2. 1 Chemical Equilibrium: “ Big K” kinetics: rate constant “little k” kinetics “little k” told us how fast a reaction proceeds and is used to indicate a possible mechanism. Eq. tells us to what extent a RXN proceeds to completion. react.  prod. @ Eq: rate forward = rate reverse H 2 CO 3 (l)  H 2 O(l) + CO 2 (g)

3. 2 Law of Mass Action: Values of K c are constant for a RXN at a given temperature. Any equilibrium mixture of the above system at that temperature should give the SAME K c value. The Equilibrium Constant: K (Temperature Dependent) (mechanism independent) aA + bB  dD + eE

4. 3 We also have K p which is sometimes used when dealing with gases with the “P” referring to the pressure of the gases. aA(g) + bB(g)  dD(g) + eE(g) Since PV = nRT (ideal gas law) so pressure is proportional to concentration.  moles of GAS temp in K 0.0821Latm molK

5. 4 Problem: PCl 3 (g) + Cl 2 (g)  PCl 5 (g) In a 5.00 L vessel (@ 230 o C) an equilibrium mixture is found to contain: 0.0185 mol PCl 3, 0.0158 mol PCl 5 and 0.0870 mol of Cl 2. Question: determine K c and K p Solution: K c = 49.08 Question: what does the value of K mean? one mol less gas on product side.

6. 5 Using I.C.E. (Initial, Change, Equilibrium) Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms.

7. 6 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms. H 2 : 0.3000 - 2x = 0.2374 moles CH 3 OH: x = 0.0313 moles Therefore E. 0.1187 0.2374 0.0313 CO(g) + 2H 2 (g)  CH 3 OH(g)

8. 7 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.30000 -x-x - 2x +x 0.1500 - x 0.3000 - 2x x Therefore E. 0.1187 0.2374 0.0313 CO(g) + 2H 2 (g)  CH 3 OH(g) Now find K c : K c = 10.52

9. 8 More I.C.E. Problem: @ 77 o C, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find K c. 2NOBr(g)  2NO(g) + Br 2 (g) I. C. E. 2.00 0 0 +x+2x - 2x 2.00 - 2x2x x Now what? Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094) and.... |x| = 0.0940 so substituting the x value into the “E. term” gives: E. 1.812 0.188 0.0940 2NOBr(g)  2NO(g) + Br 2 (g)

10. 9 More I.C.E. Problem: @ 77 o C, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find K c. 2NOBr(g)  2NO(g) + Br 2 (g) I. C. E. 2.00 0 0 +x+2x - 2x 2.00 - 2x2x x E. 1.812 0.188 0.0940 2NOBr(g)  2NO(g) + Br 2 (g) = 1.01 x 10 -3 What does the value of K c tell us?

11. 10 Treatment of Pure Solids and Liquids (as solvents) in K expressions. S(s) + O 2 (g)  SO 2 (g) would expect K c = [SO 2 ] [S(s)][O 2 ] but since M is meaningless for solids, solids are dropped out. and... Experimentally K c is found to be 4.2 x 10 52 @ 25 o C and independent of S.

12. 11 AgCl(s)  Ag + (aq) + Cl - (aq) K sp = [Ag +(aq) ][Cl - (aq) ] = 1.8 x 10 -10 @ 25 o C This is an EQUILIBRIUM value independent of the amount of solid AgCl left sitting on the bottom of the container.

