15 GENERAL CHEMISTRY Principles of Chemical Equilibrium Chemistry 140 Fall 2002 TENTH EDITION GENERAL CHEMISTRY Principles and Modern Applications PETRUCCI HERRING MADURA BISSONNETTE 15 Principles of Chemical Equilibrium In this chapter, you will learn how to deduce and write chemical formulas and how to use the information incorporated into chemical formulas. The chapter ends with an overview of the relationship between names and formulas—chemical nomenclature. PHILIP DUTTON UNIVERSITY OF WINDSOR DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Principles of Chemical Equilibrium Chemistry 140 Fall 2002 Principles of Chemical Equilibrium Contents 15-1 Dynamic Equilibrium 15-2 The Equilibrium Constant Expression 15-3 Relationships Involving Equilibrium Constants 15-4 The Magnitude of an Equilibrium Constant 15-5 The Reaction Quotient A: Predicting the Direction of Net Change 15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle 15-7 Equilibrium Calculations: Some Illustrative Examples A vital natural reaction is in progress in the lightning bolt seen here: N2(g) + O2(g) = 2NO(g). Usually this reversible reaction does not occur to any significant extent in the forward direction, but in the high-temperature lightning bolt it does. At equilibrium at high temperatures, measurable conversion of N2(g) and O2(g) to NO(g) occurs. In this chapter we study the equilibrium condition in a reversible reaction and the factors affecting it. In this chapter, our emphasis will be on the equilibrium condition reached when forward and reverse reactions proceed at the same rate. Our main tool in dealing with equilibrium will be the equilibrium constant. We will begin with some key relationships involving equilibrium constants; then we will make qualitative predictions about the condition of equilibrium; and finally we will do various equilibrium calculations. As we will discover throughout the remainder of the text, the equilibrium condition plays a role in numerous natural phenomena and affects the methods used to produce many important industrial chemicals. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Chemistry 140 Fall 2002 15-1 Dynamic Equilibrium Equilibrium –opposing processes taking place at equal rates. H2O(l) H2O(g) NaCl(s) NaCl(aq) H2O A yellow-brown saturated solution of I2 in water (top layer) is brought into contact with colorless CCl4(l) (bottom layer), I2 molecules distribute themselves between the H2O(l) and CCL4(l). When equilibrium is reached, [I2] in the CCl4 (violet, bottom layer) is about 85 times as great as in the water (colorless, top layer). I2(H2O) I2(CCl4) CO(g) + 2 H2(g) CH3OH(g) FIGURE 15-1 Dynamic equilibrium in a physical process General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-2 The Equilibrium Constant Expression The oxidation-reduction reaction of copper(II) and tin(II) in aqueous solution is reversible. k1 2 Cu2+(aq) + Sn2+(aq) 2 Cu+(aq) + Sn4+(aq) k-1 2 Cu+(aq) + Sn4+(aq) 2 Cu2+(aq) + Sn2+(aq) k1 2 Cu2+(aq) + Sn2+(aq) 2 Cu+(aq) + Sn4+(aq) k-1

General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Chemistry 140 Fall 2002 FIGURE 15-3 Three approaches to equilibrium in the reaction 2 Cu2+(aq) + Sn2+(aq) 2 Cu+(aq) + Sn4+(aq) General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

The Equilibrium Constant and Activities k-1 General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Activity Thermodynamic concept introduced by Lewis. Dimensionless ratio referred to a chosen reference state. aB = B[B] cB0 cB0 is a standard reference state = 1 mol L-1 (ideal conditions) = B[B] Accounts for non-ideal behaviour in solutions and gases. An effective concentration. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Activity A similar expression applies to gases aB = BPB PB0 PB0 is a standard reference state = 1 bar (ideal conditions) = BPB General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Reconsider the equilibrium between Cu2+ and Sn2+ Chemistry 140 Fall 2002 Reconsider the equilibrium between Cu2+ and Sn2+ Then we choose the value c°=1 mol/L ,we substitute these relationships into the equilibrium constant expression to get the expression on the next slide General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Reconsider the equilibrium between Cu2+ and Sn2+ Chemistry 140 Fall 2002 Reconsider the equilibrium between Cu2+ and Sn2+ The approximation made in writing expression (15.6) is that the values of all γ the are equal to one. This is equivalent to assuming that the ions behave ideally in solution (see page 587) General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

