Respuesta :
Answer:
(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
The value of equilibrium constant for the reverse reactions are 0.091, 81.3 & 0.641 respectively.
What is Kc?
Kc is known as equilibrium constant which defines as the ration of the concentration of products to the concentration of reactants. In the Kc expression value of pure solids and pure liquids are always one, as they don't change significantly during the equilibrium state.
(1). Given balanced chemical reaction is:
2Al(s) + 2NaOH (aq) + 6H₂O(l) ⇋ 2Na[Al(OH)₄](aq) + 3H₂(g)
Kc for this reaction is written as:
Kc = [Na[Al(OH)₄]]²[H₂]³ / [NaOH]²
Value of Kc = 11
For reverse reaction Kc = 1/11 = 0.091
(2). Given balanced chemical reaction is:
H₂O (l) + SO₃ (g) ⇋ H₂SO₄ (aq)
Kc for this reaction is written as:
Kc = [H₂SO₄] / [SO₃]
Value of Kc = 0.0123
For reverse reaction Kc = 1/0.0123 = 81.3
(3). Given balanced chemical reaction is:
P₄ (s) + 3O₂ (g) ⇋ P₄O₆ (s)
Kc for this reaction is written as:
Kc = 1 / [O₂]³
Value of Kc = 1.56
For reverse reaction Kc = 1/1.56 = 0.641
Hence, the value for the above reactions are 0.091, 81.3 & 0.641.
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