The reaction N2 (g)+ O2 (g) (equillibrium arrows)2 NO(g) contributes to air pollution when a fuel is burnt in air at a high temperature.At 1500k quillibruium constant K for it is 1.0x10-5 (no units given).Suppose [N2] =0.8 mol L-1 and [O2]=0.20 mol L-1 before any reaction occurs. Calculate the equillibrium concentraion of the reactants and the product after the mixture has been heated to 1500 k.
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Is this question out of sylabus for 12th???
Let x amount has been decayed from reactant to product then:
N2 + O2 ↔ 2NO
T =0 0.8 0.20
T=tequi. 0.8-x 0.2-x 2x
Kc = 1.0 x 10-5
Kc = 
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/ [(0.8 )(0.2)]
16 x 10-6 = 4x2
x2 = 4 x 10-6
x = 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 10-3 = 4 x 10-3