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Offline Kimyko

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Equilibrium Confusion
« on: May 09, 2007, 10:42:47 PM »
The problem I have right now is understanding this tiny concept having to do with equilibrium, and until I can get it, my mind won't let me go on to the next part. I'll probably have more questions to ask once I do get to the next parts.

Anyway, what I don't get are questions having to do with how much of [blank] chemical increases or decreases if temperature/pressure/etc increases. Or better yet, I can't tell which chemical increases or decreases when the temperature/pressure/etc increases/decreases. Sometimes, it doesn't even include pressure and whatnot, it's just if more of a certain chemical is added. Figuring out which direction equilibrium moves is not really a problem, it's just that I can never figure out what happens afterwards.

For example:

If the system 2CO(g) + O2 --> 2CO2(g) has come to equilibrium and then more CO(g) is added...

Then there are a series of four choices saying [CO2] increases and [O2] decreases and all that good stuff.

All I figured out so far is that equilibrium moves to the...right, right? (I take it back, I can't really even figure out which direction equilibrium moves in...)

And the second example, is:

If the pressure on the equilibrium system 2CO(g) + O2 --> 2CO2(g) is increased...

The quantity of CO2 increases, decreases, etc..

I've figured out nothing about this. It kind of confuses me.

Anyway, any help would be greatly appreciated. And yeah, I don't know how to make an equilibrium sign, sorry...

Offline enahs

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Re: Equilibrium Confusion
« Reply #1 on: May 09, 2007, 11:02:00 PM »
Le Chatelier's principle

Maybe you have just not heard it worded like the following:
If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure; the equilibrium will shift in order to minimize that change.

I do not really know what else to tell you without specific example of what you are trying to have trouble with.

Offline Yggdrasil

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Re: Equilibrium Confusion
« Reply #2 on: May 09, 2007, 11:07:03 PM »
If the system 2CO(g) + O2 --> 2CO2(g) has come to equilibrium and then more CO(g) is added...

Then there are a series of four choices saying [CO2] increases and [O2] decreases and all that good stuff.

All I figured out so far is that equilibrium moves to the...right, right? (I take it back, I can't really even figure out which direction equilibrium moves in...)

You are correct.  Adding more CO into the reaction mixture will make the reaction proceed forward (i.e. to the right).  Now, what will that mean in terms of the concentration of the reactants and prodcuts?

If the reaction is proceeding forward, then products are being created and reactants are being used up.  This means that the concentration of reactants (O2) will decrease and the concentration of products (CO2) will increase (the concentration of CO would also decrease from the amount present after you added CO to the reaction mixture, but it would not decrease below the original concentration of CO before you added extra CO).

On the other hand, if the reaction were to proceed backward (i.e. to the left), then the products would be converted to reactants.  This means that the concentration of the products would decrease and the concentration of reactants would increase.

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And the second example, is:

If the pressure on the equilibrium system 2CO(g) + O2 --> 2CO2(g) is increased...

The quantity of CO2 increases, decreases, etc..

I've figured out nothing about this. It kind of confuses me.

A general principle is that, when pressure is increased the reaction will proceed to the side of the equation with less gas molecules.  If the pressure is decreased, then the reaction is decreased then the reaction will proceed to the side of the reaction with more gas molecules.

In the example above, there are three molecules of gas on the left and two on the right.  Therefore, increasing the pressure will cause the reaction to proceed to the right (i.e. forward).  Now, what will happen to the concentration of CO2 in this case?  What about O2 and CO2?

Offline Kimyko

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Re: Equilibrium Confusion
« Reply #3 on: May 10, 2007, 07:39:10 AM »
Quote
And the second example, is:

If the pressure on the equilibrium system 2CO(g) + O2 --> 2CO2(g) is increased...

The quantity of CO2 increases, decreases, etc..

I've figured out nothing about this. It kind of confuses me.

A general principle is that, when pressure is increased the reaction will proceed to the side of the equation with less gas molecules.  If the pressure is decreased, then the reaction is decreased then the reaction will proceed to the side of the reaction with more gas molecules.

In the example above, there are three molecules of gas on the left and two on the right.  Therefore, increasing the pressure will cause the reaction to proceed to the right (i.e. forward).  Now, what will happen to the concentration of CO2 in this case?  What about O2 and CO2?

