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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.

Asked by Meghana Anil 6th March 2013, 12:19 AM

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

x= 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

Answered by Expert 10th March 2013, 10:37 PM
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