Chemical Equilibrium

1. Chemical Equilibrium Chapter 12

2. All chemical reactions can be described as… For Example

3. Minimizing Gibbs Free Energy The reaction proceeds to an equilbrium, with a minimum Gibbs Free Energy This is true for chemical equilibria, just as it’s true for physical equilibria

4. Equilibrium occurs at the minimum Gibbs Free Energy Gibbs Free Energy of the mixture, g, kJ/mole Equilibrium concentration Mole fraction of compound a

5. If the products and reactants form an ideal solution, the Gibbs Free Energy of the system is… Gibbs Free Energy of Mixing, which is always negative

6. Example 12.1 • Estimate the chemical equilibrium composition of a gaseous mixture of n-butane and isobutane at 298.15 K and 1 bar pressure, based on minimization of Gibbs Free Energy

7. First take the derivative Realizing that for this reaction …

8. At equilibrium, set the equation equal to 0

9. Solve for x You’ll find the pure component Gibbs Free Energies at 298.15 K on Table A.8

10. We could follow the same process for any reaction • Include terms for each product and reactant, taking stoichiometry into account • Include activity coefficients • Include fugacity coefficients • But… we usually use the law of mass action

11. Deriving the law of mass action • We know that equilibrium occurs when the Gibbs Free Energy of the system is minimized • We could use partial molal Gibbs Free Energies in our derivation (chemical potential) • It is more convenient to work with fugacities

12. For any single phase system

13. Consider the ammonia reaction In other words, the Gibbs Free Energy change caused by consuming reactants must be balanced by the products created

14. Since fugacities are easier to work with, substitute f into our equation Which with some algebra becomes…

15. Some additional manipulation (page 316) leads to… Where…

16. Remember

17. So, substituting into K … K is independent of pressure, but not of temperature For ideal solutions, γ=1

18. Example 12.2 • Find the equilibrium concentration of NO in air, at 298.15 K Use Table A.8 to find the standard state Gibbs Free Energies

19. Find ΔGreaction and K

20. For gases… The standard state fugacity for gases is 1bar, and at one bar we can assume ideal behavior

21. So for our particular reaction Since K is small, assume the concentration of nitrogen and oxygen do not change Will P cancel out in the general case?

22. Substitute in numbers At 298.15 K

23. Reaction Coordinateaka Extent of Reaction • In the example, we assumed that the reactant concentration doesn’t change • Usually that is not a good assumption • In our example the total number of moles did not change during the reaction • Usually that is not a good assumption either

24. Extent of reaction • Moles of one selected reactant consumed or product produced • For example, for our reaction Let O2 be the selected reactant

25. You can express the number of moles of each species at equilibrium using the extent and the stoichiometric coefficient It is unfortunate that the same symbol is used for the stoichiometric coefficient and the fugacity Add for the products and subtract for the reactants

26. To find the mole fraction of each species… So for our example…

28. Now we only have one unknown, and can solve for the extent Take as a basis one mole of reactants

29. When the total number of moles changes during the reaction, the calculations are a bit more complicated – but the procedure is the same

30. Factor out the pressure terms

31. Substitute in the number of moles of product and reactant Now all we need to know is the initial number of moles of each species, and we can find the equilibrium concentration

32. The effect of temperature on chemical equilibrium • Recall that And that Δg is a function of temperature

33. Take the natural log of both sides and take the derivative with respect to T

34. But Δg=Δh-TΔs

35. However, the first Gibbs relation tells us that… And at constant pressure we can delete the second term These two terms are equal

36. Of course enthalpy is also a function of temperature However, we can approximate the enthalpy as a constant for many reactions

37. Rearranging This is the format used in most chemical process design computer programs to calculate the equilibrium constant

38. However, Δh is rarely independent of temperature, making our equation good for estimates but not detailed calculations We need to back up and keep the enthalpy inside the integral

39. To find Δh of reaction at some temperature other than the standard temperature select a path where you can find the changes in enthalpy for each step… Products at T2 Reactants at T2 Reactants at T1 Products at T1

40. Constant pressure heat capacity is also a function of temperature This expression was found by simple curve fitting at pressures low enough that the gas is ideal Unfortunately, there is little consistency between sources – sometimes the temperature is in 0C and sometimes K. Sometimes it is per mole and other times it is per unit mass. Sometimes the form of the equation is even different – be very careful!!

41. The math to find the enthalpy change becomes awkward, but not very difficult I used equation 12.AS, page 330, in this derivation νis the stoichiometric coefficient

42. Since we usually call T1, 298.15 K, the second part of the expression is a constant

43. Substitute into the equilibrium constant equation This term varies depending on the Cp correlation Then all you have to do is integrate. The calculations are tedious, but not hard

44. The graph shows that K changes dramatically with temperature

45. If Δh is constant, we would expect a straight line if we plot ln(K) vs 1/T The lines show a very modest amount of curvature

46. Effect of Pressure on Chemical Reaction Equilibrium in the gas phase • for

47. Factor to give

48. Factor to give

49. The first term just makes the units work out • Knis equal to one for ideal gases. It deviates from one at high pressures • Kg is equal to one for ideal solutions – which is almost always the case for gases – it is not true for liquids

50. If the number of moles created is the same as the number of moles consumed KP does not depend on pressure Otherwise the pressure makes a big difference in K. Notice also that Kp is not necessarily dimensionless!!