Question

Consider the following equilibrium
N204(9) - 2NO2(9) Keq = 5.85 x 10-3
Which statement about this system is true?
If the equilibrium concentration of NO2 is 1.78 x 10-2 M, the equilibrium concentration of N2O4 is

Answer 1

The equilibrium concentration of N₂O₄, given that the concentration of NO₂ is 1.78×10⁻² is 5.42×10⁻²

Data obtained from the question

• N₂O₄ <=> 2NO₂
• Equilibrium constant (Keq) = 5.85×10⁻³
• Equilibrium concentration of NO₂ = 1.78×10⁻²
• Equilibrium concentration of N₂O₄ =?

How to determine the equilibrium concentration of N₂O₄

Keq = [Product] / [Reactant]

Keq = [NO₂]² / [N₂O₄]

5.85×10⁻³ = [1.78×10⁻²]² / [N₂O₄]

Cross multiply

5.85×10⁻³ × [N₂O₄] = [1.78×10⁻²]²

Divide both sides by 5.85×10⁻³

[N₂O₄] = [1.78×10⁻²]² / 5.85×10⁻³

[N₂O₄] = 5.42×10⁻²

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