For the equilibrium: 2 NO​ (g) ​ ⇌ N​ 2(g)​+ O​ 2(g)​at 300 K, the equilibrium constant, Kc, is 0.185. If 1.45 moles each of N​ 2(g)​and O​ 2(g)​are introduced in a container that has a volume of 6.00 liters and allowed to reach equilibrium at 300 K, what are the concentrations of N​ 2(g )​ , O​ 2(g)​ ,and NO​ (g)​at equilibrium?

The concentrations of at equilibrium are 0.112 M, 0.112 M  and 0.260 M

Explanation:

Moles of = 1.45 mole

Moles of   = 1.45 mole

Volume of solution = 6.00 L

Initial concentration of =

Initial concentration of =

The given balanced equilibrium reaction is,

Initial conc.          0 M             0.242 M              0.242 M  

At eqm. conc.     (2x) M      (0.242-x) M        (0.242-x) M

The expression for equilibrium constant for this reaction will be,

Now put all the given values in this expression, we get :

By solving the term 'x', we get :

x = 0.130

Thus, the concentrations of at equilibrium are :

Concentration of = (0.242-x) M  = (0.242-0.130) M = 0.112 M

Concentration of = (0.242-x) M  = (0.242-0.130) M = 0.112 M

Concentration of = 2x M = = 0.260 M