Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 15: Principles of Chemical Equilibrium

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 2 of 27 Contents 15-1Dynamic Equilibrium 15-2The Equilibrium Constant Expression 15-3Relationships Involving Equilibrium Constants 15-4The Significance of the Magnitude of an Equilibrium Constant 15-5The Reaction Quotient, Q: Predicting the Direction of a Net Change

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 3 of 27 Contents 15-6Altering Equilibrium Conditions: Le Châtelier's Principle 15-7Equilibrium Calculations: Some Illustrative Examples Focus On The Nitrogen Cycle and the Synthesis of Nitrogen Compounds

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 4 of Dynamic Equilibrium Equilibrium – two opposing processes taking place at equal rates. H 2 O(l)  H 2 O(g) NaCl(s)  NaCl(aq) H2OH2O I 2 (H 2 O)  I 2 (CCl 4 ) CO(g) + 2 H 2 (g)  CH 3 OH(g)

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 5 of The Equilibrium Constant Expression Forward:CO(g) + 2 H 2 (g) → CH 3 OH(g) Reverse:CH 3 OH(g) → CO(g) + 2 H 2 (g) At Equilibrium: R fwrd = k 1 [CO][H 2 ] 2 R rvrs = k -1 [CH 3 OH] R fwrd = R rvrs k 1 [CO][H 2 ] 2 = k -1 [CH 3 OH] [CH 3 OH] [CO][H 2 ] 2 = k1k1 k -1 = K c CO(g) + 2 H 2 (g)  CH 3 OH(g) k1k1 k -1 k1k1

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 6 of 27 Three Approaches to Equilibrium

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 7 of 27 Three Approaches to the Equilibrium

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 8 of 27 Three Approaches to Equilibrium [CH 3 OH] [CO][H 2 ] 2 K c(1) =14.5 M -2 K c(2) =14.4 M -2 K c(3) =14.2 M -2 K c = CO(g) + 2 H 2 (g)  CH 3 OH(g) k1k1 k -1

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 9 of 27 General Expressions a A + b B …. → g G + h H …. Equilibrium constant = K c = [A] a [B] b …. [G] g [H] h …. Thermodynamic Equilibrium constant = K eq = (a G ) g (a H ) h …. (a A ) a (a B ) b …. a B = [B] cBocBo c B o is a standard reference state = 1 mol l -1 (ideal conditions) =  B [B]

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 10 of Relationships Involving the Equilibrium Constant 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 11 of 27 Combining Equilibrium Constant Expressions N 2 O(g) + ½O 2 = 2 NO(g)K c = ? N 2 (g) + ½O 2 = N 2 O(g)K c(2) = 2.7 · N 2 (g) + O 2 = 2 NO(g)K c(3) = 4.7· Kc=Kc= [N 2 O][O 2 ] ½ [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O][N 2 ][O 2 ] [NO] 2 K c(2) 1 K c(3) = = 1.7· [N 2 ][O 2 ] [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O] =

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 12 of 27 Gases: The Equilibrium Constant, K P Mixtures of gases are solutions just as liquids are. Use K P, based upon partial pressures of gases. 2 SO 2 (g) + O 2 (g) = 2 SO 3 (g) K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 [SO 3 ]= V n SO 3 = RT P SO 3 [SO 2 ]= V n SO 2 = RT P SO 2 [O 2 ] = V nO2nO2 = RT PO2PO2

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 13 of 27 The Equilibrium Constant, K P 2 SO 2 (g) + O 2 (g) = 2 SO 3 (g) K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 RT P SO 3 2 RT P SO 2 RT PO2PO2 = 2 = P SO 3 2 P SO 2 PO2PO2 2 K c = K P (RT)K P = K c (RT) -1 K P = K c (RT) Δn

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 14 of 27 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) + H 2 O(g) = CO(g) + H 2 (g) K c = [H 2 O] [CO][H 2 ] = PH2OPH2O P CO P H 2 (RT) -1

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 15 of 27 Burnt Lime CaCO 3 (s) = CaO(s) + CO 2 (g) K c = [CO 2 ]K P = P CO 2 (RT)

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 16 of The Significance of the Magnitude of the Equilibrium Constant.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 17 of The Reaction Quotient, Q: Predicting the Direction of Net Change. CO(g) + 2 H 2 (g)  CH 3 OH(g) k1k1 k -1 Equilibrium can be approached various ways. Qualitative determination of change of initial conditions as equilibrium is approached is needed. At equilibrium Q c = K c Q c = [A] t a [B] t b [C]tc[D]td[C]tc[D]td

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 18 of 27 Reaction Quotient

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 19 of 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 20 of 27 Le Châtelier’s Principle Q = = K c [SO 2 ] 2 [O 2 ] [SO 3 ] 2 Q > K c 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) k1k1 k -1 K c = 2.8 · 10 2 at 1000K

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 21 of 27 Effect of Condition Changes Adding a gaseous reactant or product changes P gas. 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. K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 V n SO 3 2 V n SO 2 V nO2nO2 = 2 = V n SO 3 2 n SO 2 nO2nO2 2

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 22 of 27 Effect of change in volume K c = [A]a[B]b[A]a[B]b [G] g [H] h = V (a+b)-(g+h) nGnG a nAnA nBnB g nHnH h b V-ΔnV-Δn nGnG a nAnA nBnB g nHnH h b = 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 23 of 27 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 24 of 27 Effect of a Catalyst on Equilibrium A catalyist 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 25 of 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.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 26 of 27 Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds. N 2 (g) + O 2 (g)  2 NO(g) k1k1 k -1 K P = at 298K and 1.3 x at 1800K

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 27 of 27 Synthesis of Ammonia The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.

Prentice-Hall © 2002General Chemistry: Chapter 16Slide 28 of 27 Chapter 16 Questions Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.