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I've read many posts explaining why changing the temperature affects Kc, whereas a change in pressure/concentration keeps Kc constant, but I'm still not understanding the actual practical reason behind this!

According to Le Chatelier's principle, changing the concentration, pressure or temperature will ALL have an effect in shifting the equilibrium position. However, since Kc is a measure of how far the equilibrium position stands, how is it possible for Kc to remain constant, while the equilibrium position has changed?

Thank you!
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(Original post by I <3 WORK)
I've read many posts explaining why changing the temperature affects Kc, whereas a change in pressure/concentration keeps Kc constant, but I'm still not understanding the actual practical reason behind this!

According to Le Chatelier's principle, changing the concentration, pressure or temperature will ALL have an effect in shifting the equilibrium position. However, since Kc is a measure of how far the equilibrium position stands, how is it possible for Kc to remain constant, while the equilibrium position has changed?

Thank you!
Le Chatelier's Principle is this:

When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.
The equilibrium doesn't shift when you increase a concentration or partial pressure of a species in the system because the system is taken out of equilibrium. Due to LCP the system will counteract this increase in concentration and reestablish an equilibrium in the same position as before (unless you also change the temperature).
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(Original post by alow)
Le Chatelier's Principle is this:



The equilibrium doesn't shift when you increase a concentration or partial pressure of a species in the system because the system is taken out of equilibrium. Due to LCP the system will counteract this increase in concentration and reestablish an equilibrium in the same position as before (unless you also change the temperature).
That doesn't make sense, in all my textbooks and previous knowledge it has been written that changing the concentration - for e.g. by adding/removing product will shift the equilibrium to counteract that change?

Also do you mind clarifying what LCP means please?
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(Original post by I <3 WORK)
That doesn't make sense, in all my textbooks and previous knowledge it has been written that changing the concentration - for e.g. by adding/removing product will shift the equilibrium to counteract that change?

Also do you mind clarifying what LCP means please?
It makes no sense to say an equilibrium is shifting when the system is not in equilibrium. That is terrible wording which is used far too often. If the system was in equilibrium the whole then Le Chatelier's principle wouldn't be a thing.

This is good: http://chemistry.stackexchange.com/a/4981/33612
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(Original post by alow)
It makes no sense to say an equilibrium is shifting when the system is not in equilibrium. That is terrible wording which is used far too often. If the system was in equilibrium the whole then Le Chatelier's principle wouldn't be a thing.

This is good: http://chemistry.stackexchange.com/a/4981/33612
I've just read the post, thanks for sharing! However I still don't understand, isn't the new equilibrium the one which has shifted? What's the point of increasing/decreasing concentration if you still end up with the same amounts of reactants and products?
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(Original post by I <3 WORK)
I've just read the post, thanks for sharing! However I still don't understand, isn't the new equilibrium the one which has shifted? What's the point of increasing/decreasing concentration if you still end up with the same amounts of reactants and products?
If you have higher concentrations of the species used in your rate determining step, then the equilibrium will be established more quickly.
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(Original post by alow)
If you have higher concentrations of the species used in your rate determining step, then the equilibrium will be established more quickly.
I don't understand though, why is it that everywhere I read they say the new equilibrium will shift when changing the concentration? CGP, Edexcel Revision Guide, ChemRevise, BBC Bitesize, even Chemguide! If the equilibrium doesn't actually shift then why do they all talk about certain industrial processes like the Haber process for instance, where they increase the concentrations of H2 and N2 to produce more ammonia? Why is this some new breakthrough I've never heard of before?
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(Original post by I <3 WORK)
I don't understand though, why is it that everywhere I read they say the new equilibrium will shift when changing the concentration? CGP, Edexcel Revision Guide, ChemRevise, BBC Bitesize, even Chemguide! If the equilibrium doesn't actually shift then why do they all talk about certain industrial processes like the Haber process for instance, where they increase the concentrations of H2 and N2 to produce more ammonia? Why is this some new breakthrough I've never heard of before?
The way most people at A Level are taught with regards to "equilibrium shifts" makes it very difficult to actually understand what's going on when you change a partial pressure or concentration in a system.

It simply does not make sense to talk about equilibrium to "shift" when a system is not in equilibrium.

I posted this a while back:

(Original post by alow)
Let's say I have a reversible reaction in solution which is in equilibrium: A ⇌ B I add another 0.5mol of A to the reaction mixture, therefore increasing [A]. This means that the forward reaction will now be proceeding faster than the reverse (as there is more A being converted to B than vice versa).

The system is therefore no longer in equilibrium as the forward and reverse reactions are not proceeding at the same rate.

After some time equilibrium is reestablished as [B] "catches up" to [A], with the same ratio of [B] to [A] as before the addition of extra A.
If I define the rate constant of the forward process to be k_f and the reverse process to be k_r then the equilibrium constant K_c=\dfrac{k_f}{k_r} (Proof: http://chemistry.stackexchange.com/a/17311/33612).

A rate constant for a process is only a function of temperature, therefore the only way to change K_c (and therefore equilibrium position) is by changing the temperature, not by changing any concentrations of reactant.
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(Original post by alow)
The way most people at A Level are taught with regards to "equilibrium shifts" makes it very difficult to actually understand what's going on when you change a partial pressure or concentration in a system.

It simply does not make sense to talk about equilibrium to "shift" when a system is not in equilibrium.

I posted this a while back:



If I define the rate constant of the forward process to be k_f and the reverse process to be k_r then the equilibrium constant K_c=\dfrac{k_f}{k_r} (Proof: http://chemistry.stackexchange.com/a/17311/33612).

A rate constant for a process is only a function of temperature, therefore the only way to change K_c (and therefore equilibrium position) is by changing the temperature, not by changing any concentrations of reactant.
I think it's worthwhile noting that mark schemes can underline the "shift" and "position of equilibrium" when talking about concentration, pressure etc even if it doesn't - so you may actually lose marks by doing it properly
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(Original post by I <3 WORK)
I don't understand though, why is it that everywhere I read they say the new equilibrium will shift when changing the concentration? CGP, Edexcel Revision Guide, ChemRevise, BBC Bitesize, even Chemguide! If the equilibrium doesn't actually shift then why do they all talk about certain industrial processes like the Haber process for instance, where they increase the concentrations of H2 and N2 to produce more ammonia? Why is this some new breakthrough I've never heard of before?
I've been banging on about this for years.

Yea verily dost thou pay heed to "Alow" for he knoweth what he sayeth.

My equilibrium introductory classes always begin with a warning that many textbooks are talking out of their back covers when it comes to equilibrium.

Finally one or two examinations boards are getting the message and introducing the concept of reaction quotient.

There is one equilibrium "position" at each temperature, at which the ratio of product concentrations to reactant concentrations (the reaction quotient) is constant, and the rate of the forward reaction is equal to the rate of the back reaction.

Any change in conditions at this temperature, eg addition or removal of components, changes the reaction quotient, so that it is no longer equal to the equilibrium constant and the system is no longer at equilibrium.

The system then responds in such a way as to restore the equilibrium proportions, so that the quotient is once again equal to the equilibrium constant.

The equilibrium constant does not change UNLESS the temperature changes.

An equilibrium which is disturbed produces a system which is then no longer at equilibrium. The system then shifts to either the left-hand side or the right-hand side so as to re-establish the equilibrium proportions and restore the equality, reaction quotient = equilibrium constant.

Note: Several examinations boards do not allow Le Chatelier as an explanation, only as a rule of thumb.
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(Original post by charco)
I've been banging on about this for years.

Yea verily dost thou pay heed to "Alow" for he knoweth what he sayeth.

My equilibrium introductory classes always begin with a warning that many textbooks are talking out of their back covers when it comes to equilibrium.

Finally one or two examinations boards are getting the message and introducing the concept of reaction quotient.

There is one equilibrium "position" at each temperature, at which the ratio of product concentrations to reactant concentrations (the reaction quotient) is constant, and the rate of the forward reaction is equal to the rate of the back reaction.

Any change in conditions at this temperature, eg addition or removal of components, changes the reaction quotient, so that it is no longer equal to the equilibrium constant and the system is no longer at equilibrium.

The system then responds in such a way as to restore the equilibrium proportions, so that the quotient is once again equal to the equilibrium constant.

The equilibrium constant does not change UNLESS the temperature changes.

An equilibrium which is disturbed produces a system which is then no longer at equilibrium. The system then shifts to either the left-hand side or the right-hand side so as to re-establish the equilibrium proportions and restore the equality, reaction quotient = equilibrium constant.

Note: Several examinations boards do not allow Le Chatelier as an explanation, only as a rule of thumb.
Thank you so much alow, that final explanation makes a lot of sense, and thank you charco too for the heads up!

So basically counterintuitively, when all the textbooks write about the equilibrium shifting, whereas it's actually the quotient changing, does that mean I write the wrong answer in the exam as 'GradeA*UnderA' mentioned? My exam board is Edexcel.
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(Original post by I <3 WORK)
Thank you so much alow, that final explanation makes a lot of sense, and thank you charco too for the heads up!

So basically counterintuitively, when all the textbooks write about the equilibrium shifting, whereas it's actually the quotient changing, does that mean I write the wrong answer in the exam as 'GradeA*UnderA' mentioned? My exam board is Edexcel.
I mean, unless you intend to study chemistry at university, it's probably safer to assume what the textbooks want - but obviously, check the mark schemes of your exam boards and see what they want
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(Original post by alow)
The way most people at A Level are taught with regards to "equilibrium shifts" makes it very difficult to actually understand what's going on when you change a partial pressure or concentration in a system.

It simply does not make sense to talk about equilibrium to "shift" when a system is not in equilibrium.

I posted this a while back:



If I define the rate constant of the forward process to be k_f and the reverse process to be k_r then the equilibrium constant K_c=\dfrac{k_f}{k_r} (Proof: http://chemistry.stackexchange.com/a/17311/33612).

A rate constant for a process is only a function of temperature, therefore the only way to change K_c (and therefore equilibrium position) is by changing the temperature, not by changing any concentrations of reactant.
One more question please, does this idea also apply to pressure changes as well? When text books say that the equilibrium position moves towards the side with fewer moles of gases, why would the equilibrium position not change if that's the case?
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(Original post by I <3 WORK)
One more question please, does this idea also apply to pressure changes as well? When text books say that the equilibrium position moves towards the side with fewer moles of gases, why would the equilibrium position not change if that's the case?
For an ideal gas,  K_p is independent of pressure.
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(Original post by alow)
For an ideal gas,  K_p is independent of pressure.
Also, when you dilute a weak acid, why does the pH increase? When you add water, I know now that the equilibrium won't shift to the right but rather the quotient would. That is, the final equilibrium position would be the same as before, so therefore what causes the increase?
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(Original post by I <3 WORK)
Also, when you dilute a weak acid, why does the pH increase? When you add water, I know now that the equilibrium won't shift to the right but rather the quotient would. That is, the final equilibrium position would be the same as before, so therefore what causes the increase?
\text{pH}=-\log_{10} [\text{H}^+]

When you dilute the acid you increase the volume of the solution and, as you have the same equilibrium as before, you have the same number of hydrogen ions from the acid in a larger volume of water. Therefore [\text{H}^+] decreases, and \text{pH} increases.
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(Original post by alow)
\text{pH}=-\log_{10} [\text{H}^+]

When you dilute the acid you increase the volume of the solution and, as you have the same equilibrium as before, you have the same number of hydrogen ions from the acid in a larger volume of water. Therefore [\text{H}^+] decreases, and \text{pH} increases.
Thanks that makes sense, but also then what causes there to be a higher increase in pH for a strong acid compared to a weak acid?
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