Chemical Equilibrium A state of no net change in reactant & product concentrations. BUT There is a lot of activity at the molecular level. 1
Kinetics Equilibrium For an elementary step in the mechanism: k A + 2B f AB k 2 r rate = k [A][B] 2 f f rate r = k r [AB 2 ] 2
Kinetics Equilibrium A + 2B AB 2 If start with only reactants, forward rate is high. As products build up, reverse rate increases and forward rate decreases. 3
Kinetics Rate vs. Time rate f = k f [A][B] 2 Equilibrium equal rates rate rate r = r = k r k[ab r [AB 2 ] 2 ] Conc. vs. Time reactants prod. equilibrium, no change 4
Kinetics Equilibrium k f A + 2B AB 2 k r At equilibrium: rate f = rate r k f [A][B] 2 = k r [AB 2 ] k [AB f 2 ] = k r [A][B] 2 = K c Equilibrium Constant 5
Kinetics Equilibrium K c = k f k r = [AB 2 ] [A][B] 2 Since k f and k r are constants depending only on temp.,k c is also a constant depending on temp. K c describes the ratio of products to reactants at equilibrium. 6
Chemical Equilibrium A + B reactants C + D products Chemical equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, concentrations do not change. 7
Chemical Equilibrium N 2 O 4 (g) colorless 2NO 2 (g) brown Draw Lewis structure of NO 2 This reaction can be monitored spectroscopically. 8
N 2 O 4 (g) 2NO 2 (g) 100% N 2 O 4 time 1 time 2 time 3 time 4 no change 9
concentration N 2 O 4 (g) 2NO 2 (g) Exp.#1 Exp.#2 Exp.#3 N 2 O 4 N 2 O 4 N 2 O 4 NO 2 NO 2 NO 2 time time time 10
N 2 O 4 (g) 2NO 2 (g) Final concentrations depend on what you start with. At equilibrium, forward and reverse rates are equal. 11
Equilibrium Constant (K) aa + bb cc + dd K = [C]c [D] d [A] a [B] b Exponents are the coefficients K is independent of concentrations; it depends only on temperature. 12
Equilibrium Constant K = [C]c [D] d [A] a [B] b K<10-3 10-3 <K<10 3 K>10 3 react. prod. react. prod. react.prod. mostly similar mostly reactants mixture products 13
Equilibrium Constant Units are important even though they are not written!! K :concentration (M) c K :pressure (atm) p 14
Types of Equilibria Homogeneous: all reactants and products in same phase. Heterogeneous: reactants and products in different phases. 15
Homogeneous Equilibria N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = P 2 NO2 P N2O4 [ ] = M = mol/l Partial P (atm) (P a conc.) 16
Relationship: K c and K p aa(g) K = p P =(n RT)/V A A =[A]RT bb(g) P B b P a A P =(n RT)/V B B =[B]RT 17
Relationship: K c and K p K p = P B b P A a = ([B]RT) b ([A]RT) a [B]b K = p [A] a(rt)b-a = [B]b [A] a(rt)dn gas Dn gas = mol prod gas mol react gas 18
Relationship: K c and K p [B]b K = (RT)Dn p [A] a K = K (RT) Dn p c If Dn = 0, then K p = K c 19
K c Used when the concentration of one of the reacting chemicals doesn t change. 20
Relationship: K c and K c e.g. ionization of aqueous acetic acid (CH COOH(aq) = HAc) 3 HAc + H O Ac - + H O + 2 3 K c = [Ac- ][H 3 O + ] [HAc][H 2 O] 21
Relationship: K c and K c K c = [Ac- ][H 3 O + ] [HAc][H 2 O] But [H 2 O], the solvent, doesn t change significantly during the course of the reaction; it is ~constant. 22
Relationship: K c and K c K c = [Ac- ][H 3 O + ] [HAc][H 2 O] [Ac - ][H 3 O + ] K c [H 2 O] = K c = [HAc] Solvent is dropped from K c. 23
