7.8 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 the equilibrium mixture.

Let the concentration of N2O at equilibrium be x.
The given reaction is :
                                      2N2g           +       O2g                2N2OgInitial conc.             0.482 mol               0.933 mol                    0At equilibrium         0.48-xmol         1.933-x/2mol             x mol
Therefore, at equilibrium, in the 10L vessel:
  N2=0.482-x10,O2=0.933-x/210,N2O=x10

The value of equilibrium constant i.e. Kc=2.0×10-37 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2. Then,
     N2=0.48210=0.482 mol L-1 and O2=0.93310=0.0933 mol L-1Now,                      Kc=N2Og2N2g2O2g         2.0×10-37=x1020.048220.0933         x2100=2.0×10-37×0.04822×0.0933         x2=43.35×10-40         x=6.6×10-20N2O=x10=6.6×10-2010           =6.6×10-21