Can someone please explain to me how to derive the Gibbs-Helmholtz relationship from $G = H - TS$?

  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ May 2, 2017 at 10:37
  • $\begingroup$ @orthocresol can you please outline the chain rule step in the derivation? This is the part I don't understand: $\endgroup$ May 2, 2017 at 10:42
  • $\begingroup$ It's more of the product rule rather than chain rule. It may be more familiar to you to look at it this way: you should know that $\mathrm{d}(fg)/\mathrm{d}x = f(\mathrm{d}g/\mathrm{d}x) + g(\mathrm{d}f/\mathrm{d}x)$. Now let $x = T$, $f = G$, $g = 1/T$. $\endgroup$ May 2, 2017 at 10:43

2 Answers 2


You need to know that

$$\require{begingroup} \begingroup \newcommand{\d}[0]{\mathrm{d}} \d G = V\,\d p - S\,\d T$$

from which you can determine that

$$\newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} \pdiff{G}{p}{T} = V \qquad \pdiff{G}{T}{p} = -S$$

Therefore (by the quotient rule)

$$\begin{align}\pdiff{}{T}{p}\left(\frac{G}{T}\right) &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p} {T^2} \\[8pt] &= \frac{T(-S) - G(1)}{T^2} \\[8pt] &= \frac{-TS-G}{T^2} \\[8pt] &= -\frac{H}{T^2} \end{align}$$

as desired (since $G = H - TS$).

  • 1
    $\begingroup$ Why it also holds that $$\Delta \left(\frac{\partial{G/T}}{\partial{T}}\right)_p=-\frac{\Delta H}{T^2}$$ if initial and final temperatures are not the same? $\endgroup$
    – ado sar
    Jun 17, 2020 at 17:38

$$dG=dH-TdS-SdT=dU+VdP+PdV-TdS-SdT$$ But, $$dU=TdS-PdV$$ Adding the above two equations together, we get $$dG=VdP-SdT$$ At constant pressure, we have $$dG=-SdT$$But, from the definition of G, $$-S=\frac{G-H}{T}$$Substituting for -S yields: $$\frac{dG}{dT}=\frac{G-H}{T}\tag{constant P}$$ The rest is strightforward math.


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