A general expression for K Chemistry 140 Fall 2002 A general expression for K a A + b B …. → g G + h H …. [G]g[H]h …. Equilibrium constant = Kc= [A]m[B]n …. Thermodynamic Equilibrium constant (aG)g(aH)h …. [G]g[H]h …. 1 Δn 1 Δn Keq= ≈ = Keq (aA)a(aB)b …. c° [A]m[B]n …. c° General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-3 Relationships Involving the Equilibrium Constant Relationship of K to the Balanced Chemical Equation Reversing an equation causes inversion of K. Multiplying by coefficients by a common factor raises the equilibrium constant to the corresponding power. Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Combining Equilibrium Constant Expressions N2O(g) + ½O2 2 NO(g) Kc= ? [N2][O2]½ [N2O] = N2(g) + ½O2 N2O(g) Kc(2)= 2.710+18 [N2][O2] [NO]2 = N2(g) + O2 2 NO(g) Kc(3)= 4.710-31 Kc= [N2O][O2]½ [NO]2 = [N2][O2]½ [N2O] [N2][O2] [NO]2 Kc(2) 1 Kc(3) = = 1.710-13 General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Gases: The Equilibrium Constant, KP Mixtures of gases are solutions just as liquids are. Use KP, based upon activities of gases. KP = (aSO )2(aO ) (aSO )2 3 2 2 SO2(g) + O2(g) 2 SO3(g) = P PSO3 aSO 3 = P PSO2 aSO 2 = P PO2 aSO 3 KP = (PSO )2(PO ) (PSO )2 3 2 P General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Gases: The Equilibrium Constant, KC In concentration we can do another substitution 2 SO2(g) + O2(g) 2 SO3(g) [SO3]= V nSO3 = RT PSO3 PSO2 PO2 [SO2]= [O2] = RT RT PX [X] RT PX = [X] RT (aX ) = = c◦ c◦ General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Gases: The Equilibrium Constant, KC 2 SO2(g) + O2(g) 2 SO3(g) (aSO )2 (PSO )2 PX = [X] RT KP = = P 3 3 (aSO )2(aO ) (PSO )2(PO ) 2 2 2 2 ([SO3] RT)2 = P ([SO2] RT)2([O2] RT) ([SO3])2 KC P = = Where P = 1 bar RT ([SO2])2([O2]) RT General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

An Alternative Derivation 2 SO2(g) + O2(g) 2 SO3(g) Where P = 1 bar RT PSO3 2 PSO2 PO2 = Kc = [SO2]2[O2] [SO3] = RT PSO3 2 PSO2 PO2 Kc = KP(RT) KP = Kc(RT)-1 In general terms: KP = Kc(RT)Δn General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Pure Liquids and Solids Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids). C(s) + H2O(g) CO(g) + H2(g) Kc = [H2O]2 [CO][H2] = PH2O2 PCOPH2 (RT)1 General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Chemistry 140 Fall 2002 Burnt Lime CaCO3(s) CaO(s) + CO2(g) Kc = [CO2] KP = PCO2(RT) General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-4 The Significance of the Magnitude of the Equilibrium Constant. Chemistry 140 Fall 2002 15-4 The Significance of the Magnitude of the Equilibrium Constant. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change. k1 CO(g) + 2 H2(g) CH3OH(g) k-1 Equilibrium can be approached various ways. Qualitative determination of change of initial conditions as equilibrium is approached is needed. Qc = [A]tm[B]tn [G]tg[H]th At equilibrium Qc = Kc General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 15 Reaction Quotient General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle When an equilibrium system is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change. What happens if we add SO3 to this equilibrium? General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Le Châtelier’s Principle 2 SO2(g) + O2(g) 2 SO3(g) k1 k-1 Kc = 2.8102 at 1000K Q = = Kc [SO2]2[O2] [SO3]2 Q > Kc General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Effect of Condition Changes Adding a gaseous reactant or product changes Pgas. Adding an inert gas changes the total pressure. Relative partial pressures are unchanged. Changing the volume of the system causes a change in the equilibrium position. V nSO3 2 nSO2 nO2 = Kc = [SO2]2[O2] [SO3] = V nSO3 2 nSO2 nO2 General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Effect of Change in Volume [G]g[H]h nG g nH h Kc = = V(a+b)-(g+h) [C]c[D]d nA a nB a V-Δn nG a nA nB g nH h = When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Effect of the Change of Volume ntot = 1.16 mol gas 1.085 mol gas KP = 415 338 General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Effect of Temperature on Equilibrium Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction. Lowering the temperature causes a shift in the direction of the exothermic reaction. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

Effect of a Catalyst on Equilibrium A catalyst changes the mechanism of a reaction to one with a lower activation energy. A catalyst has no effect on the condition of equilibrium. But does affect the rate at which equilibrium is attained. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

15-7 Equilibrium Calculations: Some Illustrative Examples. Five numerical examples are given in the text that illustrate ideas that have been presented in this chapter. Refer to the “comments” which describe the methodology. These will help in subsequent chapters. Exercise your understanding by working through the examples with a pencil and paper. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.

End of Chapter Questions Problems can be solved by eliminating errors from your approach. There may be nothing wrong with your strategy, but for some reason the problem is not solving. Be willing to make errors. Be able to recognize them. General Chemistry: Chapter 15 Copyright © 2011 Pearson Canada Inc.