13. 12 PURE LIQUIDS (SOLVENTS) NH 3 (g) + H 2 O(l)  NH 4 + (aq) + OH - (aq) Note: by convention the water is ignored. HCOOH(aq) + H 2 O(l)  HCOO - (aq) + H 3 O + (aq)

14. 13 CaCO 3 (s)  CaO(s) + CO 2 (g) What is the K c expression? K c = [CO 2 ] What is the K p expression? K p = P CO2

15. 14 Reversing Equations: Reactants become Products and Products become Reactants. What is the relationship between the two K values? 1. 2H 2 (g) + O 2 (g)  2H 2 O(g) 2. 2H 2 O(g)  2H 2 (g) + O 2 (g) K 2 = 1/K 1 = K 1 -1 Relationship:

16. 15 K net for summing RXN’s: If a RXN can be obtained from the sum of RXN’s, K RXN = K 1 K 2 RXN 1: S(s) + O 2 (g)  SO 2 (g) RXN 2:SO 2 (g) + 1/2O 2 (g)  SO 3 (g) Net RXN: S(s) + 3/2O 2 (g)  SO 3 (g) K RXN = K 1 K 2

17. 16 The RXN Quotient: Q c Consider a system that may not yet be @ Equilibrium. aA + bB  dD + eE If Q c = K c ? @ Equilibrium If Q c < K c ? ratio is too small so RXN  If Q c > K c ?ratio is too large so RXN 

18. 17 Example: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) @ 250 o C K c = 4.0 x 10 - 2 If: [Cl 2 ] and [PCl 3 ] = 0.30M and [PCl 5 ] = 3.0M, is the system at Equilibrium? If not, which direction will it proceed? Q c = = 3.0 x 10 - 2 Q c < K c (not @ Eq.) Which way must the RXN go to achieve Equilbrium? Remember ratio is prod./React more products makes the number bigger RXN goes Find Q c and compare to K c to decide.

19. 18 Calculations Using K c : (1st case....Perfect Square) @ 699K H 2 (g) + I 2 (g)  2HI(g) K c = 55.17 Experiment: 1.00 mol of each H 2 and I 2 in a 0.500 L flask. Find [ ] of products and reactants @ Equilibrium. H 2 (g) + I 2 (g)  2HI(g) I. C. E. [ ] [ ] [ ] 2.00 0 -x 2x 2.00 -x 2x “perfect square” = 55.17 conc. in mol/L

20. 19 Calculations Using K c : (1st case....Perfect Square--continued) @ 699K H 2 (g) + I 2 (g)  2HI(g) K c = 55.17 H 2 (g) + I 2 (g)  2HI(g) E. [ ] [ ] [ ] 2.00 -x 2x “perfect square” = 55.17 7.428(2.00 - x) = 2x 1.58 = x

21. 20 Calculations Using K c : (1st case....Perfect Square) @ 699K H 2 (g) + I 2 (g)  2HI(g) K c = 55.17 H 2 (g) + I 2 (g)  2HI(g) E. [ ] [ ] [ ] 2.00 -x 2x 7.428(2.00 - x) = 2x 1.58 = x [H 2 ] = 2.00 - 1.58 = 0.42M [I 2 ] = 0.42M [HI] = 2(1.58) = 3.16M

22. 21 K c Problems with Quadratic Equation ax 2 + bx +c = 0 If equilibrium expression is not a perfect square must use quadratic equation.

23. 22 Problem: H 2 (g) + I 2 (g)  2HI(g) @ 458 o C K c = 49.7 Experiment: 1.00 mol H 2, 2.00 mol I 2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. H 2 (g) + I 2 (g)  2HI(g) I. C. E. [ ] [ ] [ ] 1.00 2.00 0 -x 2x 1.00 - x 2.00 - x 2x 0.920x 2 - 3.00x + 2.00 = 0

24. 23 Problem: H 2 (g) + I 2 (g)  2HI(g) @ 458 o C K c = 49.7 Experiment: 1.00 mol H 2, 2.00 mol I 2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. H 2 (g) + I 2 (g)  2HI(g) I. C. E. [ ] [ ] [ ] 1.00 2.00 0 -x 2x 1.00 - x 2.00 - x 2x 0.920x 2 - 3.00x + 2.00 = 0 2 solutions for x: 1.63  0.70 x = 2.33 or 0.93 will give positive solution for Eq. Conc.