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= 16a³(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº=-RT.In(K): (16a² (2-a)²) 27(1-a)* you can solve for a given that AG°= -32.9 kJ/mol. Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following equation: (1-1.497) a² +2 1.497a1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: Solutions: x = 0.550263 -4,Gº RT x = 5,47388 There are two answers since this is a quadratic equation. Now do the same for: -A-Gº In (16a²(2-a)²) 27(1-a)¹ to solve for a. You will also get two answers; make sure you pick the RT right one. = If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K 16a²(2-a)² p=2 27(1-a) p² and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a)-1 if a-0 and P(a)-0.5 if a-1?
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= 16a³(2-a)² 27(1-a)* When you plug this into the expression for the Gibbs energy at equilibrium A,Gº=-RT.In(K): (16a² (2-a)²) 27(1-a)* you can solve for a given that AG°= -32.9 kJ/mol. Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following equation: (1-1.497) a² +2 1.497a1.497 = 0, you would input this into the link: (1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are returned, but among them is: Solutions: x = 0.550263 -4,Gº RT x = 5,47388 There are two answers since this is a quadratic equation. Now do the same for: -A-Gº In (16a²(2-a)²) 27(1-a)¹ to solve for a. You will also get two answers; make sure you pick the RT right one. = If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression for K 16a²(2-a)² p=2 27(1-a) p² and what would be the extent of the reaction (i.e. use the Wolfram web site to recalculate a)? Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go from 4 moles of gas to 2 moles. What function of a gives P(a)-1 if a-0 and P(a)-0.5 if a-1?
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter11: Chemical Kinetics
Section: Chapter Questions
Problem 11.91PAE
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Transcribed Image Text: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-a)*
When you plug this
into the expression for the Gibbs energy at equilibrium A,Gº = -RT.In(K):
(16a² (2-a)²)
27(1-a)*
Solutions:
x = 0.550263
In
you can solve for a given that AG = -32.9 kJ/mol.
equation: (1-1.497) a² +2
Hint-hint: Use the Wolfram equation solver web site. For example, to solve the following
1.497a- 1.497 = 0, you would input this into the link:
(1-1.497)*x^2+2*1.497*x-1.497-0 and hit the equal sign. There are a lot of things that are
returned, but among them is:
-A,Gº
RT
right one.
x = 5.47388
There are two answers since this is a quadratic equation. Now do the same for:
-4,Gº
In
(16a²(2-a)²)
27(1-a)4
- to solve for a. You will also get two answers; make sure you pick the
RT
If the reactor in question 2 didn't maintain a pressure of 1 bar, what would be your expression
for K = 16a²(2-a)² poz
27(1-a) p2 and what would be the extent of the reaction (i.e. use the Wolfram web
site to recalculate a)?
Hint: Pº is 1 bar so it doesn't matter. The pressure P will drop from 1 bar to ½ bar since you go
from 4 moles of gas to 2 moles. What function of a gives P(a) = 1 if a=0 and P(a)=0.5 if a = 1?
Transcribed Image Text: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-a)
n(1-a)
4n-2na
(1-a)
4- 2a
Amount at equilibrium
Mole fractions
Partial pressures
P
K
H₂
3n(1-a)
3n(1-a)
4n-2na
3(1-a)
4-2a
P
NH3
2na
2na
4n-2na
2α
4- 2a
"₁₂
where a represents the extent of the reaction. K=II, where P is the partial pressure of species
i. As usual, Pº is just 1 bar. Now we can show:
-P
2a
(420P)²
4- 2a Po.
((1-2). P) (³(1-za). P.) ³
4 2a 4 2a
Simplification gives:
2α
4-2α P
4-2α
3
(+²2 F-)² (1-²8 - F) (+-²a - F-)³ = 16a²(2-a)³² pm²
4-2α
3(1-a)
27(1-a)*
P2¹
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