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01:48
Hi, everyone. I just joined Chemistry a little under an hour ago, and I've been browsing around meta and the main site (and answering a question!). I've seen quite a few meta posts - the faqs first and foremost - and I was wondering if there's anything else I should know about Chemistry.SE. I'm familiar with LaTeX (and now mhchem), I've read the homework policy, and I've looked at a lot of the highly-voted posts to try to see what good posts should be like. Is there anything else I should know?
Heya HDE, welcome to Chem.SE!
@NicolauSakerNeto Thanks!
You've already done quite a bit more than the average new user would when joining, so I think you're okay
Can't remember anything off the top of my head in particular to share with you
One thing you may not have read is that we avoid using TeX in question titles, in case you edit them or ask your own
I should have noted that. I've seen that issue raised elsewhere.
Because it breaks the search functionality
01:53
Indeed. Not good.
It's fine, just something to keep in mind
@NicolauSakerNeto I have to go now. Thanks for the warm welcome!
No problem, glad to have new users putting some effort in the site
Have a good day/night/whatever
Yep, night. Same to you.
 
1 hour later…
03:21
@HDE226868 Welcome! You did quite a lot of research there already, probably more than most of us who are here longer. For the rest, just give it time and vote, vote, vote! Oh and also comment, comment a lot. And have a lot of fun!
 
12 hours later…
15:33
Dumb question guys :(. Reaction quotient (Q) is undefined when the concentration of the reactants is zero correct? I'm okay with this mathematically (given the way we express Q in terms of a ratio of the concentration of reactants and products) but I'm not satisfied with the conceptual outcome. Zero reactants and some product means Q is undefined...
@HDE226868 I wish more people were like you.
15:49
That's a strange situation which I don't think is really ever chemically relevant
I like to be prepared when some smarty freshman pipes up and asks.
I haven't found a satisfying solution yet. Google has let me down. :(
I suppose you could imagine that a case where there are products but zero reactants is as if you were "doing in wrong"; invert the sides in the equilibrium to transform products into reactants, and tadaa, the new reaction quotient is zero
It's just a relative matter
Still not as satisfying as I'd like :9
I mean, you make a very good point.
But... hmmm. I guess we could assume that no real reaction ever goes to 100% completion.
You could also try to bring limits into it, the limit of the quotient with reactant concentrations approaching zero from the right would be positive infinity
So there would always be some reactant present.
Ah, I like that one. Mathematically solid.
15:52
I suppose
But I don't like that assumption because to say that no reaction ever goes to 100% completion is silly.
Start with two H atoms in a vacuum and form H2. Boom, 100% complete.
It's not silly, it's true!
Is it really true?
to say that 100% completion is impossible?
Well, depends on your definition of completion
In all practicality its true... but that doesn't make it true all the time.
Ahhh down the rabbit hole we go ;)
15:54
And it's your fault!
:(
I don't like being unprepared when I teach.
I only ask the questions I would expect to be asked of me.
For sufficiently exergonic reactions, you could imagine that at points in time, there would be exactly zero molecules of reactants left in the equilibrium
then it wouldnt be an equilibrium anymore corrrect?
But, if you consider time-averaging the equilibrium, then after years and years, due to thermal fluctuations one molecule of reactant might pop up before being quickly consumed again
OH! I LIKE THAT!
@NicolauSakerNeto you deserve some upboats. Have a star!
I'm nervous. My kids take their first exam today. I want them to do well but I think a lot of them don't really believe what they hear about how brutal I am. Only those who took me last semester really knows.
15:58
So, say, the difference between an equilibrium constant of 10^80 and 10^1000 is that the former reaction will produce one molecule of reactant that lives for 1 nanosecond per day, while the latter will produce a molecule of reactant stable for 1 nanosecond per century
Or something like that
Yes. I like this explanation very much.
I'm playing really fast and loose with the idea, but I think that's about how it works
Glad to have helped a bit
I'm satisfied. Loose and fast is what you need to be sometimes when it comes to Chemistry (unfortunately).
I think thats why so many teachers simply just say "just take my word for it" and avoid the hole.
or even better, "because I said so."
16:01
Dissenter would attest to that
To be sure, the whole concept of equilibrium kind of breaks down when you get to the level of a few molecules
The theory behind it assumes matter quantities are continuous, when they are in fact discrete
But if you have 4 octillion atoms of something, then it's essentially a continuous variable
Single out a dozen molecules and it's not quite the same
Statistical thermodynamics hooray
Ah, I have recalled something quite similar from physics
If you take a light source and measure the photon count at some distance with a sensitive detector, then start moving away, you get the expected inverse square fall with distance of light intensity
That is, until the light source is so far away that the detector starts receiving only a handful of photons
At some point, the light intensity can't get any weaker, since the detector measures individual photons at the very least
So what happens if you keep moving farther away? The light source will start to "blink", that is, the single discrete photons will start to arrive at random intervals, in such a way that the inverse square law is obeyed statistically even then
If you keep moving farther away, the photon flashes become more and more spaced out in time, even though you will always detect single photons if you wait long enough
Presumably something very similar happens with chemical equilibria
16:21
Yeah, thermo is statistical.
But H+H -> H2 done a million times.... you could stat that :)
God I love Chemistry.

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