2. Chap 14 Equilibrium Calendar 2013 M 4/8 Film B-1 4/9-10 14.1 Equil 14.2 k expression B-2 4/11-12 14.3 LeChat M 4/15 Ksp B-1 4/16-17 Lab ksp B-2 4/18-19 Review M 4/22 Test

3. Equilibrium Calendar 2014-15 M 3/30 Rev test/14.1 Equilib Syst 3/31-4/12 break M 4/13 14.1 Rev/Eq constant Keq B-1 4/14-5 14.2 Using Eq Const B-2 4/16-7 Quiz Keq 14.2 Ksp M B-1 4/20-1 Lab Ksp W = a 4/22 Quiz /rev Lab/14.3 LeChat B-2 4/23-4 Review LeChat/ M 4/27 Rev Chapter B-1 4/28-9 Test

4. Chemical Equilibrium A + B C + D Complete reactions Go to completion in one direction All reactants are converted to products Reaction proceeds in one direction only Many involve to formation of gas or ppt Many rx don’t go to completion Called reversible

5. Chemical Equilibrium A + B C + D Reversible reactions As C and D form and increase in conc, they collide and reforming A and B Opposite reactions occur at the same rate Forward reaction and reverse reaction Amounts of reactants and products are constant Can be more reactant or product

6. Equilibrium Conc of R and P

7. N 2(g) + 3H 2(g)  2NH 3(g)

8. Examples Solute and solvent in a saturated solution 2 phases at phase change temp Ice and water Liquid and vapor Acid base indicators

9. Equilibrium Constant At equilib the concentration [M] of reactants and products are constant (NOT equal) The ratio of conc of products to reactants, each raised to its coefficient as an exponent is a constant aA bB Keq = [B] b [A] a Called equilibrium constant expression Solids and liquids do not show up in the expression

10. Equilibrium Constant At equilib the concentration [M] of reactants and products are constant (NOT equal) The ratio of conc of products to reactants, each raised to its coefficient as an exponent is a constant aA + bB cC + dD [ ] means Molarity Keq = [C] c [D] d [A] a [B] b Called equilibrium constant expression Solids and liquids do not show up in the expression

11. Write the Keq Expression

12. Equilibrium Constant Exp Write the the equilibrium constant expression for: 2NO 2 (aq) N 2 (aq) + 2O 2 (aq) ZnO (aq) + CO (aq) Zn (s) + CO 2 (aq) Ni (s) + 4CO (aq) Ni(CO) 4 (aq) 4H + (aq) + 2Cl - (aq) + MnO 2 (aq) Mn 2+ (aq) + 2H 2 O (l) + Cl 2 (aq)

13. Size of Keq = At equilibrium: aA + bB cC + dD If k is about 1 there are about equal amounts of reactants and products If k is greater 1 there are more products than reactants at equilib The higher value K = more products less reactants If k is less than 1 there are more reactants than products at equilib The lower the value K = more reactants less products [C] c [D] d [A] a [B] b

14. Value of k Values of k can only be determined by experiment – run the reaction in the lab 2 types of math problems involve k 1 - given [eq] of all r and p calc k 2 - given k and [eq] of all r and p but one, solve for its [eq]

15. H 2 CO 3(aq) + H 2 O (l) H 3 0 + (aq) + HC0 3 - (aq) Determine the value of K for the reaction above if H 2 CO 3(aq) = [3.3 x 10 -2 ] H 3 0 + = [1.2 x 10 -4 ] HC0 3 - = [1.2 x 10 -4 ] 4.36 x 10 -7 Determination of K

16. Calc k if an equilib mixture has [1.2 x 10 -3 ] HCl [3.8 x 10 -4 ] O 2 [5.8x 10 -2 ] H 2 O [5.8 x 10 -2 ] Cl 2 4HCl (aq) + O 2 (aq) 2H 2 O (aq) + 2Cl 2 (aq) 1.4 x 10 10

17. H 2 and I 2 gas react to form HI. The equation for this reaction is H 2 (g) + I 2 (g) ↔ 2HI(g) If the equilibrium concentrations of gases are [H 2 ] = 0.81 M; [I 2 ] = 0.44 M; [HI] = 0.58 M, calc the value of k 0.94 M

18. Write k expression and Calc k for the eq system: 2 NO(g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O(g ) [NO] = 0.100 M[H 2 ] = 0.100 M [N 2 ] = 0.0500 M[H 2 O] = 0.100 M

19. Using K For the equilibrium: 2IBr (aq)  I 2 (aq) + Br 2 (aq) K is 4.13 x 10 -2 The equilibrium conc of is IBr [0.0124]. Calc the conc of I 2 and Br 2 at equilibrium? 0.0025

