Σν k ln (f (T,P)/f k (T, 1bar))). k (31) Now let us define

ln K o (T) = – Σν k g k o (T)/( R T), i.e.,

(32) K o (T) = exp (– Σν o (T)/( R T))= exp (– ∆G o k g k /( R T)), where

(33) the term K o (T) is conventionally called the equilibrium constant (a misnomer since it is a func-

tion of temperature and is constant only if the temperature is also held constant) and is tabu- lated for many standard reactions in Table A-28B . Using Eqs. (31) and (32), the relation be- tween K(T,P) and K o (T) is

(34) Subsequently, using Eq. (34) in Eq. (28b), as the reaction proceeds, the following inequality

K(T,P) = K o (T) Π (f (T,P)/f

(−νk)

k (T, 1bar)) .

must be satisfied: K o (T) ≥ Π (f k (T,P) ˆ α k /f k (T, 1bar)) ν k .

(35’) At chemical equilibrium K o

(35a) In order to evaluate f (T, 1bar) and the corresponding K o k (T) it is convenient to define the state

(T) = ν Π (f

k (T,P) ˆ α /f k (T, 1bar)) k k .

of each species k at a standard state corresponding to a pressure of 1 bar.

a. Standard State of an Ideal Gas at 1 Bar If one considers the state of species k to be an ideal gas at 1 bar, then f (T, 1bar) = 1. k Hence, using Eq. (30),

(35b) where g k o (T) is evaluated at the temperature T assuming the species k to be an ideal gas.

g k (T,P) = g o k (T) + R T ln (f k (T,P)/1)

Since the substance is non-ideal at the state (T,P), the second term in Eq. (35b) accounts for the correction due to non-ideal behavior.

The values of g k o (T) can be determined using the expression

(T) = h k (T) - T s k (T),

(35c) where the term h 0 k 0 (T) includes chemical and thermal enthalpies. The g k (T) can also be es-

timated using Gibbs function of formation (Chapter 11). Unless otherwise stated, Eq. (35c) will be for determining the values of 0 g

k (T).

b. Standard State of a Nonideal Gas at 1 Bar Consider the following chemical reaction

H 2 O(l) + CO(g) → CO 2 (g) + H 2 (g). The reaction involves both liquid H 2 O(l) and gaseous species. With Eq. (30), we obtain:

0 f k () l ( T , P )

g k () l ( T, P = g ) k () l () T + R T ln , where

f k () l ( T, l )

(36a)

f k(l) (T, P) = f k(l) (T,P sat ) POY k ≈f k(l) (T,P sat ),

and the Poynting correction factor is (cf. Chapter 8)

POY P

k = exp { ∫ P sat k v () l dP / RT ( ) } ≈ 1

Since POY ≈1 for most liquids and solids, Eq. (36a) assumes the form

() l ( T, P ) ≈ g k () l () T

(36b)

c. Example 3 Determine the value of g HO 2 () l (40ºC, 10 bar) and g HOg 2 () (40ºC, 10 bar) for water va-

por. Select the standard state to be at 1 bar at 40ºC. Assume ideal gas behavior at the standard state. Also determine the value of g ˆ id HO 2 () l at 40ºC and 10 bar for a salt water

mixture in which water constitutes 85% on mole basis. Solution For water,

g H2O(l) (313 K,10 bar) =

HOg 2() (313 K) + 8.314 ×313×ln ( f HO 2 () l (313 K, 10 bar)/1) = g HOg 2 () (313 K, 1 bar) + 8.314 ×313×ln ( f HO 2 () l (313 K, 10 bar)/1), where

(A)

(B) In the case of water, P sat = 0.07384 bar at 40ºC, and

f HO 2 () l (T,P)/ HO 2 f () l (T,P sat ) = POY.

POY = exp(

vdP /( RT ) ) = exp(0.00101 ∫ ×(10–0.074)×100÷(0.461×313))

= exp(0.0000695) = 1.007 ≈ 1. (C) Therefore,

f 2 () l (T,P)/ f HO sat HO 2 () l (T,P ) ≈ 1.

(D)

Since f sat H2O(l) (T,P )=f H2O(g) (T, P ) and the vapor is an ideal gas, then f H2O(g) (T, P ) =P sat = 0.07384 bar and

sat

sat

f HO 2 () l (313 K, 1 bar) = 0.074 bar. (E) Using Eqs. (A) and (E),

g o HO 2() l (313 K,10 bar) ≈ g HOg 2() (313 K, 1 bar) + 8.314 ×313×ln (0.07384/1). (F) Now,

g o HOg 2() (313 K, 1 bar) = ( h o HOg 2() –T s o HOg 2() ) 313 K, 1 bar

× 190.33 = –300894 kJ kmole = –241321 – 313 –1 . (G) Using Eqs. (A), (E), and (F),

g o HO 2() l (313 K,10 bar) = –300894 + 8.314 × 313 × ln (0.07384/1)

(H) If the liquid state is selected at 1 bar instead of an ideal gas state, Eq. (A) becomes

= –307675 kJ kmole –1 .

g HO 2 () l (T,P)= g HO 2 () l (T,1 bar) + 8.314 ln( f HO 2 () l (T,P)/ f HO 2 () l (T,1 bar)), where (I) At given T, RT d ln (f) = v dP (Chapter 8) and hence

ln( f HO 2 () l (T,P)/ f HO 2 () l (T,1 bar)) = ∫ 1 v dP/ R T

= 0.00101 ×(10–1)×100÷(0.461×313) = 0.0063, so that (J)

(K) Using Eqs. (I) and (K),

f HO 2 () l (313 K, 10 bar) ≈(f HO 2 () l (313 K, 1 bar)

HO 2() l (313 K,10 bar) = g HO 2() l (313 K,1 bar), where

(l)

o HO 2() l (313 K, 1 bar) = ( h HO 2() l –T s HO 2() l ) 313 K, 1 bar

= –285830+4.184 ×(40–25)×18.02–313×(69.95+4.184×18.02×ln (313÷298))

(M) which almost equals the previous answer (Eq. (H)).

= –307752 kJ kmole o = g

HO 2() l (313 K, 10 bar),

g ˆ id HO 2 () l (T,P) = g HO 2() l (T,P) + R T ln X HO 2 , i.e.,

g ˆ id HO 2 () l (313 K, 10 bar) = g HO 2() l (313 K, 10 bar) +

8.314 × 313 × ln 0.85 = –308159 kJ kmole –1 . Remarks

We have selected the standard state with regard to both the liquid and gaseous states. For liquids or solids, g (T,P) ≈ g o (T) f k (T,P) ≈f k (T,1) and hence K(T,P) ≈K o (T).