Chapter 15 Chemical Equilibrium

15.1

Cold Temp Hot Temp 15.1

15.2: Law of Mass Action Derived from rate laws by Guldberg and Waage (1864) For a balanced chemical reaction in equilibrium: a A + b B ↔ c C + d D Equilibrium constant expression (K eq ): K eq is strictly based on stoichiometry of the reaction (is independent of the mechanism). Units: K eq is considered dimensionless (no units) Cato Guldberg Peter Waage ( ) ( ) or

15.2 **Relates Kc to Kp

For Example Write the equilibrium expression K c for the following reactions: 15.2

Example #2 In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C: N 2 (g) + 3H 2 (g) 2NH 3 (g) Calculate Kp for this reaction at this temperature Hint..you will need to use

What does the Equilibrium constant tell us?? 15.2

Looking at reversible reactions So far we have only written the expression forwards, but it can also be written equally backwards! The constants are the recipricals of each other Kc = 0.212Kc =

15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N 2 (g) + H 2 (g) ↔ NH 3 (g) 321 Fritz Haber (1868 – 1934) 15.3

Heterogeneous: different phases CaCO 3 (s) ↔ CaO (s) + CO 2 (g) Definition:What we use: Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium. Concentrations of pure solids and pure liquids are not included in K eq expression because their concentrations do not vary, and are “already included” in K eq 15.3

For Example Write equilibrium-constant expressions for Kc and Kp for each of the following reactions: 15.3 Click for answers

15.4: Calculating Equilibrium Constants Steps to use “ICE” table: 1.“I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression 2.“C” = Determine the concentration change for the species where initial and equilibrium are known Use stoichiometry to calculate concentration changes for all other species involved in equilibrium 3.“E” = Calculate the equilibrium concentrations

Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH 1- ] is 4.64 x M. Calculate K eq at 25ºC for the reaction: NH 3 (aq) + H 2 O (l) ↔ NH 4 1+ (aq) + OH 1- (aq)

Initial Change Equilibriu m M - x M 0 M + x 4.64 x M NH 3 (aq) H 2 O (l) NH 4 1+ (aq)OH 1- (aq) X X X x = 4.64 x M

Ex: A L flask is filled with x mol of H 2 and x mol of I 2 at 448 º C. The value of K eq is What are the concentrations of each substance at equilibrium? H 2 (g) + I 2 (g) ↔ 2 HI (g)

Initial Change Equilibriu m 1.000x10 -3 M - x M (1.000x10 -3 – x) M 2.000x10 -3 M0 M - x M+ 2x M (2.000x10 -3 – x) M2x M 4x 2 = 1.33[x 2 + (-3.000x10 -3 )x x10 -6 ] 0 = -2.67x 2 – 3.99x10 -3 x x10 -6 Using quadratic eq’n: x = 5.00x10 -4 or –1.99x10 -3 ; x = 5.00x10 -4 Then [H 2 ]=5.00x10 -4 M; [I 2 ]=1.50x10 -3 M; [HI]=1.00x10 -3 M H 2 (g)I 2 (g)HI (g)

15.6 Le Chatelier’s Principle

Changes that do not affect K eq : 1.Concentration Upon addition of a reactant or product, equilibrium shifts to re- establish equilibrium by consuming part of the added substance. Addition of solids/liquids do not appreciably shift the system and can be ignored. But they are still made/consumed! Upon removal of reactant or product, equilibrium shifts to re- establish equilibrium by producing more of the removed substance. Ex: Co(H 2 O) 6 2+ (aq) + 4 Cl 1- ↔ CoCl 4 2- (aq) + 6 H 2 O (l) Add HCl, temporarily inc forward rate

Volume, with a gas present (T is constant) Upon a decrease in V (thereby increasing P), equilibrium shifts to reduce the number of moles of gas. Upon an increase in V (thereby decreasing P), equilibrium shifts to produce more moles of gas. Ex: N 2 (g) + 3 H 2 (g) ↔ 2 NH 3 (g) If V of container is decreased, equilibrium shifts right.  X N2 and X H2 dec  X NH3 inc Since P T also inc, K P remains constant.

3. Pressure, but not Volume Usually addition of a noble gas, p. 560 Avogadro’s law: adding more non-reacting particles “fills in” the empty space between particles. In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.

Catalysts Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates. Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the K eq ) Energy Rxn coordinate E a, uncatalyzed E a, catalyzed

Change that does affect K eq : Temperature: consider “heat” as a part of the reaction Upon an increase in T, endothermic reaction is favored (equilibrium shifts to “consume the extra heat”) Upon a decrease in T, equilibrium shifts to produce more heat. Effect on K eq 1.Exothermic equilibria: Reactants ↔ Products + heat Inc T increases reverse reaction rate which decreases K eq 2.Endothermic equilibria: Reactants + heat ↔ Products Inc T increases forward reaction rate increases K eq Ex: Co(H 2 O) 6 2+ (aq) + 4 Cl 1- ↔ CoCl 4 2- (aq) + 6 H 2 O (l);  H=+? Inc T temporarily inc forward rate Dec T temporarily inc reverse rate

Effect of Various Changes on Equilibrium DisturbanceNet Direction of Rxn Effect of Value of K Concentration Increase (reactant)Towards formation of product None Decrease(reactant)Towards formation of reactant None Increase (product)Towards formation of reactant None Decrease (product)Towards formation of product None

Effect of Pressure on Equilb. Pressure Increase P (decrease V) Towards formation of fewer moles of gas None Decrease P (Increase V) Towards formation of more moles of gas None Increase P ( Add inert gas, no change in V) None, concentrations unchanged None DisturbanceDirection of ReactionEffect of K

Effect of Temperature on Equilb Temperature Increase T Towards absorption of heat Increases if endothermic Decreases if exothermic Decrease TTowards release of heat Increases if exothermic Decreases if endothermic Catalyst AddedNone, forward and reverse equilibrium attained sooner None DisturbanceDirection of ReactionEffect of K

For Example For the following reaction 5 CO(g) + I 2 O 5 (s) I 2 (g) + 5 CO 2 (g) kJ for each change listed, predict the equilibrium shift and the effect on the indicated quantity. ChangeDirection of Shift (  ;  ; or no change) Effect on Quantity Effect (increase, decrease, or no change) (a)decrease in volumeKcKc No Change (b)raise temperatureamount of CO(g) Increase (c)addition of I 2 O 5 (s) No Change amount of CO(g) No Change (d)addition of CO 2 (g)amount of I 2 O 5 (s) Increase (e)removal of I 2 (g)amount of CO 2 (g) Increase