2. Reactions are reversibleReactions are reversible  A + B C + D ( forward)  C + D A + B (reverse)  Initially there is only A and B so only the forward reaction is possible  As C and D build up, the reverse reaction speeds up while the forward reaction slows down.  Eventually the rates are equal

3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

4. What is equal at Equilibrium?  Rates are equal  Concentrations are not.  Rates are determined by concentrations and activation energy.  The concentrations do not change at equilibrium.  or if the reaction is verrrry slooooow.

5. The units for KThe units for K  Are determined by the various powers and units of concentrations.  They depend on the reaction.

6. K is CONSTANTK is CONSTANT  At any temperature.  Temperature affects rate.  The equilibrium concentrations don’t have to be the same only K.  Equilibrium position is a set of concentrations at equilibrium.  There are an unlimited number.

7. Calculate KCalculate K  N 2 + 3H 2 3NH 3  Initial At Equilibrium  [N 2 ] 0 =1.000 M [N 2 ] = 0.921M  [H 2 ] 0 =1.000 M [H 2 ] = 0.763M  [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

8. Equilibrium and PressureEquilibrium and Pressure  Some reactions are gaseous  PV = nRT  P = (n/V)RT  P = CRT  C is a concentration in moles/Liter  C = P/RT

9. Equilibrium and PressureEquilibrium and Pressure  2SO 2 (g) + O 2 (g) 2SO 3 (g)  Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 )  K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

10. The Reaction QuotientThe Reaction Quotient  Tells you the directing the reaction will go to reach equilibrium  Calculated the same as the equilibrium constant, but for a system not at equilibrium  Q = [Products] coefficient [Reactants] coefficient  Compare value to equilibrium constant

11. What Q tells usWhat Q tells us  If Q<K : Not enough products : Shift to right  If Q>K : Too many products : Shift to left  If Q=K system is at equilibrium

12. Checking the assumptionChecking the assumption  The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid.  If not we would have had to use the quadratic equation  More on this later.  Our assumption was valid.

13. Using K C to determine [products] and [reactants] at equilibrium  Only known K C and initial concentrations of reactants  Changes in concentrations ( Δ C) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.

14. Steps for Complex Problems 1)Make an ICE Chart 2)Set up the equilibrium expression 3)Solve for Δ C

15. Ice ChartIce Chart “ICE”ReactantsProducts Initial Concentrations (I): Change in Concentration ( Δ C or x): Equilibrium Concentrations (E):

16. Example 3:Example 3:  H 2 O (g) is present in a rigid container at 25°C with an initial partial pressure of 0.784 atm. What are the partial pressures of H 2(g) and O 2(g) at equilibrium? (K P = 2.0 x 10 -42 )

17. Example 3: continuedExample 3: continued  If K eq < 1x10 -4, remove the “x or Δ C” value in denominator.  Δ C is very small compared to initial concentration so subtraction would not be a huge difference.  Only works when adding or subtracting Δ C  If concentrations or partial pressures are very small where their magnitude is approximately equal to K eq, CANNOT discount Δ C value.

18. Example 4:Example 4:  Sulfur trioxide decomposes to form sulfur dioxide and oxygen at 300°C°. Calculate the concentrations of all chemical compounds at equilibrium with an initial SO 3 concentration of 0.100M and K C = 1.6 x 10 -10.

19. Example 5:Example 5:  0.194 mol of COCl 2 comes to equilibrium in a 5.8L container at 25°C (K C = 7.27 x 10 -38 ). Find the equilibrium concentrations of all chemical compounds in the following equation. ( Hint: first find the initial [COCl 2 ] )  COCl 2 (g)  CO (g) + Cl 2 (g)

20. % Ionization% Ionization

22. Le Chatelier’s PrincipleLe Chatelier’s Principle  If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.  3 Types of stress

23. Change amounts of reactants and/or products  Adding product makes Q>K  Removing reactant makes Q>K  Adding reactant makes Q<K  Removing product makes Q<K  Determine the effect on Q, will tell you the direction of shift

24. Variables disrupting equilibrium  1) Addition or removal of chemical compounds from the reaction.  2) Pressure  3) Temperature/Heat

25. Basic ConceptBasic Concept  Main concept in chemical equilibrium  Change in a variable that alters the equilibrium of a system (chemical reaction) products a shift in the OPPOSITE direction  Reaction shifts to counteract the variable’s influence  Reaction tries to get back to equilibrium state-----SO reaction shifts

26. Basic Concept (cont.)Basic Concept (cont.)  What happens to a chemical reaction when equilibrium shifts  When one side of a chemical reaction is stressed, the reaction shifts to the side of LEAST stress !

27. Using K C to determine [products] and [reactants] at equilibrium  Only known K C and initial concentrations of reactants  Changes in concentrations ( Δ C) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.

28. Change PressureChange Pressure  By changing volume  System will move in the direction that has the least moles of gas.  Because partial pressures (and concentrations) change a new equilibrium must be reached.  System tries to minimize the moles of gas.

