1. Chapter 18 Chemical Equilibrium http://www.learner.org/resources/series61.html?pop =yes&pid=806#http://www.learner.org/resources/series61.html?pop =yes&pid=806# - Molecules in Action Starting at 15:30-23:52 (FYI: Co(H 2 O) 6 +2 + 4Cl - -> CoCl 4 + 6 H 2 O)

2. Reversible Reactions Reversible Reaction –Reaction that can proceed in either direction –↔ Chemical Equilibrium –Rate of forward rxn = rate of reverse direction The Equilibrium Constant (K) –Compares concentrations of products to reactants at equilibrium

3. Reversible Reactions For the reaction: aA + bB  cC + dD K = [C] c [D] d [A] a [B] b Interpretation  K = [products] coefficients [reactants] coefficients

4. Reversible Reactions Example: H 2 + I 2 ↔ 2HI Given data: Calculations of K: Experiment[H 2 ][I 2 ][HI]K 10.0004953 0.003655 20.001141 0.008410 30.0035600.0012500.01559 40.0022520.0023360.01685

5. Reversible Reactions Example: A mixture of N 2, O 2, & NO at equilibrium has [N 2 ] = 6.4x10 -3 M, [O 2 ] = 1.7x10 -3 M, and [NO] = 1.1x10 -5 M. Find K for the reaction: N 2 (g) + O 2 (g) ↔ 2NO (g)

6. Reversible Reactions FYI – SOLIDS AND LIQUIDS HAVE CONCENTRATIONS SO LARGE THEY ARE ESSENTIALLY NOT GOING TO CHANGE (a intensive property), SO YOU CAN almost always IGNORE THEM IN THE K EXPRESSION!!

7. Shifting Equilibrium Remember LeChatlier???? Changes in pressure –Affects systems with GASES involved –Move mixture to a smaller container… Creates increased pressure Rxn will shift to side with FEWER mole of gas to help alleviate that pressure Example –N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) –In a smaller container – will shift RIGHT, K stays the same!!

8. Shifting Equilibrium Changes in concentration –Add more of a substance, shifts to use it up (shifts away from an added substance), K stays the same Changes in temperature –A reversible rxn is ENDO in one direction and EXO in the other –An increase in temp causes a shift so the ENDO rxn occurs more –K does change!!

9. Shifting Equilibrium Common-ion effect –Adding a substance with an ion which is also in the rxn shifts equilibrium –Example CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -1 Adding NaCH 3 COO gives CH 3 COO -1, so causes a shift LEFT

10. Calculations involving shifts in equilibrium Reaction Quotient –Same form as K, but can be used at any point in a rxn, NOT just at equilibrium –Q = [C] c [D] d [A] a [B] b If Q = K, then –The system is at equilibrium If Q > K, then –Need more reactants (less product), so shift LEFT If Q < K, then –Need more product, so shift RIGHT

11. Calculations involving shifts in equilibrium Example: K for the rxn N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) is 2.37x10 -3. At a given point in the rxn, the concentrations are: [N 2 ] = 0.683M, [H 2 ] = 8.80M & [NH 3 ] = 3.65M. Calculate the value of Q and determine the direction of the rxn.

12. Equilibria of Acids, Bases, and Salts For weak acids: An acid losing its hydrogen ion has an equilibrium constant, K a Example: CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -1 K a = [H 3 O + ][CH 3 COO -1 ] [CH 3 COOH] ** H 2 O is not in expression, because it is a liquid so it has a concentration which essentially does not change **

13. Equilibria of Acids, Bases, and Salts Buffers –Has both the acid and its conjugate base in the solution (or base and it conjugate acid). –Example CH 3 COOH & CH 3 COO -1 NH 3 & NH 4 +1

14. Equilibria of Acids, Bases, and Salts Hydrolysis (adding water) –Anion hydrolysis Anions that come from weak acids (F -1, CH 3 COO -1 ) can react with H 2 O to remove a proton F -1 + H 2 O ↔ HF + OH -1 –A BASIC SOLUTION IS FORMED –Cation hydrolysis Cations that come from weak bases (NH 4 +1 ) can react with H 2 O to add a proton NH 4 +1 + H 2 O ↔ H 3 O +1 + NH 3 –AN ACIDIC SOLUTION IS FORMED

15. Solubility Equilibrium Solubility Product = K sp Used for slightly soluble or insoluble substance dissolving Example –AgCl (s) ↔ Ag +1 (aq) + Cl -1 (aq) {AgCl is insoluble, but still a LITTLE will ionize!!} –K = [Ag +1 ][Cl -1 ] AgCl not included because it is a solid. So concentration does NOT change.

16. Precipitation Calculations K BaSO4 = 1.1x10 -10. If [Ba +2 ] = 5.0x10 -3 M & [SO 4 -2 ] = 2.5x10 -3 M, will a precipitate form? BaSO 4 (s) ↔ Ba +2 (aq) + SO 4 -2 (aq)