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Equilibrium Expression (Keq)

Equilibrium Expression (Keq). [ ] the brackets mean “the concentration of”. 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

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Equilibrium Expression (Keq)

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  1. Equilibrium Expression (Keq) [ ] the brackets mean “the concentration of” 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

  2. So basically concentration of products over concentration reactants raised to the power of their coefficient in balanced equation Keq = [C]c x [D]d [A]a x [B]b IMPORTANT Exclude solids and pure liquids as they do not have concentration values. Products Reactants

  3. Ex: Write Keq expression for: N2(g) + 3H2(g) ↔ 2NH3(g) All gases (nothing excluded) Keq = [NH3]2 [N2] x [H2]3

  4. 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

  5. Ex: Write Keq expression for: NaCl(s) + H2SO4(l) ↔ HCl(g) + NaHSO4(s) Keq = [HCl] Take out solids and pure liquid

  6. 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.

  7. Value of Equilibrium Constant (Keq) • If Keq = 1 • Conc. products = reactants at equilibrium • If Keq > 1 • Favors Products • Large Keq = large quantities of product at equilibrium • If Keq < 1 • Favors Reactants • Small Keq = large quantities of reactant at equilibrium Keq = [Products]x [Reactants]y http://employees.oneonta.edu/viningwj/sims/equilibrium_constant_s.html

  8. 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

  9. 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

  10. Note: • The Keq value for the “reverse” reaction will be the inverse of the “forward” reaction • Products become reactants

  11. Keq interactive • http://glencoe.com/sites/common_assets/advanced_placement/chemistry_chang10e/animations/kim2s2_5.swf

  12. 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

  13. 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

  14. Try to write a dissolution equation for CaCl2(s) CaCl2(s) ↔ Ca+2(aq) + 2Cl-1(aq)

  15. 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 Ksp = [Ag+1] x [Cl-1] Answer is the “product” of the concentrations of the ions at equilibrium or “ion product”

  16. 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

  17. Value of Ksp Much more soluble! • Higher Ksp = more soluble • Lower Ksp = less soluble • Ex: Al(OH)3 Ksp = 5 x 10-33 BaCO3 Ksp = 2 x 10-9 • Large Ksp = more solid is dissolved at equilibrium • It would also indicate a higher level of conductivity since ionics are electrolytes! • Value is temperature dependant, (usually given for 25 °C)

  18. 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.

  19. 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] Ksp = [3.9 x 10-5M] x [3.9 x 10-5M] Ksp = 1.5 x 10-9 Concentration of ions is the same. (1:1 ratio)

  20. 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 Ksp = [X] x [2X]2 = [1.9 x 10-3] x [2(1.9 x 10-3)]2 = 2.7 x 10-8 Don’t know either but one is double the other

  21. 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] 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 We don’t know either concentration but they are the same

  22. 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 of1.6 x 10-5 = .016 4 [Pb+2] = .016M, [Cl-1] = .032M

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