1. Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 16 Chemical Equilibrium INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts & Connections Fifth Edition by Charles H. Corwin

2. Chapter 16 2 Most chemical reactions do not continue until all of the reactants are used up. Most reactions are ongoing, reversible processes, preceding in both the forward direction to give products and in the reverse direction to give the original reactants. We indicate an equilibrium reaction with a double arrow: reactants products → ← Equilibrium Concept

3. Chapter 16 3 In an equilibrium reaction, initially the rate of the forward reaction is very fast. As more products are formed, the rate of the reverse reaction speeds up. When the rates of the forward and reverse reactions are the same, the system is at equilibrium. reactants products forward reaction reverse reaction → ← Equilibrium Concept, continued

4. Chapter 16 4 Molecules must collide in order to react. In a successful collision, existing bonds are broken as new bonds are formed and the reactants are transformed into products. This is the collision theory of reactions. Collision Theory

5. Chapter 16 5 There are three factors that affect the rate of a chemical reaction. 1.Collision Frequency: When we increase the frequency at which molecules collide, we increase the rate of reaction. The more collisions you have, the greater the odds that a collision will be successful. 2.Collision Energy: For a reaction to occur, the molecules must collide with enough energy to form the new bonds. Factors in Successful Collisions

6. Chapter 16 6 3.Collision Geometry: For a reaction to occur, the molecules must be oriented in the proper geometry for the reaction to occur. In (a), the reactants have the correct geometry and products are formed after the collision. In (b), the reactants do not have the correct geometry and they do not react. Factors in Successful Collisions

7. Chapter 16 7 There are three factors that affect the reaction rate: 1.Reactant Concentration: –As we increase the concentration of the reactant(s), the molecules are closer together and collide more frequently. The more collisions, the faster the reaction. 2.Reaction Temperature: –As we increase the temperature, we increase the energy of the reactants. As we increase the energy of the reactants, the rate of the reaction increases because of the increased collision frequency and the collision energy. Factors That Affect Reaction Rate

8. Chapter 16 8 1.Reaction Concentration 2.Reaction Temperature 3.Addition of a Catalyst: –A catalyst increases the rate of a reaction. A catalyst increases the number of effective collisions by creating a more favorable collision geometry. –A catalyst is not consumed in a reaction. Factors That Affect Reaction Rate

9. Chapter 16 9 For a chemical reaction to occur, the reactants must collide with sufficient energy to react. This energy is required to achieve the transition state required to form the products (a). Without sufficient energy, the reaction does not occur (b). Energy Profiles of Reactions

10. Chapter 16 10 An endothermic reaction absorbs heat as the reaction proceeds. N 2 (g) + O 2 (g) + heat 2 NO(g) A reaction profile shows the energy of reactants and products during a reaction. The highest point on a reaction profile is the transition state. → ← Endothermic Reactions

11. Chapter 16 11 The energy required for reactants to achieve the transition state is the activation energy, E act. The energy difference between reactants and products is the heat of reaction,  H. The  H for an endothermic reaction is positive. Reaction Profiles

12. Chapter 16 12 An exothermic reaction releases heat as the reaction proceeds. NO(g) + O 3 (g) 2 NO 2 (g) + O 2 (g) + heat The  H for an exothermic reaction is negative. → ← Exothermic Reactions

13. Chapter 16 13 A catalyst is a substance that allows a reaction to proceed faster by lowering the activation energy. The reaction profile shows the effect of a catalyst on the reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(g) + heat A catalyst does not change  H for a reaction. A catalyst speeds up both the forward and reverse reactions. → ← Effect of a Catalyst

14. Chapter 16 14 A chemical change is a reversible process that can proceed simultaneously in both the forward and reverse directions. When the rate of the forward and reverse reactions are proceeding at the same rate, the reaction is in a state of chemical equilibrium. 3 O 2 (g) 2 O 3 (g) At equilibrium, rate f (O 2 reaction) = rate r (O 3 reaction) → ← Chemical Equilibrium Concept

15. Chapter 16 15 The rate of reaction is the rate at which the concentrations of reactants decrease per unit time. 3 O 2 (g) 2 O 3 (g) Starting with only O 2, as O 2 is consumed, the rate of the forward reaction decreases. As O 3 is produced, the rate of the reverse reaction increases. When the rates are equal, equilibrium is achieved. → ← Rates and Equilibrium

16. Chapter 16 16 The Equilibrium Process As the reaction proceeds, O 3 is formed (b). At equilibrium, the forward and reverse reactions occur at the same rate (c). Later, the amounts of O 2 and O 3 are unchanged (d). Initially, the container has only O 2 gas (a).

