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Levart [38]
3 years ago
14

2. If volume is increased, the number of collisions per second

Chemistry
1 answer:
Archy [21]3 years ago
4 0

Answer:

decreases

Explanation:

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Answer:

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 />

8 0
3 years ago
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