Chapter 13 Chemical Equilibrium REVERSE REACTION reciprocal K 1 ADD REACTIONS ADD REACTIONS Multiply Ks Multiply Ks -8.4 -8.4 LE CHATELIER’S PRINCIPLE LE CHATELIER’S PRINCIPLE CO2 + H2 H2O(g) + CO a drying agent is added to absorb H2O Shift to the right. Continuous removal of a product will force any reaction to the right H2(g) + I2(g) 2HI(g) Some nitrogen gas is added No change; N2 is not a component of this reaction system. 2 LE CHATELIER’S PRINCIPLE LE CHATELIER’S PRINCIPLE + – NaCl(s) + H2SO4(l) Na2SO4(s) + HCl(g) AgCl(s) Ag (aq) + Cl (aq) reaction is carried out in an open container some NaCl is added to the solution Because HCl is a gas that can escape from the system, Shift to left due to increase in Cl– concentration. the reaction is forced to the right. This is known as the common ion effect on solubility. This is the basis for the commercial production of hydrochloric acid. H2O(l) H2O(g) N2 + 3 H2 2 NH3 water evaporates from an open container a catalyst is added to speed up this reaction No change. Continuous removal of water vapor forces the reaction to the right, Catalysts affect only the rate of a reaction; so equilibrium is never achieved they have no effect at all on the composition of the equilibrium state LE CHATELIER’S PRINCIPLE REACTION QUOTIENT, Q hemoglobin + O2 oxyhemoglobin K is thus the special value that Q has when the reaction is at equilibrium Take up in lungs at high O2 pressure Release in cells at low O2 concentration Br2(g) 2 Br (g) Pressure increased shift to left To reduce number of molecules or atoms 3 REACTION QUOTIENT, Q REACTION QUOTIENT, Q K is thus the special value that Q has when the reaction is at equilibrium X X THERMODYNAMICS and equilibrium THERMODYNAMICS and equilibrium 1. The equilibrium constant of an endothermic reaction (ΔH° = +) increases if the temperature is raised. 2. The equilibri um constant of an exothihermic react ion (ΔH° = −) decreases if the temperature is raised. NB: understand this from Le Chatelier’s principle! 4 K IS DIMENSIONLESS! HABER-BOSCH: N +3H+ 3 H 2NH2 NH + E 2 2 3 •Concentrations in mol/liter (M) •pressures in atmospheres (atm) •ignore solids •ignore solvents Equilibrium calculation EXAMPLE Equilibrium calculation EXAMPLE X 0.001 mol Br2 0.001 mol Br2 - - 0.005 mol IO3 0.005 mol IO3 0.02 mol Br- 0.02 mol Br- First calculate Q to know the direction 1.00 mol H+ 1.00 mol H+ Solid I2 Solid I2 What will be the concentrations at equilibrium? So which way does it go? 5 Equilibrium calculation EXAMPLE SOLUBILITY PRODUCT Ksp 0.001 mol Br2 - 0.005 mol IO3 0.02 mol Br- Make an ICE table 1.00 mol H+ Ksp = equilibrium constant Solid I Solve for x 2 of a reaction that forms a precipitate SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp COMMON ION EFFECT C C + S 6 SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp COMMON ION EFFECT 2+ 3 3- 2 -26 Ksp = [Ca ] [PO4 ] = 1.0 x 10 2+ 3 3- 2 -26 Ksp = [Ca ] [PO4 ] = 1.0 x 10 3 2 -26 = (3x)3(2x)2 = 1.0 x 10-26 = (3x) (0.10 + 2x) = 1.0 x 10 X X 7 I¯ I¯ X SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp SEPARATION BY PRECIPITATION SEPARATION BY PRECIPITATION Which will form a precipitate first? Starting with 0.01 M of each, 2+ can you precipitate 99.99% of Hg2 Higher or lower Ksp? without losing any Pb2+? BrO3¯ X SOLUBILITY PRODUCT Ksp Gas – solution eaquilibrium KH SEPARATION BY PRECIPITATION Henry’s Law When (BrO —) is added to a solution containing 3 CO2 dissolves in water: + + equal concentrations of Ag and Pb2 , which will precipitate first and why? -2 CO2(g) + H2O <==> H2CO3 (aq) KH = 3.4 x 10 Ksp = 5.49 x 10-5 for AgBrO -4 3 at a CO2 pressure of 3 x 10 atmospheres, -5 Ksp = 3. 23 x 10 for Pb(BrO 3)2 whihhat is the concentrat ion o fhf the car bibonic ac idihid in the water? 10-5 M 8.