Question 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 equilibrium mixture.

Answer

Let the concentration of N2O at equilibrium be x. The given reaction is:

                                2N2 (g)         +        O2 (g)               2N2O (g)

Intial Concentration:   0.482 mol            0.933 mol                  0 mol

At equilibrium           (0.482-x) mol        (0.933 -x) mol            x mol

Therefore, at equilibrium, in the 10 L vessel:

N2 = 0.482-x / 10

O2 = 0.933-x/2  / 10

N2O = x / 10

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  =  0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1

Now,

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