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Principles of Reactivity: Chemical Equilibria Chapter 16 16.1 Nature of the Equilibrium State All chemical reactions are reversible. When both the forward and reverse reactions continue to occur at equal rates but no net change is observed. Use a double arrow to show this situation. Outside forces can affect the equilibrium! 16.2 The Reaction Quotient and Equilibrium Constant For the reaction H2 + I2 2HI the relationships among the substances has been experimentally established as [HI]2/[H2][I2] We can make an “ICE” table to summarize this relationship… The Reaction Quotient and Equilibrium Constant I2 Initial conc. 0.0175 (M) 0.0175 0 Change in -0.0138 conc. as rxn proceeds (M) -0.0138 +0.0276 Equilibrium 0.0037 conc. (M) 0.0037 0.0276 Equation H2 + 2HI The Reaction Quotient and Equilibrium Constant Putting the equilibrium concentrations into the expression … [HI]2/[H2][I2] = (0.0276)2/(0.0037)(0.0037) = 56 This quotient is always the same for all experiments under these conditions no matter what the initial concentrations. The Reaction Quotient and Equilibrium Constant The concentrations of reactants and products are always related by an expression to a value called the reaction quotient. Reaction quotient = Q = [C]c[D]d/[A]a[B]b When a reaction has reached equilibrium, the reaction quotient has a constant value referred to as the equilibrium constant, K. The Reaction Quotient and Equilibrium Constant The expression is commonly referred to as the equilibrium constant expression. Product concentrations always appear in the numerator. Reactant concentrations always appear in the denominator. Each concentration is always raised to the power of its stoichiometric coefficient in the balanced equation. When the reaction has reached equilibrium, the value of the constant K depends on the particular reaction and on the temperature. Units are never given for K! Writing Equilibrium Constant Expressions Reactions involving solids, water, and pure liquids. Concentrations of any solid reactants and products are omitted from the equilibrium constant expression. The molar concentration of water (or of any liquid reactant or product) is omitted from the equilibrium constant expression. Writing Equilibrium Constant Expressions Reactions involving gases… If reactant and product quantities are given in partial pressures, then K is given the subscript p, as in Kp. H2(g) + I2(g) 2HI(g) Kp = P2HI/(PH2PI2) Writing Equilibrium Constant Expressions For reactions concerned with substances in aqueous solution, the symbol K sometimes has the subscript c for “concentration”, as in Kc. In some cases Kc and Kp may be the same, but they are often different. Practice Problem Write the equilibrium constant expression for each of the following reactions in terms of concentrations: PCl5(g) PCl3(g) + Cl2(g) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Cu(NH3)42+(aq) Cu2+(aq) + 4NH3(aq) CH3CO2H(aq) + H2O(l) CH3CO2-(aq) + H3O+(aq) Meaning of the Equilibrium Constant, K K >> 1: Reaction is product-favored. Equilibrium concentrations of products are greater than equilibrium concentrations of reactants. “The reaction has gone to completion.” K<< 1: Reaction is reactant-favored. Equilibrium concentrations of reactants are greater than equilibrium concentrations of products. Meaning of the Reaction Quotient, Q If Q < K, the system is not at equilibrium and some reactants will be converted to products. If Q > K, the system is not at equilibrium and some products will be converted to reactants. If Q = K, the system is at equilibrium. Practice Problem Answer the following questions regarding the butane isobutane equilibrium (K = 2.50 at 298K). Is the system at equilibrium when [butane] = 0.97 M and [isobutane] = 2.18 M? If it is not at equilibrium, in which direction will the reaction proceed in order to achieve equilibrium? Is the system at equilibrium when [butane] = 0.75 M and [isobutane] = 2.60 M? If it is not at equilibrium, in which direction will the reaction proceed in order to achieve equilibrium? Practice Problem At 2000K the equilibrium constant, K, for the formation of NO(g)… N2(g) + O2(g) 2NO(g) is 4.0 x 10-4. If, at 2000K, the concentration of N2 is 0.50 M, that of O2 is 0.25 M, and that of NO is 4.2 x 10-3 M, is the system at equilibrium? If not, predict which way the reaction will proceed to achieve equilibrium. 