Brown, LeMay Ch 15 AP Chemistry Monta Vista High School Chemical Equilibrium Brown, LeMay Ch 15 AP Chemistry Monta Vista High School

15.1: 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, PX Solutes in liquids: molarity, [X]

Forward reaction: A → B Rate = kforward [A] Reverse reaction: B → A Rate = kreverse [B] or R = 0.0821 L•atm mol•K Forward reaction: Reverse reaction:

http://www. kentchemistry. com/links/Kinetics/Equilibrium/equili brium http://www.kentchemistry.com/links/Kinetics/Equilibrium/equili brium.swf ( Good link on Equilibrium) or

Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib. And if k<1, then reactants favored at equilb. K=1 (conc. Of reactants and products nearly same at equilibrium) The magnitude of Kc gives us an indication of how far the reaction has proceeded toward the formation of products, when the equilibrium is achieved. The larger the value of K, the further the reaction will have proceeded towards completion when equilibrium is reached (more products present at equilibrium).

Equilibrium Equilibrium is dynamic. The forward and reverse rxns occur at the same rate. There is a spontaneous tendency towards equilibrium. This does not mean that equilibrium will occur quickly, it simply means that there is always a drive TOWARD the equilibrium state. The amount of drive is measured as Free Energy (D G) The driving force towards equilibrium diminishes as equilibrium is approached. Thus the appearance of products actually decreases the forward impetus of the reaction, making the free energy change less negative. http://www.youtube.com/watch?v=CMs2WhGY3NE

Equilibrium The equilibrium position is the same at a given temperature, no matter from which direction it is approached. It is possible to force an equilibrium one way or the other temporarily by altering the reaction conditions, but once this “stress” is removed, the system will return to its original equilibrium.

Figure 1: Reversible reactions [A]0 or PA0 / RT PX or [X] Time → [A] or PA / RT Equilibrium is established [B] or PB / RT

Reversible Reactions and Rate Forward rate Reaction Rate Time Equilibrium is established: Forward rate = Backward rate Backward rate When equilibrium is achieved: [A] ≠ [B] and kf/kr = Keq

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 (Keq): Cato Guldberg Peter Waage (1836-1902) (1833-1900) or But Waage and Guldberg were also related through two marriages; Guldberg married his cousin Bodil Mathea Riddervold, daughter of cabinet minister Hans Riddervold, and the couple had three daughters. Waage married Bodil's sister, Johanne Christiane Tandberg Riddervold by whom he had five children, and after her death in 1869, he became Guldberg's brother-in-law a second time, in 1870, by marrying one of Guldberg's sisters, Mathilde Sofie Guldberg, by whom he had six children. Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism). Units: Keq is considered dimensionless (no units)

Relating Kc and Kp Convert [A] into PA: where Dn = = change in coefficents of products – reactants (gases only!) = (c+d) - (a+b)

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

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). Why? Q>K, reaction proceeds to left, until Q=K. Why?

Value of K For the rxn, A>B, For the reverse rxn, B >A, For the reaction, 2A > 2B For the rxn, A > C C > B K(rxn)= [B]/[A] K= 1/K(rxn) K= K(rxn)2 K (overall)= K1 X K2

15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N2 (g) + H2 (g) ↔ NH3 (g) 1 3 2 Fritz Haber (1868 – 1934) German chemist, who received the Nobel Prize in Chemistry in 1918 for his development of synthetic ammonia, important for fertilizers and explosives. He is also credited as the "father of chemical warfare" for his work developing and deploying chlorine and other poison gases during World War I; this role is thought to have provoked his wife to commit suicide. Despite his contributions to the German war effort, Haber was forced to emigrate from Germany in 1933 by the Nazis because of his Jewish background; many of his relatives were killed by the Nazis in concentration camps, gassed by Zyklon B. Though he had converted from Judaism in an effort to become fully accepted, he was forced to emigrate from Germany by the Nazis in 1933 on account of his being Jewish in their eyes. He died in the process of emigration. The Haber process now produces 500 million tons of nitrogen fertilizer per year, mostly in the form of anhydrous ammonia, ammonium nitrate, and urea. 1% of the world's annual energy supply is consumed in the Haber process (Science 297(1654), Sep 2002). That fertilizer is responsible for sustaining 40% of the Earth's population, as well as various deleterious environmental consequences.

CaCO3 (s) ↔ CaO (s) + CO2 (g) Heterogeneous: reactants and products in different phases CaCO3 (s) ↔ CaO (s) + CO2 (g) Definition: What we use: Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548). Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

15.4: Calculating Equilibrium Constants ICE tables are used to calculate equilibrium concentrations of the reactants and products from the initial concentrations. Steps to use “ICE” table: “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression “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 “E” = Calculate the equilibrium concentrations

Sample Problem 1 on Equilibrium: Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction: NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)   Initial Change Equilibrium NH3 (aq) H2O (l) NH41+ (aq) OH1- (aq) X 0.0124 M 0 M 0 M X - x + x + x X 0.0119 M 4.64 x 10-4 M 4.64 x 10-4 M x = 4.64 x 10-4 M

Sample Problem 2 on Equilibrium Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium? H2 (g) + I2 (g) ↔ 2 HI (g)

H2 (g) + I2 (g) ↔ 2 HI (g) 1.000x10-3 M 2.000x10-3 M 0 M   Initial Change Equilibrium H2 (g) I2 (g) HI (g) 1.000x10-3 M 2.000x10-3 M 0 M - x M - x M + 2x M (1.000x10-3 – x) M (2.000x10-3 – x) M 2x M 4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6] 0 = -2.67x2 – 3.99x10-3x + 2.66x10-6 Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4 Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

15.6: 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. http://www.mhhe.com/physsci/chemistry/essentialchemist ry/flash/lechv17.swf Henri Le Châtelier (1850 – 1936) French, father was a prominent aluminum industrialist, had seven children, four girls and three boys.

4 Changes that do not affect Keq: 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(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l) Add HCl, temporarily inc forward rate Add H2O, temporarily inc reverse rate

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: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) If V of container is decreased, equilibrium shifts right. XN2 and XH2 dec XNH3 inc Since PT also inc, KP 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.

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 Keq) Ea, uncatalyzed Ea, catalyzed Energy Rxn coordinate

1 Change that does affect Keq: 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 Keq Exothermic equilibria: Reactants ↔ Products + heat Inc T increases reverse reaction rate which decreases Keq Endothermic equilibria: Reactants + heat ↔ Products Inc T increases forward reaction rate increases Keq Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l); DH=+? Inc T temporarily inc forward rate Dec T temporarily inc reverse rate

Vant Hoff’s Equation Vant Hoff’s equation shows mathematically how the equilibrium constant is affected by changes in temp. ln K2 = -d H0 rxn (1 – 1) K1 R T2 T1

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

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

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