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Equilibrium Unit 4 Chapters 17, 18, 19, 20 Chapter 17 Equilibrium – when two opposite reactions occur simultaneously and at the same rate Dynamic equilibrium – equilibrium can be shifted if one or more of the reactants is changed in any way. The equilibrium will “react” Chemistry Connections Connection to Kinetics Equilibrium affects reaction rates, especially for multi-step reactions! Factors that affect equilibrium affect reaction rates Connection to Thermo If the reaction rate slows, the energy to be released or absorbed may be released more slowly or quickly….can be dangerous or take FOREVER!...when setting up a reaction MUST keep this in mind! Equilibrium Basics – Kinetics! For a reversible reaction with a one step mechanism: aA + bB cC + dD Rate = kf[A]a[B]b = kR[C]c[D]d Definition of equilibrium: Rf = Rr At Equilibrium kf[A]a[B]b = kR[C]c[D]d Kf = [C]c[D]d kR [A]a[B]b kf / kR = k c kc is the equilibrium constant (mass action expression) The constant can be determined by the ratio of the constants or the concentrations! Example 17-1, 17-2, 17-3 Equilibrium Constant kc Just as in kinetics, intermediates are not allowed, so substitution may be necessary! Must be found experimentally or by means of equilibrium concentrations from thermodynamic data. Also varies with temperature, and constant at a given temperature (just like kinetic rate constants) Independent of initial concentration Equilibrium Constant Equilibrium constants are expressed as a ratio of “activities” rather than concentrations in order to be dimensionless. Activity: Pure liquid or solid = 1 (if it is pure, amt. does not affect) Solution = the molar concentration Ideal gas = partial pressure in atm Reaction Quotient Reaction quotient – calculated the same way as kc, but the reaction is NOT YET at equilibrium. (mass action expression) If reaction quotient is calculated, comparing it to kc will allow prediction of the direction of the reaction. Comparing Quotient to K aA + bB cC + dD [C]c[D]d = Q [A]a[B]b When kc > Q, reaction moves forward When kc < Q, reaction moves reverse Reaction Quotient Forward reaction, [C]c[D]d = kc [A]a[B]b Reverse reaction, [A]a[B]b = 1/kc = kc’ [C]c[D]d Reversing the reaction, kc must be raised to -1 power. Manipulation of kc Multiply the equation by ½ a b c d A B C D 2 2 2 2 c 2 d 2 a 2 b 2 [C ] [ D ] Kc* = [ A] [ B ] [C ]c [ D] d = [ A] a [ B]b 1 2 = kc1/2 Whatever factor (n) is applied to the balanced equation, the new kc* is kc raised to the power of n Example 17-4, 17-5 What is kc used for? Generally used to calculate concentrations at equilibrium of various reactants and products Example 17-6 Example 17-7 Le Chatelier’s Principle O la la Le Chatlier’s Principle (1884) If a change in conditions (stress) is applied to a system at equilibrium, the system responds to reduce the stress and reach a new state of equilibrium. Le Chatelier’s Principle Major stresses include: 1. Change in Concentration (or pressure for gasses) PV = nRT P = (n/V) RT n/V = molar concentration if V is in L At constant T, P concentration so it can be used in its place in calculations. Example H2 (g) + I2 (g) 2HI (g) What happens when you add H2? What happens when you remove H2? What happens when you compress the container? Example 17-10 Le Chatelier’s Principle 2. Change in Volume – same moles, but smaller V, affects gas concentrations, but not liquid, solid, or solution. 2NO2 (g) N2O4 (g) What will happen if we decrease the volume of the container? You can increase the pressure by pumping in an inert gas, which does not increase the concentration. Le Chatelier’s Principle Effect of Temperature Change – depends on “thermicity” of reaction Thermicitiy is considering heat as a reactant or a product and writing it into the equation. 3. C (s) + O2 (g) CO2 (g) + 393.5kJ If I add heat, which way will equilibrium move? Le Chatelier’s Principle 4. Catalysts – Le Chatelier’s Principle Catalysts decrease the activation energy of both the forward and reverse reaction, therefore, no net effect on equilibrium, only on the rate of reaction. Example Problem 17-8 Example Problem 17-9 Example Problem 17-10 The Haber Process A Classic Example of Equilibrium Haber Process N2 (g) + 3H2 (g) 2NH3 (g) + 92.22kJ Is the enthalpy (H) Favorable? Is the entropy (S) Favorable? Will this reaction occur spontaneously? Example Problem 17-8 Mole % NH3 in Equilibrium Mixture oC Kc 25 3.6e8 209 650 51 82 98 467 0.50 4 25 80 758 0.014 0.5 5 13 10atm 100atm 1000at m near 0 near 0 near 0 The Haber Process Problems: N2 (g) + 3H2 (g) 2NH3 (g) + 92.22kJ 1. 2. Reaction is very slow at low temperatures, where Kc is highest Increasing the temperature increases the rate, but decreases the yield because heat is produced. The Haber Process Solutions: 1. 2. 3. 4. 5. Increase the Temperature (generally 450oC) Increase the Pressure (generally 200-1000atm) Use excess Reactants Remove the Ammonia as it is produced Use a catalyst (solid Iron Oxide, Potassium Oxide, and Aluminum Oxide) Partial Pressures and the Equilibrium Constant kp Partial Pressures For gas phase equilibria, the equilibrium constant (keq or kc) may be expressed in terms of the partial pressure (kp) Partial Pressure 2Cl2 (g) + 2H2O (g) 4HCl (g) + O2 (g) 4 4 [ HCl ] [O2 ] kc 2 2 [Cl2 ] [ N 2O] kp ( PHCl ) ( PO ) 2 2 ( PCl 2 ) ( PN 2 O ) 2 Partial Pressure If Kc or Kp are given and the other is wanted, you can use the relationship between them to calculate (PV = nRT) Scary Math Proof 4 PHCl PO2 4 RT RT PHCl PO2 kc 2 2 2 PCl2 PH 2O PCl2 PH 2O RT RT 2 1 RT x 1 RT 5 4 Scary Math Proof Explained Kc = Kp (1/RT) for this reaction Kc R T = Kp For every reaction: Kc = Kp(RT)-n Kp = Kc(RT)n Example Problem 17-11, 17-12, 17-13, 17-14 Heterogeneous Equilibria Heterogeneous Equilibria Remember: The activities of pure solids and liquids are 1 Solvents of very dilute solutions are treated as pure liquids with activity of 1 Heterogeneous Equilibria (Examples) Write Kc and Kp expressions: 1. 2. CaCO3 (s) CaO (s) + CO2 (g) kc = [CO2] kp = (PCO2) H2O (l) + CaF2 (s) Ca2+ (aq) + 2F1- (aq) kc = [Ca2+][F-]2 3. kp = undefined (no gas) 3Fe(s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) kc = [H2]4/[H2O]4 kp = (PH2)4/(PH2O)4 Relating Equilibrium to Thermodynamics Relation to Thermo Grxn = Gorxn + RTlnQ R = 8.314 J / k mol Can use (+ 2.303RT logQ) At equilibrium, Grxn = 0 and Q = k k can be related directly to Grxn Relation to Thermo 0 = Gorxn + RTlnk Gorxn = -RTlnk K is the thermodynamic equilibrium constant kc where solutions are involved kp where gases are involved Finding K at different Temperatures Van’t Hoff Equation Knowing Ho and kc at T1 allows the ESTIMATION of the kc at T2 ln(kT1/kT2) = Ho (T2-T1) R (T2T1) ln(k2/k1) = (H/R) ((1/T1) – (1/T2))