Chemical Equilibrium

1. Chemical Equilibrium • Chapter 13

2. Not all reactions go to completion!

3. Chemical Equilibrium • The state where the concentrations of all reactants and products remain constant with time. • On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

4. Equilibrium & Kinetics • At equilibrium, Rf = Rr • but kf is not necessarily equal to kr. • kf[Reactant] = kr[Product] • If equilibrium lies far to the right, [Products] will be large and kr will be small while[Reactants] will be small and kf will be large.

5. Reactions That Appear to Run to Completion • 1. Formation of a precipitate. • 2. Formation of a gas. • 3. Formation of a molecular substance such as water. • These reactions appear to run to completion, but actually the equilibrium lies very far to the right (favors products). All reactions in closed vessels reach equilibrium.

6. Molecular representation of the reaction 2NO2(g) ----> N2O4(g). c) & d) represent equilibrium.

7. Change in time and rates • As the concentrations of reactants decrease, the forward reaction slows down and the reverse reaction speeds up UNTIL the system reaches equilibrium

8. The equilibrium position of a rxn- left, right, or in the middle depends on many factors: • Initial Concentrations • Relative energies of the reactants and products • Organization of the reactants and products • We will focus on #2&3 more in Chapter 16

9. Concentration profile for the Haber Process which begins with only H2(g) & N2(g).

10. Haber Process • When gaseous nitrogen, hydrogen, and ammonia are mixed in a closed vessel at 25 C, concentrations don’t noticeably change over time • Why? • System is at equilbrium • Forward and reverse rxn rates are SLOW • N2 has a very strong triple bond, and unreactive • H2 is a relatively strong single bond • Under appropriate conditions, the system does reach equilibrium • H2 disappears 3x as fast as N2 • NH3 forms 2x as fast as N2 disappears

11. Shifting Gaseous Equilibrium • If the size of a container is changed, the concentration of the gases change. • A smaller container shifts the equilibrium to the right -- N2(g) + 3H2(g) ---> 2NH3(g). Four gaseous molecules produce two gaseous molecules. • A larger container shifts to the left -- two gaseous molecules produce four gaseous molecules.

12. The Law of Mass Action • For • jA + kB lC + mD • The law of mass action (Cato Guldberg & Peter Waage) is represented by the equilibrium expression: Equilibrium Constant

13. Equilibrium Expression • 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)

14. Notes on Equilibrium Expressions (EE) • The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse. • When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n • The units for K depend on the reaction being considered.

15. Equilibrium Expressions N2(g) + 3H2(g) ---> 2NH3(g) 2NH3(g) ---> N2(g) + 3H2(g) K´ = 1/K

16. Equilibrium Expressions • 1/2N2(g) + 3/2H2(g) ---> NH3(g) • K´ ´ = K1/2

17. Example: • Calculate the values of K at 127°C for the reaction shown below if the [NH3]=3.1 x10-2 M, [N2]=8.5 x 10-1, [H2]=3.1 x 10-3 • N2 + 3H2  2NH3 • Work out for solution • K=3.8 x 104 (NO UNITS)

18. Example: • Calculate the value of the equilibrium constant for the reverse reation • Work out for solution • K=2.6 x 10-5(NO UNITS)

19. Example: • Calculate the values of K at 127°C for the reaction shown below if the [NH3]=3.1 x10-2 M, [N2]=8.5 x 10-1, [H2]=3.1 x 10-3 • (1/2)N2 + (3/2)H2  1NH3 • Work out for solution • K=1.9 x 102(NO UNITS)

20. What does K value mean?

21. K Values

22. Equilibrium Position • For a given reaction at a given temperature, there is only one equilibrium constant (K), (regardless of concentrations) but there are an infinite number of equilibrium positions. • Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.

23. Three equilibrium positions but only one equilibrium constant (K).

24. Kp • K in terms of equilibrium partial pressures: • Pressure and concentration are directly proportional when the temperature is held constant.

25. K versus Kp • For jA + kBlC + mD Kp = K(RT)n • R= 0.08206 L·atm/mol·K • n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants. • n = (l+m) – (j+k)

26. K versus Kp • When n = 0 and Kp = K(RT)o, then • Kp = K • K and Kp are equal numerically but do not have the same units. • Under what situations would n = 0 ?

27. Example N2(g) + 3H2(g) 2NH3(g) Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

28. Homogeneous Equilibria • Homogeneous equilibria – involve the same phase: N2(g) + 3H2(g) 2NH3(g) HCN(aq) H+(aq) + CN-(aq)

29. Heterogeneous Equilibria • . . . are equilibria that involve more than one phase. • CaCO3(s)  CaO(s) + CO2(g) • K = [CO2] • The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. • DON’T INCLUDE SOLIDS OR LIQUIDS IN EQUILIBRIUM EXPRESSION!

