Question

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Answer 1

The following reaction:

2 NO --> N₂ + O₂

Has an equilibrium constant of 23.8 under certain conditions. If the initial concentrations are

[NO] = 0.250 M, [N₂] = 0.100 M, [O₂] = 0.100 M, what will be the concentration of N₂ at equilibrium?

Solution :

We write the K expression for above reaction

K = [N2][O2] / [NO]²

We set up ICE chart

            2 NO -- >               N₂ +                      O₂
I         0.250                    0.100                     0.100

C       -2x                         +x                            x

E      (0.250-2x)             (0.100+x)             (0.100+x)

Lets plug given value in equilibrium expression

23.8 =(0.100+x) ²/ (0.250-2x) ²

Lets take a square root of both side

4.88 =( 0.100 +x ) / (0.250-2x)

1.22 -2.44 x = 0.100+ x

1.12 =3.44 x

X = 0.326

[N2] at equilibrium = 0.100 + 0.326 = 0.426 M