Bluffer’s Guide: 15.1 - Chemical Equilibrium- occurs when opposing reactions are proceeding at equal rates - At equilibrium, the rate at which products are produced from reactants equals the rate at which reactants are produced from products - For Example: Forward Reaction: A → B Rate = kf[A] Reverse Reaction: B→A Rate = kr[B] Where kf and kr are rate constants for the forward and reverse reactions, respectively. - For gaseous substances, use the ideal-gas equation to convert between concentration (in molarity, M) and pressure (in atm): PV = nRT, so M = (n/V) = (P/RT) So for substances A and B: [A] = (PA/RT) and [B] = (PB/RT) - And the rates for the forward and reverse reactions are: Forward: Rate = kf (PA/RT) Reverse: Rate = kr (PB/RT) -At equilibrium, Kf PA P = Kr B RT RT -By rearranging this equation, a constant is established: ( P B / RT ) P B k f = = = constant ( P A / RT ) P A k r - At equilibrium, the ratio of the partial pressures of A and B equals a constant. - At equilibrium, the partial pressures of A and B don’t change, aka an equilibrium mixture, but the reactants continue to reactant; this occurs because the forward and reverse reactions are occurring at the same rate, so that there is no net change in the reactants’ amounts. - The equilibrium is dynamic. -Opposing reactions naturally lead to an equilibrium situation. - ex: the Haber Process 15.2 The Equilibrium Constant The equilibrium condition can be reached from either direction The ratio of the partial pressures in the reaction A(g) B(g) has a constant value Law of mass action: relationship between concentrations of reactions and products present in equilibrium Develops the equilibrium constant expression… Look at this reaction: aA(g) + bB(g) cC(g) + dD(g) o A,B,C,D are chemicals, and a,b,c,d are coefficients o The law of mass action states (when everything is in the gas phase): o This is the equilibrium constant expression o Keq is the equilibrium constant o The expression depends only on the stoichiometry of the reaction, NOT the mechanism o When temperature is constant, Keq doesn’t vary with initial amounts of reactants and products, but only with temperature The Magnitude of Equilibrium Constants Keq>>1 equilibrium lies to the right (toward the product side, products predominate) Keq<<1 the equilibrium lies to the left (towards the reactants, reactants predominate) The Direction of the Chemical Equation and Keq Since equilibrium can be approached from either side of a reaction, the direction that the chemical equation is written is arbitrary o RXN X: aA(g) + bB(g) cC(g) + dD(g), then: o RXN Y: cC(g) + dD(g) aA(g) + bB(g), then o Keq of X = (They are reciprocals of one another) o So, for a Keq specify reaction and temperature Other Ways to Manipulate Chemical Equations and Keq Values o Raise Keq to the power of the factor that the coefficients are multiplied by If the coefficients of a reaction are doubled, square the equilibrium constant If coefficients were divided by 2, take the square root of the original Keq o If two reactions are added, multiply the equilibrium constants together Units of Equilibrium Constants o Keq is unitless BECAUSE… The numbers entered into an equilibrium constant expression are divided by a Reference Pressure (1 atm.) or Reference Concentration (1 M) ALLOWS US to use partial pressures and concentrations in the SAME expression See p. 586 for amazing explanation of unitlessness of Keq Sample Problems: p.605 #15 - 16 15. HClO2(aq) H+ + ClO2-(aq) Keq = 1.1 a) If coefficients are divided in half, take the square root of Keq. b) If coefficients are doubled, square the original Keq = 1.2 c) If the coefficients are doubled and the reverse reaction occurs, square the inverse of the original Keq. 16. Assume all components are aqueous… RXN 1: A + B C RXN 2: C + D E + A Using Hess’s Law, the reaction simplifies to: RXN 3: B + D E (A is a catalyst and C is an intermediate, so they drop out) Keq for RXN 3 = (Keq RXN 1)(Keq RXN 2) Keq = Keq = 0.16 15.3 Heterogeneous Equilibria Homogeneous equilibria- involve substances that are all in the same phase Heterogeneous equilibria- involve substances that are in different phases Pure solids, pure liquids, and solvents are not used in the equilibrium-constant expression Only partial pressures of gases and molar concentrations of substances in solution are included in the expression Pure solids and liquids which are involved in the reaction are present at equilibrium, even though they do not show up in the equilibrium-constant expression Section 15.4 To find the