Question regarding volumes, pressures and Le Chatelier (1 Viewer)

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
668
Gender
Male
HSC
2014
Came across a Haber question:

"The volume of the reaction vessel was reduced. Explain the changes that occur"

Success One's answer:

" [formula for Haber process]
When the volume is reduced the concentrations of reactants and products immediately increase, which increases the total pressure. The equilibrium shifts forward to partially counteract this change because there are four moles of reactants and only two moles of products. A forward shift reduces the number of molecules as four react to form two. This reduces the pressure"

Is that actually the correct link between concentration and pressure?

I thought reducing the volume increased the concentration of all the substances, (since c=n/v) but increased the concentration of the reactants moreso than the products as there are more moles of reactant gases. This hence shifts equilibrium to the right. I would not have discussed pressure at all. Thoughts and explanations on why changes in volume actually shifts equilibrium to favour the side with less/more moles of gas?
 

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
668
Gender
Male
HSC
2014
volume and pressure are inveresly proportional i.e. as volume increases, pressure decreases.

Assume you have a container or something (don't ask, I'm making this up). You keep filling the container with apples. As you put in more apples, the pressure of the container increases (like come on so many more apples being forced into the same tiny space). But, now if I increased the capacity of the container i.e. the "volume" available to be filled, the apples would not be suffering from excessive pressure and would have mroe room to breathe.

Sorry for terrible analogy, but yeah they're inversely proportional. So if the question says "the volume was decreased", assume the pressure increased and by Le Chateliers principle, equilibrium shifted to the side with the least number of gaseous moles :)
Yes I understand the relationship between volume and pressure, but if you had to explain how an increase in pressure/ decrease in volume made equilibrium shift to the side with less moles, what would you say?
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Yes I understand the relationship between volume and pressure, but if you had to explain how an increase in pressure/ decrease in volume made equilibrium shift to the side with less moles, what would you say?
All you have to say is a decrease in volume results in an increase in pressure (since inversely proportional). Increase in pressure results in equilbrium shifting to the right (because of 4 to 2 mol ratio of gases) favouring the forward reaction and producing more ammonia (yield)
 
Last edited:

Kurosaki

True Fail Kid
Joined
Jul 14, 2012
Messages
1,168
Location
Tubbytronic Superdome
Gender
Male
HSC
2014
All you have to say is a decrease in volume results in an increase in pressure (since inversely proportional). Increase in pressure results in equilbrium shifting to the right favouring the forward reaction and producing more ammonia (yield)
Precisely. Repped =).

Edit: You must spread some Reputation around before giving it to faisalabdul16 again. D:
 

fatima96

Member
Joined
Sep 14, 2011
Messages
173
Location
omigod AUSTRALIA.
Gender
Undisclosed
HSC
2014
Volume of a reaction vessel is inversely proportional to pressure, so basically, if you increase the volume you decrease the pressure and vice versa.
In this case, the volume of the entire vessel has reduced, resulting in a pressure increase (because pressure is inversely proportional to volume).
When the pressure of a vessel increases, it tries to counteract the increase by producing less moles of gas.


Think of it, If I had a small room full of people, it'd be hot and stuffy and there'd be a lot more stress on the people in that room (pressure), and the only way to reduce that pressure or stress on the room (the vessel) is to get rid of some people from there :p

Likewise for this situation, pressure is increased so we need to get rid of some moles of gas to lower the pressure, thus eqm shifts right.
In terms of concentration, it is inversely proportional to volume. Reducing volume increases concentration & vice versa, and no reducing the volume / increasing conc. affects the entire vessel, doesn't affect reactants more than products :)

