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Le Chatelier's Principle

 
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msamadia1231



Joined: 24 Nov 2006
Posts: 53

PostPosted: Tue Apr 17, 2007 8:37 pm    Post subject: Le Chatelier's Principle Reply with quote

I am sure that when you take the MCAT, you will see a test question that tests if you have solid understanding of this concept. So let's make sure we've covered this.

If I push against the wall, the wall is going to push against my hand with an equal force, right? Well this principle is more of when you apply a stress, such as pressure, temp, concentration, than the system that we are analyzing is going to respond in such a way that will minimize that stress. Also keep those eyes on the eqn the MCAT gives. When talking about this above, it only applies if the reaction is in equilibrium.

How do I know if my rxn is in equilibrium? Look for the the double sided pointy arrows, one pointing to the left, the other to the right. If you see that, that means all systems go, you can proceed with Le Chatelier's Principle.

Temp increases. What will happen? Well according to above, that the system will minize, it will go to the side with less heat. Write out the rxn and see which side has heat. If it has a pos delta H thats endo and heat is a reactant. So, rxn will go to the product side.

How about pressure? This is important. Increase pressure. What happens? System will reduce the pressure by going to the side with more moles. Decrease the pressure? Goes to the side with decrease in moles. This is a neat way to remember. Increase pressure= goes to increase mole side. OK.

Increase Concentration of something? Will go to the opposite side of where that concentration was. Removing something is the same thing as decreasing the concentration of it. If I remove something, things want to replace it with more, b/c it just went missing.
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wallstng1442



Joined: 24 Jun 2007
Posts: 19

PostPosted: Sat Jul 21, 2007 2:12 pm    Post subject: Reply with quote

I think you have the pressure thing backwards. If you have an increase in pressure, the volume decreases and the system will go to the side with fewer moles in order to equilibrate.
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MANDALAM



Joined: 20 Jul 2007
Posts: 1

PostPosted: Sat Jul 21, 2007 9:00 pm    Post subject: Reply with quote

It is good to think of pressure in terms of PV=nRT. Note that Pressure is directly proportional to number of moles of gases. Now, if you get increased pressure, and you want to decrease it in one of many possible ways, how would you do it? Le Chatelier's principle is based on change in direction of reaction (of course Smile ) and therefore the best way to decrease P in PV=nRT is to decrease n. And how do you do that ? By shifting toward the side of equilibrium with fewer moles of gases.

How do you know a system goes to equilibrium?

Well, almost all chemical equations will reach an equilibrium. When a reaction goes to completion, it basically means that the Eq. constant is so high that the backward reaction's rate can be ignored. The important thing is to note that almost always the K of a reaction will be given, and of course, then you know it goes to equilibrium. A system is in equilibrium when reaction quotient (Q) is equal to reaction constant.
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kfowler_117623



Joined: 16 May 2010
Posts: 3

PostPosted: Tue Jun 08, 2010 9:58 am    Post subject: Reply with quote

wallstng1442 wrote:
I think you have the pressure thing backwards. If you have an increase in pressure, the volume decreases and the system will go to the side with fewer moles in order to equilibrate.


If pressure is increased the system will shift to decrease the pressure(LCP). So if pressure is increased the system will shift to decrease the number of moles of GAS. Therefore, shifting away from the side of the reaction with the largest number of gaseous moles towards the the side of the reaction with the fewest number of gas molecules.
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kevnm678961



Joined: 20 Nov 2010
Posts: 1

PostPosted: Sun Nov 21, 2010 11:18 am    Post subject: ? Reply with quote

The equilibrium is unaffected by solids but in the video it states an increase in solid introduces stress pushing the equilibrium to the right. Is this a mistake or am i missing something?
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mcat_premed3832



Joined: 19 Oct 2006
Posts: 413

PostPosted: Sun Nov 21, 2010 4:20 pm    Post subject: Reply with quote

If you have liquid and gas on both sides of an equilibrium, they are so incredibly more active in chemical reactions compared to solid, then indeed solid is ignored (and in fact left out of equilibrium equations).

However, from Le Chetalier's point of view, the following controlled reaction in a closed vessel of a single substance (not a chemical reaction between different compounds) was presented in the video:

solid <-> liquid + gas

This equilibrium is relevant in the context of Phase Diagrams.

www.chemguide.co.uk/physical/phaseeqia/phasediags.html
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