30. Why? • Concentrations of pure liquids and solids cannot change!

31. The position of the equilibrium CaCO3(s) ---> CaO(s) + CO2(g) does not depend upon the amounts of solid CaCO3 or CaO.

32. Magnitude of K • A K value much larger than 1 means that the equilibrium system contains mostly products -- equilibrium lies far to the right. • A very small K value means the system contains mostly reactants -- equilibrium lies far to the left. • The size of K and the time required to reach equilibrium are not directly related!

33. A physical analogy illustrating the difference between thermodynamic and chemical stability. The magnitude of K depends on E, but reaction rate depends on Ea.

34. Reaction Quotient, Q • . . . helps to determine the direction of the move toward equilibrium. • The law of mass action is applied with initial concentrations.

35. Reaction Quotient, Q (continued) • H2(g) + F2(g)  2HF(g)

36. Q versus K • Q = K System is at equilibrium. • Q > K System will shift to the left. • Q < K System will shift to the right. • See Sample Exercise 13.7 on page 625.

37. Reaction Quotient, Q • Q = K; The system is at equilibrium. No shift will occur. • Q > K; The system shifts to the left. • Consuming products and forming reactants, until equilibrium is achieved. • Q < K; The system shifts to the right. • Consuming reactants and forming products, to attain equilibrium.

38. Solving Equilibrium Problems • 1. Balance the equation. • 2. Write the equilibrium expression. • 3. List the initial concentrations. • 4. Calculate Q and determine the shift to equilibrium.

39. Solving Equilibrium Problems(continued) • 5. Define equilibrium concentrations. • 6. Substitute equilibrium concentrations into equilibrium expression and solve. • 7. Check calculated concentrations by calculating K.

40. To solve these equations…. • We set up something called RICE or ICE tables • R=reaction • I=initial concentrations • C=changed concentration • E=Equilibrium

41. Example Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) • Trial #1: 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction?

42. Set up RICE Table Fe3+(aq) + SCN–(aq) FeSCN2+(aq) Initial 6.00 10.00 0.00 Change – 4.00 – 4.00 +4.00 Equilibrium 2.00 6.00 4.00 K = 0.333

43. Example #2 • Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At a temperature of 700 K, the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00 L flask. • CO(g) + H2O (g)  CO2 (g) + H2(g)

44. Step 1: Find Q. • Concentration of everything is 1 M Q=[CO2]o[H2]o [CO]o[H2O]o =[1][1] [1][1] = 1 • Step 2: Compare this to K. • Q < K, so equilibrium shifts right • If equilibrium shifts right, then [reactant] ↓, [product] ↑

45. Step 3: Set up RICE TABLE Reaction: CO(g) + H2O(g)CO2(g) + H2(g) Initial 1.0 1.0 1.0 1.0 Change – x – x +x +x Equilib. 1.0-x 1.0-x 1.0+x 1.0+x

46. Step 4: Plug into K expression K= 5.10= [CO2][H2] = [1.0 +x][1.0+x] [CO][H2O] [1.0 -x][1.0-x] 5.10= (1.0 + x)2 (1.0 – x)2 √5.10= (1.0 + x) / (1.0 –x) 2.26=(1.0 + x) / (1.0 –x) Rearrange eqn: 0.387 M= x

47. Step 5: Plug #s in • At equilibrium: • [CO]=1-x=1-0.387=.0613 M • [H2O]= 1-x=1-0.387=.0613 M • [CO2]=1+x=1+0.387=1.387 M • [H2]= 1+x=1+0.387=1.387 M

48. Step 6: CHECK • Plug into original K expression to see if you get correct K value • Original was 5.10 • K=[1.387][1.387] =5.12 [0.387][0.387]

49. Example #3 1.000 Q < K  equilibrium shifts right. Assume for the following reaction, K=1.15x102. In a particular experiment, 3.00 mol of each component was added to a 1.5 L flask. Calculate the equilibrium concentrations of all species. H2(g) + F2(g) <---> 2HF(g) [H2]o = [F2]o = [HF]o = 3.000 mol/1.500 L = 2.000M

50. RICE • H2(g) + F2(g) <---> 2HF(g) [H2] [F2] [HF] Initial (mol/L) 2.000 2.000 2.000 Change (mol/L) - x - x + 2x Equil. (mol/L) 2.000 - x 2.000 - x 2.000 + 2x