Mg(OH) 2 (s)

1. Mg(OH)2 (s)  + 2 OH- Mg2+ [OH-]0 [Mg2+]0 K = [Mg2+]eq Q = [OH-]eq add H+ reacts with OH-

2. Le Chatelier’s Principle equilibrium balance forward reaction reverse reaction disturb balance changes in experimental conditions equilibrium shifts counteract disturbance concentration (gas phase) pressure temperature

3. Concentration Fe3+ (aq) FeSCN2+ + SCN- (aq)  add Fe(NO3)3 Q K add reactant < add NaSCN add reactant add C2O42- Q K remove Fe2+ > at equilibrium change [FeSCN2+] [FeSCN2+] K = Q = [Fe3+] [SCN-] [Fe3+] [SCN-] ratef = kf [Fe3+] [SCN-]

4. Pressure N2O4 (g) 2 NO2 (g)  at 25oC, Kc = 10.38 increase P by adding reactant or product Kc = [NO2]eq2 add N2O4 Qc = [NO2]eq2 [N2O4]eq [N2O4]i Qc Kc <

5. Pressure N2O4 (g) 2 NO2 (g)  decrease volume [N2O4] = mol N2O4 increase [N2O4] V [NO2] = mol NO2 increase [NO2] V (6.0)2 = 11.9 Qc = Kc = [NO2]2 = (3.0)2 = 10.3 (1.74) [N2O4] (0.87) Qc Kc > decrease volume decrease n increase n increase volume Δn = 0 no effect of pressure

6. PNO PNO PN O 2 2 2 4 Pressure N2O4 (g) 2 NO2 (g) add inert gas  1.00 M Ar [NO2] = 3.0 M = 3.0 mol/L increase P [N2O4] = 0.87 mol/L = 0.87 M P(1.0 L)= (4.87) (.08206) (298) PV = nRT at 298 K P (1.0 L) = (3.87) (.08206) (298) P = 119 atm P = 95 atm (95) =73 atm =(3/3.87) (119) (3/4.87) =73atm = = (.87/3.87) (95) =22atm K unchanged K = (73)2 / 22 = 242

7. Temperature changes K heat + N2O4 (g) +heat heat 2 NO2 (g) ΔH = 58.0 kJ  product reactant treat heat endothermic exothermic ΔH > 0 ΔH < 0 raising T adding heat as reactant ΔH > 0 lowering T removing heat as product ΔH < 0