Nitrogen and oxygen react to form nitrous oxide as: 2 N₂ (g) + O₂ (g) ⇌ 2 N₂O (g) Of a mixture of 0.482 mol N₂ and 0.933 mol O₂ is placed in a reaction vessel of 10.0 L and allowed to from N₂O at a temperature for which K_c = 2.0 X 10^-37, what will be the composition of the equilibrium mixture?

The given equation is:
2 N₂ (g) + O₂ (g) ⇌ 2 N₂O (g)
Therefore the equilibrium constant K_c, will be given as:
K_c = [N₂O] ² / [N₂] ² [O₂]

Initial molar concentrations are:
[N₂] = 0.482 mol / 10.0 L = 0.0482 mol L^-1
[O₂] = 0.933 mol / 10.0 L = 0.0933 mol L^-1
Let us assume that x mol of oxygen has reacted with 2x mol of N₂ to form 2x mol of N₂O, so that equilibrium concentrations are:
[N₂] = 0.0482 – 2x
[O₂] = 0.0933 – x
[N₂O] = 2x

Therefore,
K_c = (2x) ² / (0.0482 – 2x) ² (0.0933 – x)
Since, K_c is very small (2.0 X 10^-37), we can assume x to be infinitesimally small so that,
0.0482 – 2x ≈ 0.0482 and 0.0933 – x ≈ 0.0933
K_c = 4 x² / (0.0482) ² (0.0933)
i.e.
4x² = (0.0482) ² (0.0933) X (2.0 X 10^-37)
4 x² = 4.34 X 10^-412 = 4.34 X 10^-41-41-212] = 0.0482 – 2x ≈ 0.0482
[O₂] = 0.0933 – x ≈ 0.0933
[N₂O] = 2x = 2 X 3.3 X 10^-21-21