The reaction N2(g) + O2(g) 2NO(g), contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 x 10-5. Suppose in a case [N2] = 0.80 mol L-1and [O2] = 0.20 mol L-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
N2(g) + O2(g) 
2NO(g)
T=0 0.8 0.20
T=t 0.8-x 0.2-x 2x
Kc = 1.0 x 10-5
Kc = 
Or, 1.0 x 10-5 = (2x)2/ [(0.8 – x)(0.2 – x )]
If x is very small, then
0.8 – x 0.8
0.2 – x 0.2
1.0 x 10-5 = (2x)2 / [(0.80) (0.2)]
16 x (10-6) = 4 x2
X2 = 2 x 10-3
Therefore, the amount of reactant and product at equilibrium is as follows:
N2 = 0.8 - 0.002 = 0.798
O2 = 0.2 - 0.002 = 0.198
NO = 2x = 2 x 2 x 10-3 = 4 x 10-3
