1. EQUILIBRIUM CONSTANT K C 1. Equilibrium constant in terms of concentration, K c 2.For a reversible reaction at equilibrium, aP + bQ  cR + dS reactants products K c = [products] m [reactants] n = [R] c [S] d [P] a [Q] b

2. 3.For the reverse reaction, cR + dS  aP + bQ reactants products K c ’ = [P] a [Q] b [R] c [S] d = 1/K c

3. 4.K c is constant as long as temperature is constant 5.Magnitude of K c indicates extent of a chemical reaction Eg large K c [products] > [reactants] - almost complete reaction 6.K c gives no information about rate of reaction

4. 7.K c may or may not have a unit Example : CH 3 COOH + C 2 H 5 OH  CH 3 COOC 2 H 5 + H 2 O K c = [CH 3 COOC 2 H 5 ][H 2 O] [CH 3 COOH][C 2 H 5 OH] unit → (mol dm -3 ) (mol dm -3) (mol dm -3 ) (mol dm -3 ) No unit

5. N 2 O 4 (g)  2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] = (mol dm -3 ) 2 mol dm -3 Unit : mol dm -3

6. 8.Concentration of a pure solid is constant, not included in K c expressions Concentration expressed as amount per unit volume For solid – mass per unit volume → density Density is constant, concentration of solid is constant

7. Example : CaCO 3 (s)  CaO(s) + CO 2 (g) K c = [CaO(s)] [CO 2 ] [CaCO 3 (s)] constant = constant [CO 2 ] constant K c ’ = [CO 2 ] mol dm -3

8. Position of equilibrium Refers to proportion of products to reactants Eg : aP + bQ  cR + dS reactants products Position to the left, [products] < [reactants] K c = [ product ] [ reactant ] K c is small Position to the right, [products]>[reactants] K c is large

9. Examples a. K c = [Cr 2+ ] 2 [Fe 2+ ] moldm -3 [Cr 3+ ] 2 b. K c = [SO 3 ] 2 mol -1 dm 3 [SO 2 ] 2 [O 2 ]

10. Exercises (1) K c = [Sn 4+ ] [Fe 2+ ] 2 no unit [Sn 2+ ] [Fe 3+ ] 2 (2) K c = [Fe 3+ ] mol -1 dm 3 [Ag + ] [Fe 2+ ]

11. EQUILIBRIUM CONSTANT K p 1. For reactions involving gases, equilibrium usually expressed in terms of partial pressures ( K p ) K p = ( P products ) n ( P reactants ) m Note : P A = mole fraction of A x P total Mole fraction of A = n (A)/total no of moles of gases

12. Examples : a. H 2 (g) + I 2 (g)  2HI(g) K p = ( P HI ) 2 ( P H2 ) ( P I2 ) no unit where P HI = mole fraction HI x P total = n (HI ) x P total n(HI) + n(H 2 ) + n(I 2 )

13. b. 2SO 2 (g) + O 2 (g)  2SO 3 (g) K p = ( P SO3 ) 2 ( P SO2 ) 2 ( P O2 ) Unit = Pa -1 / atm -1

14. 2.K p is constant as long as temperature remains constant 3.Vapour pressure of solids are constant, therefore not included in K p expressions Eg: ZnO(s) + H 2 (g)  Zn(s) + H 2 O(g) K p = P H2O P H2

15. Exercise : write the K p expression for the following equil, including units (if any): (1) 2NOCl(g)  2NO(g) + Cl 2 (g) K p = (P NO ) 2 (P Cl2 ) atm / Pa (P NOCl ) 2 (2) 3Fe(s) + 4H 2 O(g)  4H 2 (g) + Fe 3 O 4 (s) K p = (P H2 ) 4 no unit (P H2O ) 4

16. Thermal decomposition a.Heat causes partial break-up of parent molecule b.Fraction of molecules that dissociate / decompose = degree of dissociation / decomposition (  )

17. Examples : (1) Initial amount = 1 mole,  = 20 % or 0.2  out of 1 mole, 0.2 moles will decompose Therefore at equilibrium, Amount remaining = 1 – 0.2 = 0.8 moles

18. (2) Initial amount = 0.6 mole,  = 50 % or 0.5 Amount that decompose = 50/100 x 0.6 = 0.3 moles Therefore at equilibrium, Amount remaining = 0.6 – 0.3 = 0.3 moles