Ammonia is an important industrial chemical.

Ammonia, N= Blue, H= Gray

Ammonia is synthesized using the equilibrium reaction where the reactant and products are in the gas-phase.

**N _{2} + 3 H_{2} ≈ 2 NH_{3}**

Its makes sense to use the partial pressures of the gases to define an equilibrium constant K_{p}. The reaction of forming ammonia from its elements has an equilibrium constant of K_{p} = 977 at a particular temperature.

Kp is given in terms of the partial pressures of the reactants and products in pascals where each partial pressure is divided by the standard pressure of p° = l x 10^{5} Pa = 1 bar and thus K_{p} is dimensionless. Each of the terms is raised to the power of its stoichiometric coefficient. Because the standard pressure p° = l x 10^{5} Pa each of the p° terms affect the value for K_{p} and we cannot make a simplification similar to the one we made for K_{c} but each of the K_{p} equations must be written in full with the p° values included.

For our reaction we initially have 1.00 x 10^{5} Pa pressure of N_{2} and 3.00 x 10^{5} Pa of H_{2} which are mixed, there being no NH_{3} present and the mixture is allowed to come to equilibrium. If x is the fraction of a mole of N_{2} which is lost due to the reaction then the equilibrium constant becomes

Note that each time the reaction “occurs” we form two molecules of NH_{3} hence the term 2x. Likewise, each time the reaction occurs we lose 3 molecules of H_{2} thus the term in the bracket is (3.00 — 3x). Within each of the bracketted terms the units Pa cancel out so this is written as

Firstly, calculate the fraction by solving for x using the quadratic formula. Secondly, using this value for x calculate the partial pressures of all the reactants and products. We can simplify the equation by taking out the common factors.

**Solution**

The equilibrium equation may be solved for x by:

(1) taking out the common factor of (1.00 – x) in the denominator and also squaring in the numerator;

(2) collecting the (1.00 – x) common factors together;

(4) cubing the 3.00 and rearranging the numerical factors;

**K _{p} = 4/27 [x^{2} / (1-x)^{4}] = 977 x 10^{-10} k_{p} = (27 * 977) / 4 x 10^{-10}**

(5) square rooting both sides of the equation;

**x / (1 – x) ^{2 }= √ {(27 x 977) / 4} x 10^{-10} = 81.2081 x 10^{-5}**

(6) rearranging by multiplying by the denominator on both sides of the equation, then cancelling terms; (7) squaring the bracket; (8) multiplying out the bracket;

(9) rearranging into the “standard” quadratic form, note that “162” has become “163”; and (10) dividing by 81.2081 x 10^{-5} throughout to make the coefficients much more manageable.

**(81.2081 x 10 ^{-5} x^{2}) – (162.4162 x 10^{-5} x) + 81.2081 x 10^{-5} = 0**

**x ^{2} – 2.0123x + 1 = 0**

We can solve this quadratic equation using the quadratic formula,

As the mole fraction must be positive and less than one then x = 0.8951 is the chemically correct result. This gives the three equilibrium partial pressures as