Chapter 15 Chemical Equilibrium Lecture Presentation Chapter 15 Chemical Equilibrium James F. Kirby Quinnipiac University Hamden, CT

15.1 The Concept of Equilibrium Exothermic Reaction Irreversible reactions A + B C + D Reversible reactions Chemical equilibria can be established for reversible reactions.

15.1 The Concept of Equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Physical equilibrium H2O (l) H2O (g) Chemical equilibrium N2O4 (g) 2NO2 (g) Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. At equilibrium, the conversion is still going on, but the two rates are equal.

H2O(g) + CO(g) H2(g) + CO2(g) EXERCISE #1 Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) H2(g) + CO2(g) Predict how the concentration of each chemical changes compare to its original concentration after the equilibrium is reestablished if a) You add more H2O(g) to the flask. [CO] decreases [H2] and [CO2] increase [H2O] increases but less than the total (initial + added) H2O b) You add more H2 to the flask. [CO2] decreases [H2O] and [CO] increase [H2] increases but less than the total (initial + added) H2 Copyright © Cengage Learning. All rights reserved

15.2 The Equilibrium Constant Equilibrium Constant, K Expresses the ratio between products and reactants. aA + bB cC + dD [C]c[D]d For a reaction at particular temperature, the equilibrium constant K: K = [A]a[B]b Equilibrium constant tells us whether products or reactants are favored at equilibrium. Equilibrium Will K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants K = 1 − Favor both products and reactants

Conclusions About the Equilibrium Expression 15.3 The Equilibrium Constant Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. Kf B [B] [A] A [A] [B] = 1 [B] [A] = 1 Kf Kf = Kr = Kr When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus Knew = (Koriginal)n. Korig [B] [A] A B Korig = Knew [B] 2 [A] 2 2A 2B Knew = = (Korig)2

15.3 The Equilibrium Constant K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there can be many many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. K values are usually written without units. Copyright © Cengage Learning. All rights reserved

15.4 Homogeneous Equilibria Homogeneous Equilibrium Applies to reactions in which all reacting species are in the same phase. N2O4 (g) 2NO2 (g) In most cases NO2 P2 Kc  Kp [NO2]2 Kc = Kp = [N2O4] N2O4 P aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Kc : concentration in molarity, mol/L P= (n/V)RT At constant temp, the pressure is related to the concentration, mol/L. Dn = (c + d) – (a + b) R = 0.08206 L·atm/K·mol T = temperature (in Kelvin)

15.4 Homogeneous Equilibria Homogeneous Equilibrium involving Pure Liquid Kc CH3COOH (aq) + H2O (l) CH3COO− (aq) + H3O+ (aq) [CH3COO−][H3O+] Kc = ‘ [H2O] = constant [CH3COOH][H2O] [CH3COO−][H3O+] Kc = [CH3COOH]

15.4 Heterogeneous Equilibria Heterogeneous Equilibrium Applies to reactions in which reactants and products are in different phases. Kc CaCO3 (s) CaO (s) + CO2 (g) [CaO][CO2] [CaCO3] = constant [CaO] = constant Kc = ‘ [CaCO3] Kc = [CO2] Kp = PCO 2 PCO does not depend on the amount of CaCO3 or CaO 2 Conclusion The concentrations or pressures of pure solids, pure liquids and solvent do not appear in the equilibrium constant expressions.

b) HF(aq) + H2O(l) H3O+(aq) + F‒(aq) EXERCISE #2 Write the expressions for Kc and Kp, if applicable, for the following reversible reactions at equilibrium: a) 2NO2 (g) 2NO (g) + O2 (g) [O2] [NO]2 ( P O 2 ) (P NO )2 Kc = Kp = [NO2]2 (P NO 2 )2 b) HF(aq) + H2O(l) H3O+(aq) + F‒(aq) [F−] [H3O+] Kc = No Kp [HF] c) AgCl(s) Ag+(aq) + Cl‒(aq) Kc = [Ag+] [Cl−] No Kp

15.5 Calculating Equilibrium Constants EXERCISE #3 The equilibrium concentrations for the reaction below at 74oC are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] CO [Cl2] = 0.14 0.012 0.054 Kc = = 220 Kp = Kc(RT)∆n R = 0.08206 L∙atm/K∙mol T = 74oC = 347 K ∆n = 1 – 2 = −1 Kp = Kc(RT)∆n = (220) (0.08206 x 347) −1 = 7.7

15.5 Calculating Equilibrium Constants EXERCISE #4 The equilibrium constant Kp for the reaction below is 158 at 1000 K. What is the equilibrium pressure of O2 if at equilibrium, P NO 2 = 0.400 atm and P NO = 0.270 atm? 2NO2(g) 2NO(g) + O2(g) (P NO )2 (P O 2 ) (P NO 2 )2 Kp = K p (P NO 2 )2 (P NO )2 = 158 0.400 2 0.270 2 P O 2 = = 347 atm

15.6 Applications of the Equilibrium Constant What does equilibrium constant tell us? 1. The extent of a reaction. K tells us whether products or reactants are favored at equilibrium (K>1, K=1, K<1). 2. K helps us to: calculate the equilibrium concentrations of reactants and products once the equilibrium has been achieved. predict the direction in which a reaction mixture will proceed to achieve equilibrium.

