Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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2. One of the most widely produced chemicals in the world is ammonium to be used as fertilizer. It is synthesized by the Haber process (∆G° = –32.9 kJ/mol):
N2(g) + 3H2(g) → 2NH3(g)
How much does this reaction proceeded if you start with n moles of nitrogen and 3n moles of hydrogen? Also, let’s assume that the reactor maintains a pressure of 1 bar like the reaction is in a balloon, and that the temperature is maintained at 25 °C.
Question to answer:
In question 2, why was the extent of the reaction: N2(g) + 3H2(g) → 2NH3(g) equal to α=0.968 if the pressure is maintained at 1 bar but drops slightly to =0.956 if the pressure drops to 0.5 bar?
![### Equilibrium Calculations
#### Hint:
Use the table below to calculate mole fractions and partial pressures:
| | N₂ | H₂ | NH₃ |
|---------------|------------------|------------------|---------------|
| **Amount at equilibrium** | n(1 − α) | 3n(1 − α) | 2nα |
| **Mole fractions** | \(\frac{n(1 - \alpha)}{4n - 2n\alpha}\) | \(\frac{3n(1 - \alpha)}{4n - 2n\alpha}\) | \(\frac{2n\alpha}{4n - 2n\alpha}\) |
| **Partial pressures** | \(\frac{(1 - \alpha) \cdot P}{4 - 2\alpha}\) | \(\frac{3(1 - \alpha) \cdot P}{4 - 2\alpha}\) | \(\frac{2\alpha \cdot P}{4 - 2\alpha}\) |
Where α represents the extent of the reaction. \( K = \prod \left( \frac{P_i}{P_i^{\circ}} \right)^{\nu_i} \), where \( P_i \) is the partial pressure of species *i*.
i. As usual, \( P^{\circ} \) is just 1 bar. Now we can show:
\[ K = \left(\frac{2\alpha}{4 - 2\alpha} \cdot \frac{P}{P^{\circ}} \right)^2 \bigg/ \left( \left( \frac{(1 - \alpha) \cdot P}{4 - 2\alpha} \right) \left( \frac{3(1 - \alpha) \cdot P}{4 - 2\alpha} \right) \right)^3 \]
**Simplification gives:**
\[ \left(\frac{2\alpha}{4-2\alpha} \cdot \frac{P}{P^{\circ}}\right)^2 \cdot \left(\frac{4-2\alpha}{(1-\alpha) \cdot P}\right)^3 = \frac{16\alpha^2(2-\alpha)^2}{27(1-\](https://content.bartleby.com/qna-images/question/8411f3e8-16ff-46f1-b023-d023cc4de865/240090fa-4ca9-4087-9290-a2a9eba55c6a/9gffpm_processed.jpeg)
Transcribed Image Text:### Equilibrium Calculations
#### Hint:
Use the table below to calculate mole fractions and partial pressures:
| | N₂ | H₂ | NH₃ |
|---------------|------------------|------------------|---------------|
| **Amount at equilibrium** | n(1 − α) | 3n(1 − α) | 2nα |
| **Mole fractions** | \(\frac{n(1 - \alpha)}{4n - 2n\alpha}\) | \(\frac{3n(1 - \alpha)}{4n - 2n\alpha}\) | \(\frac{2n\alpha}{4n - 2n\alpha}\) |
| **Partial pressures** | \(\frac{(1 - \alpha) \cdot P}{4 - 2\alpha}\) | \(\frac{3(1 - \alpha) \cdot P}{4 - 2\alpha}\) | \(\frac{2\alpha \cdot P}{4 - 2\alpha}\) |
Where α represents the extent of the reaction. \( K = \prod \left( \frac{P_i}{P_i^{\circ}} \right)^{\nu_i} \), where \( P_i \) is the partial pressure of species *i*.
i. As usual, \( P^{\circ} \) is just 1 bar. Now we can show:
\[ K = \left(\frac{2\alpha}{4 - 2\alpha} \cdot \frac{P}{P^{\circ}} \right)^2 \bigg/ \left( \left( \frac{(1 - \alpha) \cdot P}{4 - 2\alpha} \right) \left( \frac{3(1 - \alpha) \cdot P}{4 - 2\alpha} \right) \right)^3 \]
**Simplification gives:**
\[ \left(\frac{2\alpha}{4-2\alpha} \cdot \frac{P}{P^{\circ}}\right)^2 \cdot \left(\frac{4-2\alpha}{(1-\alpha) \cdot P}\right)^3 = \frac{16\alpha^2(2-\alpha)^2}{27(1-\
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