CO2 will increase while O2 decreases, because the O2 is being used up in order to create the CO2, right?

Enahs: I have heard of Le Chatelier's principal, and I think it's worded very similarly in my book. (I'm a horrible reader, so forgive me for my redundancy) That just basically means that it shifts in whatever direction helps it to re-establish equilibrium, right?


Oh! And there's a few other questions that just popped into my head. When you were explaining how to do the problem, Yggdrasil, I was wondering if both sides can increase or decrease at the same time. Because it seems when O2 decreases, then CO[/sub]2[/sub] increases. Can they ever both decrease/increase simultaneously? Does it work that way?

Also, regarding temperature change. I don't understand the phrasing that they used in my book. (Like I said before, I'm horrible at reading to understand)

It says, "Reversible reactions are exothermic in one direction and endothermic in another. ...The addition of heat shifts the equilibrium so that heat is absorbed. This facros the edothermic reaction."

Actually, no, I take that back, I understand what it's saying, but my problem is that I can't determine which side is endothermic and which side is exothermic.

The example I have is: If the temperature of the equilibrium system CH3OH(g) + 10kg ---> CO(g) + 2H2(g) increases...

Then the [CH3OH] increases and [CO] decreases, etc. (There are more options)

I can never figure out for sure which side is endothermic and exothermic. I try to reason it out, but then I end up confusing myself. I would think that the forward reaction is exothermic because energy is being put in in order break the first one apart. Then, does it release energy to put them back together in the reverse reaction?

Offline Yggdrasil

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Re: Equilibrium Confusion
« Reply #4 on: May 10, 2007, 05:19:01 PM »
CO2 will increase while O2 decreases, because the O2 is being used up in order to create the CO2, right?

Correct.

Quote
Oh! And there's a few other questions that just popped into my head. When you were explaining how to do the problem, Yggdrasil, I was wondering if both sides can increase or decrease at the same time. Because it seems when O2 decreases, then CO[/sub]2[/sub] increases. Can they ever both decrease/increase simultaneously? Does it work that way?

Not that I know of.  Since you need to use up O2 in order to create CO2, you can't create more CO2 without decreasing the amount of O2 (at least without adding more CO2 into the system from an external source).


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The example I have is: If the temperature of the equilibrium system CH3OH(g) + 10kg ---> CO(g) + 2H2(g) increases...

Then the [CH3OH] increases and [CO] decreases, etc. (There are more options)

I can never figure out for sure which side is endothermic and exothermic. I try to reason it out, but then I end up confusing myself. I would think that the forward reaction is exothermic because energy is being put in in order break the first one apart. Then, does it release energy to put them back together in the reverse reaction?

In general, it's hard to know which dirrection of the reaction will be endo- or exothermic.  However, I think there is a slight typo in the reaction equation you gave.  I think the right equation should be:

CH3OH(g) + 10kJ ---> CO(g) + 2H2(g)     (eq. 1)

which is equivalent to saying:

CH3OH(g)  ---> CO(g) + 2H2(g)  ΔH = +10kJ
 
For endothermic reactions, you need to add heat into your system in order to get the reaction to go forward.  Therefore, it is useful to think of heat as one of the reactants, as it is written in equation (1).  Note that if you reverse equation (1) you get the following equation:

CO(g) + 2H2(g) --> CH3OH(g) + 10kJ   (eq. 2)

which is exothermic since heat appears as a product of the reaction.  This means that the reaction releases energy.  Since exothermic reactions release heat by definition, this means that reaction (2) is exothermic.

Offline Kimyko

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Re: Equilibrium Confusion
« Reply #5 on: May 11, 2007, 01:10:16 AM »
Oh, alright, I think I understand now. So, if heat is added on the left side and it goes forward, then CH3OH would decrease while CO(g) and 2H2(g) would increase, right? And if it went the other way around, then, CO(g) and 2H2(g) would decrease while CH3OH would increase?

That brings me onto another question. I see a lot of problems in my homework having to do with either an increase in quantity and concentration, are those generally the same thing? I don't think so...right? I mean, are they sort of similar? Also, how do you find the concentration of something? That would be...molality and molarity, right?

