1. CHAPTER 13 Chemical Equilibrium

2. Equilibrium Some reactions go to completion.  All the reactants are converted into products. There are many reactions that do not go to completion:  Example: 2 NO 2 (g)  N 2 O 4 (g) Dark BrownColorless  If the reaction went to completion the container would become colorless – it doesn’t

3. EXAMPLE  Should:  2 NO 2 (g)  N 2 O 4 (g) Dark BrownColorless  Actual:  2 NO 2 (g)  N 2 O 4 (g) Light Tan

4. CHEMICAL EQUILIBRIUM Some Reactions go both way. Sometimes very few products are created. 2 CaO (s)  2 Ca (s) +O 2 (g)  Very Little appears to happen. Sometimes a lot of products are created. Cu (s) + 2 AgNO 3 (aq)  2 Ag (s) + Cu(NO 3 ) 2 (aq)  Very Little of the reactant remains.

5. CHEMICAL EQUILIBRIUM CuCl 2 (aq)  Cu +2 (aq) + 2 Cl - (aq) + Heat Green Blue Colorless Observe what happens when the following occurs:  Add Heat  Remove Heat  Add NaCl  Add AgNO 3

6. CHEMICAL EQUILIBRIUM Reaction Graph – Concentration of CuCl 2 nevers reaches zero.

7. EQUILIBRIUM Is not static. It is Highly Dynamic.  At the Macro Level everything appears to have stopped.  At the Micro Level, reactions are continuing. The reaction is travels both directions. As much Reactant is being created as Products. The Reaction Rates in both directions are in Equilibrium. Not the concentrations of Reactants and Products.

8. CHEMICAL EQUILIBRIUM EQUATION jA + kB  mC + nD j is coef for Reactant A k is coef for Reactant B m is coef for Product C n is coef for Product D Keq = [C] m [D] n [A] j [B] k

9. EXAMPLE PROBLEM The following reaction is allowed to go to equilibrium. CuCl 2 (aq)  Cu +2 (aq) + 2 Cl - (aq) + Heat The final concentrations of all the reactants and products CuCl 2 = [0.0250] Cu +2 = [1.25] Cl - = [0.625] What is the Equilibrium constant?

10. SOLUTION Keq = [C] m [D] n [A] j [B] k Keq = [1.25][0.625] 2 = 19.5 [0.0250]

11. What does it mean? If the Keq < 1, then Reactants are favored If the Keq = 1, then Products and Reactants are equal If the Keq > 1 then Products are favored Since the Keq is 19.5 and 19.5 is greater then 1, more Products will be present then reactants.

12. Another Example The Keq for the following reaction is 130, If the concentration of Nitrogen is 0.100 M and the concentration of Hydrogen is 0.200 M, what is the concentration of Ammonia? N 2 +3 H 2  2 NH 3

13. CONTINUED 130 = [ NH 3 ] 2 [N 2 ][H 2 ] 3 130 = [ NH 3 ] 2 [ 0.100 ][3 0.200 ] 3 130 = [ NH 3 ] 2 [ 0.0008 ] 130 * 0.0008 = NH 3 2 = 0.322 M/2 = 0.161 M

14. HETEROGENOUS EQUILIBRIUM Reactants or Products that are solids, and/or water are not included in the expression. H 2 SO 4 (aq) + 2 NaOH (aq)  Na 2 SO 4 (aq) + 2 H 2 O (l) H 2 SO 4 = 0.100 M NaOH = 0.200 M Na 2 SO 4 = 0.150 M What is the Keq?

15. Continued Keq = [0.150] = 37.5 [0.100][0.200] 2 In this reaction, are the products or reactants favored? How do you know?

16. A BIT HARDER 3 NaOH (aq) + H 3 PO 4 (aq)  Na 3 PO 4 (aq) + 3 H 2 O (l) Na 3 PO 4 = 0.200 M The Keq is 130 What is the concentration of H 3 PO 4 and NaOH? We have two unknowns so we must have two equations: First Equation is: Keq = [0.200] = 130 [NaOH] 3 [H 3 PO 4 ] What is the second equation?

17. A BIT HARDER CONTINUED 3 NaOH (aq) + H 3 PO 4 (aq)  Na 3 PO 4 (aq) + 3 H 2 O (l) Let’s replace NaOH with X and H 3 PO 4 with Y Let’s look at the equation now: Keq = [0.200] = 130 [X] 3 [Y] If we look at the coefficents, we can see that: 3 NaOH = 1 H 3 PO 4 We replace NaOH with X and H 3 PO 4 with Y we get: 3 X = Y Let’s look at the equation now: Keq = [0.200] = 130 [X] 3 [3X]

18. A BIT HARDER CONTINUED #2 Keq = [0.200] = [X] 3 [3X] 130 0.00154 = 3X 4 0.000513 = X 4 0.1505 = X = [NaOH] 0.1505 * 3 = [.4515] = [ H 3 PO 4 ]

19. Ksp vs Keq PbCl 2, AgCl, and HgCl 2 are considered insoluble in water. That mean they do not dissolve in water, right? Actually, a very, very, very tiny amount will dissolve in water. We use the Ksp (Constant for Solid Products) to determine the amount that will dissolve.

20. Ksp vs Keq jAB (s)  mA (aq) + nB (aq) Ksp = [A] m [B] n Notice the Ksp equation uses only the concentration of the PRODUCTS, reactants are not included.

21. EXAMPLE Lead (II) Chloride is considered insoluble in water, but experiments show that a very tiny amount will dissolve in water, if the Ksp for PbCl 2 is 3.40 x 10 -15 what is the concentration of Lead and Chlorine ions? Ksp = [A] m [B] n PbCl 2 (s)  Pb 2+ (aq) + 2 Cl - (aq) 3.40 x 10 -15 = [ Pb 2+ ] [Cl - ] 2

22. Ksp Continued Pb 2+ = 2 Cl - X = 2Y 3.40 x 10 -15 = XY 2 = 2Y*Y 2 3.40 x 10 -15 = 2Y 3 2 1.70 x 10 -15 = [Y] 3 1.19 x 10 -5 = [Y] = [Cl - ] 2 (1.19 x 10 -5 ) = [2.38 x 10 -5 ] = [Pb 2+ ]