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Question

At a particular temperature, 8.1 moles of NO2 gas is placed in a 3 L container. Over time the NO2 decomposes to NO and O2:

2NO2 (g) <--> 2NO (g) + O2(g)

At equilibrium the concentration of NO (g) was found to be 1.4 mol/L. Calculate the value of K for this reaction.

Answer 1

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 .

2 moles of NO₂ decomposes to 2 moles of NO and 1 mole of O_2.

If x be the change in the concentration of O₂,

then change in concentration of NO be 2x  and

change in concentration of NO₂ be -2x.     (negative sign indicates 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 O₂.

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.