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MexChemE

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## 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 N

_{2}when the reaction reaches equilibrium at 300 atm and 723 K. The reactor's feedstock initially contained 25% in mole of N

_{2}and 75% of H

_{2}. 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

[tex]\frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2 \rightarrow \textrm{NH}_3[/tex]

[tex]K_a = K_p K_{\gamma}[/tex]

## The Attempt at a Solution

First, I calculate the dummy constant K

_{γ}:

[tex]K_{\gamma} = \frac{\gamma_{NH_3}}{\gamma_{N_2}^{0.5} \gamma_{H_2}^{1.5}} = 0.7489[/tex]

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:

[tex]\textrm{NH}_3 \rightarrow \frac{1}{2} \textrm{N}_2 + \frac{3}{2} \textrm{H}_2[/tex]

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:

[tex]n_T = n - n\alpha + 0.5n\alpha + 1.5n\alpha = n(1+\alpha)[/tex]

Now, applying P

_{i}= n

_{i}P, the partial pressures of the components in equilibrium will be given by the following expressions:

[tex]P_{NH_3} = \left( \frac{1-\alpha}{1+\alpha} \right) P[/tex]

[tex]P_{N_2} = \left( \frac{0.5\alpha}{1+\alpha} \right) P[/tex]

[tex]P_{H_2} = \left( \frac{1.5\alpha}{1+\alpha} \right) P[/tex]

K

_{p}' will be given by:

[tex]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}[/tex]. Solving for α results in the following quadratic equation:

[tex](K_p' + 1.299P)\alpha^2 - K_p' = 0[/tex]

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 N

_{2}:

[tex]Y_{N_2} = \frac{48.31 \ atm}{300 \ atm} = 0.161[/tex]

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!

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