Avogadro's law

[Qfwfq]

When the system of gases is at equilibrium and you increase the pressure, the applied stress will drive the reaction to the left: from the side with a higher number of gas molecules – 3 – to the side with a lower number of gas molecules – 1. If you decrease the pressure when the system is at equilibrium, you will drive the reaction to the right.

Does this mean that, if I have water vapour in a cylinder and I let it expand, it will separate into hydrogen and oxygen? ;)

[Biochemist]

Does this mean that, if I have water vapour in a cylinder and I let it expand, it will separate into hydrogen and oxygen? ;)

Q- I think you ouight to stick to physics;).

 

I suspect, however, that your statement is trivially correct. The miniscule amount of water vapor that converts in equilibrium from 2H2O<->2H2 + O2 would be forced to the right under negative pressure. It is just a very small quantity on the right.

[Fishteacher73]

Ideal gas law:

PV=nRT

Pressure*Volume=Number of moles*ideal gas law constan*temp.

 

This will show that under the same conditions the same number of moles of a gas are present.

A mole can also be expressed in mass by convertying the atomic weight of the molecule to grams. (ie 1 mole of H = 1g). The gasses should have the apropriatte masses and support the hypothesis.

[TeleMad]

TeleMad: When the system of gases is at equilibrium and you increase the pressure, the applied stress will drive the reaction to the left: from the side with a higher number of gas molecules – 3 – to the side with a lower number of gas molecules – 1. If you decrease the pressure when the system is at equilibrium, you will drive the reaction to the right.

 

Qfwfq: Does this mean that, if I have water vapour in a cylinder and I let it expand, it will separate into hydrogen and oxygen?

 

First, let me support what I said. I’ll address your error in my next post (but that might to wait until tonight since I am home right now just for lunch).

 

”Consider the decomposition of N2O4(g) to NO2(g) at 298K:

 

N2O4(g) <-> 2NO2(g) Kp = 0.145

 

Figure 14.6 depicts an equilibrium mixture under an external pressure of 1 atm. Now suppose we quickly increase the external pressure to 2 atm. First, let us predict what should happen by comparing Qp and Kp.

 

…

 

Now let us consider this issue from the standpoint of Le Chatelier’s principle. When we decrease the volume of an equilibrium mixture by increasing the external pressure, we crowd the molecules more closely together. To minimize the effect of this overcrowding, a net reaction occurs, producing a smaller number of molecules. Because two moles of NO2(g) yield just one mole of N2O4(g), the reverse reaction is stimulated and equilibrium shifts to the left, just as we concluded in the assessment of Qp and Kp.

 

The following statements summarize the effect of changes in external pressure (or system volume) on an equilibrium involving gases.

 

1) When the external pressure is increased (or the system volume is reduced), an equilibrium shifts in the direction producing the smaller number of moles of gas.

 

2) When the external pressure is decreased (or the system volume is increased), an equilibrium shifts in the direction producing the larger number of moles of gas.

 

3) If there is no change in the number of moles of gas in a reaction, changes in external pressure (or system volume) have no effect on equilibrium.”

(General Chemistry: Fourth Edition, John W Hill, Ralph H Petrucci, Terry W McCreary, and Scott S Perry, Pearson/Prentice Hall, 2005, p592-593)

[TeleMad]

Does this mean that, if I have water vapour in a cylinder and I let it expand, it will separate into hydrogen and oxygen? ;)

 

Now, about your flawed logic.

 

”In principle, every reaction is reversible, at least to some extent, and can be described through an equilibrium constant expression. In many cases, however, we do not need to use equilibrium constants in calculations. How can this be? Let’s answer this question by considering three specific cases.

 

First, consider the reaction of hydrogen and oxygen gases at 298K:

 

2H2(g) + O2(g) <-> 2H2O(l)

 

Kp = 1 / [Ph2)^2(Po2)] = 1.4 x 10^83

 

Starting with a 2:1 mole ration of hydrogen to oxygen, the equilibrium partial pressure of H2(g) and O2(g) must become extremely small – approaching zero – in order of the Kp value to be so large. We conclude that, for all practical purposes, the hydrogen and oxygen are totally consumed in the reaction. We say that a reaction goes to completion if one of more reactants is totally consumed, and we do calculations by using just the principles of stoichiometry.

