Download presentation
Presentation is loading. Please wait.
Published byMelvin Carpenter Modified over 8 years ago
1
Phase changes Liquid-Vapor Equilibrium liquid gas evaporation gas liquidcondensation depends on T depends on [vapor]
2
vapor pressure = 1 atmboiling point
3
Equilibrium A + B C C rate =kfkf [A][A][B][B] C A+ B rate = krkr [C] k f [A][B] = k r [C] kfkf krkr = [C] [A] [B] = K equilibrium constant
4
Law of Mass Action aA+ bB cC + dD K =[C] c [D] d [A] a [B] b O3O3 O2O2 23at 2300 o CK = 2.5 x 10 12 = [O 2 ] 3 [O 3 ] 2 if [O 2 ] = 0.50 Mat equilibrium 2.5 x 10 12 = (0.50) 3 [O 3 ] 2 [O 3 ] = 2.2 x 10 -7 M
5
aA+ bB cC + dD K =[C] c [D] d [A] a [B] b O3O3 O2O2 23at 2300 o CK = 2.5 x 10 12 = [O 2 ] 3 [O 3 ] 2 Law of Mass Action O2O2 O3O3 32 K = = [O 3 ] 2 [O 2 ] 3 1 2.5 x 10 12 K = 4 x 10 -13 large K small K excess product excess reactant
6
Law of Mass Action CO (g) + H 2 O (g) H 2 (g) +CO 2 (g) at 830 o C equilibrium concentrations [CO] = 0.20 M [H 2 O] = 0.40 M [H 2 ] = 0.30 M [CO 2 ] = 1.36 M K = (0.30)(1.36) (0.20)(0.40) = 5.10 homogeneous equilibriumall in one phase gas K = M (mol/L) gasesP PV=nRT n V = P RT K P = Δn = mol gas reactants K(RT) ΔnΔn mol gas product - K P = K
7
K(RT) ΔnΔn 2 NO (g)+ O 2 (g) 2NO 2 (g) at 230 o C equilibrium concentrations [NO] = 0.0542 M [O 2 ] = 0.127 M [NO 2 ] = 15.5 M K = (0.0542) 2 (0.127) = 6.44 x 10 5 K P = 6.44 x 10 5 (0.08206)(503 K) = 1.56 x 10 4 K and K P are dimensionless no units (15.5) 2
8
Heterogeneous Equilibria s, l, g CaCO 3 (s) CaO (s) +CO 2 (g) 3 phases K = [CO 2 ] [CaO] [CaCO 3 ] [CaO]mol L = density L g ÷ M.W. g mol constants solids don’t appear in K = [CO 2 ] pure liquids don’t appear H 2 O as a solvent
9
Predicting the direction of a reaction 2 HI (g) H 2 (g)+ I 2 (g) at 430 o CK = 54.3 start with:0.243 mol H 2 0.146 mol I 2 1.98 mol HI 1.0 L flask [HI] 2 0 [H 2 ] 0 [I 2 ] 0 0 = initial = 1.98 2 (0.243)(0.146) = 111 111 54.3 not at equilibrium too much product reverse reaction
10
Reaction Quotient, Q Q = [products] 0 [reactants] 0 K = [products] eq [reactants] eq Q > K reverse Q = Ksystem at equilibrium Q < K forward reaction proceeds
11
Reaction Quotient, Q N 2 (g) + 3H 2 (g) 2 NH 3 (g) at 200 o C, K = 0.65 [N 2 ] 0 = 0.0711 M [H 2 ] 0 = 9.17 x 10 -3 M [NH 3 ] 0 = 1.83 x 10 -4 M Q =(1.83 x 10 -4 ) 2 (9.17 x 10 -3 ) 3 (0.0711) = 0.611 reaction proceedsforward
12
Mg(OH) 2 (s) Mg 2+ + 2 OH - K = [OH - ] eq [Mg 2+ ] eq add H + reacts with OH - Q K > < Q = [OH - ] 0 [Mg 2+ ] 0 reaction proceeds forward
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.