joonkimdds

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delta G = delta H - TdeltaS

if delta H is - and Tdelta S is - then it becomes --- => -++ so the higher T would make delta G into + which is nonspontaneous.
 

PrivateJoker

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so it will only be spontaneous at low temperatures. Your negative enthalpy value has to be larger than your tdeltaS value.
 
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joonkimdds

joonkimdds

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yeah so isn't it false to say that spontaneous rxns are thermodynamically favorable?
 

PrivateJoker

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The reaction is thermodynamically favorable if no energy is required and will proceed spontaneously.
 
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joonkimdds

joonkimdds

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well...destroyer makes it sound like it's always thermodynamically favorable.
 
Jun 14, 2009
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"Thermodynamically favorable" and spontaneity refers to the sign of deltaG, not deltaH, if that's what the confusion is about. So a spontaneous reaction is thermodynamically favorable and deltaG <0. Spontaneity implies the sign of deltaG
 
May 18, 2009
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An easy way to memorize it:

Delta H Delta S
+ - non spon
- - spon
+ + Only at high T (remember positive)
- - Only at low T ( remember negative)
 

doc3232

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"Thermodynamically favorable" and spontaneity refers to the sign of deltaG, not deltaH, if that's what the confusion is about. So a spontaneous reaction is thermodynamically favorable and deltaG <0. Spontaneity implies the sign of deltaG
I am really rusty so I might be wrong. But isn't it true that a spontaneous reaction can be thermodynamically unfavorable (while it is entropically favorable).
 

PrivateJoker

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well...destroyer makes it sound like it's always thermodynamically favorable.
A thermodynamically favorable reaction is one in which the energy state of reactants is higher than that of the products and will proceed spontaneously(without the need for added energy).
 
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I am really rusty so I might be wrong. But isn't it true that a spontaneous reaction can be thermodynamically unfavorable (while it is entropically favorable).
Hm, I always thought "thermodynamically" favorable or unfavorable referred to the (deltaH-TdeltaS) term as a whole. Anyone know for sure?
 
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joonkimdds

joonkimdds

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I thought thermodynamically refers to the high temperature which is T from TdeltaS.
 

PrivateJoker

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"Thermodynamically favorable" and spontaneity refers to the sign of deltaG, not deltaH, if that's what the confusion is about. So a spontaneous reaction is thermodynamically favorable and deltaG <0. Spontaneity implies the sign of deltaG
yes but

The sign of H is important and can determine if your delta g is positive or negative.
 
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PrivateJoker

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I thought thermodynamically refers to the high temperature which is T from TdeltaS.
If you want to get more complicated, no all thermodynamically favorable reactions may not proceed if the energy of activation is greater than the available energy. Energy would be needed to overcome the Ea or a catalyst to lower the Ea.

The other way around: Yes all spontaneous reactions are thermodynamically favorable. If it is spontaneous it will proceed so it has to be thermodynamically favorable (otherwise energy has to be put into the system or energy has to be created which goes against the first law of thermodynamics.)

To say it is spontaneous means that it does not require additional energy so again it has to be thermodynamically favorable because the reaction is going to proceed and it would not proceed without added energy it if was not spontaneous and therefore would not be thermodynamically favorable.

So Destroyer is not lying to you.
 
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Actually, it's all about ENTROPY ("entropy: don't fight it).

Spontaneity based on Gibbs free energy is simply a consequence of the second law.

Just remember, that the second law refers to the TOTAL change in entropy; the total change in entropy is the change in entropy of the system (usually the reaction) and entropy change for the surroundings....

The total change in entropy for a spontaneous process is greater than zero.

I'm not going to write out the deriviation, but....

Delta G(sys) = - TDelta S (total).....if the total change in entropy is positive, the change in gibbs free energy is negative....

It's all about entropy.
 
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^were you talking to me or someone else?
No, just my two cents....thought I would give a slightly different perspective. I think it makes more sense when you explain the connection between the second law and Gibb's free energy.

This topic has come up before, and it seems that there is a lot of confusion between entropy and spontaneity/contradiction of the second law.
 
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If you want to get more complicated, no all thermodynamically favorable reactions may not proceed if the energy of activation is greater than the available energy. Energy would be needed to overcome the Ea or a catalyst to lower the Ea.
I'm under the impression that it will still proceed under the kinetic constraints you have imposed; it's just, it'll take a lot longer.

Don't know if we're on the same page with "thermodynamically favorable" but I interpret it as spontaneity. With that in mind, the activation energy is irrelevant because spontaneity only makes the claim regarding equilibrium conditions from a given initial state.
 

PrivateJoker

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No, just my two cents....thought I would give a slightly different perspective. I think it makes more sense when you explain the connection between the second law and Gibb's free energy.

This topic has come up before, and it seems that there is a lot of confusion between entropy and spontaneity/contradiction of the second law.
Yeah. that is how they teach it in pchem. It never stuck so I had to find a different way to look at it.
 