 

A very large numerical value of Kc or Kp signifies that a reaction goes to completion, or essentially so. (In effect, the reaction is not reversible)” (General Chemistry: Fourth Edition, John W Hill, Ralph H Petrucci, Terry W McCreary, and Scott S Perry, Pearson/Prentice Hall, 2005, p586)

 

So you are trying to use a reaction that goes to completion and is not reversible to show that something stated about reversible reactions and equilibrium is wrong. Might as well go around using a hammer to try to torque in a screw. Try using the correct tools for the job.

 

 

 

PS: By the way, speaking of Le Chatelier, he was smart enough to know the difference between reversible reactions, which reach a dynamic equilibrium, and reactions that go to completion.

 

”Henri Le Chatelier was one of the first to appreciate the power of thermodynamics in dealing with chemical problems. For example, he fully understood the difference between a reaction that goes to completion and one that can only reach a state of equilibrium.” (General Chemistry: Fourth Edition, John W Hill, Ralph H Petrucci, Terry W McCreary, and Scott S Perry, Pearson/Prentice Hall, 2005, p590)

 

And one more, expressing the difference between reversible reactions, which reach a dynamic equilibrium, and reactions that go to completion.

 

”Up to now, we have limited our calculations concerning chemical reactions to reactions that go to completion and that can therefore be described by stoichiometry alone. However, not all reactions go to completion. Reversible reactions reach an equilibrium state in which forward and reverse reactions proceed at the same rate.” (General Chemistry: Fourth Edition, John W Hill, Ralph H Petrucci, Terry W McCreary, and Scott S Perry, Pearson/Prentice Hall, 2005, p575)

[Qfwfq]

Which logic? It was a question! I knew that would be, more or less, your answer TM. How about:

 

Does this mean that, if I have hydrogen and oxygen in a cylinder and I compress it, it will combine into water vapour?

 

Wouldn't it have been better to just simply say that you were leaving out a detail or two? I'm sure that, unless the compression is quite violent or there's a bit of platinum in there, they wouldn't flare up as easy as you do. Le Chatelier was also smart enough to know that, as well as mechanical work, one must also consider how endothermic/exothermic the the reaction is. Your first reply might have been slightly misleading for those not familiar with these details, quite possibly the case for anyone that had posed that question.

 

Bio, not that I claim to be a chemist, but I didn't do all that badly on the Chemistry course at Physics and neither on the course entitled Structure of Matter. Although it was many years ago, I still have a vague memory of these topics and I feared Niviene could draw too many conclusions.

[Biochemist]

Bio, not that I claim to be a chemist, but I didn't do all that badly on the Chemistry course at Physics and neither on the course entitled Structure of Matter. Although it was many years ago, I still have a vague memory of these topics and I feared Niviene could draw too many conclusions.

I was just picking on you for fun, Q. :) To me, it was obvious you were joking to make a point.

[TeleMad]

Which logic? It was a question!

 

Nope, it wasn't. I'd accept that it was a question if you had simply stated the words, but by your tacking on the silly laughing face at the end you turned it into what clearly appears to be an "in your face" counterexample.

 

If that's not what you meant, then it's your fault for not just simplly asking the question. You need to learn to communicate effectively.

 

Qfwfq: Wouldn't it have been better to just simply say that you were leaving out a detail or two?

 

Don't try to blame me again for your shortcomings. You don't know enough about chemistry so you got lost. That doesn't show a deficiency on my side, but on yours.

 

Further, if you'll compare the extensive quote I provided from the chemistry text to support what I said you'll see that I left nothing out. Did I literally leave something out? Sure. I didn't mention that the nucleus is made of protons and neutrons, which themselves are made of quarks, and a ton of other things. But things like that are irrelevant to the point I was making.

 

It's not my fault that you don't know the difference between (1) a reversible reaction that reaches a dynamic equilibrium, and (2) a reaction that goes to completeion and is irreversible. Your ignorance led you to make a blunder and now you are attempting to save face. We've seen you do it before, in our exchanges about computer programming.

 

 

Let's review.

 

1) What I said originally was correct.

2) You thought you gave a 'counterexample'

3) I supported what I had said, showing that you were wrong to try to counter me

4) I also showed that your counter was based on a misapplication of chemistry

[Qfwfq]

I was just picking on you for fun, Q. :mad: To me, it was obvious you were joking to make a point.

Just like I was pulling TM's leg! :xx:

 

I might even give a quick glance through his insults, if I get round to it. :eek: :eek: :eek: :friday: :eek:

 

:eek:

[Biochemist]

...I might even give a quick glance through his insults, if I get round to it...

