Heat Capacity temperature dependence and Gibbs energy I got this problem in school as a tool to get ready for International Chemistry Olympiad, but I have some problems with this challenge.
So, the problem is to calculate as precise as possible the Gibbs free energy for $\ce{CaCO3}$ decomposition at 1200 K.

I know that $\mathrm{d}G=\mathrm{d}H-T\cdot\mathrm{d}S$.

But $\mathrm{d}S$ and $\mathrm{d}H$ are temperature dependant, so 

$$\mathrm{d}H=\mathrm{d}H_0+\int_{T_0}^TC_p\mathrm{d}T=\mathrm{d}H_0+\mathrm{d}C_p\cdot\left(T-T_0\right)$$

and

$$\mathrm{d}S=\mathrm{d}S_0+\ln\left(T/T_0\right)\cdot\mathrm{d}C_p$$

But molar heat capacity at constant pressure is also temperature dependant, and the equation is

$C_p=a+b\cdot T+c/T^2$, where a, b, c coefficients for each compound at specific phase.

But the problem is with both solids, because this equation can not be used for them, because in 1200 K they are not gases and Einstein’s Quantum Mechanics and Debye models do not work, because Einstein’s approximates are maxed at $C_p\approx3R$, which is about 24.9, but $C_p$ at 298 K, 1 atm for both solids in tables are > 24.9.

Where is the problem? I would appreciate any possible lead to mistake/ correct answer.
At constant pressure, $dG=-SdT$.  Also, the definition of G is $G=H-TS$, so $S=\frac{H-G}{T}$.  Substituting,
$$dG=\frac{G-H}{T}dT$$
This equation can be manipulated into the form:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
So,
$$\frac{d(\Delta G/T)}{dT}=-\frac{\Delta H}{T^2}$$
I leave the rest up to you.

A math error was corrected on 10/16/15