29. Change in PressureChange in Pressure  By adding an inert gas  Partial pressures of reactants and product are not changed  No effect on equilibrium position

30. Change in TemperatureChange in Temperature  Affects the rates of both the forward and reverse reactions.  Doesn’t just change the equilibrium position, changes the equilibrium constant.  The direction of the shift depends on whether it is exo- or endothermic

31. Example 5:Example 5:  Predict the influence of certain changes to the following chemical reaction.  2SO 3(g)  2SO 2(g) + O 2(g) Δ H°= 197.84 kJ/mol  A) increasing reaction’s temperature  B) increasing reaction’s pressure  C) adding more O 2 at equilibrium  D) Removing O 2 at equilibrium

33.  Δ G°  Δ G  Concentration and partial pressures can change for reactants and products.  Δ G = Δ G° + RTlnQ What happens when we leave the standard state conditons?

34.  At equilibrium, Δ G = 0, so reaction quotient (Q) = equilibrium constant (K)  At equilibrium  Δ G° = - RTlnK  Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy ( Δ G°) Δ G = Δ G° + RTlnQ

35.  The magnitude of Δ G° indicates how far the chemical reaction in its standard state is from equilibrium.  Δ G° = 0, equilibrium  Δ G°= large value, far from equilibrium  Δ G° = small value, close to equilibrium  The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium  Positive (+) -------- shift to left, no reaction  Negative (-) -------- shift to right, reaction goes to completion What is the relationship between free energy( Δ G) and K?

36.  Calculate the equilibrium constant for the following reaction given Δ G° = -33 kJ/mol.  N 2 + 3H 2  2NH 3 Example 1Example 1

38. Reaction between H 2 O and Acid  HA + H 2 O  H 3 O + + A -  Write the equilibrium expression

39. Acid Dissociation Constant (K a )  Quantifies how much an ACID dissociates in water  Describes the acid strength  K a = [H 3 O + ] [A - ] [HA]

40. Acid Dissociation Constant (K a )  Strong Acids > 1  Weak Acids <<<< 1  Examples:  HCl Ka = 1x10 6  CH 3 CO 2 H Ka = 1.8 x 10 -5

41. Base Ionization Constant (K b )  Quantifies how much a BASE dissociates in water  Describes base strength  B + H 2 O  BH + + OH -  K b = [BH + ] [OH - ] [B]

42. Other helpful equationsOther helpful equations  pKa = -logKa  pK b = -logK b

43. Weak Acid pH CalculationsWeak Acid pH Calculations  Weak acids only partially dissociate  Calculating:  Use ICE method  Compare reaction quotient (Q a ) to equilibrium constant (Ka)----- % Ionization  When can we ignore change in initial acid concentration ( Δ C) ?

44. When can we ignore change in initial acid concentration ( Δ C)?  If Δ C < 5% of initial acid concentration  Ka < 1x10 -4 THEN, Δ C can be disregarded for equilibrium acid concentration

45. Example 1:Example 1:  Determine the H 3 O + ion concentration and pH of a 1.0M acetic acid solution with a Ka = 1.8 x 10 -5

46. Example 2:Example 2:  Calculate the pH of a 0.10M NH 3 solution with a pK b of 4.74

47. K a and K b  MUST determine K b from K a value  Use relationship between an acid and its conjugate base  K a K b = K w

48. Example 5:Example 5:  Find the pH of a 3.5M NH 3 solution with a K b value of 1.8 x 10 -5

50.  NaOAc  HOAc  Now we would have 2 sources for OAc - ion  1) Acetic Acid Dissociation  2) Salt Hydrolysis/Dissociation  **Acetate ion is COMMON to both reactions** What would happen to the pH if we added….

51.  Common ion  Ion present in both acid/base dissociation and salt ionization  Common ion effect  Seen with weak acids and bases  Hinders the ionization of a weak acid or base due to the presence of a common ion between the acid/base and a salt Common Ion EffectCommon Ion Effect

52.  HF + H 2 O (l)  H 3 O + + F -  Addition of KF  KF  K + + F -  Result is less [H 3 O + ] so pH  ----more BASIC Example 1:Example 1:

53.  NH 3 + H 2 O (l)  OH - + NH 4 +  Addition of NH 4 Cl  NH 4 Cl  NH 4 + + Cl -  Result is less [OH - ] so pH  ----more ACIDIC Example 2:Example 2:

54.  450ml of 1.0M HCO 2 H is mixed with 250ml of 0.40M NaHCO 2 at 25°C. Find the pH of the solution (Ka = 1.8 x 10 -4 ). Example 3:Example 3:

56. Salts  Ionic compound made up of CATION and ANION  Has acidic and basic properties  Based on ions produced when salts dissociate  No acid/base properties—group I/II cations (ex. Na +, Li +, K +, Ca +2 )  No basic properties—conjugate bases from monoprotic acids (ex. Cl -, Br -, NO 3 - )  Ex. NaCl, CaBr 2