For example, in one of my questions, it says, "In the equilibrium system CH3COOH(aq) + H20(l) ---> H30(aq) + CH3COO-, which species is present in the highest concentration?"

To be honest, I don't even know where to start here. Are concentrations just given to you, or are they something that you must memorize? I mean, the concentrations are different depending on certain factors, aren't they?

Uhm, and I have a few more questions (Sorry...) They're kind of general and still have to do with equilibrium, so I thought I'd just put them on here instead of making a new thread.

Okay, uhm, many times in my book, it uses the phrase "equilibrium position lies further to the right/left." What exactly does that mean? I mean, I know equilibrium goes to the left or to the right, but do some go more in those directions than others? And if so, can you figure out specifically how far to the right/left equilibrium goes? What's the use in knowing something like that?

And having to do with the question above, how do you figure out the concentration for individual particles? Does it have something to do with molality? What if the grams are not present?

Anyway, there's another thing I have about hydrolisis. In my book it says when hydrolisis occurs, but it didn't expand much on what happens when there's a weak acid/weak base It simply says that it can be both nuetral, acidic, or basic, but how can you tell when specificially?

Also it talks about how a susbstance with a low ksp is insoluble or sparingly soluable, but what constitutes a low ksp? Is it just if the number you get is low enough, like 10 to the negative really high number, then the substance is insoluable?

These questions are kind of general, I know, but I can't seem to find a direct answer to them. Sorry if they're kind of...simple.

Offline Yggdrasil

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Re: Equilibrium Confusion
« Reply #6 on: May 11, 2007, 03:06:05 AM »
Oh, alright, I think I understand now. So, if heat is added on the left side and it goes forward, then CH3OH would decrease while CO(g) and 2H2(g) would increase, right? And if it went the other way around, then, CO(g) and 2H2(g) would decrease while CH3OH would increase?

Yes.  You can think of increasing the temperature as adding heat and you can think of decreasing the temperature as taking heat away.

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That brings me onto another question. I see a lot of problems in my homework having to do with either an increase in quantity and concentration, are those generally the same thing? I don't think so...right? I mean, are they sort of similar? Also, how do you find the concentration of something? That would be...molality and molarity, right?

For equilibrium problems, concentration is generally measured in molarity.  Also, since many equilibrium problems are happening at fixed volume, molarity will be proportional to the number of moles.

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For example, in one of my questions, it says, "In the equilibrium system CH3COOH(aq) + H20(l) ---> H30(aq) + CH3COO-, which species is present in the highest concentration?"

Well, in this case, water is obviously going to have the highest concentration because everything is happening in an aqueous environment.

Quote
Okay, uhm, many times in my book, it uses the phrase "equilibrium position lies further to the right/left." What exactly does that mean? I mean, I know equilibrium goes to the left or to the right, but do some go more in those directions than others? And if so, can you figure out specifically how far to the right/left equilibrium goes? What's the use in knowing something like that?

This has to do with the equilibrium constant.  Consider the reaction:

X <--> Y

Now, the equilibrium constant for the reaction is given by:

K = [Y]/[X].

Now, what if the concentration of products is greater than the concentration of reactants that K > 1.  Similarly, if the concentrations of reactants are greater than the concentration of reactants K < 1.  Therefore, if your equilibrium constant is very small, you will have a very small amount of products present at equilibrium.  However, if your equilibrium constant is large, then you will have plenty of products at equilibrium.

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And having to do with the question above, how do you figure out the concentration for individual particles? Does it have something to do with molality? What if the grams are not present?

Anyway, there's another thing I have about hydrolisis. In my book it says when hydrolisis occurs, but it didn't expand much on what happens when there's a weak acid/weak base It simply says that it can be both nuetral, acidic, or basic, but how can you tell when specificially?

I'm not sure about these two questions.  Can you give more specific examples?

Quote
Also it talks about how a susbstance with a low ksp is insoluble or sparingly soluable, but what constitutes a low ksp? Is it just if the number you get is low enough, like 10 to the negative really high number, then the substance is insoluable?

It depends.  For a Ksp at around 10^-2 or lower, I would start calling something insoluble.

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