Larry Emme Chemeketa Community College

1. Chemical Equilibrium Chapter 16 Larry Emme Chemeketa Community College

2. Reversible Reactions

3. reversible reaction - a chemical reaction in which the products formed react to produce the original reactants.

4. cooling 2NO2(g)→ N2O4 (g) heating N2O4(g)→ 2NO2 (g) The reaction between NO2 andN2O4 is reversible. N2O4 is formed N2O4 decomposes when heated forming NO2

5. → 2NO2(g)N2O4 (g) → reaction to the right (forward) reaction to the left (reverse) Ice water Hot water

6. Rates of Reaction

7. The rate of a reaction is variable. It depends on: • concentrations of the reacting species • reaction temperature • presence or absence of catalysts • the nature of the reactants

8. Chemical Equilibrium

9. equilibrium A dynamic state in which two or more opposing processes are taking place at the same time and at the same rate. chemical equilibrium The state in which the rate of the forward reaction equals the rate of the reverse reaction in a chemical change. At equilibrium the concentrations of the products and the reactants are not changing.

10. NaCl(s) Na+(aq) + Cl-(aq) → → A saturated salt solution is in equilibrium with solid salt. salt crystalsare dissolving Na+ and Cl-are crystallizing At equilibrium the rate of salt dissolution equals the rate of salt crystallization.

11. Le Chatelier’s Principle

12. Henri LeChatelier

13. This generalization, known as LeChatelier’s Principle, states In 1888, the French chemist Henri LeChatelier set forth a far-reaching generalization on the behavior of equilibrium systems. If a stress or strain is applied to a system in equilibrium, the system will respond in such a way as to relieve that stress and restore equilibrium under a new set of conditions.

14. Effect of Concentration on Equilibrium

15. For most reactions the rate of reaction increases as reactant concentrations increase. • The manner in which the rate of reaction changes with concentration must be determined experimentally.

16. An equilibrium is disturbed when the concentration of one or more of its components is changed. As a result, the concentration of all species will change and a new equilibrium mixture will be established.

17. → → A + BC + D The system is at equilibrium results in C and D being produced faster than they are used. results in A and B being used faster than they are produced. increases the rate of the forward reaction Increasing the concentration of B

18. → → A + BC + D The system is again at equilibrium In the new equilibrium concentration of A has decreased concentrations of B, C and D have increased After enough time has passed, the rates of the forward and reverse reactions become equal.

19. Equilibrium shifts to left decrease Cl2 concentration decrease H3O+ concentration increase H2O concentration increase Cl- concentration increase HOCl concentration Equilibrium shifts to left Cl2(aq) +2H2O(l) HOCl(aq) + H3O+(aq) + Cl-(aq) → Equilibrium shifts to left Equilibrium shifts to right Equilibrium shifts to right → Effect of Concentration Changeson the Chlorine Water Equilibrium

20. Equilibrium shifts to left NaC2H3O2(aq) → Na+(aq) + C2H3O2(aq) HC2H3O2(aq) +H2O(l) H3O+(aq) + C2H3O2(aq) 1 L 0.100 M HC2H3O2 1 L 0.100 M HC2H3O2 1 L 0.100 M HC2H3O2 → → Equilibrium pH = 2.87 Equilibrium pH = 4.74 Equilibrium pH = 5.05 Effect of C2H3O2Concentration Changes on pH Add 0.100 mol NaC2H3O2 Add 0.200 mol NaC2H3O2

21. Effect of Pressure on Equilibrium

22. Changes in pressure significantly affect the reaction rate only when one or more of the reactants or products is a gas and the reaction is run in a closed container. • The effect of increasing the pressure is to increase the concentrations of any gaseous reactants or products.

23. Equilibrium shifts to left → → Increase Pressure increases CO2 concentration CaCO3(s) CaO(s) + CO2(g)

24. → Equilibrium shifts to right → Decrease Pressure decreases CO2 concentration CaCO3(s) CaO(s) + CO2(g)

25. In a system composed entirely of gases, a increase in the pressure of the container will cause the reaction and the equilibrium to shift to the side that contains the smallest number of molecules.

