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Posted

If the energy decreases, this means that energy must have been given off and so it is exothermic.

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Posted

Albert, there is something unsettling about your posts that gives me the impression that you are trolling.

You have somehow mastered asking the wrong questions and the experts are being kind and falling for it.

There is no such thing as "the energy level decreases in product from the reactants".

A reaction is exothermic if the ambient medium of the reaction gains heat.

A reaction is endothermic if the ambient medium of the reaction loses heat.

 

Now you might have believed that one should wonder, if the energy that the medium gained is not coming from the reactants as a source, then from where! We could boil a solution (adding heat) to force the reaction. This means that the product can gain heat from the flame of a Bunsen Burner. Now keeping this in mind, the product can have an enthalpy showing an increase in the energy of the product while the reactants are not even the source of the energy that was absorbed, and the reaction is endothermic. To the contrary, we might have a spontaneous combustion triggered by a nano-catalyst and it would be obviously exothermic while the starting compound loses energy to the medium. However, the energy of the product compounds could have a sum of energies that exceed the energy-difference from the combustible material as it gains back heat to trigger the spontaneous reaction.

A reaction does not necessarily have any products; surprised, right?

Dissolve some Urea in water and it will cool down the temperature of the solution, thus we have an endothermic reaction, but what are the products? Urea is still urea and water is still water but the dissolution reaction is endothermic because the hydrogen bond formation demanded some energy to form.

Now do not rush to conclusions here, because dissolving NaOH or H2SO4 in water would raise the temperature of the solution (exothermic), and here the hydrogen bond formation expels of the excess of energy.

If you wish to consider the solvent and the solute as reactants and the hydrated solution as the product, then please realise that they both define the reactants and the product and the environment as well, and here we can neither say that they gain or lose energy at all, because all is contained within the boundary.

Now we can consider some of the water as a reactant and some of the water as the environment and thus expand the boundary, but there is no way to separate the energy within the whole boundary.

So please remember that neither the reactants nor the products are exothermic or endothermic, it is the reaction that is and across the boundary that is.

Posted
Albert' date=' there is something unsettling about your posts that gives me the impression that you are trolling.

You have somehow mastered asking the wrong questions and the experts are being kind and falling for it.[/quote']

EL, I don't think Albert is trolling. I think that Albert is quite eager to learn all kinds of things and that our answers raise a thousand questions again. There is no problem with that. Of course, the questions could have somewhat more structure...

 

People having tons of basic questions should not be frightened away from a forum like this... instead, let's try and help them, pointing in the right direction (both on the subject and on how to deal with a forum like this).

 

I have seen some real trolls on Usenet (if you know sci.chem, then you'll certainly remember Dr. Mai Jaharai, KellyClarksonTV and a few other craps) and there is quite some difference between their posts and Albert's.

 

I would suggest to Albert to go out and buy a good book. That gives more structure to the subject.

The questions in this forum are scattered over many fields of chemistry. You may be able to grasp the individual (and sometimes seemingly unrelated) ideas, but getting a sound understanding of basic concepts will be hard from the replies from members of this forum. That would require even longer replies than some I have given already. A good book on the basics of chemistry fits that purpose much better.

Posted

he`s not trolling, he`s always asked many questions like these since I`ve known him, he`s perfectly safe :)

Posted
he`s not trolling, he`s always asked many questions like these since I`ve known him, he`s perfectly safe :)

 

That is OK, because I gave him the benefit of the doubt and tried to answer his "question".

For some reason he was confusing between the concept of a reaction and the reactants (substrates). There are enzyme catalyzed reactions that even a bond-energy tracing would be very difficult for a beginner. That is why I was in doubt, because if you do know that you are a beginner, you usually ask more basic questions that you feel that you know how to ask them. His curiosity is very fine, yet having no faith in our patience could be his real problem, that is why he is making short cuts and jumping to conclusions upon which he is asking the wrong questions.

To Albert: Slow down, and digest each point you get an answer for before you proceed to another point. However, as {woelen} told you, there is no replacement for a small library to cover each side of your interests professionally, then if you could not understand a passage in such books come back and ask and we will gladly help you, but show some seriousness please.

Posted

I second EL's suggestion there. Albert is not going to learn very much chemistry by shooting random questions at a forum (the answers to which, require an understanding of the fundamentals) - and not hardly even waiting for a response before the next question comes up. I repeat my recommendation of Atkins' Physical Chemistry.

 

EL : I shall respond to your previous post to me later (if you so wish it). Essentially, we are both not on the same page...and I don't want to get into a lengthy argument based on our independent interpretations of an ill-defined question. What I said in my post is not what you have understood of it, so I'll be happy to clear it up, if required.

Posted

Perhaps buying a very good book might be too demanding on a student's budget or savings.

In my country, imported hard-paperbacks in colourfully illustrated chemistry books could be as expensive as 250 dollars per issue.

Here is an excellent downloadable alternative (free of charge).

<http://www.ntu.edu.au/education/wardonli.htm >

I don't want to get into a lengthy argument based on our independent interpretations of an ill-defined question.

Now on that, we can easily agree, can't we?

 

What I said in my post is not what you have understood of it, so I'll be happy to clear it up, if required.

No need for that, because I did understand it quite clearly, and I appropriately responded, that the bond energy that forms between two reactants is "irrelevant" to the activity-coefficient of the reactants. Now you might wish to say that my correct response is not that to what you said. However, at an advanced level, there is an indirect relevancy between the newly formed bond energy and the _enthalpy-change-sum_ of the reactants, which was not asked, and you hinted at, but is not related to "reactivity".

In short, let it go. :D

Kindest regards.

Posted

Well, I am just trying to fit my chemistry knowledge with what you guys have told me, consistently.

 

I know that all reactions are either endothermic or exothermic.

 

The great things about this thread are the thermodynamics and kinetics of reactions I learned. However, I am still "shaky" on this notion, applying them to different other reactions.

 

What I am thinking now is to connect these with Chantelier's principle and exothermic/endothermic reactions, and see what theoretical conclusion I can come up whenever I do an experiment at school.

 

Albert

Posted

btw, is thermodynamics related to thermochemistry??

 

I am reading about this on a book called College Chemistry By Jerome L. Rosenberg and Lawrence M. Epstein.

 

Albert

Posted
btw' date=' is thermodynamics related to thermochemistry??

I am reading about this on a book called College Chemistry By Jerome L. Rosenberg and Lawrence M. Epstein.

Albert[/quote']

 

Very good; now keep reading. :D

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