Analysis of Gibbs Free Energy Equation

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

LBacker_2E
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Analysis of Gibbs Free Energy Equation

On the outline when it says "explain how how ∆S is related to ∆H for a change at constant temperature and pressure and explain the relationship" I am a little confused. Like I know the general equation to relate the two, but is there a trend that when you increase ∆S, ∆H will either increase or decrease? Or does it depend on what Gibbs free energy is.

705121606
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Re: Analysis of Gibbs Free Energy Equation

I believe it is all dependent on gibbs free energy because we need delta s and delta h to know if the reaction will be spontaneous or not. I would just make sure you know how to manipulate the variables and come to the conclusion of what needs to happen for the reaction to be spontaneous or non spontaneous. From a hw question we had (4.37), we can come to the conclusion about spontaneity can be found from S or G, if g is negative the reaction is spontaneous. If delta s total is positive it would also be spontaneous.

Kristina Rizo 2K
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Re: Analysis of Gibbs Free Energy Equation

Would this also be related to the equation delta S= q/T? And q = delta H?

Veronica_Lubera_2A
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Re: Analysis of Gibbs Free Energy Equation

Delta s=q/T is used when there is a constant temperature. q=DeltaH is used when there is a constant pressure. Q is heat which is separate from Gibbs free energy.

Sydney Myers 4I
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Re: Analysis of Gibbs Free Energy Equation

The relationship between delta H and delta S depends on which delta S you're referring to. If it's the entropy of the system, then as delta H increases, then delta S increases with it. If it's the entropy of the surroundings, then delta S would decrease with an increase in delta H. If its the total entropy, then delta S increases with a decrease in delta S.