Quick Qemistry Questions

[Torterrable]

Having three "Q"s in a row was too tempting.

I have some conceptual questions that I've been struggling with, and I'd love it if anyone could help me.

So, we have reaction quotient Q. It is affected by pressure, concentration of reactants and products, and volume.

How do volume and pressure affect Q given a chemical equation A+B<->C where A, B, and C are gasses?

[Kythia]

You've almost answered your own questions, to be honest.  Kinda looks like you might understand it but just be a little uncertain? 

We have 1 mole of A, 1 mole of B initially.  Let's stop splitting "volume" and "pressure" into two things and view them as flip sides of the same coin - by keeping the amount of each initial gas the same V and P are aspects of the same issue, right?  V is inversely proportional to P.

So.  What happens to the mixture when V is increased and P decreased or vice versa? In what my old chemistry teacher used to call "billiard ball terms", if that makes sense. 

[Torterrable]

The main problem is, in an explanation on a quiz or a test, I would have to relate a change in volume or pressure to Q or K (at least I think I would) and then use that as an explanation. Just saying "Le Chatelier's principle" and explaining what it is is not enough. I know what happens to K when heat is added, and I know the effect on Q when concentration is changed, but pressure and volume don't seem to directly affect either of those variables.

[Kythia]

Stick with me.  V and P both directly alter both of those variables.

Think, as I say, about the microscopic effects of a change in V or P.  What is heat on a microscopic level?  What do you actually mean when you say "concentration".

In essence, an increase in V causes the gas(es) to expand.  Which, as you know, is identical to a decrease in temperature, all else being equal.  So the equilibrium constant shifts. An increase in P is functionally identical to an increase in volume concentration for the purposes of this question - albeit an equal one for both reagents - which moves the reaction quotient.

[Torterrable]

Ah, I think I get it now. Are there any equations that can help prove this, or is this just accepted as fact?

[Kythia]

Well, the temperature aspects are just an extrapolation from the Ideal Gas Laws while the concentration aspects are kinda true by definition if you think about what is meant by "concentration" - or, rather, think about why the concentration of a reagent affects Q or K.

It's just thinking about what an increase or decrease in V and P actual means, in microscopic terms, and what Q actually stands for/represents.  Anything that affects the number of interactions between a molecule of A and a molecule of B is going to affect Q.

[Torterrable]

Thank you very much for your help; I understand now, I think.

I have one more question that just popped up; I figure I could just direct it to you since you seem to know. Under what circumstances can you use successive approximation? The professors merely say that it is "when you can ignore the x" due to the fact that it's an insignificantly small value, but what are some examples where this may occur?

[Kythia]

Ah, now you've got me I'm afraid.  Not a term I've come across before.  I think I have my notes from A-level chemistry still so I can have a look and see if anything shows up, but off the top of my head I don't have a Scooby.  Sorry.

[Torterrable]

It's okay; don't go to any trouble on my account. Honestly, I should probably just watch the recorded lectures, but since you were knowledgeable on the subject, I thought I'd simply ask first.

Thank you.

[aouser626]

The main problem is, in an explanation on a quiz or a test, I would have to relate a change in volume or pressure to Q or K (at least I think I would) and then use that as an explanation. Just saying "Le Chatelier's principle" and explaining what it is is not enough. I know what happens to K when heat is added, and I know the effect on Q when concentration is changed, but pressure and volume don't seem to directly affect either of those variables.

We can consider the shifting of the equilibrium position in the context of entropy, which, in this case, Gibbs energy, in where: ΔG = -RTInK (keeping in mind that, temperature is proportional to pressure and volume)

Considering the ratio of K, in the assumption of an equilibrium reaction where the forward reaction produces the product(s), the higher the value of K, the greater the relative concentration of the products. Hence, the lower the value of K, the lower the relative concentration of the products.

Which means, as the position of equilibrium shifts to the right according to the previously established premise, K > 1 and thus InK > 0, giving a negative value of ΔG. The opposite thus follows, whereby K < 1, InK < 0 and therefore ΔG is positive, favoring the reactants. The consideration of Gibbs energy is such that, if the value is too relatively extreme, then the position of equilibrium may be shifted so far out that relatively no reactions take place at all.

From ΔG = -RTInK and ΔG = ΔH - TΔS(sys)
-RTInK = ΔH - TΔS(sys)
InK = -(ΔH/RT) + (ΔS(sys)/R) in where RT = PV/n

Taking an exothermic reaction for the forward reaction, ΔH is negative, thus, -(ΔH/RT) is positive. Simply, as PV increases, RT increases and thus (ΔH/RT) decreases in magnitude, therefore, InK follows and K decreases in magnitude, favoring the reactants, as K is the ratio of the relative concentration of the products to the reactants.

Notice that, of a ratio of 0.999..., K < 1, InK < 0 and thus ΔG is positive, favoring the reactants as stated previously.

And from this, we saw how in changing PV, K changes and in which also changes the position of equilibrium, as considered by ΔG.

[Mathim]

Damn, ya beat me to it. I'm a chem major, going into teaching, but yeah, I'd basically have given a shorter version of what entropy said.

[aouser626]

Damn, ya beat me to it. I'm a chem major, going into teaching, but yeah, I'd basically have given a shorter version of what entropy said.

Entropy is in my name...  :D

[Sierra117]

Thank you very much for your help; I understand now, I think.

I have one more question that just popped up; I figure I could just direct it to you since you seem to know. Under what circumstances can you use successive approximation? The professors merely say that it is "when you can ignore the x" due to the fact that it's an insignificantly small value, but what are some examples where this may occur?

Hey, fantastic question. As a chemistry student-teacher, my 16-18 year olds ask this a lot.

The reason we can ignore it sometimes is due to the fact that, in order for a reaction to occur, there needs to be collisions between two or more reactants to form a product. This must happen with sufficient energy and at the correct orientation for that to happen, hence, the higher the concentration of a reactant, the more likely it is a product will be formed and therefore equilibrium shifts towards the product, affecting the equilibrium constant. As you may have noticed, even a large change in the amount of reactants doesn't massively change that number.

However, if there is so little reactant that it will make little product and therefore barely change the equilibrium number, we can ignore it. This as we use significant numbers it probably won't affect the equilibrium enough to change the significant numbers from staying the same Let me show you.

So, lets say we are making NH3 (Haber process) and the concentration is 0.15-x. However, x is .0000096. When you take away x from that, you get .1499904 which, when you round it up to two significant numbers, is still .15, so the change is so insignificant we can just ignore it. Obviously, the higher in the field you go the less this happens, but we mainly use K to get an approximate value, not an actual one. It's more a thought exercise (we can literally record the concentration of a product after a reaction through various means).