1. PHYSICAL CHEMISTRY ERT 108 Semester II 2011/2012
Huzairy Hassan
School of Bioprocess Engineering
UniMAP

2. Reaction Equilibrium in Ideal Gas Mixtures

3. Introduction
Calculate the equilibrium composition for an ideal-gas reaction from the initial composition, the temperature and pressure, and . To apply the equilibrium condition to an ideal-gas reaction, we need to relate the chemical potential of a component of an ideal gas mixture to observable properties.

4. Chemical Potentials in an Ideal Gas Mixture
Chemical Potential of a Pure Ideal Gas, µ
µ for a pure gas depends on T and P only.
Since for reaction equilibrium  mostly studied at constant T while the partial pressure of the reacting gases vary.
 variation of µ with pressure.

5. The Gibbs equation for dG for a fixed amount of substance is:
and division by the number of moles of the pure ideal gas gives;
Since the chemical potential µ of a pure substance equals (Eq 4.86), for constant T, this equation becomes;

6. If the gas undergoes an isothermal change of state from pressure P1 and P2, integration gives; )
Let P1 = standard P, P⁰ ≡ 1 bar

7. Figure 6.1 plots
at fixed T.
For a pure ideal gas;
and Hm is independent of pressure.
So, the pressure dependence of µ in the figure is due to the change of Sm with P.
In the zero pressure, S becomes infinite, and
µ goes to

8. Chemical Potentials in an Ideal Gas Mixture Definition;
- An ideal gas mixture is a gas mixture having the following properties:
(1) the equation of state is obeyed for all temperatures, pressures, and compositions (ntot is the total number of moles of gas)
(2) if the mixture is separated from pure gas i (i is any one of the mixture’s component) by a thermally conducting rigid membrane permeable to gas i only, then at equilibrium the partial pressure of gas i in the mixture is equal to the pressure of the pure- gas-i system.

9. The standard state of component i of an ideal gas mixture at temperature T is defined to be pure ideal gas i at T and pressure P⁰ ≡ 1 bar.
Let be the chemical potential of gas i in the mixture, and let be the chemical potential of the pure gas in equilibrium with the mixture through the membrane.
For phase equilibrium,
From condition (2), at equilibrium.
Therefore for ideal gas mixture; (Eq. 6.3)
xi is mole fractions of i

10. Is the fundamental thermodynamic equation for an ideal gas mixture.
From Equations 6.2 and 6.3;
Where is the partial pressure of gas i in the mixture, and
is the chemical potential of pure ideal gas i at the standard pressure of 1 bar and at the same temperature T as the mixture.
Is the fundamental thermodynamic equation for an ideal gas mixture.

11. Each of U, H, S, G, and Cp for an ideal
gas mixture is the sum of the corresponding
thermodynamic functions for the pure gases
calculated for each pure gas occupying a
volume equal to the mixture’s volume at a
pressure equal to its partial pressure
in the mixture and at a temperature
equal to its temperature in the
mixture.

12. Ideal-Gas Reaction Equilibrium
Now, we specialize to the case where all reactants and products are ideal gasses.
For the ideal-gas reaction:

15. (Eq. 6.14)
(Eq. 6.15)
is the standard equilibrium constant (or the standard pressure equilibrium constant) for the ideal-gas reaction.

16. ⨳ For example, for the ideal-gas reaction; From Eq. 6.7; (Eq. 6.18)
The standard equilibrium constant is dimensionless.

17. Example 6.1 - Finding and ΔGº from the equilibrium composition.

18. Solution: Since 0.711 mmol of CS2 was formed, so, for H2 (the same);
4(0.711 mmol) = 2.84 mmol of H2 was formed.
For CH4 (0.711 mmol reacted), 5.48 mmol – 1(0.711) mmol = 4.77 mmol CH4 present at equilibrium.
For H2S, mmol – 2(0.711) mmol = 9.60 mmol H2S present at equilibrium.
To find , we need partial pressure . We have and
P = 762 torr and xi are the mole fractions.

19. At equilibrium;
we have total moles = = 17.92
and partial pressure,

20. For the ; ⨳

21. Temperature Dependence of the Equilibrium Constant
Differentiation of with respect to T;
Use of gives;
(pressure fixed at 1 bar)

22. From Equation 6.32 & 6.33;
Where is the reaction’s standard entropy change, then;
Since ΔGº = ΔHº - T ΔSº, then;
van’t Hoff equation

23. Multiplication of Eq. 6.36 by dT and integration from T1 to T2 gives; *

24. Example 6.2 Change of with T
Find at 600 K for
Using the approximation that ΔHº is independent of T;
Using the approximation that is independent of T

25. Solution: From Appendix data, for NO2(g) and N2O4(g); = 57.20 kJ/mol
From ΔG⁰ = - RT ln =
= e – ((4730 J/mol) / ( J/mol K x 298 K))
= 0.148

26. From Eq. 6.39; ⨳

27. (b) is assumed independent of T, then;
Appendix data gives , then ⨳

28. Figure 6.6 – plots ΔHº, ΔSº, ΔGº and R ln Kºp vs. T
- plots ln Kºp vs. 1/T

29. Example 6.3 ΔH⁰ from K⁰P versus T data
Use Figure 6.6b to estimate ΔH⁰ for for temperatures in the range 300K to 500K.
Solution:
Since only an estimation, consider the line is straight.
We have 2 points;
T-1 = K-1 , ln Kºp = 20.0
T-1 = K-1 , ln Kºp = 0

30. ⨳ Hence, the slope is; (20 – 0) / (0.004 K-1 - 0.0022 K-1)
= 1.1 x 104 K
The slope of plot ln Kºp vs. K-1 is -ΔHº / R, so
ΔH⁰ = -R X slope = -(1.987 cal/mol K)(1.1 x 104 K)
= -22 kcal/mol
(in agreement with Figure 6.6a) ⨳

31. Ideal–gas Equilibrium Calculations
Aims- find the equilibrium composition of an ideal-gas reaction mixture and
Equilibrium composition is related to the initial composition by a single variable, the equilibrium extent of reaction ;
where ni,0 is the number of moles of substance i present at the start of reaction.
- ξ measures how much reaction has occurred.

32. The specific steps to find the equilibrium composition of an ideal-gas reaction mixture are as follows;

34. Example 6.4 Equilibrium composition at fixed T and P
Find the equilibrium composition.

35. Solution: Step 1 Step 2 Use Appendix data to get ;
= [ (NO2, g)] - [ (N2O4, g)]
= 2(51.31) – = 4.73 kJ/mol
Step 2

36. Step 3
Step 4

37. Step 5

38. Step 6

39. Thank you