4 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when:the rates of the forward and reverse reactions are equal andthe concentrations of the reactants and products remain constantHowever, there is a lot of activity at the molecular level!Reactants continue to form products, while products continue to yield reactants!
5 Physical vs. Chemical Equilibrium Physical equilibriumH2O (l)H2O (g)NO2Chemical equilibriumN2O4 (g)2NO2 (g)
6 N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2 Start with NO2 & N2O4
8 N2O4 (g) NO2 (g)Equilibrium ConstantK =[NO2]2[N2O4]= 4.63 x 10-3aA + bB cC + dDK =[C]c[D]d[A]a[B]bLaw of Mass Action
9 Law of Mass ActionFor a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentration has a constant value, K (the equilibrium constant)concentrations may varybut as long as temperature stays the same and the reaction is at equilibrium, K will not change.
10 Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1 Lie to the rightFavor productsK << 1Lie to the leftFavor reactants
11 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) NO2 (g)Kp =NO2P 2N2O4PKc =[NO2]2[N2O4]In most casesKc KpaA (g) + bB (g) cC (g) + dD (g)Kp = Kc(RT)DnDn = moles of gaseous products – moles of gaseous reactants= (c + d) – (a + b)
12 Homogeneous Equilibrium CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)[CH3COO-][H3O+][CH3COOH][H2O]Kc =′[H2O] = constant[CH3COO-][H3O+][CH3COOH]=Kc [H2O]′Kc =General practice not to include units for the equilibrium constant.
13 Example 1Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium:HF(aq) + H2O(l) H3O+(aq) + F-(aq)2NO(g) + O2(g) 2NO2(g)CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(l)Hints: KP applies only to gaseous reactions; the concentration of the solvent (usually water) does not appear in the equilibrium constant expression.
14 Problem 1Write Kc and KP for the decomposition of dinitrogen pentoxide: 2N2O5(g) 4NO2(g) + O2(g)
15 Example 2The following equilibrium process has been studied at 230oC: 2NO(g) + O2(g) 2NO2(g) In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO]= M, [O2]=0.127 M and [NO2]=15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature.
16 Problem 2Carbonyl chloride (COCl2), also called phosgene, was used in WWI as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride CO(g) + Cl2(g) COCl2(g) at 74oC are [CO]=1.2x10-2 M, [Cl2]=0.054 M and [COCl2]=0.14 M. Calculate the equilibrium constant (Kc).
17 Example 3At the equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorinePCl5(g) PCl3(g) + Cl2(g)is found to be 1.05 at 250oC. If the equilibrium partial pressures of PCl5 and PCl3 are atm and atm, respectively, what is the equilibrium partial pressure of Cl2 at 250oC?
18 Problem 3The equilibrium constant KP for the reaction 2NO2(g) 2NO(g) + O2(g) is 158 at 1000 K. Calculate PO2 if PNO2 = atm and PNO = atm.
19 Example 4Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) CH3OH(g) The equilibrium constant Kc for the reaction is 10.5 at 220oC. What is the value of KP at this temperature?
20 Problem 4For the reaction N2(g) + 3H2(g) 2NH3(g) KP is 4.3x10-4 at 375oC. Calculate Kc for the reaction.
21 14 Dec. 2010Objective: SWBAT write equilibrium constant expressions for reactions with heterogeneous equilibria, and calculate reaction quotients.Do now: The equilibrium constant KP for the reaction2SO3(g) ⇄ 2SO2(g) + O2(g)is 1.8x10-5 at 350oC. What is KC for this reaction?
22 Agenda Do now Homework solutions Heterogeneous equilibrium constants Reaction quotientsp. 649 #23, 25, 27, 29, 31, 35, 37
23 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.CaCO3 (s) CaO (s) + CO2 (g)[CaO][CO2][CaCO3]Kc =′[CaCO3] = constant[CaO] = constant[CaCO3][CaO]Kc x′Kc = [CO2] =Kp = PCO2The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
24 does not depend on the amount of CaCO3 or CaO CaCO3 (s) CaO (s) + CO2 (g)PCO2= KpPCO2does not depend on the amount of CaCO3 or CaO
25 Example 5Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous sytems:(NH4)2Se(s) 2NH3(g) + H2Se(g)AgCl(s) Ag+(aq) + Cl-(aq)P4(s) + 6Cl2(g) 4PCl3(l)
26 Problem 5Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) + 4CO(g) Ni(CO)4(g)
27 Example 6Consider the following heterogeneous equilibrium: CaCO3(s) CaO(s) + CO2(g) At 800oC, the pressure of CO2 is atm. Calculate a) Kp b) Kc for the reaction at this temperature.
28 Problem 6Consider the following equilibrium at 395 K: NH4HS(s) NH3(g) + H2S(g) The partial pressure of each gas is atm. Calculate KP and Kc for the reaction.
