In textbooks, I often read that the temperature dependency of the enthalpy of reaction $\Delta_\mathrm{r}H$ and the entropy of reaction $\Delta_\mathrm{r}S$ can be neglected (in a limited $T$-range).

But why is this valid? What is the intention behind this?

  • $\begingroup$ Can you please provide more context? Is this in connection with a chemical reaction or a phase change, for example? $\endgroup$ Commented Apr 25, 2019 at 11:19
  • $\begingroup$ In connection with a chemical reaction. Sorry, I am very new to this field. $\endgroup$
    – user78415
    Commented Apr 25, 2019 at 11:28
  • $\begingroup$ The standard entropy change and the standard heat of reaction are functions of temperature, and this dependence can be expressed in terms of the heat capacities of the reactants and products. However, over limited temperature ranges, these effects are typically small compared to the enthalpy and entropy changes at a specific temperature. $\endgroup$ Commented Apr 25, 2019 at 12:03
  • $\begingroup$ And what is about non-standard values? $\endgroup$
    – user78415
    Commented Apr 25, 2019 at 13:08
  • $\begingroup$ Standard refers to reaction at 1 bar pressure. $\endgroup$ Commented Apr 25, 2019 at 13:41

1 Answer 1


Three step process to evaluate temperature dependence

There is a change in enthalpy and in entropy when changing the temperature of a sample. It follows that the enthalpy of reaction and the entropy of reaction can also change when changing the temperature. The easiest way to get values for a temperature $T_2$ if you have values for a temperature $T_1$ is to do a thought experiment with three steps, starting with reactants at the temperature $T_2$:

  1. Change temperature of reaction mixture from $T_2$ to $T_1$
  2. Let the reaction proceed at a temperature of $T_1$
  3. Change temperature of product mixture from $T_1$ to $T_2$

The net process is the reaction at the temperature $T_2$, where you don't know the entropy and enthalpy change yet.

Changes depend on differences in heat capacity

If we are working at constant pressure, the enthalpy change is (with $C_P$ the heat capacity of the reaction mixture):

$$\Delta H = C_P \Delta T$$ and the entropy change is:

$$\Delta S = C_P \ln \frac{T_\text{final}}{T_\text{initial}}.$$

Because steps 1. and 3. change the temperature in opposite directions, a lot cancels out. If we define $\Delta C_P$ as the difference in heat capacity between reactant and product mixture, we get:

$$\Delta(\Delta_r H) = \Delta C_P \Delta T$$

$$\Delta(\Delta_r S) = \Delta C_P \ln \frac{T_2}{T_1}$$

Compared to other changes with temperature (e.g. Gibbs energy or equilibrium constant), these changes tend to be modest. To give an idea of the magnitude, the molar heat capacity of a diatomic ideal gas is $c_P = 3.5 R$, or about 30 J / (mol K). Given that values for $\Delta_r H$ are typically given in kJ/mol, the change in enthalpy when changing the temperature is relatively small. For the entropy, the "correction term" is not proportional to $\Delta T$ but to the logarithm of the ratio of temperatures. If you start at room temperature and increase the temperature by ten Kelvin, the value of the logarithm will be about 0.03, so the effect is also relatively small.


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