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It is not possible to confirm your derivation. If you describe the change in free energy of a system with the expression $$\Delta G=\Delta H-T\Delta S$$ then you are implicitly assuming that the initial and final temperatures of the system are equal to $T$. The more general expression for $\Delta G$, from the definition $G=H-TS$, is $$\Delta G = \Delta ...


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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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Your error occurs when you write that $$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$ since this assumes that $$\delta Q=-TdS$$ which implies reversibility and therefore that $dG=0$ at constant T and p. The more general statement is that $$dS_{tot}=\frac{\delta Q}{T_{r}}+dS$$ which means that $$dS_{tot}=-\frac{dH}{T_{r}}+dS$$ (since ...


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As an example consider an idealised/simplified model of butane, which has trans and gauche configurations as shown in the figure. We shall suppose these have different energies due to interactions between the protons on carbon $1$ and $4$ as bond C2 to C3 rotates. We take the energy of the trans state at $0^\text{o}$ to be zero, $E_0 = 0$. The energy of the ...


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You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium. For molecules in an ideal gas the molar entropy is $$S=R\ln z +RT\left(\frac{\partial z}{\partial T} \right) _V - R \ln N_A + R$$ Here $z$ is the molecular partition function $$z=\sum_i e^{\epsilon_i/kT}$$ where the summation extends over all states ...


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The gas pressure equals to the external pressure during isothermal irreversible expansion. To maintain P(gas)=P(ext), heat is supplied reversibly. By this reason delta S(surr) must be calculated above. Am I clear?


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