I am a biochemistry student and we are learning about thermodynamics. Is the Gibbs standard free energy for a reaction always constant? The equation below suggests that it changes with temperature:

$$\Delta G^{\circ\prime} = -RT\ln K'_\mathrm{eq}$$

It is important to note that we are using the biochemistry version of the Gibbs standard free energy equation. To answer my own equation, I believe that it is always constant. Changing the temperature would probably just change the equilibrium constant.


In short, no, the standard Gibbs free energy change is not constant; it is a function of temperature. The same is true for practically all other standard-state quantities.

This gets a little confusing because of how standard-state properties are often explained in lower-level (high-school and college-freshman level) textbooks. The standard Gibbs free energy change, $\Delta G^\circ$, is the value of $\Delta G$ at a certain constant pressure (usually 100 kPa or 1 atm) and constant concentration of solutions (usually 1 molal or 1 molar) and a specified temperature (usually $0~^\circ \text{C}$ or $25~^\circ \text{C}$). What is intended but not immediately obvious is that the pressure and concentration are constants but temperature is a variable; since the temperature is a variable, $\Delta G^\circ$ is a function of temperature. When the value of $\Delta G^\circ$ is given, the temperature must also be specified.

In your equation, then, $\Delta G^\circ$ is the standard Gibbs free energy change of the reaction at the specified temperature ($T$ in the equation). Change $T$, and the value of $\Delta G^\circ$ also changes; but it is still the standard Gibbs free energy change at the (new) specified temperature. (This equation cannot be used to calculate $\Delta G^\circ$ at a given temperature because $K_\text{eq}$ is also a function of temperature.)

The IUPAC "Green Book" (ed. 3) has a good explanation of what is meant by the term "standard state"; the discussion starts on p. 61 (as printed on the page; it's p. 76 in the pdf file). It explicitly expresses the standard chemical potential ($\mu^\circ$), which is very closely related to $\Delta G^\circ$, as a function of temperature.

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    $\begingroup$ Note that "standard state" is not the same as STP; this is probably where a lot of the confusion comes from. $\endgroup$ – j_foster Sep 13 '15 at 19:39
  • $\begingroup$ Thanks! I think I am getting somewhere with this. I blame most of the confusion on my textbook. My textbook does not address some of these more nuanced ideas. You mentioned "When the value of ΔG∘ is given, the temperature must also be specified." My textbook never actually does this which is ridiculous. $\endgroup$ – ctkw Sep 14 '15 at 3:11
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    $\begingroup$ See also Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions (IUPAC Recommendations 1981): “The above definitions of standard states make no reference to fixed temperature. Hence, it is possible to have an infinite number of standard states of a substance as the temperature varies.” $\endgroup$ – user7951 Sep 14 '15 at 12:06

Yes, the standard Gibbs energy of a reaction is constant. You have to keep in mind however that "standard" refers to a standard state or reference state, in which by definition, things such as temperature, pressure, phase composition, concentration of all species, etc. are fixed and unchangeable.

Changing the temperature would probably just change the equilibrium constant.

It would also mean that the system would no longer be at standard conditions.

Allowing just the temperature to vary, while holding species concentrations, phase compositions, pressure, etc. constant, the variation in $\Delta G^{\,\circ \prime}$ is given by the Gibbs-Helmholtz equation.

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    $\begingroup$ So you are saying that if I am given a (biochem) Gibbs standard free energy value, I am only able to calculate Keq for 298 K? What if I wanted Keq for a different temp (say 325 K). Could I plug 325 K into the equation I posted above and solve for Keq? Thanks for your help!!! $\endgroup$ – ctkw Sep 13 '15 at 14:09
  • $\begingroup$ Yes that is what I am saying.. (Subject to the assumptions of the Gibbs Helmholtz equation) $\endgroup$ – Curt F. Sep 13 '15 at 18:53

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