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Chemical Equilibrium and Equilibrium Law

Explore the concept of chemical equilibrium and the equilibrium law, including the study of reversible reactions, dynamic equilibrium, and the calculation of equilibrium constants. Learn how conditions affect the equilibrium position and how to interpret equilibrium constants.

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Chemical Equilibrium and Equilibrium Law

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

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

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

  4. “Chemical Eqm” can exist only if: p.04 When a condition changes, the eqm will shift to a new “eqm position”. A chemical eqm occurs when we have a reversible reaction in a closed system and the conditions do not change. A Set of [Reactant] and [Product] at eqm. (temperature / concentration / pressure)

  5. Dynamic Equilibrium: Observe the “Eqm Shift” from the Colour Change in rxn. p.05 e.g. Br2 + H2O H+ + Br- + HOBr Adding conc. OH- turns the colour lighter (paler). Adding conc. H+ turns the colour darker (deeper). [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). [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)

  6. p.06 When a condition changes, the eqm would shift to a new eqm positions. e.g. N2 + 3H2 2NH3 eqm constant (Kc) Calculation about Equilibrium: “Eqm Law” …. (1) Consider the following 3 eqm positions at temp = T:

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

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

  9. p.09 Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) Note that Kc is quite large, the FWR is more favorable. 0.05M 0.05M 0M at start 0.0122M at eqm (0.0378) Kc = (0.0122)(0.0122) (6.10/1000)(0.05) 25/1000 = 254 M-1 p. 98 Check Point 16-5A 0.0378M 0.0122M

  10. p.10 PCl5(g) PCl3(g) + Cl2(g) Note that Kc is quite small, the BWR is more favorable. 0.25 0.009 0 at start (mol dm-3) 0.007 0.252 at eqm (mol dm-3) (0.252)(0.002) Kc = (0.007) = 0.072 mol dm-3 p. 100 Check Point 16-5B 0.002

  11. Calculation about Eqm involving Gases: “Kp” …. (partial pressures) p.11 e.g. N2O4(g) 2NO2(g) = 7104 – 3.22 104 = 3.78 104 Pa P P N2O4 NO2 (3.78 104)2 Kp = = 4.44104 Pa (3.22 104) 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. = 0.46(7104) = 3.22 104 Pa

  12. p. 102 Check Point 16-6 p.12 H2(g) + I2(g) 2HI(g) 1 1 0 at start (mol) at eqm (mol) P = PT = 0.10 PT H2 0.2 1.6 (0.80)2 = P I2 P = PT = 0.80 PT 2.0 (0.10)(0.10) 2.0 HI Kp = = 64 0.2 0.2 1.6 total no. of mol = 0.2+0.2+1.6 = 2.0 mol

  13. p.13 2SO2(g) + O2(g) 2SO3(g) 0 1 3 at start (mol) (P )2 (P )2 at eqm (mol) SO3 SO2 Kp = (P ) O2 0.25 1.5 = P SO3 P = 373  = 172 kPa 3.25 3.25 SO2 (172)2 P Kp = = 373  = 28.7 kPa O2 (172)2 (28.7) = 0.035 kPa-1 p. 230 Q.6(a)(i),(ii) (1996 --- Kp) 1.5 1.5 0.25 (a) Expression for Kp: (b) total no. of mol = 1.5 + 0.25 + 1.5 = 3.25 mol

  14. p.14 N2(g) + O2(g) 2NO(g) 0 1 2 at start (mol) at eqm (mol) (2x/2)2 4x2 = Kc = 1.210-2= [(2-x)/2][(1-x)/2] (2-x)(1-x) p. 231 Q.10 (1998 --- Kc) Kc = 1.210-2 2x 2-x 1-x x = 0.077  [N2]eqm = (2 – 0.077)/2 = 0.96 mol dm-3

  15. p.15 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)]

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

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