Question

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g) ⇆ 2NO2(g)
If at equilibrium the N2O4 is 0.71 mols and NO2 is 0.78 mols, what is the value of the equilibrium constant, Kc, for the reaction under these conditions?

Answer 1

[N₂O₄] = 0.71 mol/L

[NO₂] = 0.78 mol/L

$K_c = \frac{[NO_2]^2}{[N_2O_4]} \\ K_c = \frac{[0.71]^2}{[0.78]} = 0.646$