1. Equilibrium Unit 4 Chapters 17, 18, 19, 20

2. Chapter 17 Equilibrium – when two opposite reactions occur simultaneously and at the same rate Equilibrium – when two opposite reactions occur simultaneously and at the same rate Dynamic equilibrium – equilibrium can be shifted if one or more of the reactants is changed in any way. The equilibrium will “react” Dynamic equilibrium – equilibrium can be shifted if one or more of the reactants is changed in any way. The equilibrium will “react”

3. Chemistry Connections Connection to Kinetics Connection to Kinetics Equilibrium affects reaction rates, especially for multi-step reactions! Equilibrium affects reaction rates, especially for multi-step reactions! Factors that affect equilibrium affect reaction rates Factors that affect equilibrium affect reaction rates Connection to Thermo Connection to Thermo If the reaction rate slows, the energy to be released or absorbed may be released more slowly or quickly….can be dangerous or take FOREVER!...when setting up a reaction MUST keep this in mind! If the reaction rate slows, the energy to be released or absorbed may be released more slowly or quickly….can be dangerous or take FOREVER!...when setting up a reaction MUST keep this in mind!

4. Equilibrium Basics – Kinetics! For a reversible reaction with a one step mechanism: For a reversible reaction with a one step mechanism: aA + bB  cC + dD Rate = k f [A] a [B] b = k R [C] c [D] d Definition of equilibrium: R f = R r

5. At Equilibrium k f [A] a [B] b = k R [C] c [D] d K f = [C] c [D] d k R [A] a [B] b k f / k R = k c k c is the equilibrium constant (mass action expression) k c is the equilibrium constant (mass action expression) The constant can be determined by the ratio of the constants or the concentrations! The constant can be determined by the ratio of the constants or the concentrations! Example 17-1, 17-2, 17-3 Example 17-1, 17-2, 17-3

6. Equilibrium Constant k c k c Just as in kinetics, intermediates are not allowed, so substitution may be necessary! Just as in kinetics, intermediates are not allowed, so substitution may be necessary! Must be found experimentally or by means of equilibrium concentrations from thermodynamic data. Must be found experimentally or by means of equilibrium concentrations from thermodynamic data. Also varies with temperature, and constant at a given temperature (just like kinetic rate constants) Also varies with temperature, and constant at a given temperature (just like kinetic rate constants) Independent of initial concentration Independent of initial concentration

7. Equilibrium Constant Equilibrium constants are expressed as a ratio of “activities” rather than concentrations in order to be dimensionless. Equilibrium constants are expressed as a ratio of “activities” rather than concentrations in order to be dimensionless. Activity: Activity: Pure liquid or solid = 1 (if it is pure, amt. does not affect) Pure liquid or solid = 1 (if it is pure, amt. does not affect) Solution = the molar concentration Solution = the molar concentration Ideal gas = partial pressure in atm Ideal gas = partial pressure in atm

8. Reaction Quotient Reaction quotient – calculated the same way as k c, but the reaction is NOT YET at equilibrium. (mass action expression) Reaction quotient – calculated the same way as k c, but the reaction is NOT YET at equilibrium. (mass action expression) If reaction quotient is calculated, comparing it to k c will allow prediction of the direction of the reaction. If reaction quotient is calculated, comparing it to k c will allow prediction of the direction of the reaction.

9. Comparing Quotient to K aA + bB  cC + dD [C] c [D] d = Q [A] a [B] b When k c > Q, reaction moves forward When k c < Q, reaction moves reverse

10. Reaction Quotient Forward reaction, [C] c [D] d = kc Forward reaction, [C] c [D] d = kc [A] a [B] b Reverse reaction, [A] a [B] b = 1/kc = kc’ Reverse reaction, [A] a [B] b = 1/kc = kc’ [C] c [D] d [C] c [D] d  Reversing the reaction, kc must be raised to -1 power.

11. Manipulation of kc  Multiply the equation by ½  Kc* = == kc 1/2  Whatever factor (n) is applied to the balanced equation, the new kc* is kc raised to the power of n  Example 17-4, 17-5

12. What is kc used for? Generally used to calculate concentrations at equilibrium of various reactants and products Generally used to calculate concentrations at equilibrium of various reactants and products Example 17-6 Example 17-6 Example 17-7 Example 17-7

13. Le Chatelier’s Principle O la la

14. Le Chatlier’s Principle (1884) If a change in conditions (stress) is applied to a system at equilibrium, the system responds to reduce the stress and reach a new state of equilibrium. If a change in conditions (stress) is applied to a system at equilibrium, the system responds to reduce the stress and reach a new state of equilibrium.

15. Le Chatelier’s Principle Major stresses include: 1. Change in Concentration (or pressure for gasses) PV = nRT P = (n/V) RT n/V = molar concentration if V is in L  At constant T, P  concentration so it can be used in its place in calculations.

