Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc for the reverse reaction. (1) Al (s) + NaOH (aq) + H2O (l) ⇋ Na[Al (OH) 4] (aq) + H2 (g) Kc for balanced reaction = 11 (2) H2O (l) + SO3 (g) ⇋ H2SO4 (aq) Kc for balanced reaction = 0.0123 (3) P4 (s) + O2 (g) ⇋ P4O6 (s) Kc

(1) 2Al (s) + 2NaOH (aq) + 6H2O (l) ↔2Na[Al (OH) 4] (aq) + 3H2 (g)

∴ Kc = (PH2³ * [Na[Al (OH) 4]²) / [NaOH]² = 11

(2) H2O (l) + SO3 (g) ↔ H2SO4 (aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4 (s) + 6O2 (g) ↔ 2P4O6 (s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al (s) + 2NaOH (aq) + 6H2O (l) ↔ 2Na[Al (OH) 4] (aq) + 3H2 (g)

∴ O / Al: 0 → + 2 ≡ 2e-

Na: + 1 → + 2

∴ R / H: + 1 → 0

2 - Al - 2

2 - Na - 1

8 - O - 8

14 - H - 14

⇒ Kc = (PH2³ * [Na[Al (OH) 4]²) / [NaOH]² = 11

(2) H2O (l) + SO3 (g) ↔ H2SO4 (aq)

1 - S - 1

4 - O - 4

2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4 (s) + 6O2 (g) ↔ 2P4O6 (s)

8 - P - 8

12 - O - 12

⇒ Kc = 1 / PO2∧6