20. Using k H 2 and I 2 gas react together to form HI gas. The equation for this reaction is H 2 (g) + I 2 (g) ↔ 2HI(g) The equilib constant for this reaction is 7.1 x 10 2 at 25 °C. If the final concentrations of gases are [H 2 ] = 0.42 M; [I 2 ] = 0.20 M; calculate the [HI] at equilibrium. 7.72 M

21. Sample Prob Carbonic acid is a weak acid found in carbonated beverages. Its value for keq at 25°C is 4.3 x 10 -7. Carbonic acid is added to water and creates an eq system: H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) Calculate the concentrations of [H + ] and [HCO 3 - ], if the equilibrium concentration of [H 2 CO 3 ] = 0.027 M. 1.07 x 10 -4 ??

22. Determine k Sample problems 14a, and 14b pg 526 Do practice problems 1, 2, and 3 on pg 528

23. Solubility product constant Ksp

24.  Learn how to write the solubility constant expression K sp  Learn how to determine solubility from Ksp  Learn how to determine Ksp from solubility Three parts to this section:

25. The Solubility Product Constant, K sp Many ionic compounds are only slightly soluble in water. equations are written to represent the equilibrium between the solid compound and the ions present in a saturated aqueous solution. The solubility product constant, K sp, is the product of the conc of the ions produced, each raised to their coefficient as an exponent

26. Solubility Product Constant Just like any other k value = [P]/[R] Written for the equilibrium of the solution of slightly soluble salts A a B b(s) aA + (aq) + bB - (aq) ksp = [A + ] a [B - ] b CaS (s) Ca 2+ (aq ) + S 2- (aq) ksp = [Ca 2+ ] [S 2- ] Ag 2 CO 3(s) 2Ag 1+ (aq) + CO 3 2- (aq) ksp = [Ag +1 ] 2 [CO 3 2- ] Solid and liquids do not show up in the ksp expression-like any other equilibrium equation

27. Write ksp expression for each of the following: AgOH AgOH (s)  Ag 1+ (aq) + OH 1- (aq) Cu(OH) 2 Cu(OH) 2(s)  Cu 2+ (aq) + 2OH 1- (aq) PbI 2 PbI 2(s)  Pb 2+ (aq) + 2I 1- (aq) MgCO 3 MgCO 3(s)  Mg 2+ (aq) + CO 3 2- (aq) Ca 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2(s)  3Ca 2+ (aq) + 2PO 4 3- (aq)

28. Solubility Product Constant = k sp Ksp shows the solubility of the salt High ksp = more soluble Low ksp = less soluble Types of problems: Calc ksp given molar concentrations of all salts What is the conc of one ion if know ksp and the conc of the other ion? Given ksp what is the solubility (mol/L) of a slightly soluble salt? Table 3 pg 508

29. Values For K sp At 25 o C Large Ksp = more soluble Small Ksp = less soluble

30. Relationship Between Ksp and Solubility Based on number of moles of cations and anions that dissolved (molar solubility) s = M the molarity in mol/L 1:1 Cation : anion ratio (NaCl, KF, CaSO 4, NH 4 Cl) Ksp = s 2 (s = solubility which is Molarity!) 2:1 or 1:2 ratio Cation : anion ratio (CaF 2, KS 2, Ba(NO 3 ) 2, (NH 4 ) 2 CO 3 Ksp = 4s 3 (s = solubility which is Molarity!)

31. The Solubility Equilibrium Equation and K sp For every mol NaF that breaks apart 1 Na + and 1 F - are formedNaF (s) Na + (aq) + F - (aq) s s K sp = [Na + ][F - ] or K sp = [s][s] or K sp = s 2 For every mol Ag 2 CO 3 that breaks apart 2 mol Ag + and 1 mol CO 3 2- are formed Ag 2 CO 3(s) 2Ag + (aq) + CO 3 2- (aq) 2s s K sp = [Ag + ] 2 [CO 3 2- ] or K sp = [2s] 2 [s] or K sp = 4s 3

32. Calculate Molar Solubility From K sp Calculate the solubility of AgI in water if K sp = 1.8 x 10 -10 AgI Ag + + I - s s Ksp = [Ag + ][I - ] = s 2 1.8 x 10 -10 = s 2 s = √1.8 x 10 -10 = 1.3 x 10 -5

