# Chemical Equilibrium

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1 Chemical Equilibrium Chapter 14 1 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H O (l) H O (g) Chemical equilibrium N O 4 (g) NO (g) 1

2 N O 4 (g) NO (g) K = [NO ] [N O 4 ] = 4.63 x 10-3 aa + bb K = [C]c [D] d [A] a [B] b cc + dd Law of Mass Action 3 K = [C]c [D] d aa+ bb cc + dd [A] a [B] b K >> 1 K << 1 Equilibrium Will Lie to the right Favor products Lie to the left Favor reactants 4

3 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N O 4 (g) NO (g) K c = [NO ] [N O 4 ] In most cases K p = P NO P N O 4 K c K p aa (g) + bb (g) cc (g) + dd (g) K p = K c (RT) n n = moles of gaseous products moles of gaseous reactants = (c + d) (a + b) 5 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl (g) at 74 0 C are [CO] = 0.01 M, [Cl ] = M, and [COCl ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl (g) COCl (g) K c = [COCl ] [CO][Cl ] = x = 0 K p = K c (RT) n n = 1 = -1 R = T = = 347 K K p = 0 x (0.081 x 347) -1 =

4 The equilibrium constant K p for the reaction NO (g) NO (g) + O (g) is 158 at 1000K. What is the equilibrium pressure of O if the P NO = atm and P NO = 0.70 atm? K p = P NO P O P NO P P NO O = K p P NO P O = 158 x (0.400) /(0.70) = 347 atm 7 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO (g) K c = [CaO][CO ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO ] = K c x [CaCO 3 ] [CaO] K p = P CO The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 8 4

5 CaCO 3 (s) CaO (s) + CO (g) P CO = K p P CO does not depend on the amount of CaCO 3 or CaO 9 Consider the following equilibrium at 95 K: NH 4 HS (s) NH 3 (g) + H S (g) The partial pressure of each gas is 0.65 atm. Calculate K p and K c for the reaction? K p = P NH3 P H S = 0.65 x 0.65 = K p = K c (RT) n K c = K p (RT) - n n = 0 = T = 95 K K c = x (0.081 x 95) - = 1.0 x

6 A + B C + D A + B C + D E + F E + F K c K c K c K c = [C][D] [A][B] K c = [E][F] [A][B] K c = [E][F] [C][D] K c = K c x K c If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 11 N O 4 (g) NO (g) NO (g) N O 4 (g) K = [NO ] [N O 4 ] = 4.63 x 10-3 K = [N O 4 ] = 1 [NO ] K = 16 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 1 6

7 Chemical Kinetics and Chemical Equilibrium k f A + B AB k r rate f = k f [A][B] rate r = k r [AB ] Equilibrium rate f = rate r k f [A][B] = k r [AB ] k f [AB ] = K k c = r [A][B] 13 The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14 7

8 Le Châtelier s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N (g) + 3H (g) Equilibrium shifts left to offset stress NH 3 (g) Add NH 3 15 Le Châtelier s Principle Changes in Concentration continued Remove Add Remove Add aa + bb cc + dd Change Shifts the Equilibrium Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) left right right left 16 8

9 Le Châtelier s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Increase pressure Decrease pressure Increase volume Decrease volume Shifts the Equilibrium Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas 17 Changes in Temperature Le Châtelier s Principle Change Increase temperature Decrease temperature Exothermic Rx K decreases K increases Endothermic Rx K increases K decreases colder hotter 18 9

10 Le Châtelier s Principle Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner uncatalyzed catalyzed Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. 19 Le Châtelier s Principle Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no 0 10

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Practice Problems for Chem. 1B Exam 1 F2011 These represent the concepts covered for exam 1. There may be some additional net ionic equations from chem. 1A. This is not the exact exam! Sections 16.1-16.3