2. For equilibrium to occur: System must be closed. Temperature must be constant. Reactions must be reversible (do not go to completion). H 2 (g) + Cl 2 (g)  2HCl(g) + energy No visible change…… A dynamic equilibrium exists. The rate of forward rx. = the rate of the reverse rx. Homogeneous Equilibria: all gaseous or aqueous phases.

3. Equilibrium in N 2 O 4 (g) + q 2 NO 2 (g) Initially, [N 2 O 4 ] large. It decreases rapidly then more slowly. The [NO 2 ] starts at zero; increases rapidly. Both eventually, plateau…no change Equilibrium reached. Time concentration Teq Rate f = Rate r N2O4N2O4 NO 2

4. Time Teq What happened here? N 2 O 4 (g) + q 2 NO 2 (g) Here? concentration

5. Predicting Changes Le Chatelier’s Principle: when a stress is applied to a system at equilibrium…. The reaction will “shift” in a direction to minimize the stress. 2H 2 (g) + O 2 (g)  2H 2 O(g) + energy “shift” = rxn speeds up in a particular direction. E.g. shifts right…. Forward rxn speeds up. More products “appear”. We say the reaction is “favored” to the right or the “position of the equilibrium” shifts right. Types of stress: change in concentrations; change in volume; in temperature…

6. 2H 2 (g) + O 2 (g)  2H 2 O(g) + energy [H 2 ] Make more [H 2 O] Temp Use it up![H 2 O] [H 2 ] [O 2 ] Volume Make more pressure [O 2 ] [H 2 O] [H 2 ] [O 2 ] What would a catalyst do to the equilibrium position? Nothing! It would simply be reached sooner. What would happen if we added He gas ? Nothing! It does not affect the partial press of other gases.

7. N 2 O 4 (g) + q 2 NO 2 (g) colorlessorange Describe what happens when your instructor removes the tube from the freezer, containing the system described above. Explain your observation(s) using LeChatelier’s Principle, and all of the appropriate terminology.

8. Quantitative Aspect of Equilibrium Measurements in the N 2 O 4 - NO 2 system @ 100  C If we divide Equilibrium product values by reactant, each factor raised to the correct power (explained in a moment)…. What do you notice????

9. Quantitative Aspect of Equilibrium N 2 O 4(g) + q 2 NO 2(g)

10. Mass Action Expression Given the general equation: aA (aq)* + bB (aq) cC (aq) + dD (aq) *also if g = gas [C] c [D] d [A] a [B] b K eq @ constant Temp. ( K eq can also be K p, K c, K sp, K a, K b, etc.) K eq = Product(s) Reactant(s) 0r Right Left equilibrium constant expression

11. What does the K value tell us? If K is > 1 Prod. Reactant large small At Equilibrium… there is more product than reactant! If K is < 1 Prod. Reactant Large small At Equilibrium… there is more reactant than product! If K is = 1 [Products] = [Reactants] If K is very small…..then practically no product is formed.

12. Eg. Weak Acid: HC 2 H 3 O 2 ( + H 2 O) H + (aq ) + C 2 H 3 O 2 - (aq) K a = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] = 1.8 x 10 -5 What is there more of……reactant or product?Reactant ! HC 2 H 3 O 2 only 3% ionizes. There is almost no product present. This is why it is a weak acid. Strong acids 100% ionize….K a is LARGE. Insoluble salts: PbI 2 (s) ( + H 2 O) Pb +2 (aq ) + 2 I - (aq) K sp = [Pb +2 ][I - ] 2 = 8.4 x 10 -9 What does this value tell us? This solid is very insoluble; mostly solid PbI 2 present.