Jun 24, 2009
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An easy way to memorize it:

Delta H Delta S
+ - non spon
- - spon
+ + Only at high T (remember positive)
- - Only at low T ( remember negative)
Important edit to this post with your permission:

Correction to your 2nd line:

Delta H Delta S
- - is not spon, (only at some low T) - as per your 4th line

Delta H Delta S
- + spontaneous at all temps
 

PrivateJoker

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I'm under the impression that it will still proceed under the kinetic constraints you have imposed; it's just, it'll take a lot longer.

Don't know if we're on the same page with "thermodynamically favorable" but I interpret it as spontaneity. With that in mind, the activation energy is irrelevant because spontaneity only makes the claim regarding equilibrium conditions from a given initial state.
It will not proceed at an appreciable rate. This is true in biological systems requiring a catalyst in the form of an enzyme. The catalyst or enzyme does not add energy to the system so your reactants can still be thermodynamically favorable but the reaction will not happen at a significant rate until the Ea is low enough for the reaction to proceed.
 
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It will not proceed at an appreciable rate. This is true in biological systems requiring a catalyst in the form of an enzyme. The catalyst or enzyme does not add energy to the system so your reactants can still be thermodynamically favorable but the reaction will not happen at a significant rate until the Ea is low enough for the reaction to proceed.
Just so we're on the same page....do you mean that "thermodynamically favorable" = spontaneous? That's how I interpret it...
 

PrivateJoker

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Just so we're on the same page....do you mean that "thermodynamically favorable" = spontaneous? That's how I interpret it...
I am saying all spontaneous reactions are thermodynamically favorable, but not all thermodynamically favorable reactions will proceed at an appreciable rate.
 
Jun 24, 2009
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It will not proceed at an appreciable rate. This is true in biological systems requiring a catalyst in the form of an enzyme. The catalyst or enzyme does not add energy to the system so your reactants can still be thermodynamically favorable but the reaction will not happen at a significant rate until the Ea is low enough for the reaction to proceed.
Hey bro,

Just wanted to get a little clarification on something. I was always taught that Eact has absolutely nothing to do with spontaneity of a reaction, ie., nothing to do with delta G. I'm getting a little confused by your posts cause it seems that your implying that it does. My apologies if I'm not following (both) or your posts on the topic but I just wanted to double check that which I was taught. Thanks.
 
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Correct, activation energy has nothing to do with spontaneity, just the relation of reactants to products. Activation energy determines the reaction rate of that spontaneous reaction.
 

PrivateJoker

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Hey bro,

Just wanted to get a little clarification on something. I was always taught that Eact has absolutely nothing to do with spontaneity of a reaction, ie., nothing to do with delta G. I'm getting a little confused by your posts cause it seems that your implying that it does. My apologies if I'm not following (both) or your posts on the topic but I just wanted to double check that which I was taught. Thanks.
I am saying all spontaneous reactions are thermodynamically favorable, but not all thermodynamically favorable reactions will proceed at an appreciable rate.
I was not relating spontaneity and Ea.

I was simply saying that a thermodynamically favorable reaction may not proceed at an appreciable rate.

Sorry. Didn't mean to be confusing.
 
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Jun 14, 2009
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delta G = delta H - TdeltaS

if delta H is - and Tdelta S is - then it becomes --- => -++ so the higher T would make delta G into + which is nonspontaneous.
So to get back to the original question by joonkim:

yes, thermodynamically stable means spontaneous. If you increase the T for a reaction to make deltaG positive and nonspontaneous, you're also making the reaction thermodynamically unstable. the two are synonymous.
 
Jun 24, 2009
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I was not relating spontaneity and Ea.

I was simply saying that a thermodynamically favorable reaction may not proceed at an appreciable rate.

Sorry. Didn't mean to be confusing.
Thanks for clarifying. The confusion is on my end. I've been so stressed with studying and sometimes when you read something and see terms/topics inter-related that you think shouldn't be your brain just shuts down and questions arise. Think it's time for me to close the books and embark on some mindless activities for the rest of the night.
 
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joonkimdds

joonkimdds

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1 similar question!
How does spontaneous reaction increase entropy of the universe?
that means -delta G always have positive delta S and that's not always true.
 

bennijai

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Spontaneous reactions break bonds thus creating more disorder = increase in entropy.

In order to get -delta G with with a + delta S you will need it to be in high temperatures, otherwise it will be non spontaneous. A good example is water to gas/steam.
 
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1 similar question!
How does spontaneous reaction increase entropy of the universe?
that means -delta G always have positive delta S and that's not always true.
The keyphrase is "of the universe" A single subsystem, like photosynthesis could have a negative deltaS (more order), but this correlates to an equal or greater increase of entropy somewhere else (the sun). A spontaneous reaction does not always increase the entropy of the reaction system, but if you could expand your view to the universe, the TOTAL entropy increases.

Consequently -deltaG for a single reaction in a subsystem doesnt always have to have a positive deltaS. That's why the equation of spontaneity isnt only dependent on entropy.