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Georgia [21]
3 years ago
12

exactly 1mol of n2o4 is placed in an empty 1 l container. if at equilibrium n2o4 is dissociated 20%, what is the value of equili

bruim constant
Chemistry
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

K = 0.2

Explanation:

Based on the chemical dissociation of N₂O₄:

N₂O₄ ⇄ 2NO₂

The equilibrium constant, K, of the reaction is:

K = [NO₂]² / [N₂O₄]

Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M

as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.

Replacing in K:

K = [0.4M]² / [0.8M]

<h3>K = 0.2</h3>
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