Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant

Question

The derivation shown here assumes H and U are independant of temperature, right? But what if $H(T) = H_{0}+C_V \Delta T $?

Then $dG = C_V\, dT - SdT$, right? So $ \displaystyle \left( \frac{\partial G}{\partial T}\right)_p=C_V -S$ ?

$$\begin{align} \left(\frac{\partial (G/T)}{\partial T}\right)_p &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p} {T^2} \\[8pt] &= \frac{T(C_V-S) - G(1)}{T^2} \\[8pt] &= \frac{TC_V-TS-G}{T^2} \\[8pt] &= -\frac{H_0}{T^2} \end{align}$$

as desired (since $G = H_0+C_V\Delta T - TS$).

But this way we see that the H in this equation would only be containing the constant term? Of course we insert H(T) to get thee exact G, but where in my thinking process did I go wrong? How would this derivation be made correctly, assuming that H changes with Temperature?

Answer 1

The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $G/T$ so you need to consider a polynomial expansion of $G/T$ to linear terms in $T$ around the temperature of interest (that is, an infinitely small interval dT). At the risk of redundancy:

$$\begin{align} (d(G/T))_p &= ((1/T)dG-G(dT/T^2))_p \\ &= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\ &= (-(S/T)dT-(H/T^2-S/T)dT)_p \\ &= (-(H/T^2)dT)_p \end{align} $$

The answer is therefore found in differential Calculus.

Another comment explains that, if

$G(T)=H(T) -TS(T)=H_0 +C_p \Delta T -TS(T)$

where $H(T)$ is expressed as a linear polynomial in T ($C_p$ and $p$ assumed constant) then

$dG=C_p dT -SdT-TdS$

Since for a process at constant pressure, $dq = C_p dT=TdS$ (since $C_p = (dH/dT)_p$ and $dq_p= dH$; also for a reversible process $dq= TdS$), it follows that

$\left(\frac{ \partial G}{\partial T} \right)_p=-S$

and the van't Hoff expression provided in the linked answer also follows.