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Equilibrium Expression (Keq) Also called “Mass Action Expression” Relates the concentration of products to reactants once equilibrium has been reached. For this general reaction: aA + bB ↔ cC + dD Keq = [C]c x [D]d [A]a x [B]b [ ] the brackets mean “the concentration of” So basically concentration of products over concentration reactants raised to the power of their coefficient in balanced equation Keq = [C]c [D]d x [A]a x [B]b Products Reactants IMPORTANT Exclude solids and pure liquids as they do not have concentration values. Ex: Write Keq expression for: N2(g) + 3H2(g) ↔ 2NH3(g) All gases (nothing excluded) Keq = [NH3]2 [N2] x [H2]3 Ex: Write Keq expression for: 2NO(g) + 2H2(g) ↔ N2(g) + 2H2O(l) Keq = [N2] [NO]2 x [H2]2 Take out pure liquid http://www.kentchemistry.com/links/Kinetics/EquilibriumExpression.htm Ex: Write Keq expression for: NaCl(s) + H2SO4(l) ↔ HCl(g) + NaHSO4(s) Keq = [HCl] Take out solids and pure liquid Value of Equilibrium Constant (Keq) At equilibrium if you put the concentration values (Molarity) into the Keq expression you will get a specific number (The is a unitless number and is unique to that reaction.) The only thing that can change the value of Keq is a change in temperature. Value of Equilibrium Constant (Keq) If Keq = 1 If Keq > 1 Conc. products = reactants at equilibrium Favors Products Large Keq = large quantities of product at equilibrium If Keq < 1 Keq = [Products]x [Reactants]y Favors Reactants Small Keq = large quantities of reactant at equilibrium http://employees.oneonta.edu/viningwj/sims/equilibrium_constant_s.html Plugging in Values 2A(g) + 3B(aq) ↔ 2AB(g) At equilibrium [A] = .3M, [B] = .1M, [AB] = .8M find the Keq. Keq = [.8]2 = 7111 [.3]2 x [.1]3 Favors Products Plugging in Values Find concentration of Cl2 at equilibrium if, [PCl5] = .015M, [PCl3] = .78M and Keq = 35 PCl5 (g) ↔ PCl3(g) + Cl2(g) 35 = [.78] x [“X”] [.015] [Cl2] = .67M Note: The Keq value for the “reverse” reaction will be the inverse of the “forward” reaction Products become reactants Keq interactive http://glencoe.com/sites/common_assets/a dvanced_placement/chemistry_chang10e/ animations/kim2s2_5.swf ICE Problems (Honors) Keq problems where you are given INITIAL concentrations. Use stoich ratios to find the CHANGE in concentration Subtract this from initial concentration find the EQUILIBRIUM concentration that can then go into Keq expression http://www.youtube.com/watch?v=rog8ou-ZepE&safe=active Crash Course: Equilibrium Equations (mostly Honors) http://www.youtube.com/watch?v=DP-vWN1yXrY&safe=active Solubility Equilibrium for Ionics Ionic Solids: Dissociate when placed in solution. Positive and negative ions are pulled apart. Polyatomic Ions stay together!! If an ionic solid dissolves in a polar liquid, this process is called dissolution. Dissolution Equation: AgCl(s) ↔ Ag+1(aq) + Cl-1(aq) http://programs.northlandcollege.edu/biology/Biology1111/animations/dissolve.html Try to write a dissolution equation for CaCl2(s) CaCl2(s) ↔ Ca+2(aq) + 2Cl-1(aq) Ksp Expression Equilibrium expressions for ionic solutions are called Ksp (sp = “solubility product”). Set up “K” expression as before include (g) and (aq), cross out (s) and (l) AgCl(s) ↔ Ag+1(aq) + Cl-1(aq) AgCl(s) ↔ Ag+1(aq) + Cl-1(aq) Cross out solid Answer is the “product” of the Ksp = [Ag+1] x [Cl-1] concentrations of the ions at equilibrium or “ion product” Try Writing the Ksp Expression AlPO4 Ca3(PO4)2 AlPO4 (s) ↔ Al+3 (aq) + PO4-3 (aq) Ksp = [Al+3 ] x [PO4-3 ] Ca3(PO4)2 (s) ↔ 3Ca+2 (aq) + 2PO4-3 (aq) Ksp = [Ca+2]3 x [PO4-3 ]2 Value of Ksp Higher Ksp = more soluble Lower Ksp = less soluble Ex: Large Ksp = more solid is dissolved at equilibrium Al(OH)3 BaCO3 Much more soluble! Ksp = 5 x 10-33 Ksp = 2 x 10-9 It would also indicate a higher level of conductivity since ionics are electrolytes! Value is temperature dependant, (usually given for 25 °C) Just Read Ionic compounds have different degrees of solubility (none are truly insoluble as it may indicate on your ref. tables), Ksp allows us to compare solubility. This is useful when looking at how much relatively insoluble compounds will dissolve in things like drinking water or blood plasma. Ksp Problems (Honors) Find Ksp from Solubility: A sat. solution of BaSO4 has a conc. of 3.9 x 10-5M of Ba+2 ions, find Ksp. BaSO4 (s) ↔ Ba+2 (aq) + SO4-2 (aq) Ksp = [Ba+2] x [SO4-2] Concentration of ions is the same. (1:1 ratio) Ksp = [3.9 x 10-5M] x [3.9 x 10-5M] Ksp = 1.5 x 10-9 If [Pb+2] = 1.9 x 10-3 in a saturated solution of PbF2 find Ksp. PbF2 ↔ Pb+2 + 2F-1 Ksp = [Pb+2] x [F-1]2 Don’t know either but one is double the other Ksp = [X] x [2X]2 = [1.9 x 10-3] x [2(1.9 x 10-3)]2 = 2.7 x 10-8 Ksp Problems (Honors) Find Solubility from Ksp If the Ksp of RaSO4 = 4 x 10-11 calculate its solubility in pure water. RaSO4 (s) ↔ Ra+2 (aq) + SO4-2 (aq) Ksp = [Ra+2] x [SO4-2] We don’t know either concentration but they are the same 4 x 10-11= [Ra+2] x [SO4-2] 4 x 10-11= [X] x [X] 4 x 10-11= X2 X = the square root of 4 x 10-11 = 6 x 10-6 M If Ksp of PbCl2 = 1.6 x 10-5, calculate solubility. PbCl2 ↔ Pb+2 + 2Cl-1 Ksp = [Pb+2] x [Cl-1]2 1.6 x 10-5 = [X] x [2X]2 1.6 x 10-5 = 4X3 “X” the cube root of 1.6 x 10-5 = .016 4 [Pb+2] = .016M, [Cl-1] = .032M