Let s Try Some Problems Write K c and K p for: HF(aq) +H 2 O 2NO(g) + O 2 (g) H 3 O + (aq) + F - (aq) 2NO 2 (g) CH 3 COOH(aq) + C 2 H 5 OH(aq) CH 3 COOC 2 H 5 (aq) + H 2 O 24
Some More Fun! At equilibrium at 230 o C: 2NO + O 2 2NO 2 0.0542 M 0.127 M 15.5 M Calculate K c and K p 25
Yes, There s More! PCl 5 (g) PCl 3 (g) + Cl 2 (g) K p = 1.05 at 250 o C If P PCl5 = 0.875 atm and P PCl3 = 0.463 atm What is P Cl2? 26
Heterogeneous Equilibria CaCO 3 (s) CaO(s) + CO 2 (g) K c = [CaO][CO 2 ] [CaCO 3 ] The conc. of a solid is an intrinsic property: CaCO 3 CaO [CaCO 3 ] and [CaO] = constant 27
Heterogeneous Equilibria K c = [CaO][CO 2 ] [CaCO 3 ] K c [CaCO 3 ] [CaO] = K c = [CO 2 ] (as long as CaO and CaCO 3 are present) Solids and liquids dropped from K c. 28
Try It!!! Write K c and K p for: AgCl(s) Ag + (aq) + Cl - (aq) P 4 (s) + 6Cl 2 (g) 4PCl 3 (l) 29
Keep Going CaCO 3 (s) CaO(s) + CO 2 (g) Calculate K p and K c if P CO2 is 180 mmhg at 800 o C. 30
Multiple Equilibria A + B C + D K c overall C + D E + F K c A + B E + F K c K c K c = [C][D] [E][F] x [A][B] [C][D] = K c 31
Multiple Equilibria For sequential equilibria: K c K c = K c If add reactions multiply K s (useful in many acid-base problems) 32
Try it!!! H 2 CO 3 (aq) H + + HCO 3 - K c = 4.2 x 10-7 HCO 3 - (aq) H + + CO 3-2 K c = 4.8 x 10-11 What is the expression for and value of overall K c? 33
1.For Summary reverse reaction, K c (reverse) = 1/K c (forward) N 2 O 4 (g) 2NO 2 (g) K c = 4.63 x 10-3 2NO 2 (g) N 2 O 4 (g) K c = 216 And K c K c = 1 34
Summary 2. K c depends how equation is written. N 2 O 4 (g) 2NO 2 (g) K c 2N 2 O 4 (g) 4NO 2 (g) K c K c = K c 35
Summary 3. Units of reactants & products are mol/l for K c and atm for K p. 4. K c and K p are unit-less 5. Solvents, & pure solids & liquids are not in K c 36
Summary 6. Must specify equation & T (the value of K can only change with temp.) 7. For sequence of reactions that add to a net reaction: K c = K c K c 37
Give It a Shot How are the K c values for the following equations related? 1 3 N 2 (g) + 3H 2 (g) N 2 (g) + H 2 (g) 2 3 2NH 3 (g) NH 3 (g) 38
Q c & Q p What if the concentrations of reactants and products are not at equilibrium? We define a new quantity Q, similar to K but using the actual concentrations [X] o rather than equilibrium concentrations [X]. 39
Q c & Q p For the reaction: aa + bb cc + dd Q c = [C] o c [D] o d [A] o a [B] o b The reaction direction can be determined by comparing Q vs. K. 40
Reaction Quotient: Q c Q c = [C] o c [D] o d [A] o a [B] o b If Q c > K c, reaction proceeds to left. If Q c = K c, at equilibrium. If Q c < K c, reaction proceeds to right. 41
Try It! N 2 (g) + 3H 2 (g) 2NH 3 (g) At reaction start: N 2 = 0.249 mol H 2 = 3.21 x 10-2 mol NH 3 = 6.42 x 10-4 mol in 3.50 L flask If K c = 1.2, which direction will the reaction proceed? 42
Calculating Concentrations For the reaction: A B K c = 24.0 and [A] o = 0.850 M [B] o = 0 M What are the equilibrium concentrations of A and B? 43
Calculating Concentrations using ICE Table A B Initial (M) 0.850 0 Change (M) -1x +1x Equil. (M) (0.850 x) x change controlled by stoich. Substitute into K c 44