57. 1. Salt Formation from Strong Base and Weak Acid  Salt forms a BASIC solution.  Conjugate base ion reacts with water to give hydroxide (OH - ) ions.  Ex. Potassium fluoride (KF)  KF  K + + F -  F - + H 2 O  HF + OH -

58. 2. Salt Formation from a Strong Acid and Weak Base  Salt forms an ACIDIC solution  Conjugate acid reactions with water to give hydronium ion (H 3 O + )  Ex. Ammonium nitrate (NH 4 NO 3 )

59. 3. Salt Formation from Strong Acid and Strong Base  Salt forms a NEUTRAL solution  Conjugate base resulting from salt dissociation is weak  Ex. Sodium chloride (NaCl)

60. Example 1:Example 1:  Calculate the concentration of HOAc, OAc - and OH - at equilibrium in a 0.10M NaOAc solution (Ka for HOAc = 1.8 x 10 -5 ).

62. Titration  Lab technique commonly utilized to determine an UNKNOWN concentration of a chemical compound with a KNOWN concentration of another chemical compound.  Chemical compounds combine with exact stoichiometric proportions  Analyte—  Chemical compound with unknown concentration  Titrant—  Chemical compound with known concentration  Measured with volume and concentration  Added to chemical compound with unknown concentration in titration

63. 2 Types of Acid-Base Titrations 1)Strong Acid/Strong Base Titrations 2)Weak Acid/Strong Base Titrations

64. 1. Strong Acid/Strong Base Titrations  Low initial pH value  Sharp increase in pH before equivalence point  Equivalence point is pH = 7  Rapid pH increase after equivalence point **Indicators with pH range 4-10 helpful for these titrations **Neutralization reactions

65. Strong Acid with Strong Base Titrant Strong Base with Strong Acid Titrant

66. 2. Weak Acid/Strong Base Titrations  High initial pH value  pH = pKa at half-neutralization  [weak acid] = [conjugate base]  Ka = [H 3 O + ] [A - ] SO Ka = [A - ]/[HA] is 1:1 [HA]  Ka = [H 3 O + ], SO pH = pKa

67. 2. Weak Acid/Strong Base Titrations  Equivalence point > 7 on pH scale **Indicators with pH range > 7 helpful as pH equivalence point is basic

69. Example 1: Weak Acid/Strong Base Titration Calculations  30 ml of 0.5M HC 2 H 3 O 2 titrated with 0.50M NaOH. Ka for acetic acid is 1.8x10 -5. a) Find the initial pH of 0.5M acetic acid. HC 2 H 3 O 2 + OH -  C 2 H 3 O 2 - + H 2 O (titration view) HC 2 H 3 O 2  H + + C 2 H 3 O 2 - (in detail)

70. Example 1: Weak Acid/Strong Base Titration Calculations  b) Find the pH after 15ml of NaOH were added  pH = pKa + log [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] -used when dealing with buffer solutions - buffer —mixture of weak acid/conjugate base -more on this concept later

71. Example 1: Weak Acid/Strong Base Titration Calculations  c) Find the pH at the equivalence point.  C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH -

73.  Weak acid/conjugate base mixtures OR weak base/conjugate acid mixtures  “buffers” or reduces the affect of a change in the pH of a solution  Absorbs slight changes in pH resulting from the addition of small acid/base amounts to water. Buffers

74.  HOAc + H 2 O (l)  H 3 O + + OAc –  What happens if an acid is added???  Reacts with OAc ion  [HOAc] increases slight, [OAc] decreases slightly, ratio mostly the same  No pH change Example 1: (cont.)Example 1: (cont.)

75.  HOAc + H 2 O (l)  H 3 O + + OAc –  What happens if a base is added???  Reacts with HOAc ion  More OAc ion formed, removes excess OH - from solution  No pH change Example 1: (cont.)Example 1: (cont.)

76. 1)Acidic Buffers  Formed from mixing a weak acid and its conjugate base  pH < 7  Ex. HOAc and OAc – 2)Basic Buffers  Formed from mixing a weak base and its conjugate acid  pH > 7  Ex. NH 3 and NH 4 + Types of buffersTypes of buffers

77.  Compare Ka and Kb from acid-conjugate base pair  Ka > Kb, (generally > 1x10 -7 ) -------- acidic buffer  Ka 1x10 -7 ) -------- basic buffer How do we tell acidic vs. basic buffers?

78. 1)Method applying “Common Ion Effect” 2)Henderson-Hasselbalch equation Calculating the pH of buffers

79.  Easier method  pH = pKa + log[A - ]/[HA]  pOH = pKb + log[HB + ]/[B]  [B] = molarity of weak base  [HB + ] = molarity of conjugate acid  Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium. Henderson-Hasselbalch Equation

80.  Find the pH of a buffered solution created by mixing 0.15mol NH 4 NO 3 with 0.65L of a 0.25M NH 3 solution. Assume that the volume change is negligible. (Kb = 1.8x10 -5 ) Example 1:Example 1: