Maximum Useful Work from a Process

Question

Negative of the $\Delta G$ for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way: $\Delta H$ is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently $-\Delta H$ energy is obtained from the system after doing expansion work. Since $\Delta S$ is the entropy created in the process, at the very least $-\Delta S$ entropy must be created in the surroundings - that is, at least $-T\Delta S$ energy must be lost as heat. This comes from the $-\Delta H$, and thus leaves $-\Delta H + T\Delta S$ to do useful work. No more work can be done than this. $-\Delta H + T\Delta S$ is $-\Delta G$, so $-\Delta G$ is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why $-\Delta G$ is the maximum possible useful work.

Now if $|\Delta H|$ is more than $|T\Delta S|$, with both being negative, then one can think of the $W_{max}$ or $-\Delta G$ as a part of the $|\Delta H|$; since $|\Delta G|< |\Delta H|$. Some part of $|\Delta H|$ goes as $|T\Delta S|$ to increase the surroundings' entropy, and the other part in doing useful work. But if both are positive, with $|T\Delta S|$ being more than $|\Delta H|$, again $\Delta G$ is negative, allowing useful work to be extracted. But now it seems as though $|T\Delta S|$ heat will be extracted from the surroundings, $|\Delta H|$ used up in the process, while the rest can be converted to work - in other words, useful work is a part of $|T\Delta S|$ - with the other part used by the process as $|\Delta H|$. Is this correct? Is $|T\Delta S|$ extracted from the surroundings actually?

Answer 1

"Useful work" does not include expansion work. For a proof of why $w_{\text{by,add}} \leq \Delta G$, you may want to first refer to my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work? then come back here.

In that proof, I used equalities by assuming that the process is reversible. If we drop this assumption, then by the Second Law,

$$\begin{align} \mathrm{d}S &\geq \frac{\mathrm{d}q}{T} \\ \mathrm{d}q &\leq T\mathrm{d}S \\ dU &\leq T\mathrm{d}S - p\mathrm{d}V + \mathrm{d}w_{\text{add}} \\ dG &\leq dw_{\text{add}} \\ \Delta G &\leq w_{\text{add}} \end{align} $$

Equality holds if the process is reversible, or more generally, if both the initial and final states are equilibrium states. Now this seems like an obvious contradiction of the equation in the very first paragraph. The key lies in the definition of the work - whether it is done on the system, or by the system. In my proof, I have been referring to the work done on the system; however, here we are interested in the work done by the system, since that is the work that can be "extracted". The work done by the system must be equal to the negative of the work done on the system: $w_{\text{by,add}} = -w_{\text{add}}$. This leads to the final result $w_{\text{by,add}} \leq \Delta G$.

Levine's Physical Chemistry 6th ed. writes:

In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but $w_{\text{by,add}}$ is the useful work output. The quantity $-\Delta G$ equals the maximum possible nonexpansion work output $w_{\text{by,add}}$ done by a system in a constant-$T$-and-$p$ process. Hence the term "free energy". (Of course, for a system with P-V work only, $\mathrm{d}w_{\text{by,add}} = 0$ and $\mathrm{d}G = 0$ for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses.

If you want a direct answer as to why your explanation isn't right, it's because you can't equate $\Delta H$ with the work done. Work done is given by $w = -\int\! p\,\mathrm{d}V$. You can only equate $\Delta H$ with the heat transferred at constant $p$. Apart from the First Law ($\Delta U = q + w$), heat and work are generally unrelated.