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
1 answer:
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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