5
$\begingroup$

Complete amateur here. When looking at one form of the equation deriving the change in Gibbs free energy for a given chemical reaction, the terms change in enthalpy, and the negative product of temperature (in kelvin) and change in entropy:

$$\Delta G = \Delta H - T\,\Delta S$$

This seems to double count how temperature will influence the spontaneity of the reaction if temperature is already considered when deriving entropy, $S$.

Any guidance would be appreciated.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ So what? Yes, $\Delta S$ is a function of temperature, and moreover, so is $\Delta H$, and then $\Delta G$ is exactly what you said it is. "Double counting" is not a thing of any slightest importance, much like double occurrence of the letter "S" in the text of your question. $\endgroup$
    – Ivan Neretin
    May 14, 2019 at 6:41
  • $\begingroup$ Do you think that the temperature has to change in order to get a change in enthalpy or entropy of a reacting mixture? $\endgroup$
    – Chet Miller
    May 14, 2019 at 14:02

2 Answers 2

Reset to default
7
$\begingroup$

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently.

Consider an analogous (but more intuitive) equation, the Stokes-Einstein relation for the diffusion coefficient $D$ of a particle in a liquid:

$$D = \frac{k_\mathrm{B} T}{6\pi\eta r}$$

where $\eta$ is the dynamic viscosity, $r$ is the radius of the particle, and $k_\mathrm{B}$ is Botzmann's constant. Here we see an explicit dependence of $D$ on $T$, but there is also an implicit dependence on $T$ through $\eta$ and possibly also $r$. Is this a problem? No, assuming the equation holds and we can determine the parameters accurately at each $T$ of interest. Whether that is a simple thing is another question.

Also, it is fair to wonder to what extent the parameters are fundamentally independent, or, whether by digging into the gory details of atomic structure of matter, we might not find $r$ and $\eta$ to be connected in some other way. But that is another question.

Improve this answer
$\endgroup$
4
$\begingroup$

Even if the final temperature of the products is the same as the initial temperature of the reactants, you can still have a change in entropy, enthalpy, and Gibbs free energy of products relative to the reactants. This is the result of making and breaking chemical bonds and not the result of a temperature change. Entropy and enthalpy are not just functions of temperature (and pressure), but also chemical makeup. In such a well-defined case, there would certainly not be any double counting of the temperature.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah, thank you! It only seemed that since temperature was always considered (to my very limited knowledge) when measuring the entropy, micro-states, etc. that multiplying by such a factor would be almost redundant or have some sort or ^2 squared relationship. $\endgroup$
    – Oma Höchstest
    May 15, 2019 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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