Chemical Equilibrium Forward and reverse reactions taking place at equal rates It is a dynamic state - reactions are constantly occurring

Start: 10 goldfish in the left tank and 10 guppies in the right. Equilibrium state. with 5 of each kind of fish in each tank. The equilibrium is dynamic; an averaged state and not a static condition . The fish do not stop swimming when they have become evenly mixed. If we were to observe one single fish (here a guppy among goldfish). we would find that it spends half its time in each tank .

Brightstorm videos Chemical Equilibrium Definition 5:01 http://www.youtube.com/watch?v=FYc_SoW2M40&list=PL06C3C4E3F84C6A24&index=42 Crash course chemistry http://www.youtube.com/watch?v=g5wNg_dKsYY 10:56 Equilibrium http://www.youtube.com/watch?v=DP-vWN1yXrY 9:28 Equilibrium equations You don’t need to know how to do the RICE table starting at 4:40 Isaacs Teach http://www.youtube.com/watch?v=g4TKRInLdPA 10:09 Equilibrium Good basic explanation! http://www.youtube.com/watch?v=4z4_rc6nsKU 12:46 What is the equilibrium constant, Keq? Also very good explanation

Equilibrium constant expressions aA + bB  cC + dD Keq = [C]c[D]d [A]a[B]b

General information about the Keq expression Equilibrium [ ] of products are placed in the numerator. Equilibrium [ ] of reactants are placed in the denominator. Each [ ] term is raised to an exponent equal to its coefficient in the balanced equation. If there is more than 1 product or reactant, the terms are multiplied. Solids and liquids (pure substances) are not included in the Keq expression. This is because their [ ] are their densities. The density of a substance does not change with changing temperatures.

The value of Keq is independent of the: Keq is constant for a given reaction at a given temperature. There are no units associated with the value of Keq. The value of Keq is independent of the: individual [ ] of reactants and products original [ ] of reactants and products volume of the container. The value of Keq is dependent on temperature. What does the value of Keq tell you about a reaction? Keq >1: more products than reactants at equilibrium Keq < 1: more reactants than products at equilibrium

Using equilibrium constants Calculating equilibrium concentrations: Example: At 1405 K, hydrogen sulfide, also called rotten egg gas (because of its bad odor), decomposes to form hydrogen and a diatomic sulfur molecule,S2. Keq = 2.27 x 10-3. (a) Write the balanced equation for the reaction described above. Write out the Keq expression. (b) Calculate the concentration of hydrogen gas if [S2] = 0.0540 M and [H2S] = 0.184 M.

Solving the problem – part (a) 2H2S (g)  2H2 (g) + S2 (g) Keq = [H2]2[S2] [H2S]2

Solution – (b) [H2]2= Keq [H2S]2 = [S2] (2.27 x 10-3)(0.184 M)2 = [𝐻2] 2 = 1.42 𝑥 10−3 𝑀 so [H2]  = 3.77 x 10-2 M

Le Châtelier’s principle: 1884 - Henri Le Châtelier When a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

Δ in concentration Adding more of a reactant or product: the reaction will shift in the direction to consume a portion of what was added. more reactant  shifts right more product  shifts left Removing some of a reactant or product: the reaction will shift in a direction to restore part of what was removed. reactants removed  reaction shifts left (i.e. the reverse reaction) products removed  reaction shifts right (i.e. the forward reaction).

Δ in volume Relevant when discussing gaseous equilibria and when the number of moles of gaseous reactants differ from the number of moles of gaseous products. The change in volume is a result of a change in pressure of the gaseous system. When P↓, the reaction will shift in a direction to↑ number of moles of gas. PCl5 (g)  PCl3 (g) + Cl2 (g) 1 mol  2 mol 2NH3(g)  N2(g) + 3H2(g) 2 mol  4 mol When P↑, the reaction will shift in a direction to ↓ number of moles gas. PCl5 (g)  PCl3 (g) + Cl2 (g) 1 mol  2 mol 2NH3(g)  N2(g) + 3H2(g) 2 mol  4 mol

Δ in temperature View changes in temperature as reactants or products. When the temperature of an equilibrium system is ↑ the reaction that is endothermic (ΔH>0) will take place. * forward rxn is endothermic  more product (shifts to the right). * reverse rxn is endothermic  less product (shifts to the left) When the temperature of an equilibrium system is ↓, the rxn which is exothermic (ΔH<0) will take place. * forward rxn is exothermic – more product (shifts to the right). * reverse rxn is exothermic – less product (shifts to the left) General rule: if the forward rxn is endothermic,↑K. If the forward rxn is exothermic ↓K.

