1 Chemical Equilibrium Ch 13 AP Chemistry

2 13.1: Chemical Equilibrium Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of reaction Ex: Vapor pressure: rate of vaporization = rate of condensation Saturated solution: rate of dissociation = rate of crystallization Expressing concentrations: Gases: partial pressures, P X Solutes in liquids: molarity, [X]

3 Forward reaction: A → BRate = k forward [A] Reverse reaction:B → ARate = k reverse [B] or Forward reaction: Reverse reaction: R = Latm molK

4 If equilibrium: A ↔ B forward rate = reverse rate or

5 P X or [X] Time → [B] or P B / RT Equilibrium is established Figure 1: Reversible reactions [A] or P A / RT [A] 0 or P A 0 / RT

6 Reversible Reactions and Rate Reaction Rate Time Backward rate Forward rate Equilibrium is established: Forward rate = Backward rate When equilibrium is achieved: [A] ≠ [B] and k f /k r = K eq

7 13.2: 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

8 Relating K c and K p Convert [A] into P A : where  n = = change in coefficents of products – reactants (gases only!) = (c+d) - (a+b)

9 Magnitude of K eq Since K eq  [products]/[reactants], the magnitude of K eq predicts which reaction direction is favored: If K eq > 1 then [products] > [reactants] and equilibrium “lies to the right” If K eq < 1then [products] < [reactants] and equilibrium “lies to the left”

10 Relationship Between Q and K Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient. For a reaction, A  B, Q= [B]/[A] At equilibrium, Q= K Reaction Direction: Comparing Q and K Q<K, reaction proceeds to right, until equilibrium is achieved (or Q=K) Q>K, reaction proceeds to left, until Q=K

11 Value of K For the reference rxn, A  >B, For the reverse rxn, B  >A, For the reaction, 2A  > 2B For the rxn, A  > C C  > B K(ref)= [B]/[A] K= 1/K(ref)K= K(ref) 2 K (overall)= K 1 X K 2

: 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)

13 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.

: 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

15 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)

16 NH 3 (aq) + H 2 O (l) ↔ NH 4 1+ (aq) + OH 1- (aq) Initial Change Equilibrium 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

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

18 H 2 (g) + I 2 (g) ↔ 2 HI (g) Initial Change Equilibrium 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)

: Le Châtelier’s Principle If a system at equilibrium is disturbed by a change in: Concentration of one of the components, Pressure, or Temperature …the system will shift its equilibrium position to counteract the effect of the disturbance. Henri Le Châtelier (1850 – 1936)

20 4 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. 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 Add H 2 O, temporarily inc reverse rate

21 2. 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.

22 3. Pressure, but not Volume Usually addition of a noble gas 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.

23 4. 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

24 1 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

25 Vant Hoff’s Equation Vant Hoff’s equation shows mathematically how the equilibrium constant is affected by changes in temp. ln K 2 = -d H 0 rxn (1 – 1) K 1 R T 2 T 1

26 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

27 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

28 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