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/mo

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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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?

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- 2a
-P
K =
H₂
3n(1 - α)
3n(1 - α)
4n - 2na
3(1 - α)
4 – 2α
P
In
NH3
2na
2na
4n - 2na
where a represents the extent of the reaction. K = Пi where P₁ is the partial pressure of species
pi
pou¡,
i. As usual, Pº is just 1 bar. Now we can show:
2α
4-2α
2α
(42P)²
4- 2a Po
(4-20). P) (³(1-a). P)²
4
Simplification gives:
2
2α P
3
4-2α
.
( ²2 )² (1-² F) (¹-² F) ³ = 16a²(2-a)²_p-²
4-2α
(1-α)
-
4-2α
3(1-α)
27(1-x)4
=
P
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=
When you plug this
16a²(2-a)²
27(1-α)4
into the expression for the Gibbs energy at equilibrium 4µGº = −RT · ln(K):
(16a² (2 - α)²)
27(1-x)4
you can solve for a given that AG° = -32.9 kJ/mol.
-4,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- 2a -P K = H₂ 3n(1 - α) 3n(1 - α) 4n - 2na 3(1 - α) 4 – 2α P In NH3 2na 2na 4n - 2na where a represents the extent of the reaction. K = Пi where P₁ is the partial pressure of species pi pou¡, i. As usual, Pº is just 1 bar. Now we can show: 2α 4-2α 2α (42P)² 4- 2a Po (4-20). P) (³(1-a). P)² 4 Simplification gives: 2 2α P 3 4-2α . ( ²2 )² (1-² F) (¹-² F) ³ = 16a²(2-a)²_p-² 4-2α (1-α) - 4-2α 3(1-α) 27(1-x)4 = P 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= When you plug this 16a²(2-a)² 27(1-α)4 into the expression for the Gibbs energy at equilibrium 4µGº = −RT · ln(K): (16a² (2 - α)²) 27(1-x)4 you can solve for a given that AG° = -32.9 kJ/mol. -4,Go RT
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