17. Chapter 16 17 Chemistry Connection: Ozone Hole The ozone layer is a region of the atmosphere that contains ozone, O 3. Ozone absorbs harmful UV radiation and prevents it from reaching the Earth’s surface. Molecules called chloroflurorcarbons (CFCs) were found to cause the destruction of ozone. Many nations have stopped using CFCs, and the size of the ozone hole is starting to shrink.

18. Chapter 16 18 Consider the following general reaction: a A + b B c C + d D The law of chemical equilibrium states that the molar concentrations of the products (raised to the powers c and d), divided by the molar concentrations of the reactants (raised to the powers a and b), equals a constant. → ← Law of Chemical Equilibrium

19. Chapter 16 19 Mathematically, we express the law of chemical equilibrium as follows: The constant, K eq, is the general equilibrium constant. The value of K eq varies with temperature. So a given value of K eq is valid only for a specific temperature. K eq = [C] c [D] d [A] a [B] b Equilibrium Constant K eq

20. Chapter 16 20 Let’s write the equilibrium constant expression for the reaction 2A B. Recall, K eq is product(s) over reactant(s), each raised to its coefficient in the balanced reaction. Recall that square brackets represent the molar concentration of a species. The equilibrium constant expression is: K eq = [B] [A] 2 → ← Writing Equilibrium Constants

21. Chapter 16 21 A homogeneous equilibrium is a reaction where all of the products and reactants are in the same physical state. What is the equilibrium constant expression for the homogeneous equilibrium: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K eq = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] → ← Homogeneous Equilibria

22. Chapter 16 22 A heterogeneous equilibrium is a reaction where one of the substances is in a different physical state. C(s) + H 2 O(g) CO(g) + H 2 (g) The concentrations of liquids and solids do not change, and they are therefore omitted from equilibrium constant expressions: K eq = [CO][H 2 ] [H 2 O] → ← Heterogeneous Equilibria

23. Chapter 16 23 So when we write equilibrium constant expressions, we only include the concentrations of substances in the gas or aqueous state. What is the equilibrium constant expression for the following reaction? NH 4 NO 3 (s) N 2 O(g) + 2 H 2 O(g) K eq = [N 2 O][H 2 O] 2 → ← Equilibrium Constant Expressions

24. Chapter 16 24 K eq = [HI] 2 [H 2 ][I 2 ] (1.576) 2 (0.212)(0.212) = = 55.3 → ← Experimental Determination of K eq We can calculate the numerical value of K eq if we know the concentrations of all the species in the reaction. H 2 (g) + I 2 (g) 2 HI(g) If the concentrations at equilibrium are [H 2 ] = 0.212 M, [I 2 ] = 0.212 M, and [HI] = 1.576 M, what is K eq ?

25. Chapter 16 25 It doesn’t matter how much of each substance we start with, the value of K eq is always 55.3. Values of K eq

26. Chapter 16 26 Le Chatelier’s principle states that when a reversible reaction at equilibrium is stressed by a change in concentration, temperature, or pressure, the equilibrium shifts to relieve the stress. Let’s look at the equilibrium between colorless N 2 O 4 and brown NO 2 : N 2 O 4 (g) 2 NO 2 (g) If we increase the amount of N 2 O 4, the reaction shifts to the right to produce more NO 2. If we increase the amount of NO 2, the reaction shifts to the left to produce more N 2 O 4. → ← Shifts in Gaseous Equilibria

27. Chapter 16 27 The reaction is endothermic: N 2 O 4 (g) + heat 2 NO 2 (g) If we lower the temperature, the reaction shifts to produce more colorless N 2 O 4. If we raise the temperature, the reaction shifts to produce more brown NO 2. → ← Effect of Temperature

