Chapter 15 Chemical Equilibrium

15.1 The Concept of Equilibrium Most chemical reactions are reversible. reversible reaction = a reaction that proceeds simultaneously in both directions Examples: Double arrows ( ) denote an equilibrium reaction. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Equilibrium Consider the reaction At equilibrium, the forward reaction: N2O4(g)  2 NO2(g), and the reverse reaction: 2 NO2(g)  N2O4(g) proceed at equal rates. Chemical equilibria are dynamic, not static – the reactions do not stop. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Equilibrium Let’s use 2 experiments to study the reaction each starting with a different reactant(s). Exp #1 pure N2O4 Exp #2 pure NO2 Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Equilibrium Experiment #1 Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Equilibrium Experiment #2 Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Equilibrium Are the equilibrium pressures of NO2 and N2O4 related? Are they predictable? Copyright McGraw-Hill 2009

15.2 The Equilibrium Constant At equilibrium, or where Kc is the equilibrium constant Copyright McGraw-Hill 2009

The Equilibrium Constant This constant value is termed the equilibrium constant, Kc, for this reaction at 25°C. Copyright McGraw-Hill 2009

The Equilibrium Constant For the NO2 / N2O4 system: equilibrium constant expression equilibrium constant Note: at 100°C, K = 6.45 Copyright McGraw-Hill 2009

The Equilibrium Constant reaction quotient = Qc = the value of the “equilibrium constant expression” under any conditions. For, Q > K  reverse reaction favored Q = K  equilibrium present Q < K  forward reaction favored Copyright McGraw-Hill 2009

The Equilibrium Constant For a reaction: For gases: P in atm For solutions: [ ] = mol/L The Law of Mass Action: Cato Maximilian Guldberg & Peter Waage, Forhandlinger: Videnskabs-Selskabet i Christiana 1864, 35. Copyright McGraw-Hill 2009

The Equilibrium Constant Note: The equilibrium constant expression has products in the numerator, reactants in the denominator. Reaction coefficients become exponents. Equilibrium constants are temperature dependent. Equilibrium constants do not have units. (pg. 622) If K >>> 1, products favored (reaction goes nearly to completion). If K <<< 1, reactants favored (reaction hardly proceeds). Copyright McGraw-Hill 2009

15.3 Equilibrium Expressions homogeneous equilibria = equilibria in which all reactants and products are in the same phase. heterogeneous equilibria = equilibria in which all reactants and products are not in the same phase. Ex: The equilibrium constant expression is, K = [CO2] [CaO] and [CaCO3] are solids. Pure solids and liquids are omitted from equilibrium constant expressions. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise: Write the expressions for Kp for the following reactions: Solution: Copyright McGraw-Hill 2009

Equilibrium Expressions A. Reverse Equations For, For, Conclusion: Copyright McGraw-Hill 2009

Equilibrium Expressions B. Coefficient Changes For, For, Conclusion: Copyright McGraw-Hill 2009

Equilibrium Expressions C. Reaction Sum (related to Hess’ Law) For, For, Add [1] + [4], Copyright McGraw-Hill 2009

Equilibrium Expressions Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise: At 500ºC, KP = 2.5  1010 for, (a) At 500ºC, which is more stable, SO2 or SO3? Compute KP for each of the following: 1 ¾ ¾ ® (b) SO (g) + O (g) ¬ ¾ ¾ SO (g) 2 2 3 2 3 ¾ ¾ ® (c) 3 SO (g) + O (g) ¬ ¾ ¾ 3 SO (g) 2 2 3 2 1 ¾ ¾ ® (d) SO (g) ¬ ¾ ¾ SO (g) + O (g) 3 2 2 2 Copyright McGraw-Hill 2009

