3. (12) The expression of K in terms of fugacity coefficient is: The standard state for a gas is the ideal-gas state of the pure gas at the standard-state pressure P 0 of 1 bar. Since the fugacity of an ideal gas is equal to its pressure, f i 0 = P 0 for each species i. Thus for gas-phase reactions, and Eq. (12) becomes: (26)

4. The equilibrium constant K is a function of temperature only. However, Eq. (26) relates K to fugacities of the reacting species as they exist in the real equilibrium mixture. These fugacities reflect the nonidealities of the equili- brium mixture and are functions of temperature, pressure, and composition. This means that for a fixed temperature the composition at equilibrium must change with pressure in such a way that remains constant

5. The fugacity is related to the fugacity coefficient by Substitution of this equation into Eq. (26) provides an equilibrium expression displaying the pressure and the composition: Where =  i i and P 0 i s the standard-state pressure of 1 bar, expressed in the same units used for P. (27)

6. If the assumption that the equilibrium mixture is an ideal solution is justified, then each becomes  i, the fugacity coefficient of pure species i at T and P. In this case, Eq. (27) becomes: (27) For pressures sufficiently low or temperatures sufficiently high, the equilibrium mixture behaves essentially as an ideal gas. In this event, each  i = 1, and Eq. (27) reduces to: (28)

7. Although Eq. (28) holds only for an ideal-gas reaction, we can base some conclusions on it that are true in general:  According to Eq. (20), the effect of temperature on the equilibrium constant K is determined by the sign of  H 0 : o  H 0 > 0 (the reaction is endothermic)  T>>  K >>. Eq. (28) shows that K >> at constant P  >>   >> o  H 0 >  K <<. Eq. (28) shows that K << at constant P  <<   e <<

8.  If the total stoichiometric number (   i i ) is negative, Eq. (28) shows that am increase in P at constant T causes an increase in, implying a shift of the reaction to the right.  If the total stoichiometric number (   i i ) is positive, Eq. (28) shows that am increase in P at constant T causes a decrease in, implying a shift of the reaction to the left, and a decrease in  e.

9. For a reaction occurring in the liquid phase, we return to (12) For the usual standard state for liquids f 0 i is the fugacity of pure liquid i at the temperature of the system and at 1 bar. The activity coefficient is related to fugacity according to: (29) The fugacity ratio can now be expressed (30)

10. (31) Gibbs free energy for pure species i in its standard state at the same temperature: (7) Gibbs free energy for pure species i at P and the same temperature: (7.a) The difference between these two equations is:

11. Fundamental equation for Gibbs energy: (32) For a constant-temperature process: (32) For a pure substance undergone a constant-temperature process from P 0 to P, the Gibbs free energy change is: (33) (34)

12. Combining eqs. (31) and (34) yields: (35) (36) or Combining eqs. (31) and (35) yields:

13. Combining eqs. (36) and (12) yields: (37)

14. For low and moderate pressure, the exponential term is close to unity and may be omitted. Then, (38) and the only problem is determination of the activity coefficients. An equation such as the Wilson equation or the UNIFAC method can in principle be applied, and the compositions can be found from eq. (38) by a complex iterative computer program. However, the relative ease of experimental investigation for liquid mixtures has worked against the application of Eq. (38).

15. If the equilibrium mixture is an ideal solution, then  i is unity, and Eq. (38) becomes: (39) This relation is known as THE LAW OF MASS ACTION. Since liquids often form non-ideal solutions, Eq. (39) can be expected in many instances to yield poor results.

17. Suppose a single reaction occurs in a homogeneous system, and suppose the equilibrium constant is known. In this event, the calculation of the phase composition at equilibrium is straightforward if the phase is assumed an ideal gas [Eq. (28)] or an ideal solution [Eq. (27) or (39)]. When an assumption of ideality is not reasonable, the problem is still tractable for gas-phase reactions through application of an equation of state and solution by computer. For heterogeneous systems, where more than one phase is present, the problem is more complicated and requires the superposition of the criterion for phase equilibrium developed in Sec. 11.6

18. Single-Phase Reactions The water-gas shift reaction, CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) Is carried out under the different set of conditions below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas. The reactants consist of 1 mol of H 2 O vapor and 1 mol of CO. The temperature is 1100 K and the pressure is 1 bar. Example

19. Solution COH2OH2OCO 2 H2H2 – 1 + 1 A3,3763,4705,5473,249 B  10 3 0,5571,4501,0450,422 C  10 6 – 4,392000 D  10 -5 – 0,0310,121– 1,1570,083  H 0 f,298 – 110.525– 241.818– 393.5090  G 0 f,298 – 137.169– 228.572– 394.3590

20. 20

21. 21

22. Since the reaction mixture is an ideal gas: The number of each species at equilibrium is: While total number of all species at equilibrium is:

24. Therefore the fraction of the steam that reacts is 0.5

25. Estimate the maximum conversion of ethylene to ethanol by vapor-phase hydration at 523.15 K and 1.5 bars for an initial steam-to-ethylene ratio of 5. Example Solution Reaction: C 2 H 4 (g) + H 2 O (g)  C 2 H 5 OH (g)

26. C2H4C2H4 H2OH2OC 2 H 5 OH – 1 + 1 A1,4243,4703,518 B 14,394  10 -3 1,450  10 -3 20,001  10 -3 C – 4,392  10 -6 0 – 6,002  10 -6 D0 0,121  10 5 0  H 0 f,298 52.510– 241.818– 235.100  G 0 f,298 68.460– 228.572– 168.490

27. 27

29. For hign temperature and sufficiently low pressure:

30. The number of each species at equilibrium is: While total number of all species at equilibrium is:

32. The gas-phase oxidation of SO 2 to SO 3 is carried out at a pressure of 1 bar with 20% excess air in an adiabatic reactor. Assuming that the reactants enter at 298.15 K and that equilibrium is attained at the exit, determine the composition and temperature of the product stream from the reactor. Example Solution Reaction: SO 2 (g) + ½ O 2 (g)  SO 3 (g)

33. Basis: 1 mole of SO 2 entering the reactor: moles of O 2 entering = (0.5) (1.2) = 0.6 moles of N 2 entering = (0.6) (79/21) = 2.257 The amount of each species in the product stream is:

34. Total amount of all species: Mole fraction of each species:

35. Energy balance: Reactant T = 298.15 K SO 2 = 1 O 2 = 0.6 N 2 = 2.257 Product T = 298.15 K SO 2 = 1 –  e O 2 = 0.6 – 0.5  e N 2 = 2.257 Reaction  e Product T SO 2 = 1 –  e O 2 = 0.6 – 0.5  e N 2 = 2.257

36. (a) (b)

37. SO 2 O2O2 SO 3 – 1 + 1 A5.6993.6398.060 B 0.801  10 -3 0.506  10 -3 1.056  10 -3 C000 D – 1.015  10 5 – 0.227  10 5 – 2.028  10 5  H 0 f,298 – 296 8300– 395 720  G 0 f,298 – 300 1940– 371 060

38. 38

39. (c) (d) (e)

40. For hign temperature and pressure of 1 bar: (f)

41. Algorithm: 1.Assume a starting value of T 2.Evaluate K 1 [eq. (c)], K 2 [eq. (d)], and K [eq. (e)] 3.Solve for  e [eq. (f)] 4.Evaluate T [eq. (a)] 5.Find a new value of T as the arithmatic mean value just calculated and the initial value; return to step 2. The scheme converges on the value  e = 0.77 and T = 855.7 K

42. The composition of the product is: