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3 January, 19:04

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

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  1. 3 January, 19:15
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    (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
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