Question

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, the reaction below can proceed to a measurable extent. N2(g) + O2(g) ⇔ 2 NO(g) At 3000 K, the reaction above has Keq = 0.0153. If 0.2702 mol of pure NO is injected into an evacuated 2.0-L container and heated to 3000K, what will be the equilibrium concentration of NO?

Answer 1

Solution:

            2 NO(g) ⇔ N₂(g) + O₂(g)

            0.2702          0           0                        (initial)

(0.2702-x)      0.5x       0.5x                             (Equilibrium, moles)

            (0.2702-x)/2      0.5x/2       0.5x/2       (Equilibrium, moles/L)

Equilibrium constant for the above reaction can be given as:

K_eq = [N₂][O₂]/ [NO]² = 1/0.0153

(0.5x/2)( 0.5x/2)/ [(0.2702-x)/2]² = 65.35

0.25x² / (0.2702-x)² = 65.35

0.25x² = 4.77                [neglecting ‘x’ since it is very small at the equilibrium]

x² = 19.08

x = 4.36