[wanabBScN]

Equilibrium Equation:
2NH3 ::equil::N2+3H2

Equilibrium Expression:
Ke= [N2][H2]3/[NH3]
Ke=[N2](0.523)3/(0.731)2=0.612
[N2]= 2.29 mol/L

To get this answer I have to divide (0.731)2 by (0.523)3 and multiply it by the answer 0.612. Is this the correct new formula?

[Joseph]

2NH3 ::equil::N2+3H2

Ke = [N2][H2]^3 / [NH3]^2

Take the [NH3]^2 to the other side to give:

Ke * [NH3]^2 = [N2][H2]^3

Now take the [H2]^3 to the other side to give:

Ke * [NH3]^2 / [H2]^3 = [N2]

Now, substitute in the known values:

0.612 * [0.731]^2 / [0.523]^3 = [N2]

Therefore [N2] = 2.29

Assuming the numbers you supplied are correct, this is how the problem is solved