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User Buck Thorn writes here that

In chemistry we are most often interested in the effect of changes in $T$ or $P$ on $\Delta G$ for some process such as a phase change or chemical reaction which is otherwise carried out at constant $T$ and $P$.

Does this mean that the derivations using $\Delta G$ for a reaction need not be consistent with entropy interpretations?

I have been trying to prove that for a homogeneous phase at constant composition and all the processes taking place reversibly:

$V\,\mathrm{d}P - S\,\mathrm{d}T <0$ is a criterion for "favourable change" using entropy arguments only but without any success. If Gibbs energy and entropy arguments are compatible with each other can anyone showing the above only using entropy statements?

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  • $\begingroup$ Can you explain what is not consistent? $\endgroup$
    – Greg
    May 22 at 11:04
  • $\begingroup$ I have been trying to prove that for a homogeneous phase at constant composition and all the processes taking place reversibly: $VdP−SdT<0$ is a criterion for "favourable change" using entropy arguments only but without any success. This makes me feel that Gibbs energy interpretation is not consistent with entropy. Can you please prove that with entropy arguments only? $\endgroup$
    – hsawhsiv
    May 22 at 11:08
  • $\begingroup$ Indeed is hard to see the question. Or it negates the very need for a free energy function. $\endgroup$
    – Alchimista
    May 22 at 11:14
  • $\begingroup$ @NisargBhavsar By entropy arguments, I mean entropy of the whole universe. Can you arrive at same result by showing that Entropy of the universe increases whenever VdP−SdT< 0. If no, why one cannot do so? Any condition for "favorable change" must trace its root back to entropy of universe trying to increase. Right? Side Note: Sorry for editing the question after being answered since that was a very fundamental error. $\endgroup$
    – hsawhsiv
    May 22 at 11:20
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    $\begingroup$ $V\,\mathrm{d}P - S\,\mathrm{d}T <0$ for a spontaneous process, but reversible processes are not spontaneous, they're reversible and then $V\,\mathrm{d}P - S\,\mathrm{d}T =0$. If you are interested in the effect of p or T on a phase transition then you write $V_1\,\mathrm{d}P - S_1\,\mathrm{d}T =V_2\,\mathrm{d}P - S_2\,\mathrm{d}T$ which leads to the Clausius equation. $\endgroup$
    – Buck Thorn
    May 22 at 13:48
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Entropy alone cannot decide if a process is spontaneous.

Haber's process is an good example where entropy decrease but still the process on a whole is spontaneous.

The whole idea of defining free energy was to take into account both enthalpy and entropy arguments. You cannot derive the conclusion you want using only entropy interpretations. Both enthalpy and entropy need to be considered to predict the spontaneity of a process.


Coming to the result you want to prove:

We know that a process is spontaneous when $\mathrm dG<0$.

And Gibbs free energy is defined as:

$$G = H - TS$$

Now differentiating this we can get:

$$ \begin{align} \mathrm dG &= \mathrm dH - \mathrm d(TS) \\ \mathrm dG &= \mathrm d(U+PV) - \mathrm d(TS) \\ \mathrm dG &= \mathrm dU+ P\,\mathrm dV + V\,\mathrm dP - T\,\mathrm dS - S\,\mathrm dT\\ \mathrm dG &= (\mathrm dU+ P\,\mathrm dV) + V\mathrm dP - (T\,\mathrm dS) - S\,\mathrm dT\\ \mathrm dG &= \mathrm dq + V\,\mathrm dP - \mathrm dq - S\,\mathrm dT\\ \mathrm dG &= V\,\mathrm dP - S\,\mathrm dT\\ \end{align} $$

Therefore for a process to be spontaneous:

$$ \begin{align} \mathrm dG&<0\\ V\,\mathrm dP - S\,\mathrm dT &<0\\ \end{align} $$

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  • $\begingroup$ "The whole idea of defining free energy was to take into account both enthalpy and entropy arguments. You cannot derive the conclusion you want using only entropy interpretations." This isn't quite correct. The idea behind the Gibbs free energy was to identify a state function that, under the common conditions of constant T & p, could provide a criterion for spontaneity and thus for equilibrium. This is entirely determined by the sign/value of dS_universe (0 for equilibrium; >0 for spontaneity). However, it's not typically possible to measure this directly, [continued] $\endgroup$
    – theorist
    May 23 at 7:47
  • $\begingroup$ so Gibbs designed a function that, at const. T and P, acts as a surrogate for this. Yes, to construct such a function he needed to include the enthlapy, but the reason is not fundamentally because enthlapy plays some role in spontaneity. It's all about the entropy, and the enthalpy happens to play a role here only to the extent it contributes to the entropy change of the universe. See my answer here: chemistry.stackexchange.com/questions/124412/… $\endgroup$
    – theorist
    May 23 at 7:50
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    $\begingroup$ @theorist This entropy mentioned in the answer is the entropy of the system and not the universe. Surely the entropy of the universe is "the" deciding factor. $\endgroup$ May 23 at 7:51
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    $\begingroup$ It's still incorrect to say that balancing entropy and enthalpy is the fundamental idea behind Gibbs free energy. Rather, those are a consequence of the fundamental idea behind it, not the fundamental idea itself, which was to determine a state function surrogate for the entropy change of the universe. $\endgroup$
    – theorist
    May 23 at 8:00

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