equilibrium concentration and Keq value complete the following: o Tabulate the known initial and equilibrium concentrations of all the species in the equilibrium-constant expressions. o For those species for which both the initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium. o Use the stoichiometry of the reaction to calculate the changes in concentration for all the other species in the equilibrium. o From the initial concentrations and the changes in concentration, calculate the equilibrium concentrations. These are used to evaluate the equilibrium constant. Example Problem. A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 degrees Celsius. The value of the equilibrium constant for the reaction at this temperature is 50.5. What are the partial pressures of H2, I2, and HI in the flask? H2(g) + I2 (g) First, calculate the initial pressures of the reactants. P =( nRT) / V For H2: For I2: Initial Change Equilibrium H2 59.19 atm -x atm 59.19-x atm I2 118.4 atm -x atm 118.4-x atm HI 0 atm +2x atm 2x atm Keq = (PHI)2 / (PH2) (PI2) Reject 137.6. Thus, x=55.3 Partial Pressures: H2= 59.19 – 55.3 = 3.9 atm I2 = 118.4 – 55.3 = 63.1 atm HI = 2(55.3) = 110.6 atm Section 15.5 The equilibrium constant allows us to… o Predict the direction in which the reaction mixture will proceed to achieve equilibrium o Calculate the concentrations of reactants and products when equilibrium has been reached -When the reactant and products partial pressures or concentrations are substituted into the equilibrium-constant expression, the result is the reaction quotient. o The letter Q represents the reaction quotient. o ***The reaction quotient is equal to the Keq value ONLY at equilibrium. o When Q > Keq substances on the right side of the chemical equation will react to form substances on the left; the reaction moves right to left to approach equilibrium. o When Q < Keq substances on the left side of the chemical equation will react to form substances on the right; the reaction moves left to right to approach equilibrium. Bluffer’s Guide for 15.6: Le Chatelier’s Principle -Le Chatelier’s principle- if a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, then the system will shift its equilibrium position so as to counteract the effect of the disturbance Change in Reactant or Product Concentrations -The addition of either a reactant or a product to a chemical system will shift to reestablish equilibrium by consuming part of the added substance -Removing either a reactant or product will cause the reaction to move in the direction that forms more of that substance -increase concentration of reactants, equilibrium shifts to right -increase concentration of products, equilibrium shifts to left -decrease concentration of reactants, equilibrium shifts to left -decrease concentration of products, equilibrium shifts to right Effects of Volume and Pressure Changes -At constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas -Pressure-volume changes do not change the value of Keq, but change the concentrations of the substances -Consider the reaction: A(g) 2B(g) -A decrease in volume (increase in pressure) shifts the reaction in the direction that produces a smaller number of moles of gas (shifts to left in above reaction) -An increase in volume (decrease in pressure) shifts reaction in direction that produces a larger number of moles of gas (shifts to right in above reaction) -If there’s no change in the number of moles of gases in a reaction, a volumepressure change does not affect the position of equilibrium Effect of Temperature Changes -When temperature is increased, it’s as if a reactant or product has been added to the system at equilibrium. The equilibrium shifts in the direction that consumes the excess reactant or product, namely heat -Endothermic: reactants + heat products; increasing the heat in an endothermic reaction results in an increase in Keq -Exothermic: Reactants products + heat; increasing the heat in an exothermic reaction results in a decrease in Keq -In an endothermic reaction, heat is absorbed as reactants are converted to products, thus increasing the temperature will cause equilibrium to shift to the right, and Keq increases. Opposite for exothermic reactions -Cooling reactions has opposite effect of heating them: cooling an endothermic reaction shifts the equilibrium to the left, decreasing Keq. Opposite for exothermic reactions The Effect of Catalysts -A catalyst increases the rate at which equilibrium is achieved, but does not change the composition of the equilibrium mixture