Calculating Concentrations K c = 24.0 = [B] [A] = x (0.850 x) and x =.816 M = [B] 0.850 x = 0.034 M = [A] Substitute into K c to check. 45
Give It A Shot! H 2 (g) + I 2 (g) 2HI(g) K c = 54.3 at 430 o C If start with 0.500 mol each of H 2 and I 2 in a 1.00 L flask, what are equilibrium concentrations of all species? 46
Try Another 2NO 2 (g) 2NO(g) + O 2 (g) At 1000 K, K p = 158 Starting with pure NO 2, it is found that partial pressure of O 2 at equilibrium is 0.25 atm. Calculate P NO and P NO2. 47
Quadratics are Possible! H 2 (g) + I 2 (g) 2HI(g) Set up ICE table and substitute into K c : K c = 54.3 0.062 M = [H 2 ] o 0.041 M = [I 2 ] o 0.22 M = [HI] o (Don t solve!) 48
Factors Affecting Equilibrium Changes in external variables (T, P, concentrations, etc.) can shift equilibrium to right or left. right products reactants left 49
Le Chatelier s Principle If a stress (change) is applied to a system at equilibrium, the system will adjust to partially offset the stress. Stresses: concentration & dilution pressure/volume (gases) temperature 50
Le Chatelier s Principle Haber synthesis of ammonia: N 2 (g) + 3H 2 (g) 2NH 3 (g) If additional N 2 is added at equilibrium, what happens to: rate of forward reaction? rate of reverse reaction? amounts of each substance? 51
concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) initial equilibrium new equilibrium H 2 NH 3 obey K N 2 N 2 added time 52
Concentration Stress Consider the equilibrium: FeSCN +2 Fe +3 + SCN - red yellow clear stress : Add NaSCN left red Add Fe(NO 3 ) 3 left red 53
Concentration Stress FeSCN +2 Fe +3 + SCN - red yellow clear stress Add H 2 C 2 O 4 Fe +3 + 3C 2 O 4-2 Fe(C 2 O 4 ) 3-3 Thus right 54
Concentration Stress If concentrations of reactants or products change,the system will respond in accordance with K c. K c = products reactants 55
Concentration Stress N 2 (g) + 3H 2 (g) 2NH 3 (g) 0.683M 8.80M 1.05M (at eq.) At 720 o C, K c = 2.37 x 10-3 If [NH 3 ] is increased to 3.65 M, then Q c > K c, so reverse reaction predominates. 56
Concentration: Dilution HF(aq) + H 2 O H 3 O + (aq) + F - (aq) What happens to the ionization of HF if the system is diluted? This is just a special case of Le Chatelier s principle. Adding H 2 O shifts reaction right. 57
Pressure & Volume Stress P & V have no effect on solids & liquids, but for gases: P = (n/v) RT As P increases or V decreases the conc. of gases increase. 58
P & V Stress N 2 O 4 (g) 2NO 2 (g) At equilib, if volume is decreased, product conc. Is affected more than reactant side: Q c = [NO 2 ] o 2 > Kc (shift left) [N 2 O 4 ] o 59
P & V Stress In general, increased P or decreased V shifts reaction to side with lesser moles of gas. Exception: if P is increased by adding an inert gas, it does not affect equilibrium. 60
Y all Be Stressin Predict the direction of the reaction resulting from increased P (or decreased V): PCl 5 (g) H 2 (g) + CO 2 (g) PCl 3 (g) + Cl 2 (g) H 2 O(g) + CO(g) 61
Temperature Stress Unlike concentration, P, or V, temperature changes the value of K c. Increasing temperature will favor the reaction that is endothermic ( uses heat ). 62
Temperature Stress N 2 O 4 (g) 2NO 2 (g) colorless brown DH = 58 kj T=-78 o C T=-10 o C T=+20 o C 63