Animation demonstration http://www.learnerstv.com/animation/animation.php?ani=120&cat=chemistry This has some of the actual equilibria we will be investigating later this week. Le Chatelier’s Principle – Bozeman Science http://www.youtube.com/watch?v=PciV_Wuh9V8 7:00 good explanations with visuals and excellent discussion on how to increase yield of a reaction Equilibrium Disturbances – Bozeman Science http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36 This one will help you with the lab we’re doing. He also discusses the effect of disturbances (changes) in an equilibrium system and how they affect the value of K (the equilibrium constant)

Reactions that go to completion Formation of a gas H2CO3 (aq)  H2O (l) + CO2 (g) Formation of precipitate (remember Double displacement reactions) Formation of a slightly ionized product; often times H2O (i.e. in a neutralization reaction) H3O+ + OH-  2H2O (l)

Solubility equilibria http://www.youtube.com/watch?v=YJ-dyEtB66A&feature=topics Brightstorm 4:17

The Solubility Product Constant, Ksp Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution. The solubility product constant, Ksp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.

The Solubility Equilibrium Equation And Ksp CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp = [Ca2+][F-]2 Ksp = 5.3x10-9 As2S3 (s) 2As3+ (aq) + 3S2- (aq) Ksp = [As3+]2[S2-]3 Ksp = 6 x 10-51

Ksp And Molar Solubility The solubility product constant is related to the solubility of an ionic solute, but Ksp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing. Calculating solubility equilibria fall into two categories: determining a value of Ksp from experimental data calculating equilibrium concentrations when Ksp is known.

Calculating Ksp From Molar Solubility It is found that 1.2x10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25 oC. What is the Ksp at this temperature? Solution: PbI2 (s) Pb2+ (aq) + 2I- (aq) Ksp = [Pb2+] [I-]2 Ksp = (1.2 x 10-3 M) (2 x 1.2 x 10-3 M)2 Ksp = 6.9 x10-9

Calculating Molar Solubility From Ksp Calculate the molar solubility of silver chromate, Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4. Solution: Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq) Ksp = [Ag+]2 [CrO4 2-] Ksp = (2x)2(x) = 1.1 x 10-12 4x3 = 1.1 x 10-12 X = 6.5 x 10-5 M

The Common Ion Effect In Solubility Equilibria The common ion effect also affects solubility equilibria. Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.

Solubility Equilibrium Calculation -The Common Ion Effect What is the solubility of Ag2CrO4 in 0.10 M K2CrO4? Ksp = 1.1x10-12 for Ag2CrO4. Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq) Ksp = [Ag+]2 [CrO4 2-] Ksp = (2x)2(0.10) = 1.1 x 10-12 x = 1.65 x 10-6 M Comparison of solubility of Ag2CrO4 In pure water: 6.5 x 10-5 M (prior slide) In 0.10 M K2CrO4: 1.7 x 10-6 M The common ion effect!!

Determining Whether Precipitation Occurs Q is the ion product reaction quotient and is based on initial conditions of the reaction. Q can then be compared to Ksp. To predict if a precipitation occurs: - Precipitation should occur if Q > Ksp. - Precipitation cannot occur if Q < Ksp. - A solution is just saturated if Q = Ksp. DR lab: unexpected PPT according to solubility rules! Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq) Ksp = [Ca2+][OH-]2 Ksp = 6.5 x 10-6

Determining Whether Precipitation Occurs – An Example The concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate to precipitate? Ksp = 2.3x10-9. Three steps: Determine the initial concentrations of ions. Evaluate the reaction quotient Q. Compare Q with Ksp.

Solution CaC2O4 (s) Ca2+ (aq) + C2O42- (aq) Ksp = [Ca2+] [C2O42-] = 2.3x10-9 Qsp = (2.5 x 10-3 M) (1.0x10-7 M) = 2.5 x 10-10 2.5 x 10-10 < 2.3x10-9 Q < Ksp therefore no ppt will be formed

Summary The solubility product constant, Ksp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution. The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound. The solubilities of some slightly soluble compounds depends strongly on pH.

Equilibrium lab Fe(OH)3 (s) Fe3+ (aq) + 3OH- (aq) Ksp = [Fe3+][OH-]3 = 4 x 10-38 Q vs. Ksp Q = [Fe3+][OH-]3 = (0.2M)(6.0M)3 = 43.2 Q >Ksp so a PPT forms to take the Fe3+ out of solution

Qualitative Inorganic Analysis Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis. “Qualitative” signifies that the interest is in determining what is present, not how much is present. Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.