# Exercise 2 | Chemical Equilibrium

2 NO (g) + O(g) $⇄$ 2 NO(g)

Initially [NO]0 = 7.50 M, [O2]0 = 3.50 M, and [NO2]0 = 6.50 M.

If [O2] = 0.750 M at equilibrium, calculate the equilibrium values of [NO] and [NO2].

From the stoichiometry of the chemical equation, 2 moles of NO react with 1 mole of O2 to form 2 moles of NO2.

Change in mol.L-1 of O2 = [O2]0 - [O2] = 3.50 – 0.750 = 2.75 M

[NO] = [NO]0 – 2 x Change of O2 = 7.50 – 5.50 = 2.00 M

[NO2] = [NO2]0 + 2 x Change of O2 = 6.50 + 5.50 = 12.0 M