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.
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,
