# Why is my conclusion inconsistent with the van't Hoff equation?

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 $$\ln K = -∆G^\circ/(RT)$$?

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 $$\frac{\Delta S}{R}$$.

• 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? Commented Apr 16, 2019 at 23:37
• @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.
– Karsten
Commented Apr 17, 2019 at 0:31
• If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.
– 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

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

Finally, do you have a reference where it asserts that the intercept at $$(1/T) \to 0$$ is supposed to be $$\Delta S^\circ/R$$?
• Actually, if you take the tangent line to the curve of ln(K) vs 1/T at the point (T*, K*), the intercept is $$\ln{K^*}+\frac{\Delta S^0(T^*)}{R}$$ Commented Apr 17, 2019 at 17:17