Reaction between nitrogen and oxygen takes place as following: 2N2(g) + O2 ⇌ 2N2O(g) If a mixture of 0.482 mole ofN2 and 0.933 mole of O2 is placed in a reaction vessel of volume 10 litre and allowed to form at a temperature for which KC = 2.0 x 10-37 litre mol-1 . Determine the composition of the equilibrium mixture.

The reaction given in the question –

2N2(g)+O2(g)⇌2N2O(g)

The value for equilibrium constant (Kc) of this reaction will be –

Kc=[N2O]2[N2]2[O2]

Given in the question,

Equilibrium constant = Kc = 2.0 x 10-37 litre mol-1

Volume = 10 L

Initial moles of N2 (dinitrogen gas) = 0.482 mol

Initial moles of O2 (dioxygen gas) = 0.993 mol

– We can calculate concentration by calculating the ratio of mass and volume i.e.

concentration=mass/volume

Concentration of N2 (dinitrogen gas) = 0.48210molL−1

Concentration of O2 (dioxygen) = 0.99310molL−1

– Now, we need to find the equilibrium concentration of all compounds – nitrogen gas, oxygen gas and nitrous oxide.

Let us assume that ‘x’ oxygen is reacted with ‘2x’ moles of nitrogen. Therefore –

2N2(g) + O2 ⇌ 2N2O(g)

The value of Kc by substituting above concentration –

Kc=[2x 10]2/[0.0482-2x /10]2[0.933-x/10]

As we can see, the value of Kc is very small, therefore, the concentration of nitrogen and oxygen that reacted will be very small. So, we will ignore ‘x’ from molar concentration of nitrogen and oxygen.

Therefore, we get –

2.0 x 10-37=[2x 10]2/[0.0482]2[0.0933]

Solving the above equation –

4×2100=[0.0482]2[0.0933][2.0 x 10-37]

4x2 = 4.34 x 10−39

x2 = 1.08 x 10−39

x = 3.3 x 10-20

Now, substituting the value of ‘x’ in equilibrium concentrations, we get –

Equilibrium concentration of N2 = 0.0482-2x /10 = 0.00482molL−1

Equilibrium concentration of O2 = 0.933-x/10 = 0.0099molL−1

Equilibrium concentration of N2O = 2x/10 = 6.6×10−21molL−1

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