Graphs of H and S

$\Delta S = \frac{q_{rev}}{T}$

K Stefanescu 2I
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am

Graphs of H and S

I understand why S is at a minimum at equilibrium, but why is H at a maximum when this occurs?

Chem_Mod
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Re: Graphs of H and S

At equilibrium, the energy (i.e. enthalpy) is at a minimum, whereas the entropy (i.e. delta S is) at a maximum. For example, if you observed a gas in a cylinder, over time it would occupy the entire cylinder volume and reach equilibrium; thus, at maximum volume it will have the greatest number of positions and entropy is at maximum. For enthalpy at a minimum, you can think of a ball on a hill. The ball will naturally reach equilibrium by losing energy to reach the bottom of the hill. Thus, enthalpy is at a minimum at equilibrium.

Mike Vinci 2B
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Re: Graphs of H and S

I'm also a little confused with exactly why this occurs, however I did look at the formula for delta G=0, and found that delta S= delta H/ T. If we maximize delta H, and considering that delta S is proportional to it, wouldn't that also maximize delta S? I could be analyzing the wrong equation, but if that is not the case, how would you also relate the question asked to the one I presented above?

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