Let's say I hypothesize that a graph of $\ln K$ vs. $1/T$ has a slope of $-∆G^\circ/R$ and a $y$-intercept of $0$. I prove it simply:

$$∆G^\circ = -RT\ln K \quad\to\quad \ln K = -\frac{∆G^\circ}{RT}$$

This matches the linear form $y = mx + b$. Thus, plotting $\ln K$ vs. $1/T$ would have a slope $m = -∆G^\circ/R$ and a $y$-intercept $b = 0$.

However, I understand that a van't Hoff plot defines a graph of $\ln K$ vs. $1/T$ to have a slope of $-ΔH^\circ/R$ and a $y$-intercept of $∆S^\circ/R$. It is clear from the relation $∆G^\circ = ∆H^\circ - TΔS^\circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that

$$\ln K = -\frac{∆H^\circ}{RT} + \frac{∆S^\circ}{R},$$

but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^\circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^\circ/R$ instead of $-ΔG^\circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $\ln K = -∆G^\circ/(RT)$?


2 Answers 2


In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $\Delta G^\circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -\frac{\Delta H}{R} + \frac{T \Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-\frac{\Delta H}{R} \times \frac{1}{T}$ tends to zero and $\frac{T \Delta S}{R} \times \frac{1}{T}$ cancels to be just $\frac{\Delta S}{R}$.

  • $\begingroup$ Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well? $\endgroup$ Commented Apr 16, 2019 at 23:37
  • 2
    $\begingroup$ @MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree. $\endgroup$
    – Karsten
    Commented Apr 17, 2019 at 0:31
  • $\begingroup$ If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller. $\endgroup$
    – Karsten
    Commented Apr 17, 2019 at 4:33

The fact of the matter is that the differential version of your equation

$$\frac{\mathrm{d}\ln{K}}{\mathrm{d}\left(\frac{1}{T}\right)} = -\frac{\Delta G^\circ}{R}$$

is not exact (because $\Delta G^\circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials

$$\frac{\mathrm{d}\ln{K}}{\mathrm{d}\left(\frac{1}{T}\right)} = -\frac{\Delta H^\circ}{R}$$

is exact. Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $\Delta G^\circ$ are constrained to be at 1 bar. So, in the van't Hoff development, the temperature derivative of $\Delta G^\circ$ is exactly given by

$$\frac{\Delta G^\circ}{\mathrm{d}T} = -\Delta S^\circ$$

This important constraint is not even addressed in your approach.

Finally, do you have a reference where it asserts that the intercept at $(1/T) \to 0$ is supposed to be $\Delta S^\circ/R$?

  • $\begingroup$ I took the intercept to be ∆Sº/R as 1/T approaches 0 just from the linear form of the van't Hoff equation, ln(K) = -∆Hº/RT + ∆Sº/R. If this is in y=mx+b form, b is ∆Sº/R. I guess my reference is the Wikipedia page for the van't Hoff equation, which mentions it a few times (such as in the "van't Hoff plot" section). Is this not an accurate assertion to make? $\endgroup$ Commented Apr 17, 2019 at 14:26
  • $\begingroup$ Actually, if you take the tangent line to the curve of ln(K) vs 1/T at the point (T*, K*), the intercept is $$\ln{K^*}+\frac{\Delta S^0(T^*)}{R}$$ $\endgroup$ Commented Apr 17, 2019 at 17:17

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