N2 + O2 2NO K1 = 2.0 x 10-25 2NO + O2 2NO2 K2 = 6.4 x 109 N2 + 2O2 2NO2
K3 = ???? 2 [NO]2 [NO2]2 [NO2] K1K2= x = =K 2 [N2] [O2] [NO] [O2] [N2][O2] 2 3 2.0 x 10-25 x 6.4 x 109 = 1.28 x 10-15
2NO + O2 2NO2 2NO2 2NO + O2 Kc = 6.4 x 109 Kc = ? K prime 1 = 1.6 x 10-10 Kc =
Kc 2NO + O2 2NO2 Kc = 6.4 x 109 4NO + 2O2 4NO2 Kc = ? K double prime
2 Kc = Kc = 4.1 x 1019 2NO + O2 2NO2 NO + O2 NO2 Kc = 6.4 x 109 Kc = ?
K triple prime Kc= 4 = 8.0 x 10 Kc O3 + CO CO2 + O2
Rxn has K=16 Starts with 9 O3 & 12 CO [CO2][O2] [CO [x] 2][O [x]2] K= = 16 K= [O3] [CO] [O [9]3] [CO] [12]
Values NOT at equilibrium!! - O3 + CO 9 12 CO2 + O2 0
0 Initially NO PRODUCTS Rxn MUST move FORWARD 9-x 12-x +x +x (x) (x) = 16 (9-x) (12-x) + Reactants Products
[x] [x] 16 = [9-x] [12-x] 0 = 15x2 - 336x + 1728 x = 8 or 14.4 [NOCl]o = 1.0 M 2 NOCl Kc = 1.6 x 10-5 2NO + Cl2
Initial 1.0 M 0 0 Change -2x +2x +x
If amount is not stated assumed ZERO in most cases 1.0 2x 2x x [NO]2 [Cl2] (2x)2 (x) -5 = = 1.6
x 10 K= [NOCl]2 (1.0 2x)2 Equilibrium (2x) (x) -5 = 1.6 x 10 (1.0 2x)2 [ ] is 100 times
2 o larger (4x) = 1.6 x 10-5 1.0 3 x = 0.016 with small K values, you can
cross xs that are + or - @ equilibrium [NO] = 2x = 2(0.016) = 0.032 M [Cl2] = x = 0.016 M [NOCl] = 1.0 2x = 1.0 2(0.016) = 0.97M [NOCl]o = 2.0 M [Cl2]o = 1.0 M Kc = 1.6 x 10-5 2 NOCl 2NO + Cl2 2.0 M
0 1.0 M -2x +2x +x 2.0 2x 2x 1.0 + x [NO]2 [Cl2] (2x)2(1.0 + x)
-5 = = 1.6 x 10 K= [NOCl]2 (2.0 2x)2 (2x) (1.0 + x) -5 = 1.6 x
10 (2.0 2x)2 2 (2x)2 (1.0) -5 =1.6 x 10 2 (2.0) x = 0.0040 @ equilibrium [NO] = 2x = 2(0.0040) = 0.0.0080 M [Cl2] = 1.0 + x = 1.0 + 0.0040 1.0 M
[NOCl] = 2.0 2x = 2.0 2(0.0040) = 1.992 M can predict how certain changes in a reaction will affect the position of equilibrium REMOVES STRESS from the RXN Rxn will SHIFT AWAY from the ADDITION and SHIFT TOWARDS the SUBTRACTION Rxn will SHIFT AWAY from the ADDITION and SHIFT TOWARDS the SUBTRACTION CuCl2(H2O)2 + 2H2O Cu(H2O)42+ + 2Cl-
system will shift away from the added component or towards a removed component N2 + 3H2 2NH3 Ex: if more N2 is added, then equilibrium position shifts to right if some NH3 is removed, then equilibrium position shifts to right adding or removing gaseous reactant or product is same as changing conc. adding inert or uninvolved gas increase the total pressure doesnt effect the equilibrium position
changing the volume decrease V decrease in # gas molecules shifts towards the side of the reaction with less gas molecules increase V increase in # of gas molecules shifts towards the side of the reaction with more gas molecules all other changes alter the concentration at equilibrium position but dont actually change value of K value of K does change with temperature
if energy is added, the reaction will shift in direction that consumes energy treat energy as a reactant: for endothermic reactions product: for exothermic reactions N2(g) + 3H2(g) 2NH3(g) + Energy increase add NH3 temp to left decrease add H2 volume
to to right right add Ne gas remove N2 no shift to left Increase P Must produce LESS gas
P4(s) + 6Cl2(g) 4PCl3(l) decrease remove Clvolume 2 to right left increase volume add Kr gas to noleft shift add P PCl
4 3 to right left energy + N2(g) + O2(g) 2NO(g) endo increase or exo? volume no H=181 shift kJ decrease endothermic temp
increase to left temp to right N2(g) + 3H2(g) 2NH3(g) 2 NOCl 2NO + Cl2 9 12 0
0 -x -x +x +x 9-x 12-x
x x Video Quiz 1) Did the lecturer have Glasses or a Mustache? 2) There was a clipart object during a problem House Explosion Plant Bonus Element Symbol of element that he pronounces really oddly.