Consider N2O4 (g) ---> 2 NO2 (g) . 3.00 X 10-2 mol of N2O4 are placed...

Consider N₂O₄ (g) ---> 2 NO₂ (g) . 3.00 X 10^-2 mol of N₂O₄ are placed in a 1.0 L flask. AT equilibrium, 2.36 X 10^-2 mol of N₂O₄ remain. What is the Keq for this reaction?

Expert Solution

N₂O₄ (g) ---> 2 NO₂ (g)

Initial 0.03 molar 0

Change -x x

At equilibrium 0.03-x x

Keq = [NO2]^2 / [N2O4]

Given :

[N2O4] = x = 0.0236 molar

so [NO2] = 0.03- 0.0236 = 0.0064 molar

so Keq = 0.0064 / (0.0236)^2 = 11.49

queen_honey_blossom answered 1 year ago

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