Equilibrium of ammonia synthesis

[MexChemE]

Homework Statement

In order to obtain commercial conversion levels in the production of ammonia, reactors operate between a pressure range from 150 to 300 atm, and a temperature range from 700 to 750 K. Calculate the mole fraction of N₂ when the reaction reaches equilibrium at 300 atm and 723 K. The reactor's feedstock initially contained 25% in mole of N₂ and 75% of H₂. Under these conditions we have the following parameters:
K_a = 0.0066
γ_NH3 = 0.91
γ_H2 = 1.09
γ_N2 = 1.14

Homework Equations

$$\frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2 \rightarrow \textrm{NH}_3$$
$$K_a = K_p K_{\gamma}$$

The Attempt at a Solution

First, I calculate the dummy constant K_γ:
$$K_{\gamma} = \frac{\gamma_{NH_3}}{\gamma_{N_2}^{0.5} \gamma_{H_2}^{1.5}} = 0.7489$$
Now I can get K_p which is K_a/K_γ; 0.0088129. Now, I figured out it was easier working with the reversed reaction:
$$\textrm{NH}_3 \rightarrow \frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2$$
The reversed reaction has K_p' = 113.47. Now, I applied the following method. Suppose we have the following initial conditions:
NH3 - n moles
N2 - 0 moles
H2 - 0 moles
When the reaction reaches equilibrium we will have:
NH3 - n (1 + α) moles
N2 - 0.5n α moles
H2 - 1.5n α moles
The total moles found in equilibrium are given by:
$$n_T = n - n\alpha + 0.5n\alpha + 1.5n\alpha = n(1+\alpha)$$
Now, applying P_i = n_iP, the partial pressures of the components in equilibrium will be given by the following expressions:
$$P_{NH_3} = \left( \frac{1-\alpha}{1+\alpha} \right) P$$
$$P_{N_2} = \left( \frac{0.5\alpha}{1+\alpha} \right) P$$
$$P_{H_2} = \left( \frac{1.5\alpha}{1+\alpha} \right) P$$
K_p' will be given by:
$$K_p' = \frac{\left( \frac{0.5\alpha}{1+\alpha} \right)^{0.5} P^{0.5} \left( \frac{1.5\alpha}{1+\alpha} \right)^{1.5} P^{1.5}}{\left( \frac{1-\alpha}{1+\alpha} \right) P}$$. Solving for α results in the following quadratic equation:
$$(K_p' + 1.299P)\alpha^2 - K_p' = 0$$
Plugging in K_p' = 113.47 and P = 300 atm and solving the equation results in α = 0.475. Now we can calculate the partial pressure of each of the components in equilibrium, which are 106.78 atm for ammonia, 48.31 atm for nitrogen and 144.91 atm for hydrogen. Finally, we calculate the mole fraction of N₂:
$$Y_{N_2} = \frac{48.31 \ atm}{300 \ atm} = 0.161$$
This result is apparently correct, although the procedure is a bit laborious; is there another method for solving this problem without involving the reversed reaction and α?

Thanks in advance for any input!

 

[Chestermiller]

Sure. Start out with 0.25 moles of N2 and 0.75 moles of H2, and let them react until x moles of NH3 are formed. So, in the final equilibrium,

moles of N2 = 0.25 - 0.5x

moles of H2 = 0.75 - 1.5x

total moles = 1 - x

Then get the mole fractions and partial pressures.

Chet

 

[MexChemE]

Thanks, Chet! I got the same result using that method, and it sure is a bit shorter. I think I'll be working with this method for synthesis reactions and the other method for dissociation reactions. Is there a set of restrictions for using each method? Like knowing the initial proportion of reactants? In this case 25% nitrogen and 75% hydrogen.

 

[Chestermiller]

Thanks, Chet! I got the same result using that method, and it sure is a bit shorter. I think I'll be working with this method for synthesis reactions and the other method for dissociation reactions. Is there a set of restrictions for using each method? Like knowing the initial proportion of reactants? In this case 25% nitrogen and 75% hydrogen.

Yes, there's a restriction. You can't do it by the "reverse" method that you used unless you know in advance that the reactants were in stochiometric proportions.

Chet