# What value of temperature does T represent in Gibbs Free Energy Equation?

Temperature before, or after reaction? Temperature change? Or what is it?

• Welcome to chemistry.SE! If you had any questions about the policies of our community, take your time to visit the help center or take a tour of the site. – It's Over Jan 12 '15 at 9:48
• Wow, Graham, this is turning out to be a long debate! Would you mind showing us that what you meant by "Gibbs Free Energy Equation"? Graham: Of course not. :) – It's Over Jan 12 '15 at 12:39

We've got:$$G_{(p,T)}= U+pV-TS$$ Where G is Gibbs free energy, U is the thermodynamic system's internal energy, p is pressure, V is volume, T is temperature and finally, S in entropy. In a chemical reaction, we can change the equation above to have:$$\Delta{G}_{(p,T)}= \Delta{U} +p\Delta{V}- T\Delta{S}$$ Which is then "interpreted" to:$$\Delta{G}_{(p,T)}= \Delta{H} - T\Delta{S}$$ This is what learners like me learn during the high school. Note that Gibbs free energy (measured in here) works well when deciding about the spontaneity of processes that occur in constant temperatures and pressures.

So if you're to use the equation, know that you oblige the poor poor system to be in a constant temperature! Think about it. If you're being asked a question from things you've studied, you will definitely run into phrases like STP or RTP in the question content. There is, however, a "formula expansion" to where we have variable temperatures; but of course, that wouldn't use the equations above.

If it's still vague for you, you can comment me.

• I think the question is really clear. It seems you're just answering a high school question. It's a conceptual question not a high school exam. – Mohebifar Jan 12 '15 at 12:00
• Nah @Mohebifar, I just wanted to, or maybe tried to, look at the Q from his/her point of view. I just could have answered "the T of system which is constant."; instead, I took an approach a learner with some research on the topic would take, a position s/he is in. Plus, the "stp or rtp" question example is just an evidence for the fact, nothing more. – It's Over Jan 12 '15 at 12:05
• It is definitely false and that's why I am telling you that. The equation you wrote, is just an special case, not a universal equation. This is what you can find in high school books. – Mohebifar Jan 12 '15 at 12:19
• That's exactly what I say! Read the whole answer. The equation only works in constant p and T, if this is the especial case you're talking about. Graham hasn't provided the equation in his mind; but since the Net is full of this equation, and since this question can rise when talking about this equation, I assumed this was what that was in his mind. – It's Over Jan 12 '15 at 12:25
• Sorry I should have clarified, yes this is a question for AP chem so basically high school, and the equation I had in mind is dG=dH-TdS (d meaning delta) – Graham Jan 14 '15 at 7:11

We can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation: $$\Delta G = \Delta H - \Delta TS$$

• Temperature can be constant or not,if it is constant we can change the equation to deltaG = deltaH - TdeltaS – Fatima Sajadi Jan 12 '15 at 11:26
• Yes it can. But if it would be inconstant then the equation needs expansion. By the way, visit this page to make better posts. – It's Over Jan 12 '15 at 11:30
• @MARamezani The answer seems almost correct because the equation deals with partial derivatives. Your answer is just a simplified equation. – Mohebifar Jan 12 '15 at 12:23

The temperature must remain constant so it is the value both at the start and the end. I will try to explain this simply as, from your question, I suspect that you have just been introduced to $G$ with the equation $G=H-TS$.

The Gibbs free energy helps to indicate whether a process/reaction will be spontaneous at conditions where the pressure and temperature remain constant. This is useful because these are the conditions that occur in lab when we do an experiment. As I am sure you know the Gibbs free energy must decrease for a spontaneous process; in other words $\Delta G<0$.

Notice that the equation $G=H-TS$ does not involve $\Delta G$ is only gives a relation between Gibbs free energy and enthalpy and entropy. It is clearly useful to consider changes in $G$ so lets say $G$ changes by an infinitesimally small amount: $\mathrm{d}G$ (this basically means the same as $\Delta G$ but the change is much smaller) we can change the expression so that $\mathrm{d}G$ is written in terms of infinitesimal changes in entropy temperature and enthalpy:

$$(G+\mathrm{d}G)=(H+\mathrm{d}H)-(T+\mathrm{d}T)(S+\mathrm{d}S)$$ Expanding the brackets in the last expression: $$G+\mathrm{d}G=H+\mathrm{d}H-TS+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$ I will now rearrange to make the following step clear: $$\mathrm{d}G+[G=H-TS]+\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$ Notice the $G=H-TS$ in the equation (highlighted in square brackets) – this can now cancel because it’s the same as having $G$ on each side. This leaves: $$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S+\mathrm{d}S\mathrm{d}T$$ The last term is the product of two infinitesimally small changes (very very small numbers) so is basically zero and can be ignored. This leaves: $$\mathrm{d}G=\mathrm{d}H+S\mathrm{d}T+T\mathrm{d}S$$ Now, remember towards the start of my answer I said that this is useful at a constant temperature and pressure. Well, if temperature is constant $\mathrm{d}T=0$ so the $S\mathrm{d}T$ term vanishes leaving:

$$\mathrm{d}G=\mathrm{d}H-T\mathrm{d}S$$ and for larger changes: $$\Delta G =\Delta H -T\Delta S$$

It is crucial that you know, and hopefully now understand after my explanation, that the above equation is only valid at a constant pressure and temperature.

So in short $T$ is the constant value for temperature.