1. Chapter 13 Chemical Equilibrium

4. REVERSE REACTION  reciprocal K

5. ADD REACTIONS  Multiply Ks

6. ADD REACTIONS  Multiply Ks -8.4

7. LE CHATELIER’S PRINCIPLE

8. CO 2 + H 2  H 2 O(g) + CO a drying agent is added to absorb H 2 O Shift to the right. Continuous removal of a product will force any reaction to the right H 2 (g) + I 2 (g)  2HI(g) Some nitrogen gas is added No change; N2 is not a component of this reaction system.

9. LE CHATELIER’S PRINCIPLE NaCl(s) + H 2 SO 4 (l)  Na 2 SO 4 (s) + HCl(g) reaction is carried out in an open container Because HCl is a gas that can escape from the system, the reaction is forced to the right. This is the basis for the commercial production of hydrochloric acid. H 2 O(l)  H 2 O(g) water evaporates from an open container Continuous removal of water vapor forces the reaction to the right, so equilibrium is never achieved

10. LE CHATELIER’S PRINCIPLE AgCl(s)  Ag + (aq) + Cl – (aq) some NaCl is added to the solution Shift to left due to increase in Cl– concentration. This is known as the common ion effect on solubility. N 2 + 3 H 2  2 NH 3 a catalyst is added to speed up this reaction No change. Catalysts affect only the rate of a reaction; they have no effect at all on the composition of the equilibrium state

11. LE CHATELIER’S PRINCIPLE hemoglobin + O2  oxyhemoglobin Take up in lungs at high O 2 pressure Release in cells at low O 2 concentration Br 2 (g)  2 Br (g) Pressure increased shift to left To reduce number of molecules or atoms

12. REACTION QUOTIENT, Q K is thus the special value that Q has when the reaction is at equilibrium

13. REACTION QUOTIENT, Q K is thus the special value that Q has when the reaction is at equilibrium

14. REACTION QUOTIENT, Q

15. THERMODYNAMICS and equilibrium X

16. 1. The equilibrium constant of an endothermic reactionequilibrium constantendothermic reaction (ΔH° = +) increases if the temperature is raised. 2. The equilibrium constant of an exothermic reactionequilibrium constantexothermic reaction (ΔH° = −) decreases if the temperature is raised. NB: understand this from Le Chatelier’s principle! X

17. HABER-BOSCH: N 2 + 3 H 2  2 NH 3 + E

18. K IS DIMENSIONLESS! Concentrations in mol/liter (M) pressures in atmospheres (atm) ignore solids ignore solvents

19. What will be the concentrations at equilibrium? 0.001 mol Br 2 0.005 mol IO 3 - 0.02 mol Br - 1.00 mol H + Solid I 2 Equilibrium calculation EXAMPLE X

20. First calculate Q to know the direction 0.001 mol Br 2 0.005 mol IO 3 - 0.02 mol Br - 1.00 mol H + Solid I 2 So which way does it go? Equilibrium calculation EXAMPLE

21. Make an ICE table 0.001 mol Br 2 0.005 mol IO 3 - 0.02 mol Br - 1.00 mol H + Solid I 2 Solve for x Equilibrium calculation EXAMPLE

22. SOLUBILITY PRODUCT Ksp Ksp = equilibrium constant of a reaction that forms a precipitate

23. SOLUBILITY PRODUCT Ksp

24. COMMON ION EFFECT C C + S

25. SOLUBILITY PRODUCT Ksp Ksp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 1.0 x 10 -26 = (3x) 3 (2x) 2 = 1.0 x 10 -26

26. SOLUBILITY PRODUCT Ksp COMMON ION EFFECT Ksp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 1.0 x 10 -26 = (3x) 3 (0.10 + 2x) 2 = 1.0 x 10 -26

27. X

28. X

29. Which will form a precipitate first? Higher or lower Ksp? SOLUBILITY PRODUCT Ksp SEPARATION BY PRECIPITATION I¯ X

30. Starting with 0.01 M of each, can you precipitate 99.99% of Hg 2 2+ without losing any Pb 2+ ? SOLUBILITY PRODUCT Ksp SEPARATION BY PRECIPITATION I¯

31. When (BrO 3 — ) is added to a solution containing equal concentrations of Ag + and Pb 2 +, which will precipitate first and why? Ksp = 5.49 x 10 -5 for AgBrO 3 Ksp = 3.23 x 10 -5 for Pb(BrO 3 ) 2 SOLUBILITY PRODUCT Ksp SEPARATION BY PRECIPITATION Stoichiometry! BrO 3 ¯ X

32. Gas – solution eaquilibrium K H Henry’s Law CO 2 dissolves in water: CO 2 (g) + H 2 O H 2 CO 3 (aq) K H = 3.4 x 10 -2 at a CO 2 pressure of 3 x 10 -4 atmospheres, what is the concentration of the carbonic acid in the water? 10 -5 M