Reaction between N2 and O2– takes place as follows:
2N2 (g) + O2 (g) ⇌2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture.
Equilibrium constant of reaction
2N2 (g) + O2 (g) ⇌2N2O (g)
is given as,
Kc =
= 2.0 × 10–37
Initial concentration of substances,
[N2O] = 0
[N2] = number of mole of N2/Volume of container
= 0.482/10 = 0.0482
[O2] = number of mole of O2/Volume of container
= 0.933/10 = 0.0933
2N2 (g) + O2 (g) ⇌2N2O (g)
where x is concentration of O2 that has reacted.
Thus, equilibrium constant,
Kc = 
= 
Since value of equilibrium constant is small, the amount of O2 and N2 reacted is very less and thus can be neglected.
∴ Kc =
= 2 × 10-37
i.e., (2x)2 = (2 × 10-37) ×(0.0482)2× 0.0933
2x= 4.335 × 10-41
x = 2.168 × 10-41
Thus, Equilibrium concentrations of substances are,
[N2] = 0.0482 – 4.335 × 10-41 = 0.0482 mol L-1
[O2] = 0.0933 – 2.168 × 10-41 = 0.0933 mol L-1
[N2O] = 4.335 × 10-41mol L-1