Equilibrium Expression (Keq) Also called “Mass Action Expression” Also called “Mass Action Expression” Relates the concentration of products to reactants once equilibrium has been reached. 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 x [D] d [A] a x [B] b IMPORTANT Exclude solids and pure liquids as they do not have concentration values. Products Reactants

Ex: Write Keq expression for: N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g) All gases (nothing excluded) Keq = [NH 3 ] 2 [N 2 ] x [H 2 ] 3 [N 2 ] x [H 2 ] 3

Ex: Write Keq expression for: 2NO(g) + 2H 2 (g) ↔ N 2 (g) + 2H 2 O(l) Keq = [N 2 ] [NO] 2 x [H 2 ] 2 [NO] 2 x [H 2 ] 2 Take out pure liquid

Ex: Write Keq expression for: NaCl(s) + H 2 SO 4 (l) ↔ HCl(g) + NaHSO 4 (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.

If Keq = 1 If Keq = 1 Conc. products = reactants at equilibrium Conc. products = reactants at equilibrium If Keq > 1 If Keq > 1 Favors Products Favors Products Large Keq = large quantities of product at equilibrium Large Keq = large quantities of product at equilibrium If Keq < 1 If Keq < 1 Favors Reactants Favors Reactants Small Keq = large quantities of reactant at equilibrium Small Keq = large quantities of reactant at equilibrium Value of Equilibrium Constant (Keq) Keq = [Products] x [Reactants] y

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 [.3] 2 x [.1] 3 Favors Products Plugging in Values

Find concentration of Cl 2 at equilibrium if, [PCl 5 ] =.015M, [PCl 3 ] =.78M and Keq = 35 Find concentration of Cl 2 at equilibrium if, [PCl 5 ] =.015M, [PCl 3 ] =.78M and Keq = 35 PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g) 35 = [.78] x [“X”] [.015] [.015] [Cl 2 ] =.67M Plugging in Values

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

Keq interactive Keq interactive dvanced_placement/chemistry_chang10e/ animations/kim2s2_5.swf dvanced_placement/chemistry_chang10e/ animations/kim2s2_5.swf dvanced_placement/chemistry_chang10e/ animations/kim2s2_5.swf dvanced_placement/chemistry_chang10e/ animations/kim2s2_5.swf

ICE Problems (Honors) Keq problems where you are given INITIAL concentrations. Keq problems where you are given INITIAL concentrations. Use stoich ratios to find the CHANGE in concentration 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 Subtract this from initial concentration find the EQUILIBRIUM concentration that can then go into Keq expression Crash Course: Equilibrium Equations (mostly Honors)

Solubility Equilibrium for Ionics Ionic Solids: Dissociate when placed in solution. Dissociate when placed in solution. Positive and negative ions are pulled apart. Positive and negative ions are pulled apart. Polyatomic Ions stay together!! Polyatomic Ions stay together!! If an ionic solid dissolves in a polar liquid, this process is called dissolution. If an ionic solid dissolves in a polar liquid, this process is called dissolution. Dissolution Equation: AgCl(s) ↔ Ag +1 (aq) + Cl -1 (aq)

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

Ksp Expression Equilibrium expressions for ionic solutions are called Ksp (sp = “solubility product”). Equilibrium expressions for ionic solutions are called Ksp (sp = “solubility product”). Set up “K” expression as before Set up “K” expression as before include (g) and (aq), cross out (s) and (l) include (g) and (aq), cross out (s) and (l) 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”

Try Writing the Ksp Expression AlPO 4 Ca 3 (PO 4 ) 2 AlPO 4 (s) ↔ Al +3 (aq) + PO 4 -3 (aq) Ksp = [Al +3 ] x [PO 4 -3 ] Ca 3 (PO 4 ) 2 (s) ↔ 3Ca +2 (aq) + 2PO 4 -3 (aq) Ksp = [Ca +2 ] 3 x [PO 4 -3 ] 2

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

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. 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. 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 BaSO 4 has a conc. of 3.9 x M of Ba +2 ions, find Ksp. BaSO 4 (s) ↔ Ba +2 (aq) + SO 4 -2 (aq) Ksp = [Ba +2 ] x [SO 4 -2 ] Ksp = [3.9 x M] x [3.9 x M] Ksp = 1.5 x Concentration of ions is the same. (1:1 ratio)

If [Pb +2 ] = 1.9 x in a saturated solution of PbF 2 find Ksp. PbF 2 ↔Pb F -1 Ksp = [Pb +2 ] x [F -1 ] 2 Ksp = [X] x [2X] 2 = [1.9 x ] x [2(1.9 x )] 2 = 2.7 x Don’t know either but one is double the other

Ksp Problems (Honors) Find Solubility from Ksp If the Ksp of RaSO 4 = 4 x calculate its solubility in pure water. If the Ksp of RaSO 4 = 4 x calculate its solubility in pure water. RaSO 4 (s) ↔ Ra +2 (aq) + SO 4 -2 (aq) Ksp = [Ra +2 ] x [SO 4 -2 ] 4 x = [Ra +2 ] x [SO 4 -2 ] 4 x = [X] x [X] 4 x = X 2 X = the square root of 4 x = 6 x M We don’t know either concentration but they are the same

If Ksp of PbCl 2 = 1.6 x 10 -5, calculate solubility. PbCl 2 ↔Pb Cl -1 Ksp = [Pb +2 ] x [Cl -1 ] x = [X] x [2X] x = 4X 3 “X” the cube root of 1.6 x = [Pb +2 ] =.016M, [Cl -1 ] =.032M