1. 1 Chemical Equilibrium Chapter 13 AP CHEMISTRY

2. 2 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.

3. 3 The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

4. 4 The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) A mixture of N 2 O 4 (initially present) and NO 2 (initially formed) appears brown. When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4

5. 5 The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

6. 6 A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

7. 7 Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

8. 8 Notes on Equilibrium Expressions (EE) For a given equation: jA + kB ↔ lC + mD K does not include any pure solids or liquids The expression shows products divided by reactants Like the rate constant, k, the units of K depend on the experiment being performed For the reverse reaction K’ = 1/K (reactants and products switch) Sometimes you will see K written as K c Law of mass action K = [C] l [D] m [A] j [B] k

9. 9 Equilibrium Expression Write the equilibrium expression for: 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g) Complete sample problems #1-4. See sample problems 13.1 and 13.2 in your textbook for more worked examples. Complete sample problems # 1-4.

10. 10 Significance of Equilibrium Expression The inherent tendency for a reaction is occur is indicated by the magnitude of K. A K value much larger than 1 means that at equilibrium the reaction system will consist of mainly product – equilibrium lies to the right A very small K means that the system at equilibrium will consist mainly of reactants – equilibrium position is far to the left The size of K and the time required to reach equilibrium are NOT directly related. Complete sample problem #5.

11. 11 Notes on Equilibrium Expressions (EE) For a reaction multiplied by an integer, “n”, K new = (K orig ) n See sample exercise 13.2(c) on page 614. For a given reaction, K is dependent only on temperature K = [C] l [D] m [A] j [B] k

12. 12 Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

13. 13 Practice Problem: Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq)  FeSCN 2+ (aq) In trial #1, you start with 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq), and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value of the equilibrium constant for this reaction? Fe 3+ (aq) + SCN - (aq)  FeSCN 2+ (aq) Initial Change Equilibrium

14. 14 Practice Problem Fe 3+ (aq) + SCN - (aq)  FeSCN 2+ (aq) Equilibrium 2.00 6.00 4.00 Complete sample problems 6 & 7 for more practice using ICE charts.

15. 15 Equilibrium Constant in Terms of Pressure Equilibria involving gases can be described in terms of either pressure or concentrations. The relationship can be seen using the Ideal Gas Law, PV=nRT. Rearranging this equation gives n/V represents concentration in M. Complete sample problem # 8 now.

16. 16 K vs. K p For any reaction: K p = K(RT)  n  n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants. Complete Sample Problem #9.

17. 17 Reaction Quotient H 2 (g) + F 2 (g)  2HF(g)  After the equilibrium constant (K) is known, we can use it to determine if a reaction is at equilibrium.  The reaction quotient, Q, has the same form as the equilibrium constant expression EXCEPT initial concentrations are used instead of equilibrium concentrations.

18. 18 Predicting the Direction of a Reaction Using Reaction Quotient  If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K).  If Q < K then the forward reaction must occur to reach equilibrium.  If Q = K then the reaction is at equilibrium.

19. 19 Practice Problem Part 2: Using the previous reaction: Fe 3+ (aq) + SCN - (aq)  FeSCN 2+ (aq) and the K value we determined: K = 0.33 determine if the following concentrations are at equilibrium: Initial:10.0 M Fe 3+ (aq), 8.00 M SCN - (aq), and 2.00 M FeSCN 2- Complete sample problems 10 & 11.

20. 20 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.

21. 21 Solving Equilibrium Problems (continued ) 5.Define the change needed to reach equilibrium. 6.Substitute equilibrium concentrations into equilibrium expression and solve. 7.Check calculated concentrations by calculating K. Complete more complex sample problems #12-15.

22. 22 Le Châtelier’s Principle... if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

23. 23 Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance. Le Châtelier’s Principle

24. 24 Le Châtelier’s Principle

25. 25 Effects of Changes on the System 1.Concentration: The system will shift away from the added component. 2.Temperature: K will change depending upon the temperature (treat the energy change as a reactant).

26. 26 Increase of Pressure to an Equilibrium.

27. 27 Effects of Changes on the System (continued) 3.Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

28. 28 Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases. Le Châtelier’s Principle

29. 29 Change in Reactant or Product Concentrations Le Châtelier’s Principle

30. 30 The Haber Process Le Châtelier’s Principle

31. 31 N 2 and H 2 are pumped into a chamber. The pre-heated gases are passed through a heating coil to the catalyst bed. The catalyst bed is kept at 460 - 550  C under high pressure. The product gas stream (containing N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies but not N 2 or H 2. The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized The Haber Process for producing NH 3 Le Châtelier’s Principle