15.4 Using Equilibrium Expressions to Solve Problems Predicting the direction of a reaction Compare the computed value of Q to K Q > K  reverse reaction favored Q = K  equilibrium present Q < K  forward reaction favored Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #1: At 448°C, K = 51 for the reaction, Predict the direction the reaction will proceed, if at 448°C the pressures of HI, H2, and I2 are 1.3, 2.1 and 1.7 atm, respectively. Solution: 0.47 < 51  system not at equilibrium Numerator must increase and denominator must decrease. Consequently the reaction must shift to the right. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #2: At 1130°C, K = 2.59  102 for At equilibrium, PH2S = 0.557 atm and PH2 = 0.173 atm, calculate PS2 at 1130°C. Solution: PS2 = 0.268 atm Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #3: K = 82.2 at 25°C for, Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm. What are the equilibrium pressures of I2, Cl2, and ICl? Solution: Initial 2.00 atm 2.00 atm 0.00 atm Change x  x +2x Equilibrium (2.00 – x) (2.00 – x) 2x perfect square Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #3: (cont.) square root  2 x = 18.132 – 9.066 x 11.066 x = 18.132 x = 18.132 / 11.066 = 1.639 PI2 = PCl2 = 2.00 – x = 2.00 – 1.639 = 0.36 atm PICl = 2x = (2)(1.639) = 3.28 atm Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #4: At 1280°C, Kc = 1.1  103 for Initially, [Br2] = 6.3  102 M and [Br] = 1.2  102 M. What are the equilibrium concentrations of Br2 and Br at 1280°C? Solution: Initial 6.3  102 M 1.2  102 M Change -x +2x Equilibrium (6.3  102) - x (1.2  102) + 2x 4 x2 + 0.0491x + (7.47  105) = 0 Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 4 x2 + 0.0491x + (7.47  10-5) = 0 quadratic equation: a x2 + b x + c = 0 solution: x =  1.779  103 and  1.050  102 Q: Two answers? Both negative? What’s happening? Equilibrium Conc. x =  1.779  103  1.050  102 [Br2] = (6.3  102) – x = 0.0648 M 0.0735 M [Br] = (1.2  102) + 2x = 0.00844 M  0.00900 M [Br2] = 6.5  102 M [Br] = 8.4  103 M impossible Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #5: A pure NO2 sample reacts at 1000 K, KP is 158. If at 1000 K the equilibrium partial pressure of O2 is 0.25 atm, what are the equilibrium partial pressures of NO and NO2. Solution: Initial ? 0 atm 0 atm Change Equilibrium 0.25 atm 0.50 +0.50 +0.25 PNO2 +0.50 atm rearrange and solve Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #5: (cont.) = 3.956  104 PNO2 = 0.020 atm PNO = 0.50 atm see ICE table Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise #6: The total pressure of an equilibrium mixture of N2O4 and NO2 at 25°C is 1.30 atm. For the reaction: KP = 0.143 at 25°C. Calculate the equilibrium partial pressures of N2O4 and NO2. two equations and two unknowns – BINGO! PNO2 + PN2O4 = 1.30 atm Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 PNO2 + PN2O4 = 1.30 atm Exercise #6: (cont.) PN2O4 = 1.30 atm - PNO2 PNO22 + 0.143 PNO2  0.1859 = 0 Use the quadratic formula, PNO2 = +0.366 atm and 0.509 atm PN2O4 = 1.30 atm - PNO2 = 1.30  0.366 = 0.934 atm PN2O4 = 0.93 atm Copyright McGraw-Hill 2009

15.5 Factors That Affect Chemical Equilibrium Le Châtelier’s Principle “If an equilibrium system variable is changed, the equilibrium will shift in the direction (right or left) that tends to reduce the change.” Example: N2, H2, and NH3 are at equilibrium in a container at 500°C. kJ 92 H ) ( NH 2 3 N rxn = D ¾ ¬ ® + o g (continued on next 5 slides) Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Case I : Change: N2 is added Shift: ??? to the right Q: Why? Ans: [N2] has increased. Which direction will decrease [N2]? N2 decreases N2 increases right left Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Case II: Change: compress the system Shift: ??? to the right N2 H2 NH3 Q: Why? Ans: Total pressure has increased. Which direction will decrease the total pressure? Recall: P  n (4 moles gas) (2 moles gas) less gas less pressure more gas more pressure Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Case III: Change: increase the temperature Shift: ??? to the left Q: Why? Ans: Temperature has increased. Which direction decreases the temperature? Recall, the reaction is exothermic. endothermic heat absorbed right left exothermic heat evolved Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Case IV: Change: add helium at constant volume Shift: ??? none Q: Why? Ans: Helium is not a reactant or product. Adding helium (at constant V) does not change PN2, PH2 or PNH3. Hence the equilibrium will not shift. Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Case V: Change: add helium at constant total pressure Shift: ??? to the left Q: Why? Ans: If the total pressure is constant, PN2 + PH2 + PNH3 must decrease. Which direction increases this sum? Recall: P  n (4 moles gas) (2 moles gas) less gas less pressure more gas more pressure Copyright McGraw-Hill 2009

Copyright McGraw-Hill 2009 Exercise: Hydrogen (used in ammonia production) is produced by the endothermic reaction, Ni 750C Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if H2O(g) is removed. The temperature is increased. The quantity of Ni catalyst is increased. An inert gas (e.g., He) is added. H2(g) is removed. The volume of the container is tripled. Left Right None Copyright McGraw-Hill 2009