Answer:
[N₂] = 1.1M
Explanation:
Based on the chemical reaction:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Equilibrium constant, K, is defined as:
K = 5.93 = [NO]² / [N₂] [O₂]
<em>Where [] are equilibrium concentrations of each specie</em>
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As initial concentrations are:
N₂ = 0.40mol / 0.500L = 0.8M
NO = 1mol / 0.500L = 2M
The equilbrium concentrations are:
[NO] = 2M - 2X
[N₂] = 0.8M +X
[O₂] = X
Replacing:
5.93 = [2 - 2X]² / [0.8+X] [X]
5.93 = 4 - 8X + 4X² / 0.8X + X²
4.744X + 5.93X² = 4 - 8X + 4X²
1.93X² + 12.744X - 4 = 0
Solving for X:
X = -6.9M → False solution. There are no negative concentrations
X = 0.3M. Real solution.
[N₂] in equilibrium is:
[N₂] = 0.8M +0.3M
[N₂] = 1.1M
