Step 1:

Initially 8.1 moles of NO₂ is placed in 3.0 L of container.

Initial amount of NO₂ is

Concentration of [NO] at equlibrium is 1.4 m/L

The chemical equation is .

Let x be the concentration of O₂.

Then change in concentration of NO be 2x.

Then change in concentration of NO₂ be -2x.     ( negative sign indicates that decomposition)

Step 2:

(ii)

Find the change in amount of NO.

Initially the amount of NO is 0 M.

Equilibrium Amount of NO is 2x.

Equilibrium Amount = Initial amount + Change in amount.

Equilibrium Amount = 0 + 2x.

1.4 = 2x

x = 0.7.

(ii)

Find the Equilibrium Amount of Cl₂.

Initially the amount of O₂ is 0 M.

Change in concentration of O₂ is x

Equilibrium Amount = Initial amount + Change in amount.

Equilibrium Amount = 0 + x.

Equilibrium Amount = 0.7.

(iii)

Find the equilibrium amount of NO₂.

Initial amount of NO₂ is 2.7 M.

Equilibrium Amount = Initial amount + Change in amount.

Step 3:

Definition of equilibrium constant:

Equilibrium constant  is the ratio of the concentrations of products to the concentration of reactants.

.

The chemical reaction is .

Solution:

Equilibrium constant is K = 0.8118.