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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.

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.

Introduction to Chemical Engineering Thermodynamics
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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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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. 

Hint: We use the table below to calculate mole fractions and partial pressures: N₂ n(1 - α) n(1 - α) 4n - 2na Amount at equilibrium Mole fractions Partial pressures (1 - α) 4 – 2α P K = H₂ 3n(1 - α) 3n(1-a) 4n - 2na 3(1-x) 4 – 2α P pºi where a represents the extent of the reaction. K = [₁ pou¡, where P; is the partial pressure of species i. As usual, Pº is just 1 bar. Now we can show: NH3 2na 2na 4n - 2nd In 2α P 4- 2α (42P)²2 2α Р 4 – 2a P° 3 (4-20 P) (³4²--2a P) (3(1-α). 2α Simplification gives: 2 3 2α Р 4-2α 4-2α Po . - (1²2 P)² (1-² F) (+-²a F-)³ = 16a²(2-a)² p-² 4-2α (1-α) 3(1-α) 27(1-x)4 P2 Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K= 16a²(2-a)² 27(1-α)4 When you plug this into the expression for the Gibbs energy at equilibrium ArGº = −RT · ln(K): (16a² (2-a)²) 27(1-α)4 you can solve for a given that AG° = -32.9 kJ/mol. - -A-Gº RT
Transcribed Image Text:Hint: We use the table below to calculate mole fractions and partial pressures: N₂ n(1 - α) n(1 - α) 4n - 2na Amount at equilibrium Mole fractions Partial pressures (1 - α) 4 – 2α P K = H₂ 3n(1 - α) 3n(1-a) 4n - 2na 3(1-x) 4 – 2α P pºi where a represents the extent of the reaction. K = [₁ pou¡, where P; is the partial pressure of species i. As usual, Pº is just 1 bar. Now we can show: NH3 2na 2na 4n - 2nd In 2α P 4- 2α (42P)²2 2α Р 4 – 2a P° 3 (4-20 P) (³4²--2a P) (3(1-α). 2α Simplification gives: 2 3 2α Р 4-2α 4-2α Po . - (1²2 P)² (1-² F) (+-²a F-)³ = 16a²(2-a)² p-² 4-2α (1-α) 3(1-α) 27(1-x)4 P2 Since the pressure is maintained at 1 bar and the standard pressure is 1 bar, we can eliminate the pressure component on the right side and are thus left with: K= 16a²(2-a)² 27(1-α)4 When you plug this into the expression for the Gibbs energy at equilibrium ArGº = −RT · ln(K): (16a² (2-a)²) 27(1-α)4 you can solve for a given that AG° = -32.9 kJ/mol. - -A-Gº RT
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