What is the value of Kc for this equilibrium reaction?

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SUMMARY

The equilibrium constant (Kc) for the reaction involving 1.0 mol of N2O4(g) and 1.0 mol of NO2(g) in a 1.0 dm3 vessel, where 0.75 mol of N2O4(g) remains at equilibrium, is definitively calculated to be 3.0. The correct approach involves using the equilibrium concentrations of the reactants and products in the expression Kc = [NO2]2 / [N2O4]. The balanced chemical equation is N2O4(g) ⇌ 2NO2(g), confirming the stoichiometry used in the calculation.

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Homework Statement


A 1.0 dm3 reaction vessel contains initially 1.0 mol of N2O4(g) and 1.0 mol of NO2(g).
At equilibrium, 0.75mol of N2O4(g) are present.
What is the value of Kc?
A. 0.33
B. 0.50
C. 2.0
D. 3.0

The Attempt at a Solution


Right answer = D, but I do not why. My attempt for Kc is (3/4)^2 / (9/8) = 0.5, so wrong answer. Please, help me with this stricky one.
 

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Show the equilibrium equation in terms of the agents and also the actual balanced chemical equation.
 
GCT said:
Show the equilibrium equation in terms of the agents and also the actual balanced chemical equation.

Yes indeed. That is the right method to solve this.

Thank you!
 

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