16. Example H 2 (g) + I 2 (g)  2HI (g) H 2 (g) + I 2 (g)  2HI (g) What happens when you add H 2 ? What happens when you add H 2 ? What happens when you remove H 2 ? What happens when you remove H 2 ? What happens when you compress the container? What happens when you compress the container? Example 17-10 Example 17-10

17. Le Chatelier’s Principle 2. Change in Volume – same moles, but smaller V, affects gas concentrations, but not liquid, solid, or solution. 2NO 2 (g)  N 2 O 4 (g) What will happen if we decrease the volume of the container? You can increase the pressure by pumping in an inert gas, which does not increase the concentration.

18. Le Chatelier’s Principle 3. Effect of Temperature Change – depends on “thermicity” of reaction Thermicitiy is considering heat as a reactant or a product and writing it into the equation. C (s) + O 2 (g)  CO 2 (g) + 393.5kJ If I add heat, which way will equilibrium move?

19. Le Chatelier’s Principle 4. Catalysts –

20. Le Chatelier’s Principle Catalysts decrease the activation energy of both the forward and reverse reaction, therefore, no net effect on equilibrium, only on the rate of reaction. Catalysts decrease the activation energy of both the forward and reverse reaction, therefore, no net effect on equilibrium, only on the rate of reaction. Example Problem 17-8 Example Problem 17-8 Example Problem 17-9 Example Problem 17-9 Example Problem 17-10 Example Problem 17-10

21. The Haber Process A Classic Example of Equilibrium

22. Haber Process N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92.22kJ Is the enthalpy (  H) Favorable? Is the entropy (  S) Favorable? Will this reaction occur spontaneously? Example Problem 17-8

23. Mole % NH 3 in Equilibrium Mixture oCoCoCoC KcKcKcKc10atm100atm 1000at m 253.6e8 near 0 209650518298 4670.5042580 7580.0140.5513

24. The Haber Process Problems: Problems: N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92.22kJ 1. Reaction is very slow at low temperatures, where Kc is highest 2. Increasing the temperature increases the rate, but decreases the yield because heat is produced.

25. The Haber Process Solutions: Solutions: 1. Increase the Temperature (generally 450 o C) 2. Increase the Pressure (generally 200-1000atm) 3. Use excess Reactants 4. Remove the Ammonia as it is produced 5. Use a catalyst (solid Iron Oxide, Potassium Oxide, and Aluminum Oxide)

26. Partial Pressures and the Equilibrium Constant kpkpkpkp

27. Partial Pressures For gas phase equilibria, the equilibrium constant (k eq or k c ) may be expressed in terms of the partial pressure (k p ) For gas phase equilibria, the equilibrium constant (k eq or k c ) may be expressed in terms of the partial pressure (k p )

28. Partial Pressure 2Cl 2 (g) + 2H 2 O (g)  4HCl (g) + O 2 (g)

29. Partial Pressure If Kc or Kp are given and the other is wanted, you can use the relationship between them to calculate (PV = nRT) If Kc or Kp are given and the other is wanted, you can use the relationship between them to calculate (PV = nRT)

30. Scary Math Proof

31. Scary Math Proof Explained Kc = Kp (1/RT) for this reaction Kc = Kp (1/RT) for this reaction Kc R T = Kp Kc R T = Kp  For every reaction:  For every reaction: Kc = Kp(RT) -  n Kc = Kp(RT) -  n Kp = Kc(RT)  n Kp = Kc(RT)  n Example Problem 17-11, 17-12, 17-13, 17-14 Example Problem 17-11, 17-12, 17-13, 17-14

32. Heterogeneous Equilibria

33. Remember: Remember: The activities of pure solids and liquids are 1 The activities of pure solids and liquids are 1 Solvents of very dilute solutions are treated as pure liquids with activity of 1 Solvents of very dilute solutions are treated as pure liquids with activity of 1

34. Heterogeneous Equilibria (Examples) Write Kc and Kp expressions: Write Kc and Kp expressions: 1. CaCO 3 (s)  CaO (s) + CO 2 (g) kc = [CO 2 ] kp = (P CO2 ) 2. H 2 O (l) + CaF 2 (s)  Ca 2+ (aq) + 2F 1- (aq) kc = [Ca2+][F-]2kp = undefined (no gas) 3. 3Fe(s) + 4H 2 O (g)  Fe 3 O 4 (s) + 4H 2 (g) kc = [H 2 ] 4 /[H 2 O] 4 kp = (P H2 ) 4 /(P H2O ) 4

35. Relating Equilibrium to Thermodynamics

36. Relation to Thermo  G rxn =  G o rxn + RTlnQ R = 8.314 J / k mol Can use (+ 2.303RT logQ) At equilibrium,  Grxn = 0 and Q = k  k can be related directly to  Grxn

37. Relation to Thermo 0 =  G o rxn + RTlnk  G o rxn = -RTlnk K is the thermodynamic equilibrium constant K is the thermodynamic equilibrium constant kc where solutions are involved kc where solutions are involved kp where gases are involved kp where gases are involved

38. Finding K at different Temperatures

39. Van’t Hoff Equation Knowing  H o and kc at T1 allows the ESTIMATION of the kc at T2 Knowing  H o and kc at T1 allows the ESTIMATION of the kc at T2 ln(k T1 /k T2 ) =  H o (T 2 -T 1 ) R (T 2 T 1 ) ln(k 2 /k 1 ) = (  H/R) ((1/T 1 ) – (1/T 2 ))