A study of the following system, 4 NH3(g) + 7 O2(g) <--> 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [N2O4] = [H2O] = 3.60 M as the only components initially. At equi­librium, [H2O] is 0.600 M. Calculate the equilibrium concentration of O2(g).

3.50 M

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

Step 2: Make an ICE chart

        4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)

I              0               0              3.60            3.60

C         +4x            +7x               -2x             -6x

E            4x              7x           3.60-2x      3.60-6x

Step 3: Calculate the value of x

The concentration of water at equilibrium is 0.600 M. Then,

3.60-6x = 0.600 M

x = 0.500 M

Step 4: Calculate the concentration of O₂ at equilibrium

The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M