33. Ksp of MgCO 3 is 6.8 x 10 -6. Calculate the solubility of MgCO 3. MgCO 3(s)  Mg 2+ (aq) + CO 3 2- (aq)

34. Calculate K sp From Molar Solubility It is found that 1.2 x 10 -3 mol of lead (II) iodide, PbI 2, dissolves in 1.0 L of aqueous solution at 25 o C. What is the K sp at this temperature? PbI 2 (s) Pb 2+ (aq) + 2I 1- (aq) s 2s Ksp = [Pb 2+ ][I - ] 2 = 4s 3 Ksp = 4(1.2 x 10 -3 ) 3 Ksp = 6.9 x 10 -9

35. The solubility of silver bromide (Ag 2 S) in water is 0.0072 M. Calculate it’s Ksp.

36. Lab Write equation for dissolution of Sr(OH) 2 Write ksp expression If 49.0 ml of saturated Sr(OH) 2 leaves 0.77 g of solid upon drying, calc ksp

37. Ksp and Solubility – Equation Form s = solubility of the salt

38. Common Ion Effect

39. Methanol can be synthesized by the reaction of CO and hydrogen gas according to the reaction: CO (g) + 2H 2 (g) CH 3 OH (g) Keq = 290 Calculate the concentration of hydrogen gas when [CO] = 0.0098 M, [CH 3 OH] = 0.0098 M.

40. LeChatelier’s Principle When an equilibrium system’s conditions are changed, the equilibrium system shifts to the right or left to relieve the stress One reaction occurs more than the other Equil shifts toward (produces more) reactant or product 3 types of changes affect equilibrium  [concentration] [ ] means molarity (mol/L) Increase or decrease  temperature  pressure – for gas systems only

41. Changes in k Changes in conc DO NOT change k adding or removing reactant or product [R] and [P] change but k remains the same Changes in the pressure of the system DO NOT change k Placing in a bigger or smaller container Changes in temp DO cause k to change [R] or [P] gets bigger or smaller Depends on the way the equil shifts

42. Increase in Concentration A + B C + D Increase in the amt of reactants in number of collision between reactants Forward rx occurs more than the reverse rx Conc of all r and p change K remains the same An increase in conc of a subst pushes the eq away from the side of the increase Same thought process for an increase in the conc of the products

43. Decrease in Concentration A + B C + D Decrease in the amt of reactants in number of collision between reactants Reverse rx occurs more than the forward rx Conc of all r and p change K remains the same Decreases in conc of a substance pulls the equilib towards the side of the decrease Same thought process for a decrease in the conc of the products

44. Examples 2NO 2(g) N 2 O 4(g) What happens to k if the conc of NO 2 is increased? What happens to conc of N 2 O 4 if the conc of NO 2 is decreased? What happens to k if the conc of N 2 O 4 is increased? What happens to conc of NO 2 if the conc of N 2 O 4 is increased?

45. LeChatelier’s Principle : Changing the pressure on the container. Increase in the pressure. Rx will shift toward the side with the fewer moles of gas. Decreasing the pressure. Rx will shift toward the side with more moles of gas. Changing the pressure will make no difference if there are = number of moles of gas on each side k does NOT change

46. LeChat Pressure N 2(g) + 3H 2(g) 2NH 3(g) Which way will the eq shift if the pressure is increased? Which direction will the eq shift if the container size is increased?

47. Change in Temperature Do you know the forms for endo and exo reactions??? A + B C + D + heat heat + A + B C + D Heat, energy, or a number of KJ on reactant side = endo Heat, energy, or a number of KJ on product side = exo

48. Changes in Temperature When temp changes, write “heat” into the eq as a reactant or product (if not given) Which side is heat written? Based on whether rx is endo or exo Treat “heat” as a reactant or product in heat push rx away from the side “heat” is located in temp pull rx towards the side “heat” is located K will get bigger or smaller depending on which direction the equil rx shifts k get bigger k gets smaller

49. Change in Temperature The rx2NO 2(g) N 2 O 4(g) gives off 57.2 KJ of energy. What happens to the conc of N 2 O 4 if the equilibrium system is heated? What happens to the value of K? Rx is exo therefore 2NO 2(g) N 2 O 4(g) + heat (57.2 KJ) increase in heat will push rx away from the side with heat

50. Change in Temperature Co(H 2 O) 6 2+ (aq) + 4Cl - (aq) CoCl 6 2- (aq) + 6H 2 O (aq) Pink Blue Based on the demo, is the forward reaction endothermic or exothermic?

51. 5. Methanol can be synthesized by the reaction of CO and hydrogen gas according to the reaction: CO (g) + 2H 2 (g) CH 3 OH (g) At 700 K the value of Keq = 290. Calculate the concentration of hydrogen gas when [CO] = 0.0098 M, [CH 3 OH] = 0.0098 M. (0.094 M)