1. Chemical Equilibrium Chapter 15

2. Equilibrium - state in which there are no observable changes with time Achieved when: rates of the forward and reverse reactions are equal and concentrations of the reactants and products remain constant dynamic equilibriumdynamic equilibrium Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) H 2 O (g) 2NO 2 (g) colorlessred-brown

3. Fig 15.1 The N 2 O 4 (g) ⇌ 2 NO 2 (g) equilibrium colorlessred-brown

4. Since both reactions are elementary, we use kinetics, Forward reaction:N 2 O 4 (g)  2 NO 2 (g) Rate Law:Rate = k f [N 2 O 4 ] Reverse reaction:2 NO 2 (g)  N 2 O 4 (g) Rate law:Rate = k r [NO 2 ] 2 At equilibrium:Rate f = Rate r or k f [N 2 O 4 ] = k r [NO 2 ] 2

5. N 2 O 4 (g) 2NO 2 (g) Fig 15.2 (a) Achieving chemical equilibrium for: Start with N 2 O 4 Equilibrium occurs when concentrations no longer change

6. N 2 O 4 (g) 2NO 2 (g) Fig 15.2 (b) Achieving chemical equilibrium for: Equilibrium occurs when k f and k r are equal

7.  Start with N 2 and H 2 or with NH 3  Result: same proportions of all three substances at equilibrium N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) e.g., the Haber process to form ammonia

8. aA + bB cC + dD K C = [C] c [D] d [A] a [B] b Law of Mass Action Must be caps! Equilibrium constant Lies to the rightLies to the left

9. Table 15.1 Initial concentrations in gas phase at 100 o C N 2 O 4 (g) 2NO 2 (g) Fig 15.1 Concentration changes approaching equilibrium constant

10. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO2 P2P2 N2O4 P In most cases K c  K p K p = K c (RT) Δn Δn = moles of gaseous products – moles of gaseous reactants The equilibrium constant may be expressed in terms of concentration or in terms of pressure

11. In the synthesis of ammonia from nitrogen and hydrogen, K c = 9.60 at 300 °C. Calculate K p for this reaction at this temperature N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Sample Exercise 15.2 p 634

12. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74°C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 (0.012) · (0.054) = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = 0.0821 (L·atm)/(mol·K)T = 273 + 74 = 347 K K p = 220 ·[0.0821 · 347] -1 = 7.7