Question

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M and [ NO ] = 0.500 M . N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re‑established?

Answer 1

Answer:

The answer to the question is the final concentration of NO after equilibrium is re‑established = 0.7143 moles

Explanation:

The reaction is given by

O₂(g) + N₂(g) → 2NO(g)

Kc = $\frac{[NO]^{2} }{[O_{2}][N_{2} ]}$

= (0.5^2)/(0.1×0.1) = 25

However, initially, y moles each of O and N gave off x moles each to form 2·x moles of NO, therefore.

Hence we have 2·x = 0.5 or x = 0.25

and y = 0.1+0.25= 0.35

When the concentration is increased to 0.8, we have each mole of NO giving out 2·a moles while each O₂ and N₂  increases by a moles which gives

(0.8-2·a)^2/(0.1+a)^2 = 25

Taking square roots of both sides and cross multiplying, we have 7×a=0.3

Which gives a = 0.04286 moles.

Therefore the number of moles of NO at equilibrium

= 0.8-2×0.04286 = 0.7143 moles