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### Topic: Confusion over position of equilibrium shifting.  (Read 398 times)

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#### etotheipi

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##### Confusion over position of equilibrium shifting.
« on: September 03, 2019, 10:40:03 AM »
Consider a reaction A + B  C + D occuring at constant temperature.

If we decrease [C], the reaction quotient Q decreases. The net rate of reaction increases in the forward direction to restore Q to the constant value of Kc (i.e. more A and B react forward than C and D backward). After the system has once again reached equilibrium, we find that Q is the same as it was initially. This must be the case since the value of Kc has not changed (though all of the concentrations will have decreased a little bit, but this does not affect the ratio).

Le Chatelier's principle implies that if we reduce [C], the position of equilibrium shifts to the right. I don't understand this, since the quotients at the beginning and end are equal. Does the 'position of equilibrium' in this case refer to the position of the reaction (i.e. the position of the reaction shifts to the right if [C] is reduced), or am I missing something here?

The only context where the phrasing 'equilibrium shifts to the left/right' makes sense to me is when there is a change in temperature, because in that case the new value of the quotient at equilibrium has changed to the new value of Kc.

I believe I'm getting confused over what is meant by the position of equilibrium. I had a thought that it could it be equivalent to the value of Q, however this doesn't make that much sense considering that when it is 'shifting', the system isn't in equilibrium!

Thank you very much in advance
« Last Edit: September 03, 2019, 11:20:33 AM by etotheipi »

#### Corribus

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##### Re: Confusion over position of equilibrium shifting.
« Reply #1 on: September 03, 2019, 01:48:43 PM »
Le Chatelier's principle implies that if we reduce [C], the position of equilibrium shifts to the right. I don't understand this, since the quotients at the beginning and end are equal. Does the 'position of equilibrium' in this case refer to the position of the reaction (i.e. the position of the reaction shifts to the right if [C] is reduced), or am I missing something here?
Your thinking is more or less correct. Although this is the way it's frequently stated in casual discussion, I don't think it's a very good word choice. The only time "equilibrium shifts" is if new conditions actually change the value of the equilibrium constant (e.g., temperature or pressure change). Removing some product C doesn't change the actual equilibrium point (for reasonably small changes, anyway). It does drive the net consumption of products to form more reactants, so in this sense it "moves the reaction to the right". I guess there's a bit of sloppy language going on, where words "reaction" and "equilibrium" are used interchangeably. But we should make a more rigorous distinction about when changes to the system result in changes to the actual equilibrium constant and when they don't, even though Le Chatelier's Principle applies (in different ways) to both situations.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### etotheipi

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##### Re: Confusion over position of equilibrium shifting.
« Reply #2 on: September 03, 2019, 03:17:34 PM »
Le Chatelier's principle implies that if we reduce [C], the position of equilibrium shifts to the right. I don't understand this, since the quotients at the beginning and end are equal. Does the 'position of equilibrium' in this case refer to the position of the reaction (i.e. the position of the reaction shifts to the right if [C] is reduced), or am I missing something here?
Your thinking is more or less correct. Although this is the way it's frequently stated in casual discussion, I don't think it's a very good word choice. The only time "equilibrium shifts" is if new conditions actually change the value of the equilibrium constant (e.g., temperature or pressure change). Removing some product C doesn't change the actual equilibrium point (for reasonably small changes, anyway). It does drive the net consumption of products to form more reactants, so in this sense it "moves the reaction to the right". I guess there's a bit of sloppy language going on, where words "reaction" and "equilibrium" are used interchangeably. But we should make a more rigorous distinction about when changes to the system result in changes to the actual equilibrium constant and when they don't, even though Le Chatelier's Principle applies (in different ways) to both situations.

Thank you, that's really helpful

#### Vidya

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##### Re: Confusion over position of equilibrium shifting.
« Reply #3 on: September 05, 2019, 01:42:50 AM »
Try to understand this shifting and difference in Q and K using this image which is attached here.

#### etotheipi

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##### Re: Confusion over position of equilibrium shifting.
« Reply #4 on: September 05, 2019, 05:06:14 AM »
Try to understand this shifting and difference in Q and K using this image which is attached here.

This is a nice example. I did a rough calculation and as expected the value of the quotients before and after the change are equal, presumably equal to the value of Kc.

My confusion is that in this sense the position of equilibrium hasn't shifted (in terms of the value of Q at equilibrium), though the relative concentrations of each reactant have.

I think it might come down to terminology: should we think of the position of equilibrium as an arbitrary descriptor of the amounts of products and reactants?

The same thing occurs with gases; if we decrease the volume of a container with a reaction consisting of unequal amounts of moles on either side, the value of K must remain constant although the mole fractions change in order to keep K constant. So in a sense the composition of the reaction mixture has shifted, although the position of equilibrium as defined by the value of K has not.
« Last Edit: September 05, 2019, 05:50:15 AM by etotheipi »

#### Vidya

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##### Re: Confusion over position of equilibrium shifting.
« Reply #5 on: September 06, 2019, 01:11:10 AM »
From  image you can clearly see  position of equilibrium is changing to maintain the value of K constant.
It is K which remains  constant and  to keep it constant if some external changes are done equilibrium shifts.
Add more product it will give value of Q greater than K  so equilibrium shifts towards reactants to reduce the product which increases the amount of reactant and Q again become equal to K

It is K which is constant and equilibrium is dynamic .