# How does work from other sources affect Gibbs free energy of the system?

In most physical chemistry textbooks, Gibbs free energy is given like this:

$$\Delta_\mathrm rG = \Delta_\mathrm rG^\circ + RT\ln Q\tag{1}$$

and Gibbs free energy is part of this three-way relationship:

$$\Delta_\mathrm rG^\circ = -RT\ln K = -nFE^\circ\tag{2}$$

which gives rise to the Nernst equation.

However, in many cases, $$\Delta_\mathrm rG > 0,$$ so it is not thermodynamically favourable. In these cases, there would be an external source of work.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $$-VIt$$ term or something else which would cause $$\Delta_\mathrm rG < 0?$$

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $$\Delta_\mathrm rG < 0?$$

I will treat the first and second question as if they were part one and part two of a single question. I think the OP wants to know how non-PV work plays a role when looking at the Gibbs energy of a reaction, and its interpretation.

To start off, the Gibbs energy for a given reaction is a state function, so it does not depend on how much work or how much heat are exchanged between system and surrounding. The total change in entropy, however, is affected by what is going on in the system and in the surrounding, so work vs heat is relevant for the second law of thermodynamics and the direction of the reaction.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra $$-VIt$$ term or something else which would cause $$\Delta_\mathrm rG < 0?$$

The easiest way to deal with the power source is to place the system/surrounding boundary such that the power source is in the surrounding. This shows it does not affect the Gibbs energy of the reaction at all. The Gibbs energy is positive, but the reaction can go forward because it is offset by the work done on the system. As a result, the combined entropy increases, and we are not breaking the second law.

In general, the criterion is not that the Gibbs energy of reaction has to be negative, but that

$$\Delta_r G < w$$

When there is no work done on the system or by the system, this simplifies to the requirement that the Gibbs energy be negative.

As an aside, when the Gibbs energy of reaction is negative, we can not have the reaction do an arbitrary amount of work on the surrounding, but that amount is limited to the Gibbs energy. That's why the Gibbs energy is sometimes called the maximal work (this is a good term for cases when it is negative; when it is positive, it is the minimal work that needs to be done on the system for it to be possible for the reaction to go forward).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause $$\Delta_\mathrm rG < 0?$$

The hydrolysis of ATP usually does not do work on the surroundings. Instead, the uncatalyzed reaction has a positive Gibbs energy of reaction. To use up the reactants and make the products, the cell runs a different reaction that includes the hydrolysis of ATP. It is not sufficient that ATP be hydrolized somewhere while the desired reaction goes on somewhere else. Instead, the two reactions are coupled so that ATP hydrolysis can not occur without the other part of the reaction happening. Here is an example how that might look for the reaction $$\ce{A + B -> C}$$:

$$\ce{A + ATP -> A-P + ADP}\tag{1}$$ $$\ce{A-P + B + H2O -> C + P_i}\tag{2}$$

In this scheme, A-P refers to a phosphate group bound to A, and $$\ce{P_i}$$ refers to phosphate. Both steps can have a negative Gibbs energy of reaction, even thought the Gibbs free energy of reaction for $$\ce{A + B -> C}$$ is positive. As net result, A and B turn in to C, without work done on the system.

There are other cases where there is work done on the system or by the system. These include proton pumps, and myosin hydrolyzing ATP to contract muscle fibers. You could also discuss photosynthesis in terms of work done to drive the synthesis of ATP.

This is what I think. Please correct me if there is any wrong notion.

First Question: In the case of an electrolytic cell, in the system, there is also a power source. How would this external voltage source affect the Gibbs free energy equation? Would it just be an extra −VIt term or something else which would cause ΔrG<0?

At first ΔrG > 0. You plug in external voltage, and the process happens. That does not mean ΔrG somehow becomes < 0 though. You did the work, the process is still not spontaneous, it is your work that makes the process happen. So, the external voltage source has no direct effect on ΔrG (Of course you can say that voltage causes change in composition, and the composition will affect ΔrG but that is the effect of composition on ΔrG).

Second Question: In the case of a biological cell, the work usually comes from the hydrolysis of ATP. How would this alter the equation and cause ΔrG<0?

Same reasoning as above, no effect on ΔrG. In this case, the work from hydrolysis drives non-spontaneous process. Just because a non-spontaneous process happens does not mean it is spontaneous, you are doing work.

• I disagree with the term work in the case of a coupled reaction. There is no mechanical, electrical or other work. Also, the ATP hydrolysis reaction would have to be running in the surrounding in order to do work on the system. In most cases, ATP is not hydrolysed in an actual reaction. Instead, it undergoes a multi-step reaction, with a net reaction that is the sum of ATP hydrolysis and the reaction you want to drive. I like the first part of the answer. – Karsten Theis Sep 6 at 2:33