13. K eq is used to calculate concentrations of species at equilibrium. Given: N 2(g) + O 2(g) = 2NO (g). At 25°C the K c = 1.0 x 10 -30. [N 2 ] =0.040 & [O 2 ] = 0.010. What is the concentration of NO? M.A.E. : K c = [NO] 2 [N 2 ] [O 2 ] = 1.0 x 10 -30 [NO] 2 = [N 2 ] [O 2 ] 1.0 x 10 -30 = (0.040) (0.010) (1.0 x 10 -30 ) = 4.0 x 10 -34 [NO] = 2.0 x 10 -17 mol/L Small K value…….very little product!

14. For the system: CO 2 + H 2 = CO + H 2 O K c = 0.64 @ 900°C. The initial concentrations of reactants are both 0.100mol/L. When the system reaches equilibrium what are the concentrations of reactants and products? Orig. Conc (mol/L) Change in Conc. Equilib. Conc. CO 2 H 2 CO H 2 O 0.100 0.000 -x 0.100-x -x 0.100-x +x x K c = 0.64 = [CO] [H 2 O] [CO 2 ] [H 2 ] = x 2 (0.100-x) 2 x = 0.044; [CO] = [H 2 O] = 0.044mol/L [CO 2 ] = [H 2 ] = 0.100 - 0.044 = 0.056 mol/L

15. Heterogeneous equilibrium Reactions in which one or more of the substances involved is a pure liquid or solid. CO 2 (g) + H 2 (g) CO(g) + H 2 O (l) Experimentally, we find, that the position of the equilibrium is independent of the amount of solid or liquid. Adding or removing a liquid or a solid has no effect on the equilibrium. We do not need to include terms for solids or liquids in the expression for K eq. They are totally ignored. Thus, K eq = [CO] [CO 2 ] [H 2 ]

16. Solubility & K: (read pgs.597-612) Calculating solubility from K sp Given, PbSO 4, K sp = 1.6 x 10 -8. Calculate its solubility, [Pb +2 ] & [SO 4 -2 ]., PbSO 4 (s) = Pb +2 (aq) + SO 4 -2 (aq) K sp = [Pb +2 ] [SO 4 -2 ] = 1.6 x 10 -8 x 2 = 1.6 x 10 -8 (since x = [Pb +2 ] & [SO 4 -2 ]) x = 1.3 x 10 -4. [Pb +2 ] & [SO 4 -2 ] = 1.3 x 10 -4 M

17. Try this problem: Mg(OH) 2 (s) = Mg +2 (aq) + 2OH - 1 (aq) K sp = 1.5 x 10 - 11. Calculate the [Mg +2 ], [OH -1 ], and the [Mg(OH) 2 ] at equilibrium. If X = [Mg +2 ][OH -1 ] =2x So… how do we set up the mass action expression? (x) (2x) 2 =1.5 x 10 -11 x = 1.6 x10 -4 = [Mg +2 ]; [OH -1 ] = 3.2 x10 -4 [Mg(OH) 2 ] = 1.6 x10 -4

18. More Problems: 1. Bromine(I) chloride gas is formed in an endothermic reaction. At 400°C, after the reaction reaches equilibrium, the mixture contained 0.82M BrCl, 0.20M Br 2 (g) and 0.48M Cl 2 (g). a. write the equation for this reaction: 1 Cl 2 (g) + 1Br 2 (g) + q = 2 BrCl (g) b. Write the equilibrium Expression, the MAE [BrCl] 2 [Cl 2 ][Br 2 ] = K eq c. Calculate the value for K eq : [.82] 2 [.20] [.48] = 7.0 d. What direction is favored?

19. 2. Will a precipitate form when the following two solutions are combined? (assume volumes are additive.) 0.0025 M Pb(NO 3 ) 2 + 0.000036 M NaI The K sp for PbI 2 = 8.4 x 10 -9 [Pb +2 ] = 2.5 x 10 -3, [I - ] = 3.6 x 10 - 5 Reaction Quotient: (2.5 x 10 -3 ) (3.6 x 10 - 5 ) 2 = 3.24 x 10 -12 Since the rx. Quotient < the K sp, No ppt. Forms!