The equilibrium constant for the reaction N₂O₂(g) → N2(g) + O₂(g) is 0.456 at 322 k. Suppose we put 1.00 mole of N₂O₂ into a 1.00-L container and heat it to 322 k. At time zero, the concentration of N₂O₂ would be 1.00 M, and the concentration of products N₂ and O₂ would both be zero. By the time equilibrium is reached, an unknown amount of N₂O₂ (call it x) would have been converted into N₂ and O₂. Since the stoichiometry of the reaction is 1 to 1 to 1 (all balancing coefficients are equal to 1), if x was the amount of N₂O₂ used up, then x amount of N₂ and x amount of O₂ would be produced and the amount of N₂O₂ left would be 1 – x. Thus, at equilibrium, we can write (remember, square brackets mean “concentration at equilibrium”):

This means we can write the equilibrium constant as:

The value for x has to be somewhere between the starting concentration of N₂O₂ (that is, 1.00 M) and zero. Work together to plug in values for x until you find a value of x that comes close to solving this equation. Then tell what the concentrations of all three species would be once the reaction reaches equilibrium.

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[N₂O₂] = 1.00 x [N₂] = x [0₂] = x