Successive approximations for the problem:

Calculate the equilibrium pressure for N2, H2 and NH3 for the Haber process at 400oC
if one starts with 10.0 atm each of N2 and H2 in a closed volume.

The equilibrium equation is:  N2 + 3H2 ⇌ 2NH3

The equilibrium constant at 400oC = 2.0 x 10-4

Assume x atms of reaction occures to come to equilibrium.  This means, x atm of N2
is consumed, 3x atm of H2 is consumed and 2x of NH3 is produced.  OR:
 

P(N2) P(H2) P(NH3)
initially  10.0 10.0 0.0
equilibrium 10.0 - x 10.0 - 3x 0.0 + 2x

The equilibrium equation is: 
Substituting in: 

Assume x << 10.0.  If this works, then the answer will show that x << 10.0.
 
 

P(N2) P(H2) x P(NH3)
1st Approximation 10.0 10.0 0.71  1.41

Note that x << 10.0 and the approximation works.  One can continue by using the
approximation that:  10.0 - x = 10.0 - 0.7 and 10.0 - 3x = 10.0 - 2.2.  This gives
the 2nd approximation.
 
2nd Approx 9.29 7.88  0.48   .95
3rd Approx 9.52 8.57  0.55 1.09
4th Approx 9.45 8.36 0.53  1.05
5th Approx 9.47  8.42 0 .53 1.06
6th Approx 9.47 8.40 0 .53  1.06