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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 $$\Delta G = \Delta ... 3 Your textbook's derivation is done under the assumption of constant T, which means T_{sys} = T_{surr} =T. However, this does not mean dG_{sys} is always zero. Let's start with the following:$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$Since heat flow always affects the surroundings ... 1 For a microscopic step at constant T and p$$\mathrm dG=0\tag{constant $T$ and $p$}$$implies: reversibility (equilibrium) \mathrm dS_\mathrm{univ} = 0 \mathrm dH_\mathrm{sys} = T\,\mathrm dS_\mathrm{sys} since \mathrm dG = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys} \tag{constant T and p} The derivation you suggest seems strange. ... 2 Your error occurs when you write that$$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$$(since ... 2 As an example consider an idealised/simplified model of butane, which has trans and gauche configurations as shown in the figure. We shall suppose these have different energies due to interactions between the protons on carbon 1 and 4 as bond C2 to C3 rotates. We take the energy of the trans state at 0^\text{o} to be zero, E_0 = 0. The energy of the ... 2 You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium. For molecules in an ideal gas the molar entropy is$$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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