16.3 Determining an Equilibrium Constant If the experimental values of the concentrations of all the reactants and products at equilibrium are known, an equilibrium constant can be calculated by substituting the data into the equilibrium constant expression. Determining an Equilibrium Constant If not all concentrations are known, use an ICE table to show the initial concentrations, how they change on proceeding to equilibrium, and the concentrations at equilibrium. Practice Problem a. b. A solution is prepared by dissolving 0.050 mol of diiodocyclohexane, C6H10I2, in the solvent CCl4. The total solution volume is 1.00 L. When the reaction C6H10I2 C6H10 + I2 has come to equilibrium at 35oC, the concentration of I2 is 0.035 M. What are the concentrations of C6H10I2 and C6H10 at equilibrium? Calculate Kc, the equilibrium constant. Homework After reading sections 16.1-16.3, you should be able to do the following… P. 752 (1-10) 16.4 Using Equilibrium Constants The value of K and the initial amounts of reactants are often known, and you are asked to calculate the quantities present at equilibrium. Practice Problem At some temperature, K = 33 for the reaction H2(g) + I2(g) 2HI(g) Assume the initial concentration of both H2 and I2 are 6.00 x 10-3 mol/L. Find the concentration of each reactant and product at equilibrium. Using Equilibrium Constants in Calculations If the solution is a quadratic expression… K= 1.20 = x2/(1.60-x) x2 + 1.20x -1.92 = 0 Use appdx A to find roots to equation Negative values aren’t chemically meaningful! So, answer is x = 0.91. PCl5 PCl3 + Cl2 I 1.60 0 0 C -x +x +x E 1.60-x x x Practice Problem Graphite and carbon dioxide are kept at 1000 K until the reaction C(graphite) + CO2(g) 2CO(g) has come to equilibrium. At this temperature, K = 0.021. The initial concentration of CO2 is 0.012 mol/L. Calculate the equilibrium concentration of CO. Is there an easier way? If K is very small, then you can use another expression… If 100K < [A]o the approximate expression will give acceptable values of equilibrium concentrations. In general, when K is about 1 or greater, the approximation cannot be made. This means you can drop the “-x” off of your expression! Practice Problem The reaction N2 + O2 2NO contributes to air pollution whenever a fuel is burned in air at a high temperature, as in a gasoline engine. At 1500K, K=1.0x10-5. Suppose a sample of air has [N2] = 0.80M and [O2] = 0.20M before any reaction occurs. Calculate the equilibrium concentrations of reactants and products after the mixture has been heated to 1500K. 16.5 Balanced Equations and Equilibrium Constants When the stoichiometric coefficients of a balanced equation are multiplied by some factor, the equilibrium constant for the new equation (Knew) is the old equilibrium constant (Kold) raised to the power of the multiplication factor. The equilibrium constants for a reaction and its reverse are the reciprocals of each other. When two or more chemical equations are added to produce a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants for the added equations. Practice Problem a. b. The conversion of oxygen to ozone has a very small equilibrium constant. 3/ O (g) O (g) -29 K = 2.5 x 10 2 2 3 What is the value of K when the equation is written using whole number coefficients? What is the value of K for the conversion of ozone to oxygen? Practice Problem The following equilibrium constants are given at 500 K: H2(g) + Br2(g) 2HBr(g) Kp = 7.9x1011 H2(g) 2H(g) Kp = 4.8x10-41 Br2(g) 2Br(g) Kp = 2.2x10-15 Calculate Kp for the reaction of H and Br atoms to give HBr. H(g) + Br(g) HBr(g) Kp = ? 16.6 Disturbing a Chemical Equilibrium A change in any factor that determines the equilibrium conditions of a system will cause the system to change to counteract the effect of the change. (Le Chatelier’s principle) Change temperature Change concentrations of reactants or products Change volume (for gas system) Disturbing a Chemical Equilibrium When temp of a system at equilibrium increases, rxn shifts in direction that absorbs heat energy (endothermic direction) If temp decreases, rxn shifts in direction that releases heat energy (exothermic direction) Either way, equilibrium composition will change and therefore K will change Practice Problem Consider the effect of temperature changes on the following equilibria. Does the concentration of NOCl increase or decrease at equilibrium as the temperature of the system is increased? 2NOCl(g) 2NO(g) + Cl2(g) ΔH = +77.1kJ Does the concentration of SO3 increase or decrease when the temperature increases? 2SO2(g) + O2(g) 2SO3(g) ΔH = -198 kJ Effect of Addition or Removal of Reactant/Product If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually. The value of K will still be the same! Practice Problem Equilibrium exists between butane and isobutane when [butane] = 0.20M and [isobutane] = 0.50M. What are the concentrations of butane and isobutane, after equilibrium has been established, if 2.00 mol/L of isobutane is added to the original mixture? K = 2.50 Effect of Volume Changes on GasPhase Equilibria The stress of a volume decrease (pressure increase) will be counterbalanced by a change in the equilibrium composition to one having a smaller number of gas molecules. For a volume increase (pressure decrease), the equilibrium will shift toward the side of rxn with the greater number of gas molecules. If there is no change in the number of gas molecules in the reaction, then a volume change will have no effect. Practice Problem The formation of ammonia from its elements is an important industrial process. 3H2(g) + N2(g) 2NH3(g) How does the equilibrium composition change when extra H2 is added? When extra NH3 is added? What is the effect on the equilibrium when the volume of the system is increased? Does the equilibrium composition change or is the system unchanged? Homework After reading sections 16.4-16.6, you should be able to do the following… P. 753 (15-20,25-28)