## Gibbs at Constant Pressure

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

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

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

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Shrita Pendekanti 4B
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Joined: Fri Sep 25, 2015 3:00 am

### Gibbs at Constant Pressure

Is Gibbs Free Energy at a constant pressure always 0? I figured since ΔH= q at a constant pressure and ΔS = q/t that substituting these terms into the Gibbs free equation ΔG = ΔH - TΔS would become ΔG = q-T(q/t) = q - q = 0.

I am wondering if someone can explain if this is correct, why this is the case (a conceptual explanation), and how this compares to an isothermal situation.

Thanks so much!

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### Re: Gibbs at Constant Pressure

This is not correct, because dS = dqrev/T. Since T is not constant during a constant pressure process, this does not integrate to deltaS = q/T (it DOES for isothermal process). Instead, it integrates to deltaS = Cpln(Tf/Ti).

The Gibbs energy at constant pressure has a concrete meaning: it is the maximum amount of non-expansion work that can be done.

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