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Your textbook's derivation is done under the assumption of constant $T$, which means $T_{sys} = T_{surr} =T$. However, this does not mean $dG_{sys}$ is always zero. Let's start with the following: $$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$ Since heat flow always affects the surroundings ...


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For a microscopic step at constant $T$ and $p$ $$\mathrm dG=0\tag{constant $T$ and $p$}$$ implies: reversibility (equilibrium) $\mathrm dS_\mathrm{univ} = 0$ $\mathrm dH_\mathrm{sys} = T\,\mathrm dS_\mathrm{sys}$ since $\mathrm dG = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys} \tag{constant $T$ and $p$}$ The derivation you suggest seems strange. ...


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