Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 1 of 33 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 15: Principles of Chemical Equilibrium

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 2 of 33 Contents 15-1Dynamic Equilibrium 15-2The Equilibrium Constant Expression 15-3Relationships Involving Equilibrium Constants 15-4The Magnitude of an Equilibrium Constant 15-5The Reaction Quotient, Q: Predicting the Direction of a Net Change 15-6Altering Equilibrium Conditions: Le Châtelier’s Principle 15-7Equilibrium Calculations: Some Illustrative Examples

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 3 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 CO(g) + 2 H 2 (g) CH 3 OH(g)

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 4 of 33 Dynamic Equilibrium

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 5 of The Equilibrium Constant Expression  Methanol synthesis is a reversible reaction. CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1 CH 3 OH(g) CO(g) + 2 H 2 (g) CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 6 of 33 Three Approaches to the Equilibrium

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 7 of 33 Three Approaches to Equilibrium

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 8 of 33 Three Approaches to Equilibrium CO(g) + 2 H 2 (g) CH 3 OH(g) k1k1 k -1

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 9 of 33 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) k 1 k -1 k1k1

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 10 of 33 General Expressions a A + b B …. → g G + h H …. Equilibrium constant = K c = [A] m [B] n …. [G] g [H] h …. Thermodynamic Equilibrium constant = K eq = (a G ) g (a H ) h …. (a A ) a (a B ) b …. [A] m [B] n …. [G] g [H] h …. (  G ) g (  H ) h …. (  A ) a (  B ) b ….  =  1 under ideal conditions

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 11 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 © 2007 General Chemistry: Chapter 15 Slide 12 of 33 Combining Equilibrium Constant Expressions N 2 O(g) + ½O 2 2 NO(g)K c = ? 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] = 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 

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 13 of 33 Gases: The Equilibrium Constant, K P  Mixtures of gases are solutions just as liquids are.  Use K P, based upon activities of gases. K P = (a SO ) 2 (a O ) (a SO ) = PP P SO 3 a SO 3 = PP P SO 2 a SO 2 = PP PO2PO2 3 K P = (P SO ) 2 (P O ) (P SO ) PP 2 SO 2 (g) + O 2 (g) 2 SO 3 (g)

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 14 of 33 Gases: The Equilibrium Constant, K C  In concentration we can do another substitution 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) [SO 3 ]= V n SO 3 = RT P SO 3 [SO 2 ]= RT P SO 2 [O 2 ] = RT PO2PO2 (a X ) = [X] c◦c◦ RT PXPX c◦c◦ = PXPX = [X] RT

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 15 of 33 Gases: The Equilibrium Constant, K C 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K P = (a SO ) 2 (a O ) (a SO ) = (P SO ) 2 (P O ) (P SO ) PP PXPX = [X] RT = ([SO 2 ] RT) 2 ([O 2 ] RT) ([SO 3 ] RT) 2 PP = ([SO 2 ]) 2 ([O 2 ]) ([SO 3 ]) 2 PP RT = KCKC Where P  = 1 bar

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 16 of 33 An Alternative Derivation 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 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 In general terms: Where P  = 1 bar

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 17 of 33 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). K c = [H 2 O] 2 [CO][H 2 ] = PH2O2PH2O2 P CO P H 2 (RT) 1 C(s) + H 2 O(g) CO(g) + H 2 (g)

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 18 of 33 Burnt Lime CaCO 3 (s) CaO(s) + CO 2 (g) K c = [CO 2 ]K P = P CO 2 (RT)

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 19 of The Significance of the Magnitude of the Equilibrium Constant.

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 20 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 m [B] t n [G] t g [H] t h

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 21 of 33 Reaction Quotient

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 22 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. What happens if we add SO 3 to this equilibrium?

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 23 of 33 Le Châtelier’s Principle Q = = K c [SO 2 ] 2 [O 2 ] [SO 3 ] 2 Q > K c K c = 2.8  10 2 at 1000K 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) k1k1 k -1

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 24 of 33 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 ] V n SO 3 2 V n SO 2 V nO2nO2 = 2 = V n SO 3 2 n SO 2 nO2nO2 2

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 25 of 33 Effect of Change in Volume K c = [C] c [D] d [G] g [H] h = V (a+b)-(g+h) nGnG a nAnA nBnB g nHnH h a V-ΔnV-Δn nGnG a nAnA nBnB g nHnH h a =  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 © 2007 General Chemistry: Chapter 15 Slide 26 of 33 Effect of the Change of Volume mol gas1.16 mol gas n tot = K P =

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 27 of 33 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 © 2007 General Chemistry: Chapter 15 Slide 28 of 33 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.

Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 29 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.  Exercise your understanding by working through the examples with a pencil and paper.