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Andreyy89
3 years ago
13

The equilibrium fraction of lattice sites that are vacant in silver (ag) at 500°c is 0.5 x 10-6. calculate the number of vacanci

es (per meter cubed
Chemistry
1 answer:
Sedbober [7]3 years ago
6 0
Missing question: <span>Assume a density of 10.35 g/cm3 for Ag, A(Ag) = 107.87 g/mol.
N(Ag) = Na </span>· d(Ag) ÷ A(Ag).
N(Ag) = 6,023·10²³ atoms/mol · 10,35 g/cm³ · 10⁶ cm³/m³ ÷ 107,87 g/mol.
N(Ag) = 5,78·10²⁸ atoms/mol.
Nv = 5,78·10²⁸ atoms/mol · 5·10⁻⁵.
Nv = 2,89·10²².
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At 850 K, the equilibrium constant for the reaction
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Answers are in the explanation.

Explanation:

<em>Given concentrations are:</em>

  • <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
  • <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
  • <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>

<em />

In the reaction:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

Kc is defined as:

Kc = 15 = [SO₃]² / [O₂] [SO₂]²

<em>Where concentrations of each species are equilbrium concentrations.</em>

<em />

Also, you can define Q (Reaction quotient) as:

Q = [SO₃]² / [O₂] [SO₂]²

<em>Where concentrations of each species are ACTUAL concentrations.</em>

<em />

If Q > Kc, the reaction will shift to the left until Q = Kc;

If Q < Kc, the reaction will shift to the right until Q = Kc

If Q = Kc, there is no net reaction because reaction would be en equilibrium.

Replacing with given concentrations:

  • Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
  • Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
  • Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right

<em />

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