Principles of Reactivity: Chemical Equilibria

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1 Principles of Reactivity: Chemical Equilibria This chapter addresses the principle of equilibrium equilibrium What can you do to reestablish equilibrium? non-equilibrium Whose principle supports this? 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 Le Châtelier s Principle N 2 (g) + 3H 2 (g) Equilibrium shifts left to offset stress 2NH 3 (g) Add NH

2 Le Châtelier s Principle Changes in Concentration continued aa + bb cc + dd Change Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) Shifts the Equilibrium left right right left 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 14.5 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

3 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 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 14.5 The Nature of the Equilibrium State Ca 2+ (aq) + 2HCO 3- (aq) CaCO 3 (s) + CO 2 (g) + H 2 O(l) Chemical equilibria are dynamic When the system is at equilibrium, the forward and reverse reactions continue, but at the same rate. 3

4 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 2 O (l) H 2 O (g) Chemical equilibrium N 2 O 4 (g) 2NO 2 (g) 14.1 N 2 O 4 (g) 2NO 2 (g) equilibrium equilibrium equilibrium Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O constant

5 Fig 15-2 Pg 709 Changes in the concentrations of NO 2 and N 2 O 4 with time. The system reaches equilibrium before either substance is fully consumed. The Equilibrium Constant (K) The equilibrium constant relates concentrations of reactants and products at equilibrium at a given temperature to a numerical constant H 2 (g) + I 2 (g) 2 HI (g) K = [products] raised to the stoich. coefficient [reactants] raised to the stoich. coefficient H 2 (g) + I 2 (g) 2 HI (g) K = [HI] 2 [H 2 ] [I 2 ] The equilibrium expression will tell you the value of the equilibrium constant as well as the concentrations of the reactants and products at equilibrium 5

6 You may write you equilibrium expression solving for either Kc or Kp. What do you think is the difference? Kc is used in expressions for solution concentrations. Include only gases and aqueous solutions. Kp is used in expression for pressure so the units are in pressure instead of concentration ***Remember to only include gases in your expression for Kp*** Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. K c = [NO 2 ]2 [N 2 O 4 ] N 2 O 4 (g) 2NO 2 (g) K p = P 2 NO 2 P N2 O 4 In most cases K c K p aa (g) + bb (g) cc (g) + dd (g) K p = K c (RT) Δn Δn = moles of gaseous products moles of gaseous reactants = (c + d) (a + b) 14.2 Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO- ][H 3 O + ] [CH 3 COOH] = K c [H 2 O] General practice not to include units for the equilibrium constant

7 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x [CaCO 3 ] [CaO] K p = P CO2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant The equilibrium constant K p for the reaction 2NO 2 (g) 2NO (g) + O 2 (g) is 158 at 1000K. What is the equilibrium pressure of O 2 if the equilibrium pressures of P NO = atm and P NO = atm? 2 P NO P O2 K p = 2 P NO 2 2 P P NO O2 = K p 2 2 P NO P O2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm 14.2 How do you solve the expression for either K or the equilibrium concentrations? ICE Table Initial concentration change Equilibrium concentration 7

8 How to set up an ice table There are two ways to solve this: either for the equilibrium constant (K) or the equilibrium concentrations H 2 (g) + I 2 (g) 2 HI (g) I C E Initial concentration Change Equilibrium concentration H 2 (g) + I 2 (g) 2 HI (g) K = [HI] 2 I C E x -x +2x x x 2x [H 2 ] [I 2 ] K = 56 K = [2x] 2 [.0175 x] [.0175-x] How do you solve this? quad 8

9 9

10 10

11 Consider the following equilibrium at 295 K: NH 4 HS (s) NH 3 (g) + H 2 S (g) The partial pressure of each gas is atm. Calculate K p and K c for the reaction? K p = P NH3 P H 2 S = x = K p = K c (RT) Δn K c = K p (RT) -Δn Δn = 2 0 = 2 T = 295 K K c = x ( x 295) -2 = 1.20 x A + B C + D A + B C + D E + F E + F K c K c K c [C][D] K c = [A][B] K c = [E][F] K c = [A][B] [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

12 What About Solving for Kp? Remember: K forward = 1/K reverse In addition: If you add two equations together, the Knet is the product of the two Ks. K net = K 1 K 2 12

13 Helpful Hints Do not include concentrations of solids in your equilibrium expression. It is a fixed amount. Do not include water in your expression unless you are solving for Kp which has H 2 O (g) in the equation. Water is present in such a large amounts that it essentially state unchanged throughout the reaction. If the value of K is very large, what does that tell you about the reaction? Fig Pg 742 Schematic representation of the relationship between K eq and the extent of reaction. (a) When K eq <<1, the difference from initial conditions, x, is small. (b) When K eq >>1, the difference from completion, y, is small. What happens if you are not at equilibrium? Instead of solving for K, you solve for Q, the reaction quotient A + B C + D Q = [C] [D] [A] [B] Looks like K, but the concentrations in the expression are not the concentrations at equilibrium 13

14 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 14.4 What would be the purpose of knowing the value of Q? If Q< K, the reaction will move to the right To re-establish equilibrium If Q> K, the reaction will move to the left To re-establish equilibrium 14

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