CHEMISTRY question!!! - desperately need help!!! i don't get this?

Question

The equilibrium constant for the Haber reaction is 3.82 x 10-5 atm-2 at a certain temperature. If a reaction starts with 2.70 atm of H2 and 5.40 atm of N2 at this temperature, what will be the equilibrium pressure of NH3? Enter in atmospheres.


please explain - thank you

Answer 1

Let X be the equilibrium pressure of NH3.
From the Haber-Bosch process:
3 H2 + N2 → 2 NH3, K = 3.82x10^-5 /atm^2
we know that to form every 2 moles of NH3, 3 moles of H2 and 1 mole of N2 must be consumed. We also know that in ideal gas model the partial pressure of any gas is directly proportional to its number of moles. Hence:
.........3 H2 .... + .... N2 .. → .. 2 NH3
Initial: 2.70..............5.40.........0.00
Final: {2.70-(3/2)X}..(5.40-X/2)...X
and K = 3.82x10^-5 = X^2 /{(2.70-(3/2)X)^3*(5.40-X/2)}
Or: 3.82x10^-5*{(2.70-(3/2)X)^3*(5.40-X/2)} = X^2
This is a fourth-order single variable equation. The numerical positive solution is: X = 0.0604 (atm)