I was studying Ellingham Diagram in Metallurgy. The following line is given from my textbook:
It is interesting to note that $ΔH$ (enthalpy change) and $ΔS$ (entropy change) values for any chemical reaction remain nearly constant even on varying temperature. So the only dominant variable in the equation $ΔG=ΔH-TΔS$ becomes $T$.
My question is why do the values of enthalpy change and entropy change remain nearly constant for a chemical reaction even on varying temperature?
Reflecting the typical context of metallurgy and oxidation/reduction, let's consider constant-pressure reactions occurring at room temperature and above; we'll work in molar quantities since the number of moles is conserved in chemical reactions.
The temperature dependence of the molar enthalpy $H$ and molar entropy $S$ are related to the constant-pressure molar heat capacity $C_P$ as
$$\left(\frac{\partial H}{\partial T}\right)_P=C_P\qquad \left(\frac{\partial S}{\partial T}\right)_P=\frac{1}{T}C_P.$$
Thus, if we calculate or are given $\Delta H$ or $\Delta S$ between two materials at one temperature, to work at a different temperature is to consider whether $\Delta C_P$, the difference in $C_P$ between the materials, is substantial or not. (Make sure not to confuse a difference between materials with a difference between temperatures; the delta symbol always means the former in this answer.)
As described by the Dulong-Petit Law, the molar heat capacity $C_P$ of condensed matter at room temperatures and higher is generally around $3R$ (where $R$ is the ideal gas constant), and those of monatomic and diatomic gases are $\frac{5}{2}R$ and $\frac{7}{2}R$, respectively.
Because these values are all similar, $\Delta H$ and $\Delta S$ don't vary much over a temperature range where a reaction occurs between consistent phases.
Equivalently, the Ellingham diagram slope is essentially constant for consistent reactants, with its slope representing the comparatively large entropy change from generating or eliminating (high-entropy) gas in the oxidation/reduction process.
Simply put, the enthalpy and entropy changes for a reaction are not constant over different temperatures! They are only approximately so over relatively small temperature ranges, where this range depends greatly on the details of your system of interest. To show this with an example that I give my students, we can estimate the boiling point of water with and without temperature dependence. We will first use the reference data for the standard molar enthalpy of formation and standard molar entropy of liquid and gaseous water at 25$^\circ$C (taken from the CRC Handbook), \begin{gather*} \begin{array}{ c c c } & H_2O_{(l)} & H_2O_{(g)} \\ \hline \Delta \bar{H}_f^\circ \text{ (kJ/mol)} & -285.83 & -241.83 \\ \bar{S}^\circ \text{ (J/mol}\cdot \text{K)} & 69.95 & 188.83 \end{array} \end{gather*} to estimate the boiling point of water from the Van't Hoff equation, \begin{align*} \ln(K_{eq}) &= - \frac{\Delta \bar{H}_{vap}^\circ}{RT} + \frac{\Delta \bar{S}^\circ_{vap}}{R} \\ \ln(P_{H_2O}) &= - \frac{\Delta \bar{H}_{vap}^\circ}{RT} + \frac{\Delta \bar{S}^\circ_{vap}}{R} \\ \implies \frac{\Delta \bar{H}_{vap}^\circ}{RT_{BP}} &= \frac{\Delta \bar{S}^\circ_{vap}}{R} \\ \implies T &= \frac{\Delta \bar{H}^\circ_{vap}}{\Delta \bar{S}_{vap}^\circ} = 370 \text{ K} = 97^\circ \text{C} \end{align*} where we have used the fact that at the boiling point the vapor pressure of the water must match the ambient pressure (1 atm) and the condensed phase water concentration is omitted from the equilibrium constant. We also related the temperature to $T_{BP}$ to indicate the boiling point. Notice that ignoring the temperature dependence of $\Delta H_{vap}^\circ$ and $\Delta S_{vap} ^\circ$ made us underestimate the boiling point. The CRC Handbook also has some temperature dependent data for the molar Gibbs free energy of formation for these two phases. By linearly fitting this data over the region of interest, we obtain the following, \begin{gather*} \begin{array}{ c c c } & H_2O_{(l)} & H_2O_{(g)} \\ \hline \Delta \bar{G}_f^\circ \text{ (kJ/mol)} & 0.160T - 284.72 & 0.0498T-243.64 \end{array} \end{gather*} Once again, we can use the thermodynamic data to determine, \begin{gather*} \Delta \bar{G}_{vap}^\circ = -0.1102T + 41.08 \end{gather*} which lets us find the boiling point, \begin{align*} \Delta \bar{G}_{vap}^\circ &= -RT\ln(K_{eq}) \\ \implies -0.110T_{BP} + 41.08 &= 0 \\ \implies T_{BP} &= 372.8 \text{ K} = 99.6^\circ\text{C} \end{align*} which is the correct boiling point of water. This simple example shows that the values are really not constant. But the values for a given system, like a solid, for instance, might vary extremely minutely due to the small variation of volume with temperature. This will depend both on the substances and the phases of the substances in question.
