Physical Equilibrium ratevaporization ratecondensation = equilibrium Go = Ho - TSo = 0 Hof products - Hof products T = Soproducts - Soproducts molecules with high K.E. leave liquid vaporization gas phase molecules collide with surface condensation

Napoleon - 1812 Snwhite Sngrey tin buttons ΔHof (kJ/mol) So (J/mol K) white tin 0.0 51.55 grey tin -2.1 44.14 ΔHo = -2.1 - 0.0 = -2.1 kJ Snwhite Sngrey  ΔSo = 44.14 - 51.55 = -7.4 J/mol K T = -2100 J = 283 K = 10oC -7.4 J/K -20 F ∆G298 = .105 kJ ∆G244 = -.294 kJ

Chemical Equilibrium kf [A][B] = kr [C] A + B  C rate = kf [A] [B] kf [C] = = K A + B rate = kr [C] C  kr [A] [B] equilibrium constant

Law of Mass Action aA + bB  cC + dD K = [C]c [D]d [A]a [B]b 2 O3 3 O2 at 300 K Kc = K = 2.6 x 1055 = [O2]  3 [O3] 2 if [O2] = 0.50 M at equilibrium 2.6 x 1055 = (0.50)3 [O3] = 6.9 x 10-29 M [O3]2

Law of Mass Action aA + bB  cC + dD K = [C]c [D]d [A]a [B]b 2 O3 3 O2 at 300 K Kc = K = 2.6 x 1055 = [O2]  3 [O3] 2 1 3 O2  2 O3 Kc = = [O3] 2 2.6 x 1055 [O2] large K excess product 3 Kc = 3.5 x 10-56 small K  excess reactant

Law of Mass Action aA + bB  cC + dD K = [C]c [D]d K = PCc PDd PAa PBb [A]a [B]b 3 O2  O3 2 K = 3.5 x 10-56 (0.08206 x 300 K)-1 Kc = 3.5 x 10-56 K = 1.4 x 10-57 Δn K = M (mol/L) gases P K = Kc (RT) n = P PV=nRT Δn = mol gas product - V RT mol gas reactants

Law of Mass Action CO (g) + H2O (g)  H2 (g) + CO2 (g) at 830oC equilibrium concentrations [CO] = 0.20 M (0.30) (1.36) K = = 5.10 [H2O] = 0.40 M (0.20) (0.40) [H2] = 0.30 M [CO2] = 1.36 M KC = K homogeneous equilibrium all in one phase gas

Δn K = Kc (RT) 2 NO (g) + O2 (g)  2NO2 (g) at 230oC equilibrium concentrations [NO] = 0.0542 M [O2] = 0.127 M Kc = (15.5)2 = 6.44 x 105 [NO2] = 15.5 M (0.0542)2(0.127) K = 6.44 x 105 (0.08206) (503 K) -1 = 1.56 x 104 K and Kc are dimensionless no units

Heterogeneous Equilibria s, l, g 3 phases CaCO3 (s)  CaO (s) + CO2 (g) Kc = [CO2] [CaO] = [CO2] [CaCO3] L g g mol [CaO] mol L density M.W. = ÷ constants solids don’t appear in K pure liquids don’t appear H2O as a solvent

Predicting the direction of a reaction H2 (g) + I2 (g)  2 HI (g) at 430oC K = 54.3 start with: 0.243 mol H2 0.146 mol I2 1.0 L flask 1.98 mol HI [HI]20 = 1.982 = 111 = Q [H2]0 [I2]0 (0.243) (0.146) 111  54.3 not at equilibrium 0 = initial

Reaction Quotient, Q Q = [products]0 K = [products]eq [reactants]0 [reactants]eq Q > K reverse reaction proceeds Q = K system at equilibrium Q < K forward reaction proceeds

Reaction Quotient, Q N2 (g) + 3H2 (g)  2 NH3 (g) at 200o C, Kc = 1085 [N2]0 = 0.0711 M [H2]0 = 9.17 x 10-3 M [NH3]0 = 1.83 x 10-4 M Q = (1.83 x 10-4)2 = 0.611 (9.17 x 10-3)3 (0.0711)

“insoluble” salts Mg(OH)2 (s)  Mg2+ + 2 OH- Ksp = [Mg2+]eq [OH-]2eq = 1.02 x 10-11 1.02 x 10-11 = (x) (2x) 2 = 4x3 x = 1.37 x 10-4 Q = [OH-]20 [Mg2+]0 add H+ reacts with OH-