1. Learning objective: To be confident working out equilibrium amounts, Kc and units 23/11/2016 Write an expression (with units) for the equilibrium constant (Kc) for the reaction: 2 SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g)

2. Kc – equilibrium constant concentration CH 3 COOH + C 2 H 5 OH CH 3 COOCH 2 CH 3 + H 2 O Kc = [CH 3 COOCH 2 CH 3 ][H 2 O]/[CH 3 COOH][C 2 H 5 OH] Units: none N 2 + 3 H 2 2 NH 3 Kc = [NH 3 ] 2 /[N 2 ][H 2 ] 3 Units: mol -2 dm 6

3. Whiteboards – Kc & Kp equations CH 4 (g) + H 2 O(g) ⇋ CO(g) + 3 H 2 (g) initial moles 1 1 0 0 equilibrium moles 0.4 N 2 (g) + 3 H 2 (g) ⇋ 2 NH 3 (g) initial moles 0.1 0.1 0 equilibrium moles0.07 H 2 (g) + I 2 (g) ⇋ 2 HI(g) initial moles 10 10 0 equilibrium moles 3 CO 2 (g) + 3 H 2 (g) ⇋ CH 3 OH(g) + H 2 O(g) initial moles 1 1 0 0 equilibrium moles 0.2 2 SO 2 (g) + O 2 (g) ⇋ 2 SO 3 (g) initial moles 2 2 0 equilibrium moles 0.7 0.6 1.2 0.01 0.06 314 0.80.40.2 1.31.65

4. PCl 5 (g) ⇋ PCl 3 (g) + Cl 2 (g) initial moles 1 1 1 equilibrium moles 1.8 CH 4 (g) + H 2 O(g) ⇋ CO(g) + 3 H 2 (g) initial moles 1 1 1 1 equilibrium moles 1.6 N 2 (g) + 3 H 2 (g) ⇋ 2 NH 3 (g) initial moles 10 0 10 equilibrium moles 9 PCl 5 (g) ⇋ PCl 3 (g) + Cl 2 (g) initial moles 2 0 1 equilibrium moles 1.4 PCl 5 (g) ⇋ PCl 3 (g) + Cl 2 (g) initial moles 2 3 1 equilibrium moles 1.3 0.21.8 0.8 1.2 10.51.5 1.60.4 1.73.3

5. Calculation Practice Some PCl 5 was heated in a sealed container at 250  C, resulting in the following equilibrium. Analysis of the equilibrium mixture showed that it contained 0.0042 moles of PCl 5, 0.040 moles of PCl 3 and 0.040 moles of Cl 2. The total volume was 2.0 dm 3. Calculate the concentration of each species at equilibrium and then Kc. PCl 5 (g) ⇋ Cl 2 (g) + PCl 3 (g) ΔH +ve 1.Calculate Kc 2.What effect will: an increase in temperature; a decrease in pressure; addition of a catalyst have on (i) the equilibrium position, (ii) value of Kc and (iii) the rate of attainment of equilibrium? 3.Kc = [Cl 2 ][PCl 3 ]/[PCl 5 ] = (0.040/2)x(0.040/2)/(0.0042/2) = 0.190 moldm -3 4.i) shifts right; shifts right; none ii) increases; no effect; no effect iii) faster; slower; faster

6. Calculation Practice 2 moles of sulphur dioxide and 1 mole of oxygen were mixed and allowed to reach equilibrium in the presence of a suitable catalyst under a total pressure of 500000 Pa. At equilibrium, 0.67 moles of oxygen were present. Find Kp. 2 SO 2 (g) + O 2 (g) ⇋ 2 SO 3 (g) Equilibrium amount of SO 2 = 2.00 – 0.66 = 1.34 moles Equilibrium amount of SO 3 = 0.66 moles Mole fraction SO 2 = 1.34/(1.34+0.67+0.66) = 0.502 Mole fraction O 2 = 0.67/2.67 = 0.251 Mole fraction SO 3 = 0.66/2.67 = 0.247 Partial pressures: SO 2 = 0.502 x 500 = 250.9 kPa; O 2 = 0.251 x 500 = 125.5 kPa; SO 3 = 0.247 x 500 = 123.6 kPa Kp = p(SO 3 ) 2 /p(SO 2 ) 2 (pO 2 ) = 0.00193 kPa -1