I try to ignore them. Responding just generates more of them.

[TeleMad]

Responding just generates more of them.

 

 

Just like my pointing out the lowlife tactics of you and Qfwfq just seems to generate more of them. Ain't life funny!

[niviene]

Which logic? It was a question! I knew that would be, more or less, your answer TM. How about:

 

Does this mean that, if I have hydrogen and oxygen in a cylinder and I compress it, it will combine into water vapour?

 

Wouldn't it have been better to just simply say that you were leaving out a detail or two? I'm sure that, unless the compression is quite violent or there's a bit of platinum in there, they wouldn't flare up as easy as you do. Le Chatelier was also smart enough to know that, as well as mechanical work, one must also consider how endothermic/exothermic the the reaction is. Your first reply might have been slightly misleading for those not familiar with these details, quite possibly the case for anyone that had posed that question.

 

Bio, not that I claim to be a chemist, but I didn't do all that badly on the Chemistry course at Physics and neither on the course entitled Structure of Matter. Although it was many years ago, I still have a vague memory of these topics and I feared Niviene could draw too many conclusions.

 

Q, I thought it was funny. ;) I know nothing about this stuff, and I would have asked these questions seriously... just because I wouldn't know and would like to... ;)

[Qfwfq]

More or less, Niviene, given a mix of hydrogen and oxygen at a high enough temperature you would find the equilibrium to gradually vary with both pressure and temperature, according to Le Chatelier. At ordinary temperature two things get in the way:

 

1) The reaction from H + O to H2O is very strongly exothermic so the equilibrium is just about completely on the H2O side.

 

2) The kinetic barrier becomes important, so it isn't enough for H and O to simply be in contact. This is the same reason why many substances, which are combustible with oxygen, may remain in presence of it without burning until you put a match to them or somehow start the combustion.

[niviene]

More or less, Niviene, given a mix of hydrogen and oxygen at a high enough temperature you would find the equilibrium to gradually vary with both pressure and temperature, according to Le Chatelier. At ordinary temperature two things get in the way:

 

1) The reaction from H + O to H2O is very strongly exothermic so the equilibrium is just about completely on the H2O side.

 

2) The kinetic barrier becomes important, so it isn't enough for H and O to simply be in contact. This is the same reason why many substances, which are combustible with oxygen, may remain in presence of it without burning until you put a match to them or somehow start the combustion.

 

So, are you saying it needs some kind of catalyst - heat, apparently - that changes the kinetic barrier? The molecules have to be spread out or be moving at a certain speed for that to happen? This is awesome to think about... it's nice that someone will explain it in a way that I actually understand, rather than turning into some kind of equation or other jargon I don't understand, either... but I'm understanding a little of what you're saying. It takes a lot of heat to get the two together, apparently. Of all the sciences, I understand chemistry the least. Sorry. I do remember some common sense things - what different levels of heat do to things... and I understand that it might take more than letting water "expand" to separate the hydrogen and oxygen, because of how they are bound, but I don't understand why at a superhigh temp they wouldn't eventually separate again... or is that what you are saying? If not, what would it take to get them apart?

[Qfwfq]

Kinetic energy is what heat is, while a catalyst helps the molecules get over the kinetic barrier without needing as great a kinetic energy. In short, a catalyst causes he reaction to occur much faster in the same conditions, this can even cause the combustion to begin at lower temperatures.

 

Very high temperatures, according to Le Chatelier, will move the equilibrium to less H2O and more H and O. This has a quite different reason: a reaction that absorbs heat (endothermic) will be favoured by higher temperatures while one that gives out heat (exothermic) will be less favoured, this is what changes the balance between the two opposite reactions.

 

Le Chatelier also applies with pressure and volume and this is where the number of moles comes in. If the reaction, going one way, at a given temperature, decreases the total volume then it will be favoured by higher pressure. This also changes the balance between the two opposite reactions.

[TeleMad]

In short, a catalyst causes he reaction to occur much faster in the same conditions, this can even cause the combustion to begin at lower temperatures.

 

No, that's not what a catalyst is: there's more to it. A catalyst is a substance that speeds up a chemical reaction, by lowering the activation energy, and is not permanently changed in the process.

 

You know, maybe I wouldn't point out such a slight oversight, but hey, turn about is fair play.

[Qfwfq]

IOW