Jtdcftul

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:

$$∆G^circ = -RTln K quadtoquad ln K = -frac{∆G^circ}{RT}$$

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 $frac{Delta S}{R}$.

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 $Delta S^circ/R$?