Does the free energy of the universe decrease for every process?

Question

The second law of thermodynamics states that $dS_{universe}>0$ for all processes.

In addition to the second law, gibbs free energy is used to consider the feasibility of a reaction. $dG=dH_{syst}-T*dS_{syst}$. If $dG_{syst}<0$, this means that the reaction is spontaneous, but a reaction with $dG_{syst}$ can also occur if energy is input into the system (for example electrolysis).

Therefore, $dS_{univ}>0$ does not imply $dG_{syst}<0$.

However, is there such a quantity as $dG_{univ}$, and if so is the relationship $dG_{univ}<=>dS_{univ}$ true?

Answer 1

Didn't you ask the last question about this, too?

The issue here is that when $\Delta G > 0$, the reaction is indeed non-spontaneous as written. If you wish to add energy to the system, for example, via electrolysis, you should include the additional changes in the computation of the free energy change. You will find that after you sum up everything $\Delta G < 0$.

Also, note that $\Delta S_{\mathrm{universe}} < 0$ does not mean that the process will not happen; it only means that it's (fairly) unlikely to happen.

Answer 2

The relationship is: $\Delta G = - T \Delta S_{univ} $

Also, if you want to consider electrolysis in this context, your system needs to include the power input in some form. Your entropy increase from obtaining that power is large enough to offset the entropy penalty of an unfavorable process making it spontaneous.

Answer 3

I think the answer is no, there is no such thing as the Gibbs energy of the universe, and thus no way to say that it only decreases. The reason is that thermodynamics does not take account of how things got to be the way they are. If a raised weight happens to be conveniently waiting there attached to a pulley, which is attached to a mechanism that allows it to drive the pistons of a Van't Hoff equilibrium box, in which a chemical reaction with $\Delta G > 0$ takes place, thermodynamics does not ask how all that stuff got there. Unless I am mistaken, the potential energy, $mgh$, of the weight does not figure into its Gibbs energy because it is not internal energy (though I would be excited to be corrected on this if I am mistaken). Thus it is entirely possible for the weight to fall very slowly and reversibly drive the increase of free energy of the chemical system without any compensating decrease of free energy elsewhere. In the reversible limit, the falling weight does not generate any compensating entropy, either, which is frequently claimed to be necessary to avoid violating the second law. The position of the weight has no entropy associated with it.

So does this net $\Delta G > 0$ process mean that the total entropy decreases, violating the second law? No, it doesn't. The formula $\Delta G = -T\Delta S_{total}$ only applies if there is no work being done on the system other than expansion or compression work. That formula is derived by assuming that the change in enthalpy is equal to the heat absorbed from the surroundings, or $\Delta H = Q$, which is the usual situation for a reaction in a lab. In that case $$\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$.

But if work is also done on the system, then the enthalpy change is not just the heat absorbed. You also have to add the work done.

$$ \Delta H = \Delta(U+PV) = Q + W + P\Delta V = Q + W - W_{PV} = Q + W_{non-PV}$$.

Now when you calculate the total entropy change, you get

$$ \Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - (\Delta H-W_{non-PV})/T = (W_{non-PV}-\Delta G)/T \geq 0$$.

The last inequality comes from $W_{non-PV} \geq \Delta G$, which follows from Clausius' inequality, $\Delta S \geq Q/T$. So the second law checks out, even though it seems like some free energy was created out of nowhere (but actually was converted from the potential energy of the raised weight).