29 For which of the following reactions is Kc equal to KP? 4NH3(g) + O2(g) ⇄ 4NO(g) + 6H2O(g)2H2O2(aq) ⇄ 2H2O(l) + O2(g)PCl3(g) + 3NH3(g) ⇄ 3HCl(g) + P(NH2)3(g)
30 What if the product molecules of one reversible reaction are involved in a second reaction as the products?A + B C + DC + D E + F
31 ′ ′ ′ ′ ′′ Kc = [C][D] [A][B] Kc = [E][F] [C][D] A + B C + D Kc Kc ′′ C + D E + F[E][F][A][B]Kc =A + B E + FKcKc =Kc′′′xIf a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
32 ′ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] = 216When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
33 Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.The equilibrium constant is a dimensionless quantity.In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.
34 15 Dec. 2010Objective: SWBAT calculate reaction quotient and calculate an equilibrium concentration.Do now: If the equilibrium constant for the reactionN2O4 (g) ⇄ 2NO2 (g) is 4.63 x 10-3calculate the equilibrium constant for the reaction2NO2(g) ⇄ N2O4(g)
35 Agenda Do now Homework solutions Reaction quotient Equilibrium concentrationsHomework: p. 650 #23, 30, 36, 39
36 The Relationship between Chemical Kinetics and Chemical Equilibrium ratef = kf [A][B]2A + 2B AB2kfkrrater = kr [AB2]Equilibriumratef = raterkf [A][B]2 = kr [AB2]kfkr[AB2][A][B]2=Kc =
37 What does the Equilibrium Constant tell us? We can use the equilibrium constant to calculate unknown equilibrium concentrations of products or reactants (at a constant temperature)We can predict the direction in which a reaction will proceed to achieve equilibrium
38 What if the reaction is not yet at equilibrium? Calculate reaction quotient (Qc) by substituting the initial concentrations into the equilibrium constant expression.
39 The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.IFQc > Kc system proceeds from right to left to reach equilibriumQc = Kc the system is at equilibriumQc < Kc system proceeds from left to right to reach equilibrium
40 Example 7At the start of the reaction, there are mol N2, 3.21x10-2 mol H2 and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant Kc for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.
41 Problem 7The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl2(g) ⇄ 2NOCl(g) is 6.5x104 at 35oC. In a certain experiment, 2.0x10-2 mole of NO, 8.3x10-3 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0 L flask. In which direction will the system proceed to reach equilibrium?
42 What do you do with the Kc and KP? If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations!
43 Calculating Equilibrium Concentrations Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.Having solved for x, calculate the equilibrium concentrations of all species.
44 Example 8 At 1280oC the equilibrium constant (Kc) for the reaction Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.Br2 (g) Br (g)
45 ICE At 1280oC the equilibrium constant (Kc) for the reaction Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.Br2 (g) Br (g)Let x be the change in concentration of Br2Br2 (g) Br (g)Initial (M)0.0630.012ICEChange (M)-x+2xEquilibrium (M)xx[Br]2[Br2]Kc =Kc =( x)2x= 1.1 x 10-3Solve for x
46 Kc =( x)2x= 1.1 x 10-34x x = – x4x x = 0-b ± b2 – 4ac2ax =ax2 + bx + c =0x =x =Br2 (g) Br (g)Initial (M)Change (M)Equilibrium (M)0.0630.012-x+2xxxAt equilibrium, [Br] = x = Mor MAt equilibrium, [Br2] = – x = M
47 Example 9A mixture of mol H2 and mol I2 was placed in a 1.00 L stainless steel flask at 430oC. The equilibrium constant KC for the reaction H2(g) + I2(g) ⇄ 2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2 and HI at equilibrium.
48 Example 10For the reaction below, the initial concentrations of each species are H2 = M, I2 = M and HI = M at 430oC. The equilibrium constant KC for the reaction is still Calculate the concentrations of these species at equilibrium. H2(g) + I2(g) ⇄ 2HI(g)
50 Problem 8Consider the reaction below: H2(g) + I2(g) ⇄ 2HI(g) with an equilibrium constant of Starting with a concentration of M for HI, calculate the concentrations of HI, H2 and I2 at equilibrium.
51 Problem 9At 1280oC the equilibrium constant (Kc) for the reaction Br2(g) ⇄ 2Br(g) is 1.1x10-3. If the initial concentrations are [Br2] = 6.3x10-2 M and [Br] = 1.2x10-2 M, calculate the concentrations of these species in equilibrium.
52 3 January 2011 Take Out AP Problem Set: Equilibrium Objective: SWBAT review equilibrium concentration calculations and describe factors that affect chemical equilibrium.Do now: What does a very large Kc indicate?What does it mean if Qc is much smaller than Kc?
