Equilibrium constant

  • Thread starter 97lmn
  • Start date
  • #1
9
0

Main Question or Discussion Point

N2(g) + 2O(g) <--> 2NO2(g)

At a certain temperature, the follwoing are the equilibrium concentrations for the above reaction:

[N2] = 8.0 mol/L
[O2] = 2.0 mol/L
[NO2] = 4.0 mol/L

Determine the equilibrium constant.

My answer..

Ke = [NO2]^2 / [N2][O2]^2

Ke = [4.0]^2 / [8.0][2.0]^2

Ke = 0.5

Correct?
 

Answers and Replies

  • #2
59
0
I believe your calculations are correct. Latex is friendly though =)
 
  • #3
9
0
Another question having to due with equilibrium...

The Claus Process is an industrial process used to remove toxic hydrogen sulfide gas during the processing of crude oil. The chemicle equation for this process is:

2 H2S(g) + SO2(g) <--> 3 H2O(g) + Heat

Use Le Chatliers Principle to describe why the following changes favour the removal of H2S.

a) Removing sulfur as soon as it forms
b) Cooling the reaction chamber

I think it has something to due with the temperature..and either the products or reactants being favored...not sure tho.
 
  • #4
59
0
97lmn said:
Another question having to due with equilibrium...

The Claus Process is an industrial process used to remove toxic hydrogen sulfide gas during the processing of crude oil. The chemicle equation for this process is:

2 H2S(g) + SO2(g) <--> 3 H2O(g) + Heat

Use Le Chatliers Principle to describe why the following changes favour the removal of H2S.

a) Removing sulfur as soon as it forms
b) Cooling the reaction chamber

I think it has something to due with the temperature..and either the products or reactants being favored...not sure tho.
I'm not sure what you mean by a); since there is no sulfur on the right side of the equation.

I don't want to give you the answer, but for b), look at the equation. Is it exothermic or endothermic? What would it mean? Try to think of temperature as an actual "molecular species", or in this case, a product. Try to apply your knowledge dealing with the removal or addition of a product in this case to predict which way the system would shift. I hope that leads you to the right answer.
 
  • #5
9
0
oh sorry got the equation wrong.

2 H2S(g) + SO2(g) <--> 3 S(s) + 2 H2O(g) + Heat
 
  • #6
59
0
97lmn said:
oh sorry got the equation wrong.

2 H2S(g) + SO2(g) <--> 3 S(s) + 2 H2O(g) + Heat
yes well, I was too lazy to actually fix your equation haha :P
do you understand or do you have further questions?
 
  • #7
GCT
Science Advisor
Homework Helper
1,728
0
Use Le Chatliers Principle to describe why the following changes favour the removal of H2S.

a) Removing sulfur as soon as it forms
b) Cooling the reaction chamber
The best way to explain these changes is to offer an explanation in relevance to relative rate constants.
 
  • #8
97lmn said:
Another question having to due with equilibrium...

The Claus Process is an industrial process used to remove toxic hydrogen sulfide gas during the processing of crude oil. The chemicle equation for this process is:

2 H2S(g) + SO2(g) <--> 3 H2O(g) + Heat

Use Le Chatliers Principle to describe why the following changes favour the removal of H2S.

a) Removing sulfur as soon as it forms
b) Cooling the reaction chamber

I think it has something to due with the temperature..and either the products or reactants being favored...not sure tho.

remember that heat can be considered as one of the products in this reaction. so cooling the reaction chamber means that you are absorbing heat from the right side of the reaction, and so the reaction will shift to the right by L'Chatlier's principle in order to restore the heat in equilibrium.
 
  • #9
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
17
I don't particularly favor quetzal's and atremis' approach of "treating heat as a reactant/product" as a means of first understanding Le Chatelier - this approach provides no physical insight. And I strongly recommend GCT's approach of deriving the effects on the forward and reverse reaction rates. Once you have completely understood the reasons though, you may use the first method, as a quick 'n' easy tool.
 

Related Threads on Equilibrium constant

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
6K
Replies
16
Views
5K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
0
Views
3K
Replies
3
Views
2K
Top