Temperature Stress 58kJ + N 2 O 4 (g) 2NO 2 (g) Raising temperature (adding heat) shifts equilibrium to use up the heat (Le Chatelier). Reaction shifts right, producing more NO 2. Value of K c increases. 64
Temperature Stress CoCl -2 + 6H O 4 2 Co(H O) +2 + 4Cl - 2 6 Based on the following clip, predict whether this reaction is endo- or exothermic. 65
Effect of Catalyst Catalysts affect the rate of reaction, but NOT the equilibrium. DH E a with catalyst is lower so rate increases. There is no effect on DH, so equilibrium doesn t change. 66
Le Chatelier: Try It! N 2 F 4 (g) 2NF 2 (g) DH = 38.5 kj What happens if: NF 2 is added V is increased He gas if added T is decreased 67
Physical Equilibria Systems can also be in physical equilibria, where rate of forward and reverse processes are equal. H 2 O(s) H 2 O(l) (at 0 o C) CO 2 (g) CO 2 (aq) (in close container) KCl(s) KCl(aq) (in saturated soln) (demo) 68
Le Chatelier: Physical Change H 2 O(s) H 2 O(l) What happens at equilibrium if: heat is added? pressure is increased? 69
Solubility Equilibria Application of equilibria to insoluble ionic compounds. Even insoluble ionic compounds are very slightly soluble! 70
Soluble Compounds Soluble Compounds Exceptions Alkali metals & NH + 4 KNOW!!! NO - 3,HCO - 3, ClO - 3 KNOW!!! Halides Ag + & Hg 2+ 2 & Pb 2+ Sulfates (SO 2-4 ) Ag + & Ca 2+ & Sr 2+ & Ba 2+ & Pb 2+ 71
Insoluble Compounds Insolubles CO 3 2- & PO 4 3- Exceptions KNOW!!! & CrO 4 2- & S 2- Alkali metals & NH 4 OH - Alkali metals & Ba 2+ + 72
Solubility Equilibria Now let s get quantitative. AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.6E-10 = [Ag + ][Cl - ] Why isn t AgCl(s) included in K sp? Ca 3 (PO 4 ) 2 3Ca +2 + 2PO -3 4 K sp = 1.2E-26 = [Ca +2 ] 3 [PO -3 4 ] 2 73
Solubility and K sp For a saturated solution: Molar solubility: Sol. in mol/l Solubility: Solubility in g/l The solubility of CaSO 4 is 0.67g/L. What is K sp? 0.67 g L x 1 mol 136.2 g = 4.9 x 10-3 M 74
K sp of CaSO 4 : ICE Table CaSO 4 Ca +2 + SO 4-2 I 0 0 C -4.9E-3 4.9E-3 4.9E-3 E 4.9E-3 4.9E-3 K sp = [Ca +2 ][SO 4-2 ] = (4.9 x 10-3 ) 2 = 2.4 x 10-5 75 75
Solubility and K sp Given K sp of Cu(OH) 2 is 2.2E-20, what is its solubility? (s = molar solubility of Cu(OH) 2 ) Cu(OH) 2 Cu +2 + 2OH - I 0 0 C +s +2s E s 2s 76
Solubility and K sp K sp = 2.2x10-20 = [Cu +2 ][OH - ] 2 = (s)(2s) 2 = 4s 3 Thus solubility in g/l is: s = 1.8x10-7 M 1.8E-7M x 97.57 g/mol = 1.8E-5 g/l 77
Solubility and K sp Table 16.3 shows the relationship between s and K sp for several different cases. K sp = s 2 K sp = 27s 4 AgCl Al(OH) 3 K sp = 4s 3 K sp = 108s 5 PbF 2 Ca 3 (PO 4 ) 2 78
Solubility and K sp Your turn! Calculate the K sp of Bi 2 S 3 given its solubility is 1.0E-15 M. Calculate the solubility of Al(OH) 3 given K sp = 1.8E-33 79 79
Solubility Equilibria & Q As with K eq problems, calculate reaction quotient, Q, for unsaturated or supersaturated solutions to determine the direction of reaction toward equilibrium. 80
Will a Precipitate Form? Mix 200.mL of 1.0E-4 M Sr(NO 3 ) 2 & 100.mL of 1.0E-4 M NaF. The possible precipitate is SrF 2 (K sp = 2.0E-10) Do the ion concentrations of the mixture exceed K sp? Final solution volume is 0.300 L. 81
