Reaction between N2 and O2 takes place as follows:
2N2(g) +O2(g) ↔2N2O(g)
If a mixture of 0.482 mol of 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)Initial conc.0.482 mol0.933 mol0At equilibrium(0.482−x)mol(1.933−x) molx mol
Therefore, at equilibrium, in the 10 L vessel:
[N2]=0.482−x10,[O2]=0.933−x210,[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=0.0482 mol L−1 and [O2]=0.93310=0.0933 mol L−1
Now,
Kc=[N2O(g)]2[N2(g)][O2(g)]⇒2.0×10−37=(x10)2(0.0482)2(0.0933)⇒x2100=2.0×10−37×(0.0482)2×(0.0933)⇒x2=43.35×10−40
[N2O]=⇒x=6.6×10−20 mol L−1
[N2]=0.0482 mol L−1 and [O2]=0.0933 mol L−1