15.6 Applications of the Equilibrium Constant Reaction Quotient, Q Q tells us the current ratio between products and reactants. Helps predict the direction of a reaction. Used when all of the initial concentrations are nonzero. aA + bB cC + dD Kc = [C]c[D]d [A]a[B]b [C]0[D]0 c d [A]0, [B]0, [C]0, [D]0 are initial concentrations ≠ 0 Qc = [A]0[B]0 a b Qc > Kc system shifts to the left consuming products and forming reactants, until equilibrium is achieved. Qc = Kc the system is at equilibrium. No shift will occur. Qc < Kc system shifts to the right consuming reactants and forming products, to attain equilibrium.

EXERCISE #5 A 1.00 mol sample of N2O4(g) is placed in a 10.0-L vessel and allowed to reach equilibrium according to the equation: Calculate the equilibrium concentrations of N2O4(g) and NO2(g). N2O4(g) 2NO2(g) Kc = 4.00 × 10‒4

EXERCISE #5 [N2O4]eq? [NO2]eq? N2O4(g) 2NO2(g) 2 Kc = 4.00 x 10‒4 Initial (M) Change (M) Equilibrium (M) n (N2O4) = 1.00 mol 0.1 V = 10.0 L −x +2x = 1.00 mol 10.0 L [N2O4]0 = n V = 0.1 M 0.1 – x 2x Kc << [N2O4]0  the change in [N2O4] is negligible. [N2O4]eq ≈ 0.1 M [NO 2 ]2 [N 2 O 4 ] ≈ 2x 2 (0.1) Kc = = 4.00 x 10‒4 x = 0.0032 M % error = x [N 2 O 4 ]o x 100% = 0.0032 0.1 x 100% = 3.2% < 5% ok [N2O4] = 0.1 – x = 0.1 M – 0.0032 M = 0.097 M [NO2] = 2x = 2(0.0032 M) = 6.32 x 10‒3 M

EXERCISE #6 Consider the formation of ammonia from nitrogen and hydrogen: The equilibrium constant Kc for the reaction is 1.0 at 350oC. At the start of a reaction, there are 8.0 mol N2, 4.0 mol H2, and 2.0 mol NH3 in a 2.0-L reaction vessel. N2(g) + H2(g) 2NH3(g) a) Predict the direction of the reaction. Is Q less than, greater than, or the same as K? M = n V Qc = [NH 3 ]o2 [N 2 ]o H 2 o = 1.0 2 (4.0)(2.0) = 0.13 2.0 mol 2.0 L [NH3]0 = = 1.0 M Qc << Kc Proceeds from left to right 8.0 mol 2.0 L [N2]0 = = 4.0 M 4.0 mol 2.0 L [H2]0 = = 2.0 M

15.7 Factors that Affect Chemical Equilibrium H2O(g) + CO(g) H2(g) + CO2(g) 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 concentration/s volume and pressure temperature Adding a catalyst Copyright © Cengage Learning. All rights reserved

15.7 Factors that Affect Chemical Equilibrium Changes in Concentration/s Remove Add Remove Add aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

15.7 Factors that Affect Chemical Equilibrium Changes in Pressure and Volume A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas Change of pressure will not affect the concentration of reacting species in condensed phase, e.g. liquid and solid. Addition of inert gas does not affect the equilibrium position.

a) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) b) PCl5(g) PCl3(g) + Cl2(g) EXERCISE #7 Predict the direction of the net reaction in each of the following reversible reactions if the pressure of the system is increased (by decreasing the volume) at constant temperature. a) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) b) PCl5(g) PCl3(g) + Cl2(g) c) H2(g) + CO2(g) H2O(g) + CO(g) nreac > nprod nreac < nprod nreac = nprod nothing happens P shifts to the side with fewest moles of gas

15.7 Factors that Affect Chemical Equilibrium Changes in Temperature Change Exothermic Rxn Endothermic Rxn Increase temperature K decreases K increases Decrease temperature K increases K decreases A (g) B (g) + energy K Exothermic rxn: At equilibrium, there will be more B than A At equilibrium, there will be more A than B energy + A (g) B (g) K Endothermic rxn: At equilibrium, there will be more A than B At equilibrium, there will be more B than A A temperature increase favors an endothermic reaction, whereas a temperature decrease favors an exothermic reaction.

15.7 Factors that Affect Chemical Equilibrium Adding a Catalyst Catalyst lowers Ea for both forward and reverse reactions. does not change equilibrium constant K does not shift the position of an equilibrium system system will reach equilibrium sooner uncatalyzed catalyzed

15.7 Factors that Affect Chemical Equilibrium Summary of Changes and Le Châtelier’s principle Change Equilibrium Constant Change Shift Equilibrium Concentration yes no Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products

END OF CHAPTER 15