Will a Precipitate Form? SrF Sr 2+ + 2F - 2 Q = [Sr +2 ][F - ] 2 mol Sr +2 = 0.200 L x 1.0E-4 M = 2.0E-5 mol mol F - = 0.100 L x 1.0E-4 M = 1.0E-5 mol 82
Will a Precipitate Form? Q = [Sr +2 ][F - ] 2 = 2.0E-5mol 0.300 L = 7.3 x 10-4 x 1.0E-5mol 0.300 L 2 Since Q > K sp of 2.0E-10, a precipitate will form. 83
Will a Precipitate Form? For this problem, what are [Sr +2 ] and [F - ] at equilibrium? limiting SrF 2 Sr 2+ + 2F - I 2.0E-5/.300 1.0E-5/.300 = 6.67E-5 = 3.33E-5 C(stoich) -1.67E-5-3.33E-5 E 5.00E-5 ~0 detn. by 84K sp
Will a Precipitate Form? Next plug into K sp : K sp = 2.0E-10 = [Sr +2 ][F - ] 2 detn. 2.0E-10 = (5.00E-5) [F - ] 2 by K sp [F - ] = 2.0E-3M and [Sr +2 ] = 5.0E-5M 85
Will a Precipitate Form? Try It!! 200.mL of 0.200 M NaOH is mixed with 1.00 L of 0.100 M CaCl 2. Will a precipitate form? What are [Ca +2 ] and [OH - ] at equilibrium? For Ca(OH) 2 K sp = 8.0E-6 86
Common Ion Effect E.g., AgCl is less soluble in aqueous AgNO 3 than in water. AgCl(s) Ag + (aq) + Cl - (aq) As Ag + increases, Cl - must decrease in accordance with K sp. Thus less AgCl will dissolve. Le Chatelier 87
Common Ion Effect What is the solubility of AgCl in 0.0065 M AgNO 3? Let s = molar solubility of AgCl AgCl Ag + + Cl - I 0.0065 0 C +s +s E 0.0065 +s +s ~E 0.0065 +s 88
Common Ion Effect K sp = [Ag + ][Cl - ] 1.6 x 10-10 = (0.0065)(s) s = 2.5 x 10-8 M Thus the amount of dissolved AgCl is 2.5 x 10-8 M or 3.6 x 10-6 g/l Note s is small compared to 0.0065M so approximation was valid. 89
Common Ion Effect Your turn!! What is the molar solubility of silver carbonate: in water in 0.0030 M sodium carbonate? K sp (Ag 2 CO 3 ) = 8.1 x 10-12 90
One More What is the K sp of bismuth sulfide (Bi 2 S 3 ) given its solubility is 8.8E-13 g/l? (M = 514.2 g/mol) What is the molar solubility of bismuth sulfide in a 0.020 M Na 2 S solution? 91
The End See K sp summary in problem set. 92 92
93
Warm-up For the reaction: 2CO(g) + O 2 (g) 2CO 2 (g) Write the algebraic expressions for K c and K p 94
Warm-up Write K c for PbCl 2 (s) Pb +2 (aq) + 2Cl - (aq) For the reaction at 170 o C, 2H 2 (g) + O 2 (g) 2H 2 O(g) Calculate the ratio of K p /K c. Are these constants large (>>1) or small (<<1)? 95
Warm-up 2NOCl(g) = 2NO(g) + Cl 2 (g) 1.0 mol NOCl was placed in a 2.0 L flask at 200 o C. At equilibrium it was found that 40.% of NOCl had decomposed. What is K p? 96
Warm-up N 2 (g) + O 2 (g) 2NO(g) If 1.0 mol each of N 2, O 2, and NO are placed in a 2.0 L vessel, what is the equilibrium concentration of NO? (K c = 3.4E-2) 97
Warm-up The solubility of SrF 2 (s) in water is 5.8E-4M. Calculate K. c 98
Warm-up Are they soluble in water? silver nitrate calcium carbonate copper(ii) sulfide barium hydroxide calcium hydrogen carbonate 99
Warm-up Write the expression for K of PbF. sp 2 What is the K of PbF, sp 2 given solubility is 0.54 g/l? 100
Warm-up What is molar solubility of PbF 2 in water and in 0.020 M Pb(NO 3 ) 2? (given K sp = 4.1E-8) 101
Warm-up Silver sulfide has K sp = 6.0E-51. What is its molar solubility in water? What is its molar solubility in 1.0E-7 sodium sulfide solution? If 100.mL each of 4.0E-8M silver nitrate and 1.0E-7M sodium sulfide solutions are mixed, what is [Ag + ] and [S -2 ]? 102