Hope this helped and correct me if im wrong
 

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
668
Gender
Male
HSC
2014
I think if you were asked to explain it though, you'd have to link it to concentration I.e. if you have 2mols of gas on one side and 1 mol of gas on the other, halving the volume will increase the concentration of one side by a factor of 4 and the other by a factor of 2. Thus eq will shift to the side with less moles I.e. a lower concentration.
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,282
Location
Lords
Gender
Female
HSC
2014
All you have to say is a decrease in volume results in an increase in pressure (since inversely proportional). Increase in pressure results in equilbrium shifting to the right favouring the forward reaction which has less gaseous moles, in order to reduce the pressure in the system by Le Chatelier's principleand thus producing more ammonia product as a result (yield)
^Just added some stuff as the bolded part for a complete answer
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Say that the decrease in volume corresponded to an increase in pressure, since they are inversely proportional. By Le Chatelier's Principle, the increase in pressure will drive the reaction to the side with the least number of gaseous moles in order to partially alleviate this imposed change in conditions.
Alright State ranker in english, let us feeble english students get some marks in chem :(
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,282
Location
Lords
Gender
Female
HSC
2014
I think if you were asked to explain it though, you'd have to link it to concentration I.e. if you have 2mols of gas on one side and 1 mol of gas on the other, halving the volume will increase the concentration of one side by a factor of 4 and the other by a factor of 2. Thus eq will shift to the side with less moles I.e. a lower concentration.
not really, I've never seen it related to concentration and it's best not to incase your marker is a bitch and thinks that you're talking about concentration instead :/

Perhaps you're overcomplicating it just a bit.
 

zhertec

Active Member
Joined
Aug 1, 2012
Messages
217
Gender
Male
HSC
2015
volume and pressure are inveresly proportional i.e. as volume increases, pressure decreases.

Assume you have a container or something (don't ask, I'm making this up). You keep filling the container with apples. As you put in more apples, the pressure of the container increases (like come on so many more apples being forced into the same tiny space). But, now if I increased the capacity of the container i.e. the "volume" available to be filled, the apples would not be suffering from excessive pressure and would have mroe room to breathe.

Sorry for terrible analogy, but yeah they're inversely proportional. So if the question says "the volume was decreased", assume the pressure increased and by Le Chateliers principle, equilibrium shifted to the side with the least number of gaseous moles :)
This and I believe if you decrease the volume yes all the concetration of chemicals in the system increase but I think it is to a very similar extent for all substances, hence pressure will be the main cause of the equilibrium shift.
 

harrypotterfan

Active Member
Joined
Nov 5, 2012
Messages
77
Gender
Female
HSC
2014
I think if you were asked to explain it though, you'd have to link it to concentration I.e. if you have 2mols of gas on one side and 1 mol of gas on the other, halving the volume will increase the concentration of one side by a factor of 4 and the other by a factor of 2. Thus eq will shift to the side with less moles I.e. a lower concentration.
Enigma is right, I avoid personally avoid bringing up concentration too because some markers can be st00pid.

But if you were to, I would probably explain with n=cv. As volume decreases, concentration increases. Hence the system will drive to the other side (the side without the increased concentration product) in order to minimise the disturbance to the equilibrium.

However, this is assuming the volume of one reactant or product increased. Theoretically its fine, but in reality (plus in 99% of questions), they'll say the total volume of the system. At this point, you can't just consider one reactant anymore, and total volume represents pressure...

So this is how it works, but you don't want a marker to think you're talking about total volume (the more likely method), especially if the question specificially asks for total volume (as I said before highly likely). So I would personally avoid concentration at all costs and focus only on pressure.
 

emilios

Well-Known Member
Joined
Jan 31, 2013
Messages
668
Gender
Male
HSC
2014
Ok here's the actual explanation, starting at 1:20 and it'll make more sense for those of you who have done Industrial.

http://www.youtube.com/watch?v=V0AdJJc478E

Lol this is why I hate HSC Science. I learnt all this Le Chatelier stuff 10 months ago but simply saying "By LCP, an increase in pressure shifts equilibrium to favour the side producing less moles of gas, in order to counteract the imposed change" doesn't actually tell me anything, even though it's the answer I'd mindlessly reproduce in a test. The link is sorta kinda there but nobody actually told me why equilibrium shifts to favour the side with less moles (and no, saying 'by LCP' is not an explanation).

So TLDR; the difference I'm stressing here is the difference between an "Identify" and "Explain" question. For an identify q you'd obviously just do the normal LCP routine with the pressures and whatnot. But for an Explain q, is simply saying "BY LCP...counteract the imposed change blah blah blah" enough?
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top