In the equation for Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$, does $T$ refer to the temperature of the system or the surroundings?

I know we have to calculate Gibbs free energy of the system but the criterion for spontaneity says $\Delta S_\mathrm{total}$ should be greater than zero. When we relate it to Gibbs free energy to show that Gibbs free energy change should be always negative, we keep both system and surrounding temperature same with pressure also constant. Then how can the process occur? Wikipedia says that it is chemical potential that undergoes changes there but what about Gibbs free energy change?

From Wikipedia:

Thus, Gibbs free energy is most useful for thermochemical processes at constant temperature and pressure: both isothermal and isobaric. Such processes don't move on a P—V diagram, such as phase change of a pure substance, which takes place at the saturation pressure and temperature. Chemical reactions, however, do undergo changes in chemical potential, which is a state function. Thus, thermodynamic processes are not confined to the two dimensional P—V diagram. There is a third dimension for n, the quantity of gas

  • $\begingroup$ Actually, it is $\Delta G$ that needs to be negative for a spontaneous reaction. $\endgroup$
    – LDC3
    Mar 30 '14 at 14:28

If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.

You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this must occur equally in both directions, otherwise a temperature gradient would settle between system and environment. We could describe thermal equilibrium as the result of "a zero balance of rate of heat transfer between system and environment". That's what goes on during a phase transition (e.g. solid to liquid, where entropy of the system increases due to heat given by the environment, whereas temperature stays constant).

Now, applying the second law of thermodynamics to an ensemble system + environment, undergoing an infinitesimal transformation, we write:

$$\mathrm dS_\text{syst} + \mathrm dS_\text{env} \geq 0 \tag1$$

As said above, the entropy of the environment changes if there's a reversible heat exchange with the system:

$$\mathrm dS_\text{env} = \frac{\delta Q_{env}}{T}=-\frac{\delta Q_{syst}}{T} \tag2$$

Merging eq. $\text{(1)}$ and $\text{(2)}$, you write:

$$\mathrm dS_\text{syst} - \frac{\delta Q_{syst}}{T} \geq 0 \tag3$$

Now the first law allows you to write:

$$\delta Q - P\mathrm dV_\text{syst}=\mathrm dE_\text{syst} \tag4$$

and then

$$T\mathrm dS_\text{syst} - \left(\mathrm dE_\text{syst}+P\mathrm dV_\text{syst}\right) \geq 0 \tag5$$

Now (and this is crucial), assuming both pressure and temperature constant, you end up with

$$\mathrm d{\left(E_\text{syst}+PV_\text{syst}-TS_\text{syst}\right)_{T,P}} \leq 0 \tag6$$

The expression within the brackets is a state function of the only system and corresponds to the free energy. Quoting Wikipedia:

A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state.

Considering that $E+PV=H$, we write:

$$\left(\mathrm dG_\text{syst}\right)_{T,P} = \left(\mathrm dH_\text{syst}-T\mathrm dS_\text{syst}\right)_{T,P} \leq 0 \tag7$$

Now, while equation $\text{(1)}$ is much more general, equation $\text{(7)}$ applies only to the situation where the work is expansion or contraction and for constant temperature and pression. Nonetheless, since for most situations of chemical interest, those conditions are satisfied, eq $\text{(7)}$ maintains a well deserved relevance and is very convenient to establish whether a process is spontaneous or not, within the limitation of constant temperature and pressure.

A chemical reaction occurs down to the point where $\left(\Delta G\right)_{T, P}=0$.

For a general reaction:

$$\ce{$a$A + $b$B <=> $l$L + $m$M}$$

For constant temperature and pressure:

$$\left(\Delta G\right)_{T, P}=l\mu_{\ce L} + m\mu_{\ce M} - a\mu_{\ce A} - b\mu_{\ce B} \tag8$$

being $\mu$ the chemical potential for each of the species involved. The chemical potential relates to activity, according to:

$$\mu=\mu^\circ+RT\ln a \tag9$$

Equation $\text{(8)}$ and $\text{(9)}$ shows that free energy varies with the composition (expressed as activity) of our system: that's exactly what happens during a chemical reaction.

Proceeding further, you will end up with Van't Hoff isotherm:

$$\left(\Delta G\right)_{T, P}=\left(\Delta G^\circ\right)_{T}+RT\ln\frac{a_{\ce L}^{l}a_{\ce M}^{m}}{a_{\ce A}^{a}a_{\ce B}^{b}}$$

Activities vary until a chemical equilibrium is established, so that $\left(\Delta G\right)_{T, P}=0$.

  • 1
    $\begingroup$ I already know everything that you wrote .My weakness is that I am strong in explaining my question and answer while speaking and not when writing because question here is much theoretical and I am not from English Speaking Country.Well My Question is actually regarding meaning of Paragraph from wikipedia in second edit.Please Look EDIT2 $\endgroup$ Mar 30 '14 at 15:37
  • $\begingroup$ Above wikipedia passage tells that system and surrounding temperature are same and not isothermal for each of system and surrounding. $\endgroup$ Mar 30 '14 at 15:47
  • $\begingroup$ I can't just respond to you if you respond to me.Please consider me if made response after 12 hours .I am in GMT+5:30 zone. $\endgroup$ Mar 30 '14 at 15:49
  • $\begingroup$ @VishvajeetPatil edited, hope that helps $\endgroup$
    – mannaia
    Mar 30 '14 at 21:50
  • $\begingroup$ @mannaia Why system and surroundings must be in thermal equilibrium when a reaction takes place? Is it because in such a way we can maintain a constant temperature for the system? Also when we say chemical, mechanical or thermal equilibrium should we say with respect to surroundings or not? Is equilibrium a property of the system or a property of both system and surroundings? $\endgroup$
    – ado sar
    Feb 3 '21 at 0:11

In the equation for Gibbs free energy change, ΔG=ΔH−TΔS, does T refer to the temperature of the system or the surroundings?

The temperature $T$ of the Gibbs energy of a system is the temperature of the system. There is a related thermodynamic state function called the "exergy" or "availability" and sometimes given the symbol $B$. This function, unlike the Gibbs energy, is defined relative to the temperature of the surroundings.

$dB = dH - T_0 dS$

The Gibbs energy change for a process represents how much useful work could have been extracted from that process, if it were run at constant temperature.

The "exergy" change for a process represents how much useful work could have been extracted from that process, if after running it at some constant system temperature, a series of reversible heat engines was used to cool the system down to the temperature of the surroundings.


It depends on what you want to calculate. For the Standard Free Energy, $\Delta G°$, the temperature is 25°C. If you want to calculate the Free Energy at another temperature, $\Delta G$, it is usually provided. https://www.chem.tamu.edu/class/majors/tutorialnotefiles/gibbs.htm

  • 1
    $\begingroup$ You aren't clear $\endgroup$ Mar 30 '14 at 13:51
  • $\begingroup$ In essence, you choose the temperature you want to do the calculation at. $\endgroup$
    – LDC3
    Mar 30 '14 at 13:55
  • $\begingroup$ Please check Edit $\endgroup$ Mar 30 '14 at 14:24
  • $\begingroup$ I would if you would point out what you think is wrong. I'm not a mind reader. $\endgroup$
    – LDC3
    Mar 30 '14 at 14:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.