26. N2(g) + 3H2(g) 2NH3(g) 1 mol 3 mol 2 mol 6.02 x 1023 molecules 1.81 x 1024 molecules 1.20 x 1024 molecules 2.41 x 1024 molecules → → Equilibrium shifts to the right towards fewer molecules. Increase Pressure

27. N2(g) + O2(g) 2NO(g) 1 mol 1 mol 2 mol 6.02 x 1023 molecules 6.02 x 1023 molecules 1.20 x 1024 molecules 1.20 x 1024 molecules → → Increase Pressure Equilibrium does not shift. The number of molecules is the same on both sides of the equation.

28. Effect of Temperature on Equilibrium

29. When the temperature of a system is raised, the rate of reaction increases. The rate of the reaction that absorbs heat is increased to a greater extent, and the equilibrium shifts to favor that reaction. When the process is endothermic, the forward (left to right) reaction is increased. When the process is exothermic, the reverse (right to left) process is increased. In a reversible reaction, the rates of both the forward and the reverse reactions are increased by an increase in temperature.

30. At 1000oC moles CO2moles CO C(s) + CO2(g) + heat 2CO(g) → Equilibrium shifts to right → Heat may be treated as a reactant in endothermic reactions. At room temperature very little CO forms.

31. Effect of Catalystson Equilibrium

32. A catalyst is a substance that influences the rate of a reaction and can be recovered essentially unchanged at the end of the reaction. A catalyst does not shift the equilibrium of a reaction. It affects only the speed at which the equilibrium is reached.

33. Energy Diagram for an Exothermic Reaction Activation energy: the minimum energy required for a reaction to occur. A catalyst speeds up a reaction by lowering the activation energy. A catalyst does not change the energy of a reaction.

34. 2KClO3(s) → 2KCl + 3O2(l) The laboratory preparation of oxygen uses manganese dioxide as a catalyst to increase the rate of the reaction. AlCl3 PCl3(l)+ S(s) → PSCl3(l) Very little thiophosphoryl chloride is formed in the absence of a catalyst because the reaction is so slow. In the presence of a catalyst the reaction is complete in a few seconds. MnO2 Δ

35. Equilibrium Constants

36. At equilibrium the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products are constant.

37. The equilibrium constant (Keq) is a value representing the unchanging concentrations of the reactants and the products in a chemical reaction at equilibrium.

38. aA + bB cC + dD → → For the general reaction at a given temperature

39. 3H2 + N2 2NH3 → → For the reaction

40. 4NH3 + 3O2 2N2+ 6H2O → → For the reaction

41. The magnitude of an equilibrium constant indicates the extent to which the forward and reverse reactions take place. H2 + I2 2HI At equilibrium more product than reactant exists. At equilibrium more reactant than product exists. COCl2 CO+ Cl2 → → → →

42. When the molar concentrations of all species in an equilibrium reaction are known, the Keq can be calculated by substituting the concentrations into the equilibrium constant expression.

43. Calculate the Keq for the following reaction on concentrations of PCl5 = 0.030 mol/L, PCl3 = 0.97 mol/L and Cl2 = 0.97 mol/L at 300oC. PCl5(g )PCI3(g) + Cl2(g) → →

44. Ionization Constants

45. In addition to Kw, several other ionization constants are used.

46. Ka

47. HC2H3O2(aq) H+ + C2H3O2 → → When acetic acid ionizes in water, the following equilibrium is established: Ka is the ionization constant for this equilibrium. Ka is called the acid ionization constant. Since the concentration of water is large and does not change appreciably, it is omitted from Ka.

48. At 25oC, a 0.100 M solution of HC2H3O2 is 1.34% ionized and has an [H+] of 1.34 x 10-3 mol/L. Calculate Ka for acetic acid. HC2H3O2(aq) H+ + C2H3O2 Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal. → → The moles of unionized acetic acid per liter are 0.100 mol/L – 0.00134 mol/L = 0.099 mol/L

49. Substitute these concentrations into the equilibrium expression and solve for Ka. [HC2H3O2] = 0.099 mol/L

50. What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5. HC2H3O2(aq) H+ + C2H3O2 Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal. → → The equilibrium expression and Ka for HC2H3O2 are: