Question

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 10L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37 determine the composition of equilibrium mixture.

Answer 1

Let x moles of N₂(g) take part in the reaction. According to the equation, \(\frac{x}{2}\) moles of O₂(g)  will react to form x moles of N₂(g). The molar concentration per litre of difference species before the reaction and at the equilibrium point as:

2N₂(g) + O₂(g) ⇌ 2N₂O(g) (zero)

Initial moles\litre \(\frac{0.482}{10}\)\(\frac{0.933}{10}\)

Moles per litre at eqn. point

The value of equilibrium constant (2.0 x 10^-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, x is extremely small and can be omitted as far as the reacts are concerned.

Thus, in the equilibrium mixture

Molar conc. of N₂ = 0.0482 mol L^-1

Molar conc. of O₂ = 0.0933 mol L^-1

Molar conc. of N₂O = 6.6 x 10^-20 mol L^-1