# How are reaction enthalpy and entropy affected by temperature

The equation for Gibbs Free Energy demonstrates how the standard change in Gibbs Free Energy for a reaction is affected by temperature.

$$\Delta G^\circ = \Delta H^\circ - T\,\Delta S^\circ$$

However, to my understanding, the equation makes the inherent assumption that the standard change and enthalpy and the change in entropy for the reaction is unaffected by temperature. I have heard that this is not true and that reaction enthalpy and entropy are indeed dependent on temperature. However, I do not know how exactly temperature affects the two and why.

Would a reaction occurring at a higher temperature be more or less exothermic? Or does it perhaps depend on the reaction itself?

I know that the entropy of a substance increases at higher temperatures but I do not know how this correlates to the situation of a reaction since I'd assume that both the reactants and the products would be at a higher entropy.

• There is nothing to know here. $\Delta H$ may change either way, depending on the reaction. Ditto for $\Delta S$. – Ivan Neretin Feb 7 at 8:06

For each species you could use Kirchoff's equation to calculate the change in enthalpy with temperature, in its integrated form (at constant pressure) this is $$\displaystyle \Delta H_2-\Delta H_1=\int_{T_1}^{T_2}\Delta C_p dT$$. The heat capacity $$C_p$$ changes with temperature also, but for small temperature changes it may be considered to be effectively constant, then $$\displaystyle \Delta H_2-\Delta H_1=\Delta C_p(T_2-T_1)$$, if not, the heat capacity can be approximated by $$C_p=a+bT+cT^2 \cdots$$ and integrated but the constants should be determined experimentally.

The entropy is zero at zero Kelvin for a pure substance so must increase with temperature and as $$\displaystyle S_T=\int_0^TC_p/T dT$$ you can calculate the change in entropy between two temperature.

However, there are phase changes as the temperature increases (solid, liquid, gas) then there are discontinuities in the heat capacity and the entropy becomes $$\displaystyle S_T=\int_0^TC_p/T dT+\Delta H_{fus}/T_{fus} + \Delta H_{vap}/T_{vap}$$. The terms with phase changes cancel out in $$\Delta S$$ if your species are in one phase, i.e. all reactants are in the gas phase.

(If you are familiar with partition functions in Statistical Mechanics then thermodynamic properties of molecules can, in principle, be calculated directly.)

The standard Gibbs free energy, enthalpy, and entropy changes, $$\Delta G^\circ, \Delta H^\circ,\text{and }\Delta S^\circ$$, are all formally temperature-dependent. Indeed, one could write them as: $$\Delta G^\circ (T), \Delta H^\circ (T),\text{and }\Delta S^\circ (T)$$.

"Standard" simply means that the pressure is 1 bar. See "standard thermodynamic quantities", https://goldbook.iupac.org/terms/view/S05927

Most commonly, standard thermodynamic quantities are tabulated at $$298.15 \text{K}$$, but they can be given for any temperature. Indeed, the reason why one needs to specify the temperature is that they are temperature-dependent.

However, while the following equation does not assume the temperature-independence of any thermodynamic quantity, it does assume that the temperature is constant for the change in state:

$$\Delta G^\circ = \Delta H^\circ - T\,\Delta S^\circ$$

This is because, by defintion:

$$G = H - TS \Rightarrow dG = dH - TdS - SdT$$

Then, under the assumption of constant T,

$$dG = dH - TdS \Rightarrow \Delta G = \Delta H - T \Delta S$$

As to the specific temperature-dependence of each of these thermodynamic quantities, we have:

$$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$$

$$\pd {\Delta G}Tp = -\Delta S$$ $$\pd {\Delta H}Tp = \Delta C_p$$ $$\pd {\Delta S}Tp = \dfrac{\Delta C_p}{T}$$

Note the above hold for any change in state in which the pressure is constant, which of course includes when the pressure is the standard pressure.

In case you are confused by $$\Delta C_p$$, note that it is just the difference between the constant-pressure heat capacitity of the products and reactants. I.e., just as $$\Delta G$$ is $$G_{prod} - G_{react}$$, $$\Delta C_p$$ is $$C_{p, prod} - C_{p, react}$$.

Hence, as the temperature increases, if (say) $$C_{p, prod} > C_{p, react}$$, then:

$$\pd {\Delta H}Tp > 0,$$

which means the reaction will become less exothermic/more endothermic as the temperature increases. Note, however, that $$C_p$$ is itself also temperature-dependent.

$$\endgroup$$