Thermodynamic Spontaneity

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David Vanden Bout

David Vanden Bout

11 жыл бұрын

Пікірлер: 15
@AmritBedi47
@AmritBedi47 11 жыл бұрын
i love this channel!!! makin studying chem so easy!!! keep makin videos guys!!!
@alecdodge2814
@alecdodge2814 9 жыл бұрын
Shout-out to all the chem 2 students watching this for Dr M class in the morning
@108.LordShivaYT
@108.LordShivaYT 3 жыл бұрын
He Pronounces Spontaneity Perfectly😍😍😍
@VagabondMorrison
@VagabondMorrison 9 жыл бұрын
First of all, GREAT channel! I'm learning so much and you articulate it so well! Okay, so I have question which was the question that first lead me to your channel. I saw an article about a car that runs on water and upon further investigation I learned the explanation for this is it has "water spark plugs" which separate the hydrogen and oxygen. Right away I thought of thermodynamics which admittedly I still only have a very cursory knowledge of (hopefully I'll have a better grasp when I finish your course), but as I understand it thermodynamics explains why "free energy" is more or less impossible. The claim for this car is that the power it takes to separate the hydrogen from oxygen is less than the power it creates... but is that even possible?
@YawnGod
@YawnGod 11 жыл бұрын
Ha! The Party Boy song. Someone making this video had a sense of humor.
@InnerFireme
@InnerFireme 7 жыл бұрын
Brilliant. Thanks so much for helping us understanding the practical applications of Chemistry
@BrenoSimas
@BrenoSimas 9 жыл бұрын
sanapathi prasad i reckon it's been a while since you made your question, but if it hasn't been answered so far, I hope it will be from now on: Indeed you are right about the fact that the system's entropy decreases, which is actually kind of intuitive since the number of moles is decreasing (even though that's no guarantee, if water is also a gas. See what I wrote down there about it). But when you say that entropy is decreasing, beware that you are talking about the system's entropy only, and that the second principle (a lot of people call it "the second law", but i feel kind of uncomfortable with that) talks about the entropy of the universe. This way, you'd have to evaluate the variation of entropy of the system (which we did, even though not in the most formal way) and the variation of entropy of the surroundings. By doing so, you'd see that, since the reaction we're talking about is a combustion, it's surely exothermic. Exothermic changes will always increase the entropy of the surroundings, since heat will be delivered to it (if the process occurs at constant temperature, we can use a something called "Clausius' Postulate, which, in a less general form than the usual one, can define variation of entropy as ΔS=-ΔH/T). This way, although the system's entropy is decreasing, the entropy of the surroundings is increasing, and if the increase in the entropy of the surroundings overcomes the decrease of entropy in the system, we'll have a spontaneous process. I apologize for the long text, but I hope I've made myself clear... The safest way to see the decrease in entropy is by doing an analysis as follows: I took this data from the book "Chemical Principles: The Quest for Insight" (authors Atkins and Jones): the molar entropys are: for water (presumably in it's gas state, which makes me tell you to always put the physical states of the reactants when talking about physical chemistry), 188,83 J/molK, for hydrogen (gas), 130,68 and for oxygen (gas), 205,14. thinking of the reaction as H2 + 1/2O2----->H2O, we'd have: ΔS= S(products)-S(reactants)= 188,83 - (130,68 + 0,5 x 205,14) = -44,42 J/molK (I left it per mol because it's a combustion reaction. Please have in mind that the " per mol" here means "per mol of hydrogen", which is the combustible). The result we found is coherent with what you expected, but this is the way to be sure about it... If you need anything else or if something wasn't as clear as you'd like, I'll be glad to answer you again (:
@akashshete1695
@akashshete1695 6 жыл бұрын
Thanks a lot for this great explanation sir.
@Phoenixggs123
@Phoenixggs123 Жыл бұрын
Chemistry never gets old, today is my exam studying thermo wish me luck 😂
@mannyefule3500
@mannyefule3500 9 жыл бұрын
Thanks! Your videos are actually pretty great. I'm a chemistry grad student browsing through better ways of explaining things like gibbs free energy and I think these are the best explanations I've seen so far. Do you happen to have a web page or videos on crystal growth and interfacial free energy?
@sleepnessuk1476
@sleepnessuk1476 8 жыл бұрын
It is useful for me
@prof.fabioflamigni1972
@prof.fabioflamigni1972 4 жыл бұрын
Free Energy is related to both Hentalpy and Enthropy changes.. non only Entrophy.. :)
@19sumanth
@19sumanth 11 жыл бұрын
Hi, 2H2 + O2 ---> 2H2O here left side has 3 moles, and right side 2 moles. here entropy is decreasing and hence is non spontaneous?? can u please clarify. i m confused...plz..
@fuckthisshiat
@fuckthisshiat 11 жыл бұрын
nice video. although i find it misleading when you say that "non-spontaneous processes can never happen" because if you do some "work" they will.
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