A + B C + D Chemical Equilibrium and Equilibrium Law p.01 Chemical Equilibrium and Equilibrium Law It is the study of Reversible Rxns! A + B C + D The rxn makes products which themselves react to give back the reactants. The rxn never stops. C. Y. Yeung (CHW, 2009)

How does the Concentrations change with time in a reversible rxn? p.02 How does the Concentrations change with time in a reversible rxn? conc. time [reactant] eqm reached: no further change [reactant]  0 i.e. Not all reactants converted to products. [product] t’ At time = t’, an equilibrium is established. No further change in the concentrations. At eqm, [reactant] and [product] may not be the same.

How does the Rate change with time in a reversible rxn? p.03 How does the Rate change with time in a reversible rxn? rate time forward rate t’ rate  0 i.e. The reaction is Dynamic! backward rate At time = t’, an equilibrium is established. Rate of FWR = Rate of BWR  0 FWR and BWR do not stop, but have the same rate.

A Set of [Reactant] and [Product] at eqm. “Chemical Eqm” can exist only if: A chemical eqm occurs when we have a reversible reaction in a closed system and the conditions do not change. (temperature / concentration / pressure) When a condition changes, the eqm will shift to a new “eqm position”. A Set of [Reactant] and [Product] at eqm.

p.05 Dynamic Equilibrium: Observe the “Eqm Shift” from the Colour Change in rxn. e.g. Br2 + H2O H+ + Br- + HOBr Adding conc. OH- turns the colour lighter (paler). [If no eqm is involved, no colour change would be observed.] (OH- decreases [H+], eqm shifts forward to increase [H+], and decreases [Br2] at the same time). Adding conc. H+ turns the colour darker (deeper). [If no eqm is involved, no colour change would be observed.] ([H+] increases, eqm shifts backward to decrease [H+], and increases [Br2] at the same time). (ref. p. 92 for another example)

Calculation about Equilibrium: “Eqm Law” …. (1) p.06 Calculation about Equilibrium: “Eqm Law” …. (1) When a condition changes, the eqm would shift to a new eqm positions. e.g. N2 + 3H2 2NH3 Consider the following 3 eqm positions at temp = T: [H2]/mol dm-3 [N2]/mol dm-3 [NH3]/mol dm-3 0.50 1.00 0.087 1.35 1.15 0.412 2.44 1.85 1.27 [NH3]2/[N2][H2]3 0.060 eqm constant (Kc)

Kc = [NH3(g)]2 [N2(g)][H2(g)]3 Kc = [FeSCN2-(aq)] [Fe3+(aq)][SCN-(aq)] p.07 Calculation about Equilibrium: “Eqm Law” …. (2) e.g.1 N2(g) + 3H2(g) 2NH3(g) Kc = [NH3(g)]2 [N2(g)][H2(g)]3 e.g.2 Fe3+(aq) + SCN-(aq) FeSCN2-(aq) Kc = [FeSCN2-(aq)] [Fe3+(aq)][SCN-(aq)] e.g.3 CaCO3(s) CaO(s) + CO2(g) Kc = [CO2(g)] p. 95 Check Point 16-4

Position lies to the LEFT Position lies to the RIGHT Implication of “Kc” ? p.08 e.g.1 CH3COOH + H2O CH3COO- + H3O+ Kc = 1.0010-5 Position lies to the LEFT i.e. reactants dominate. ( CH3COOH is a weak acid) e.g.2 HCl + H2O H3O+ + Cl- Kc = 5.56108 Position lies to the RIGHT i.e. products dominate. ( HCl is a strong acid)

p. 98 Check Point 16-5A Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) Kc = at start at eqm 0.05M 0M 0.0122M (6.10/1000)(0.05) 25/1000 0.0122M 0.0378M Kc = (0.0378) (0.0122)(0.0122) = 254 M-1 Note that Kc is quite large, the FWR is more favorable.

p. 100 Check Point 16-5B PCl5(g) PCl3(g) + Cl2(g) Kc = at start (mol dm-3) at eqm (mol dm-3) 0.009 0.25 0.007 0.252 0.002 Kc = (0.252)(0.002) (0.007) = 0.072 mol dm-3 Note that Kc is quite small, the BWR is more favorable.

Calculation about Eqm involving Gases: “Kp” …. (partial pressures) e.g. N2O4(g) 2NO2(g) At 350K, the eqm pressure is 7104 Pa in a closed container. The mixture is pale brown in colour, and mole fraction of N2O4(g) = 0.46 at eqm. Find Kp. P N2O4 = 0.46(7104) = 3.22 104 Pa = 7104 – 3.22 104 = 3.78 104 Pa P NO2 (3.22 104) Kp = (3.78 104)2 = 4.44104 Pa

p. 102 Check Point 16-6 H2(g) + I2(g) 2HI(g) at start (mol) at eqm (mol) 1 0.2 0.2 1.6 total no. of mol = 0.2+0.2+1.6 = 2.0 mol P H2 = PT  0.2 2.0 = P I2 = 0.10 PT P HI = PT  1.6 2.0 = 0.80 PT Kp = = 64 (0.80)2 (0.10)(0.10)

p. 230 Q.6(a)(i),(ii) (1996 --- Kp) 2SO2(g) + O2(g) 2SO3(g) at start (mol) at eqm (mol) 3 1 1.5 0.25 1.5 Kp = (P )2 SO3 (P ) O2 SO2 (a) Expression for Kp: (b) total no. of mol = 1.5 + 0.25 + 1.5 = 3.25 mol P SO2 = 373  1.5 3.25 = P SO3 = 172 kPa Kp = (172)2 (28.7) = 0.035 kPa-1 P O2 = 373  0.25 3.25 = 28.7 kPa

p. 231 Q.10 (1998 --- Kc) N2(g) + O2(g) 2NO(g) Kc = 1.210-2 at start (mol) at eqm (mol) 2 1 2-x 1-x 2x Kc = 1.210-2 = (2x/2)2 [(2-x)/2][(1-x)/2] 4x2 (2-x)(1-x) = x = 0.077  [N2]eqm = (2 – 0.077)/2 = 0.96 mol dm-3

Assignment Study the examples in p. 98-104, p.104 Check Point 16-7 p.127 Q.1-5, 7 p. 231 Q.9(a) [due date: 9/3(Wed)] Quiz on Chemical Kinetics (Ch. 13-15) [9/3(Mon)]

p.16 Next …. Significances of Eqm Constant and Le Chatelier’s Principle (p. 110 - 123) [Tuesday]