Increasing Yield of a System with Le Chatelier's Principle

[Physics345]

Homework Statement

I am given 2NH3(g) + 2.5O2(g) ⇌ 2NO(g) + 3H2O(g)+heat. By using Le Chatelier's principle, describe and explain four ways in which an industrial chemist could increase the yield of nitrous oxide.

Homework Equations

The Attempt at a Solution

For 2NH3(g) + 2.5O2(g) ⇌ 2NO(g) + 3H2O(g) + heat
This is an unbalanced equation first off let’s balance it. If we multiply both sides by two it will be balanced giving us:
4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) + heat
Now to make it easier let:
A=4NH_3(g) B=5O_2(g) C=4NO(g) D=6H2O(g)
Giving us:
A+B⇌C+D+heat

1. If heat is removed from the product by cooling it down, the system will shift right in order to make up for the missing product (the heat which can be viewed as a product in the system), in turn adding more products which will yield to an increase in the products molecules. This can be viewed as a reverse reaction.
For example: ↓A + ↓B ⇌ ↑C + ↑D

2. If we add more A there will be a forward shift causing an increase in C and D there will be a shift forward (right) in turn making more product. This occurs because the system is trying to re-establish equilibrium by adding re balancing it’s self by adding more product.
For example: A + ↓B ⇌ ↑C + ↑D+heat

If we add more A and more B there will be even more of an increase in C and D there will be a shift forward (right) in turn making more product. This occurs because the system is trying to re-establish equilibrium by adding re balancing it’s self by adding more product.
For example: A + B ⇌ ↑↑C + ↑↑D+heat

3. If we decrease the pressure on the system it will respond by going right since there are 9 moles on the reactants side and 10 moles on the products side.

For example: ↓A + ↓B ⇌ ↑C + ↑D+ heat


4. If we remove D the system will respond by shifting forward (right) to counteract the removal of D the system will create more products resulting in more C in order to re-establish equilibrium.
For example: A + B ⇌ ↑C + ↑heat

Do I have a good understanding of Le Chatelier's principle? or am I missing something here. For some reason I feel like there is something missing it could be due to my lack of practice. Any advice would be helpful, thanks.

 

[Borek]

This is an unbalanced equation

It is balanced, just not with integer coefficients.

If we add more A and more B there will be even more of an increase in C and D there will be a shift forward (right) in turn making more product.

I wouldn't say this is because of the shift, you just react more substances, but the equilibrium doesn't change at all.

Other than that it looks OK, but being stuffed after an Easter breakfast I could miss something.

 

[Physics345]

It is balanced, just not with integer coefficients.



I wouldn't say this is because of the shift, you just react more substances, but the equilibrium doesn't change at all.

Other than that it looks OK, but being stuffed after an Easter breakfast I could miss something.

First off happy Easter!
Wouldn't the equilibrium favor the right side in this case though in order to relieve the stress, wouldn't that be worded as a shift forward (right)?

 

[Physics345]

Also I updated that part of it to say this instead:
If we add more A and more B there will be even more of an increase in C and D the reaction will favour the right side in order to relieve the stress caused by adding more of A and B, in turn making more product. This occurs because the system is trying to re-establish equilibrium by adding re balancing it’s self by adding more product.

 

[Borek]

No, just because you double amounts of reactants and you get more products doesn't mean there was a shift of equilibrium.

Think about it this way: you start with one mole of ammonia and you express amount of oxide produced per this one mole of ammonia used. Let's say you produced 0.1 mole of oxide. As long as the conversion doesn't change, there is no shift. You take 5 moles of ammonia and produce 0.5 moles of the oxide - no shift. You use 100 kmoles and get 10 kmoles of the oxide - same thing. But when you use 0.1 mole of ammonia and you change other things in such a a way you get not 0.01 moles but 0.02 moles of oxide - equilibrium was shifted.

 

[Physics345]

No, just because you double amounts of reactants and you get more products doesn't mean there was a shift of equilibrium.

Think about it this way: you start with one mole of ammonia and you express amount of oxide produced per this one mole of ammonia used. Let's say you produced 0.1 mole of oxide. As long as the conversion doesn't change, there is no shift. You take 5 moles of ammonia and produce 0.5 moles of the oxide - no shift. You use 100 kmoles and get 10 kmoles of the oxide - same thing. But when you use 0.1 mole of ammonia and you change other things in such a a way you get not 0.01 moles but 0.02 moles of oxide - equilibrium was shifted.

You honestly just answered my main confusion in this topic. Basically it's the moles conversion that determines whether there was a shift or not. This makes a lot more sense to me now. It's kind of awesome how the world works, it kind of feels like a pattern. The more I learn, the more I feel as though everything is relating to each other.
I really appreciate your help.
Thanks, Borek