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Author Topic: Relationship between gibbs, enthalpy, and entropy  (Read 3520 times)

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Foobarz

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Relationship between gibbs, enthalpy, and entropy
« on: February 06, 2012, 03:44:35 PM »

So Gibbs free energy is  :delta:G= :delta:H-T:delta:S

And Gibbs free energy is "the capacity of a system to do non-mechanical work."

The more negative the  :delta:G, the more energy the system gives off

If so, then why is the sign before T:delta:S a minus sign? Is increasing entropy (positive :delta:S) equate to the release of more free energy (more negative  :delta:G ?) due to the minus sign? But how does that make sense?
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Rutherford

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Re: Relationship between gibbs, enthalpy, and entropy
« Reply #1 on: February 07, 2012, 12:43:23 AM »

State of disorder is much more likely than the state of order, use for example the experiment with the numbered balls in a box, when you shake the box the probability that the balls after that are in the same sequence like at the beginning is very low. It is the same with chemical equations, the reaction will happen easier when a bigger disorder is going to be made (dS>0). If you know this, and that the lower the Gibbs energy is (dG<0) the equation is more likely going to happen then. I think that's why the ''-'' is there. Bigger disorder, entrophy more positive, Gibbs energy lower and the reaction will happen easier.
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Foobarz

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Re: Relationship between gibbs, enthalpy, and entropy
« Reply #2 on: February 10, 2012, 06:13:10 AM »

AHHH I GET IT NOW

So it's why reactions are much less likely to happen in the solid state than in aqueous or gaseous states because of the lower entropy in the solid state!

THANK YOU
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juanrga

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Re: Relationship between gibbs, enthalpy, and entropy
« Reply #3 on: February 10, 2012, 08:52:11 AM »

So Gibbs free energy is  :delta:G= :delta:H-T:delta:S

And Gibbs free energy is "the capacity of a system to do non-mechanical work."

The more negative the  :delta:G, the more energy the system gives off

If so, then why is the sign before T:delta:S a minus sign? Is increasing entropy (positive :delta:S) equate to the release of more free energy (more negative  :delta:G ?) due to the minus sign? But how does that make sense?

U = U(S,V,N)

dU = TdS - pdV + mu dN

but for laboratory chemists is more easy to measure (T,p,N). Therefore we apply a change of variables.

Using d(xy) = xdy + ydx, the above expression can be rewritten like

dU = d(TS) - SdT - d(pV) + Vdp + mu·N

or

dU + d(pV) - d(TS)  = - SdT + Vdp + mu·N

The left hand side is a new function and we denote it by G = U + pV - TS

dG = - SdT + Vdp + mu·N

This explains why there is a minus sign before TdS.

Notice that G=G(T,p,N).
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