Chapter 15 Chemical Equilibrium

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

Copyright McGraw-Hill 2009 Equilibrium Consider the reaction At equilibrium, the forward reaction: N 2 O 4 (g)  2 NO 2 (g), and the reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) proceed at equal rates. Chemical equilibria are dynamic, not static – the reactions do not stop.

Copyright McGraw-Hill 2009 Equilibrium Let’s use 2 experiments to study the reaction each starting with a different reactant(s). Exp #2 pure NO 2 Exp #1 pure N 2 O 4

Copyright McGraw-Hill 2009 Equilibrium Experiment #1

Copyright McGraw-Hill 2009 Equilibrium Experiment #2

Copyright McGraw-Hill 2009 Equilibrium Are the equilibrium pressures of NO 2 and N 2 O 4 related? Are they predictable?

Copyright McGraw-Hill The Equilibrium Constant At equilibrium, or where K c is the equilibrium constant

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

Copyright McGraw-Hill 2009 The Equilibrium Constant For the NO 2 / N 2 O 4 system: equilibrium constant expression equilibrium constant Note: at 100°C, K = 6.45

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

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 Equilibrium Expressions homogeneous equilibria = equilibria in which all reactants and products are in the same phase. [CaO] and [CaCO 3 ] are solids. Pure solids and liquids are omitted from equilibrium constant expressions. Ex: The equilibrium constant expression is, K = [CO 2 ] heterogeneous equilibria = equilibria in which all reactants and products are not in the same phase.

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

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

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

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

Copyright McGraw-Hill 2009 Equilibrium Expressions

Copyright McGraw-Hill 2009 Exercise:At 500ºC, K P = 2.5  for, Compute K P for each of the following: (a) At 500ºC, which is more stable, SO 2 or SO 3 ? (g)O 2 1 SO (g)SO (d) 223    (g)SO (g)O SO (b) 322    2 1 (g)SO 3 (g)O SO 3 (c) 322    2 3

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

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, H 2, and I 2 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 Exercise #2: At 1130°C, K = 2.59  10  2 for At equilibrium, P H 2 S = atm and P H 2 = atm, calculate P S 2 at 1130°C. Solution: P S 2 = atm

Copyright McGraw-Hill 2009 Exercise #3:K = 82.2 at 25°C for, Initially, P I 2 = P Cl 2 = 2.00 atm and P ICl = 0.00 atm. What are the equilibrium pressures of I 2, Cl 2, and ICl? Solution: Initial 2.00 atm 2.00 atm0.00 atm Change  x  x +2x Equilibrium(2.00 – x)(2.00 – x) 2x perfect square

Copyright McGraw-Hill 2009 square root  2 x = – x x = x = / = P I 2 = P Cl 2 = 2.00 – x = 2.00 – = 0.36 atm P ICl = 2x = (2)(1.639) = 3.28 atm Exercise #3:(cont.)

Copyright McGraw-Hill 2009 Exercise #4: At 1280°C, K c = 1.1  10  3 for Initially, [Br 2 ] = 6.3  10  2 M and [Br] = 1.2  10  2 M. What are the equilibrium concentrations of Br 2 and Br at 1280°C? Initial6.3  10  2 M1.2  10  2 M Change -x +2x Equilibrium (6.3  10  2 ) - x(1.2  10  2 ) + 2x Solution: 4 x x + (7.47  10  5 ) = 0

Copyright McGraw-Hill x x + (7.47  ) = 0 quadratic equation:a x 2 + b x + c = 0 solution: x =   10  3 and   10  2 Q:Two answers? Both negative? What’s happening? Equilibrium Conc.x =   10  3   10  2 [Br 2 ]= (6.3  10  2 ) – x = M M [Br]= (1.2  10  2 ) + 2x = M  M impossible [Br 2 ]= 6.5  10  2 M [Br]= 8.4  10  3 M

Copyright McGraw-Hill 2009 Exercise #5:A pure NO 2 sample reacts at 1000 K, K P is 158. If at 1000 K the equilibrium partial pressure of O 2 is 0.25 atm, what are the equilibrium partial pressures of NO and NO 2. Solution: Initial ?0 atm0 atm Change Equilibrium0.25 atm atm  0.50 rearrange and solve P NO 2

Copyright McGraw-Hill 2009 Exercise #5:(cont.) =  10  4 P NO 2 = atm P NO = 0.50 atm see ICE table

Copyright McGraw-Hill 2009 Exercise #6: The total pressure of an equilibrium mixture of N 2 O 4 and NO 2 at 25°C is 1.30 atm. For the reaction: K P = at 25°C. Calculate the equilibrium partial pressures of N 2 O 4 and NO 2. P NO 2 + P N 2 O 4 = 1.30 atm two equations and two unknowns – BINGO!

Copyright McGraw-Hill 2009 P NO P NO 2  = 0 P N 2 O 4 = 1.30 atm - P NO 2 P NO 2 + P N 2 O 4 = 1.30 atm Exercise #6:(cont.) Use the quadratic formula, P NO 2 = atm and  atm P N 2 O 4 = 1.30 atm - P NO 2 = 1.30  = atm P N 2 O 4 = 0.93 atm

Copyright McGraw-Hill Factors That Affect Chemical Equilibrium “If an equilibrium system variable is changed, the equilibrium will shift in the direction (right or left) that tends to reduce the change.” Example: N 2, H 2, and NH 3 are at equilibrium in a container at 500°C. (continued on next 5 slides) Le Châtelier’s Principle kJ  92 H )(NH 2 )(H 3 )(N rxn322      ggg

Copyright McGraw-Hill 2009 Case I:Change: N 2 is added Shift: ??? to the right Q:Why? Ans:[N 2 ] has increased. Which direction will decrease [N 2 ]? N 2 decreases N 2 increases right left

Copyright McGraw-Hill 2009 Case II:Change: compress the system Shift: ??? to the right Q:Why? Ans:Total pressure has increased. Which direction will decrease the total pressure? Recall: P  n N 2 H 2 NH 3 (4 moles gas) (2 moles gas) less gas less pressure more gas more pressure

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 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 P N 2, P H 2 or P NH 3. Hence the equilibrium will not shift.

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, P N 2 + P H 2 + P NH 3 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 Exercise:Hydrogen (used in ammonia production) is produced by the endothermic reaction, 750  C Ni Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if (a)H 2 O(g) is removed. (b)The temperature is increased. (c)The quantity of Ni catalyst is increased. (d)An inert gas (e.g., He) is added. (e)H 2 (g) is removed. (f)The volume of the container is tripled. Left Right None Right