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g Determine the value of the equilibrium constant for this reaction if an initial concentration of N2O4(g) of 0.040 mol/L is reduced to 0.0055 mol/L at equilibrium. There is no NO2(g) present at the start of the reaction.

Respuesta :

Answer:

Explanation:

N₂O₄(g) ⇄ 2 NO₂

N₂O₄ reacted = .04 - .0055 = .0345 mole

NO₂ formed = 2 x .0345 = .069 moles

equilibrium constant = [ NO₂ ] ² / [ N₂O₄]

= .069² / .0055

= 0.865 .

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