Chapter 14: Chemical EquilibriumSome Reactions “Proceed to Completion”Dynamic EquilibriumEquilibrium ConstantConnection Between Kinetics and EquilibriumSlide 6Pressures or ConcentrationsExamplesHeterogeneous Equilibria: Reactions involving solids or liquidsRelationships Between K’s of Related ReactionsSlide 11Slide 12The Reaction QuotientSlide 14ExampleCalculations Involving Equilibrium Constants: ICE TablesExample “Calculate Equilibrium Constant When Given Initial Concentrations and one equilibrium concentration...”Slide 18Slide 19Change this examleSimplifying Assumption when K is smallLe Châtelier’s PrincipleAdding or Removing Reactant or ProductChange in Pressure/VolumeChange in VolumeChanging the TemperatureChange in TemperatureDriving Reactions to Completion/Increasing YieldChapter 14 1Chapter 14: Chemical EquilibriumGeorgia Gwinnett CollegeChem 1212KFall 2013(B. Shepler)Some Reactions “Proceed to Completion”•“Proceed to Completion”: All of the reactants are converted to products.–Many reactions do proceed essentially to completion.–Example: Combustion of propane•Many other reactions do not proceed to completion. Only some of the reactants are converted to products. We are left with a mixture of reactants and products.–Example: Vaporization of water below its boiling point (vapor pressure)2Chapter 14Dynamic Equilibrium•Reaction rates depend on the concentration of the reacting species.–If we start with all reactants, we start to form products. As products are formed the reverse reaction starts to occur•More products  Reverse reaction gets faster•Less reactants  Forward reaction gets slower–Dynamic Equilibrium is achieved when the rate of forward and reverse reactions are equal.•No net change in concentrations of reactants or products. However, the forward and reverse reactions continue to occur.3Chapter 14Equilibrium Constant•The equilibrium constant is a way to quantify the amount of reactants and products that are present at equilibrium.•Equilibrium constant is the ratio of the concentrations of products to reactants AT EQUILIBRIUM:•Example, write the equilibrium constant expression for:Chapter 14 4 K =[C]c[D]d[A]a[B]bC2H6O(g) + O2(g) Û CO2(g) + H2O(g)Connection Between Kinetics and Equilibrium•Write the equilibrium constant expression for the above reactions.•If I tell you the above reaction is an elementary step:–Write the rate law for the forward reaction–Write the rate law for the reverse reaction•If we are at equilibrium then the rates of the forward and reverse reactions are equal.–Rearrange•Note: the equilibrium constant expression can be written for any reaction, not just elementary steps.Chapter 14 5Equilibrium Constant•Small K would be significantly less than 1. At Eq more reactants than products.•Large K would be significantly greater than 1. At Eq more products than reactants•Intermediate K would be ≈ 1. Significant amount of both reactants and products at Eq.6 Kc=C[ ]cD[ ]dA[ ]aB[ ]bChapter 14K=1x10380K=5K=1 x 10-30aA+bBÛ cC +dDPressures or Concentrations•The equilibrium expression and equilibrium constant for a gas phase reaction can be expressed using partial pressures rather than concentrations.7Chapter 14Examples•Consider the following reactionCO(g) + 2 H2(g)  CH3OH(g)An equilibrium mixture of this reaction at a certain temperature was found to have [CO] = 0.105 M, [H2] = 0.114 M, and [CH3OH] = 0.185 M. What is the value of the equilibrium constant (Kc) at this temperature.•At 775 (K) the reaction below has an equilibrium constant K=0.0584. If [N2]=0.120 M and [H2]=0.140 M at equilibrium, what is the equilibrium concentration of NH3?N2(g) + 3 H2(g)  2 NH3(g)Chapter 14 8Heterogeneous Equilibria: Reactions involving solids or liquids•Concentration = moles/Volume–This is a constant for pure liquids and solids (It is the density)–Pure liquids and solids DO NOT APPEAR in the equilibrium expression (they are given a value of 1)9CaCO3(s) ↔ CaO(s) + CO2(g) Amount of CO2Does not dependOn relative amountsOf the two solidsAmount of CO2Does not dependOn relative amountsOf the two solidsRelationships Between K’s of Related Reactions10#1: Forward and Reverse Reactions2 H2 (g) + O2 (g) ↔2 H2O (g)2 H2O (g) ↔ 2 H2 (g) + O2 (g) #1#2Relationships Between K’s of Related Reactions11#2 When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor.2 H2 (g) + O2 (g) ↔2 H2O (g)4 H2 (g) + 2 O2 (g) ↔ 4 H2O (g)Relationships Between K’s of Related Reactions#3 When chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation.122 BrCl (g) ↔ Br2 (g) + Cl2 (g)Br2 (g) + I2 (g) ↔ 2 IBr (g)2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g)(PBr2)(PCl2)(PBrCl)2= K1 = 0.45 @ 25oC(PIBr)2(PBr2) (PI2)= K2 = 0.051 @ 25oCThe Reaction Quotient•When a reaction is not at equilibrium the ratio of the product and reactant concentrations is called the reaction quotient, Qc.•Q is calculated with the same equation as K.–If the reaction is at equilibrium we call the ratio K–If the reaction is not a equilibrium we call the ratio Q–At equilibrium Q=K13aA+bBÛ cC +dDThe Reaction Quotient•The reaction quotient tells us in which direction the reaction will go.14Example•The equilibrium constant (Kp) for the reaction P4(g) ↔ 2 P2(g) is 1.39 at 400oC. Suppose that 2.75 atm of P4(g) and 1.08 atm of P2(g) are mixed in a closed 25.0 L container at 400oC. Compute Q and state the direction in which the reaction proceeds to attain equilibrium.15Calculations Involving Equilibrium Constants:ICE Tables16A + 2B C + 3DI =C = E =Example“Calculate Equilibrium Constant When Given Initial Concentrations and one equilibrium concentration...”17Consider the equilibrium 4 NO2(g)  2 N2O(g) + 3 O2(g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature.NOTE: Don’t think about the different “types” of equilibrium calculations, as labeled by the book, as being different