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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Agreetodisagree on January 10, 2022, 04:48:53 PM

Title: Why Gibbs free energy isn't always 0
Post by: Agreetodisagree on January 10, 2022, 04:48:53 PM

Hello, I would like to ask if dG=dH-TS (Gibbs free energy = enthalpy - thermodynamic temperature * entropy), dH=q(enthalpy heat) and dS=q/T (entropy-heat/ thermodynamic temperature) than why it isn't always 0. I would assume (probably it's wrong but I can't see why) that dG=q-q.

Title: Re: Why Gibbs free energy isn't always 0
Post by: Borek on January 10, 2022, 06:32:10 PM

Quote from: Agreetodisagree on January 10, 2022, 04:48:53 PM

dS=q/T

That would be ideal, reversible process. In reality dS > q/dT.

Title: Re: Why Gibbs free energy isn't always 0
Post by: joyalin on February 06, 2022, 02:25:24 AM

dS=qrev/T. For irreversible processes, dS>qirrev/T. Thermodynamically, reversible processes are those that remain infinitely close to equilibrium. Delta G=0 indicates a process at equilibrium.