28. Chapter 16 28 In a gaseous equilibrium, increasing the pressure will shift the reaction to the side with fewer gas molecules. In the reaction N 2 O 4 (g) 2 NO 2 (g), increasing the pressure will shift the reaction to the left, producing more N 2 O 4. → ← Effect of Pressure

29. Chapter 16 29 If we add an inert gas to a gaseous reaction at equilibrium, what will happen? The volume of the container does not change, therefore the concentration (and the partial pressure) of the substances do not change. Adding an inert gas has no effect on a system at equilibrium. Effect of an Inert Gas

30. Chapter 16 30 The equilibrium constant for the ionization of a weak acid or base is the ionization equilibrium constant, K i. What is K i constant for the ionization of hydrofluoric acid? HF(aq) H + (aq) + F – (aq) K i = [H + ][F - ] [HF] → ← Ionization Equilibrium Constant, K i

31. Chapter 16 31 We can also write a K i expression for the ionization of ammonium hydroxide: NH 4 OH(aq) NH 4 + (aq) + OH – (aq) K i = [NH 4 + ][OH - ] [NH 4 OH] → ← Ionization of a Weak Base

32. Chapter 16 32 We can calculate the numerical value of K i if we know the concentrations of all the species in the reaction. HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 – (aq) If the concentrations at equilibrium are [HC 2 H 3 O 2 ] = 0.100 M, [H + ] = 0.00134 M, and [C 2 H 3 O 2 – ] = 0.00134 M, what is K i ? K i = [H + ][C 2 H 3 O 2 – ] [HC 2 H 3 O 2 ] (0.00134)(0.00134) (0.100) = = 1.80 × 10 -5 → ← Experimental Determination of K i

33. Chapter 16 33 Let’s look at the following weak acid equilibrium: HF(aq) H + (aq) + F – (aq) If we increase the amount of HF, the reaction shifts to the right to produce more H + and F -. If we raise the pH (by adding base, for example), we decrease [H + ], and the reaction shifts to the right. If we add some soluble NaF, we increase the [F - ], and the reaction shifts to the left. If we add some soluble NaCl, nothing happens. → ← Weak Acid-Base Equilibria Shifts

34. Chapter 16 34 Insoluble salts are really very slightly soluble. If we add Ag 2 SO 4 to water, some of the slightly soluble Ag 2 SO 4 dissolves: Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2- (aq) We can write the solubility product equilibrium constant, K sp, for the reaction: K sp = [Ag + ] 2 [SO 4 2- ] Recall, we don’t include pure solids or liquids in equilibrium constant expressions. → ← Solubility Product Equilibria

35. Chapter 16 35 We can calculate the numerical value of K sp if we know the concentrations of all the species in the reaction. Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) If the concentrations at equilibrium are [Mg 2+ ] = 0.00016 M, and [OH - ] = 0.00033 M, what is K sp ? K sp = [Mg 2+ ][OH - ] 2 = (0.00016)(0.00032) 2 K sp = 1.6 × 10 -11 → ← Experimental Determination of K sp

36. Chapter 16 36 Let’s look at the following solubility equilibrium: AgCl(s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] What happens if we add more AgCl? –Nothing, since AgCl does not appear in K sp. What happens if we add some soluble NaCl? –We increase [Cl - ], and the equilibrium shifts to the left producing more solid AgCl. → ← Solubility Equilibria Shifts

37. Chapter 16 37 According to collision theory we can speed up a reaction in three ways: –increasing the concentration of reactants –raising the temperature of the reaction –adding a catalyst An endothermic reaction absorbs heat energy, and an exothermic reaction releases heat energy. Chapter Summary

38. Chapter 16 38 The amount of energy necessary to achieve the transition state is the activation energy, E act. The difference in the energy of the reactants and products is the heat of reaction,  H. A catalyst speeds up a reaction by lowering the activation energy. A catalyst speeds up both the forward and reverse reactions. Chapter Summary, continued

39. Chapter 16 39 We can write an equilibrium expression for reactions at equilibrium. Chapter Summary, continued

40. Chapter 16 40 According to Le Chatelier’s principle, a reaction at equilibrium shifts in order to relieve a stress. Chapter Summary, continued