53 Agenda Do now AP Problem Set Questions? Review ICE boxes and equilibrium concentrationsFactors that affect chemical equilibrium: DemoNotesHomework: p. 651 #45, 49, 51, 52, odds
54 DemoSolid copper metal reacts with nitric acid to produce dinitrogen tetraoxide gas, water vapor and a solution of copper (II) nitrate.Dinitrogen tetraoxide gas forms nitrogen dioxide gas.
55 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.Changes in ConcentrationN2 (g) + 3H2 (g) NH3 (g)AddNH3Equilibrium shifts left to offset stress
56 Le Châtelier’s Principle Changes in Concentration continuedRemoveAddAddRemoveaA + bB cC + dDChangeShifts the EquilibriumIncrease concentration of product(s)leftDecrease concentration of product(s)rightIncrease concentration of reactant(s)rightDecrease concentration of reactant(s)left
57 Le Châtelier’s Principle Changes in Volume and PressureA (g) + B (g) C (g)ChangeShifts the EquilibriumIncrease pressureSide with fewest moles of gasDecrease pressureSide with most moles of gasIncrease volumeSide with most moles of gasDecrease volumeSide with fewest moles of gas
58 Example 1At 720oC, the equilibrium constant Kc for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) is 2.37x10-3. In a certain experiment, the equilibrium concentrations are [N2]=0.683 M, [H2]=8.80M and [NH3]=1.05 M. Suppose some NH3 is added to the mixture so that its concentration is increased to 3.65 M. a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. b) Confirm your prediction by calculation the reaction quotient Qc and comparing its value to Kc.
59 Problem 1At 430oC, the equilibrium constant (KP) for the reaction 2NO(g) + O2(g) ⇄ 2NO2(g) is 1.5x105. In one experiment, the initial pressures of NO, O2 and NO2 are 2.1x10-3 atm, 1.1x10-2 atm and 0.14 atm respectively. Calculate QP and predict the direction that the net reaction will shift to reach equilibrium.
60 Le Châtelier’s Principle Changes in Temperature (gases only!)ChangeExothermic RxEndothermic RxIncrease temperatureK decreasesK increasesDecrease temperatureK increasesK decreasesN2O4 (g) NO2 (g)colderhotter
61 Example 2 Consider the following equilibrium systems: 2PbS(s) + 3O2(g) ⇄ 2PbO(s) + 2SO2(g)PCl5(g) ⇄ PCl3(g) + Cl2(g)H2(g) + CO2(g) ⇄ H2O(g) + CO(g)Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature.
62 Problem 2Consider the equilibrium reaction involving nitrosyl chloride, nitric oxide and molecular chlorine: 2NOCl(g) ⇄ 2NO(g) + Cl2(g) Predict the direction of the net reaction as a result of decreasing the pressure (increasing the volume) on the system at a constant temperature.
63 5 January 2011Objective: SWBAT describe how stress on a system shifts equilibrium.Do now: Complete this table:Stress to system (exothermic)EffectIncrease in temperatureIncrease in pressureincrease in concentration of reactantincrease in volume
64 Agenda Do now Homework answers Finish Le Chatlier’s examples Equilibrium problem set (work time tomorrow, too)Test Monday, 2nd half of class
65 Le Châtelier’s Principle Adding a Catalystdoes not change Kdoes not shift the position of an equilibrium systemsystem will reach equilibrium soonerCatalyst lowers Ea for both forward and reverse reactions.Catalyst does not change equilibrium constant or shift equilibrium.
66 Example 3Consider the following equilibrium process between dinitrogen tetrafluoride and nitrogen difluoride:N2F4(g) ⇄ 2NF2(g) ∆Ho=38.5 kJ/molPredict the changes in the equilibrium ifthe reacting mixture is heated at a constant volume.some N2F4 gas is removed from the reacting mixture at constant temperature and volumethe pressure on the reacting mixture is decreased at constant temperaturea catalyst is added to the reacting mixture.
67 Problem 3 Consider the equilibrium between molecular oxygen and ozone: 3O2(g) ⇄ 2O3(g) ΔHo=284 kJ/molWhat would be the effect ofincreasing the pressure on the system by decreasing the volume?adding O2 to the system at constant volume?decreasing the temperature?adding a catalyst?
68 Life at High Altitudes and Hemoglobin Production Chemistry In ActionLife at High Altitudes and Hemoglobin ProductionHb (aq) + O2 (aq) HbO2 (aq)Kc =[HbO2][Hb][O2]
69 Chemistry In Action: The Haber Process N2 (g) + 3H2 (g) NH3 (g) DH0 = kJ/mol
70 Le Châtelier’s Principle - Summary ChangeShift EquilibriumChange EquilibriumConstantConcentrationyesnoPressureyes*noVolumeyes*noTemperatureyesyesCatalystnono*Dependent on relative moles of gaseous reactants and products