1. Introductory Chemistry: Concepts & Connections Introductory Chemistry: Concepts & Connections 4 th Edition by Charles H. Corwin Chemical Equilibrium Christopher G. Hamaker, Illinois State University, Normal IL © 2005, Prentice Hall Chapter 16

2. 2 Equilibrium Concept 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

3. Chapter 163 Equilibrium Concept Continued 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 speed up. When the rates of the forward and reverse reactions are the same, the system is at equilibrium. reactants ⇆ products forward reaction reverse reaction

4. Chapter 164 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 165 Factors in Successful Collisions 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.

6. Chapter 166 Factors in Successful Collisions 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.

7. Chapter 167 Factors that Effect Reaction Rates There are three factors that effect 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.

8. Chapter 168 Factors that Effect Reaction Rates 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.

9. Chapter 169 Energy Barriers in Reactions 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).

10. Chapter 1610 Endothermic Reactions 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.

11. Chapter 1611 Reaction Profiles 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.

12. Chapter 1612 Exothermic Reactions 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.

13. Chapter 1613 Effect of a Catalyst 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 + heat A catalyst does not change  H for a reaction. A catalyst speeds up both the forward and reverse reactions.

14. Chapter 1614 Chemical Equilibrium Concept 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)

15. Chapter 1615 Rates and Equilibrium 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.

16. Chapter 1616 Law of Chemical Equilibrium 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.

17. Chapter 1617 Equilibrium Constant, K eq 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

18. Chapter 1618 Writing Equilibrium Constants Lets 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

19. Chapter 1619 Homogeneous Equilibria A homogeneous equilibrium is a reaction where all of the products and reactants are in the physical same 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 ]

20. Chapter 1620 Heterogeneous Equilibria 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]

21. Chapter 1621 Equilibrium Constant Expressions 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

22. Chapter 1622 Experimental Determination of K eq We can calculate the numerical value of K eq if we know the concentrations of all of 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 ? K eq = [HI] 2 [H 2 ][I 2 ] (1.576) 2 (0.212)(0.212) = = 55.3

23. Chapter 1623 Values of K eq It doesn’t matter how much of each substance we start with, the value of K eq is always 55.3.

24. Chapter 1624 Shifts in Gaseous Equilibria Le Chatleir’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. Lets 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.

25. Chapter 1625 Effect of Temperature 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 N 2 O 4. If we raise the temperature, the reaction shifts to produce more NO 2.

26. Chapter 1626 Effect of Pressure 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.

27. Chapter 1627 Effect of an Inert Gas 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.

28. Chapter 1628 Ionization Equilibrium Constant 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]

29. Chapter 1629 Ionization of a Weak Base 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]

30. Chapter 1630 Experimental Determination of K i We can calculate the numerical value of K i if we know the concentrations of all of 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

31. Chapter 1631 Weak Acid-Base Equilibria Shifts Lets 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.

32. Chapter 1632 Solubility Product Equilibria 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.

33. Chapter 1633 Experimental Determination of K sp We can calculate the numerical value of K sp if we know the concentrations of all of 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

34. Chapter 1634 Solubility Equilibria Shifts Lets 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.

35. Chapter 1635 Conclusions 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 release heat energy.

36. Chapter 1636 Conclusions Continued 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.

37. Chapter 1637 Conclusions Continued We can write an equilibrium expression for reactions at equilibrium.

38. Chapter 1638 Conclusions Continued According to Le Chatlier’s principle, a reaction at equilibrium shifts in order to relieve a stress.