1. KpKp

2. The equilibrium constant in terms of partial pressures KpKp

3. Mole fraction

4. Partial pressures Partial pressure, p The contribution of a gas towards the total pressure Partial pressure = mole fraction x Total pressure A gas mixture with a total pressure of 320 kPa contains 2 mol of N 2 (g) and 3 mol of O 2 (g). Mole fractions Partial pressures Sum of partial pressures = Total pressure p(N 2 ) + p(O 2 ) = 128 + 192 = 320 kPa x(N 2 ) = = 0.4x(O 2 ) = = 0.6 p(O 2 ) = x(O 2 )P = 0.6 x 320 = 192 kPa p(N 2 ) = x(N 2 )P = 0.4 x 320 = 128 kPa 2525 3535

5. What is K p Similar to K c but partial pressures used in place of concentration Equilibrium: 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g) Units: K p =

6. Calculating K p Equilibrium: 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g) Partial pressures: SO 2 (g), 74 kPa; O 2 (g), 23 kPa; SO 3 (g), 142 kPa K p = x x = 0.160 kPa –1

7. Heterogeneous equilibria Equilibrium contains different phases Equilibrium: CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) K p expression contains only gaseous species K p = p(CO 2 ) Solid species are omitted (solids have no gas pressure)