In high school we learn that when a reaction has both negative $\Delta{}S$ (entropy change) and negative $\Delta{}H$ (enthalpy change) it occurs spontaneously at lower temperatures, but becomes non-spontaneous at higher temperatures.

I know that water freezing below $\pu{293 K}$ is an example of such change, but I'm trying to find an example of a chemical reaction with such a property. Does it even exist?

  • $\begingroup$ Something like carbonation of oxides, eg CaO + CO2 $\endgroup$
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    Oct 21, 2021 at 13:48
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5 Answers 5


A very simple reaction which is readily accessible for study is the dimerisation of nitrogen dioxide:

$$\ce{2 NO2(g) -> N2O4 (g)}$$

This process is has a standard reaction enthalpy change of $\rm{\Delta_r H^o=-57.2\ kJ\ mol^{-1}}$ and a standard reaction entropy change of $\rm{\Delta_r S^o=-175.83\ J\ mol^{-1}\ K^{-1}}$. With these values, the standard reaction Gibbs free change $\rm{\Delta_r G^o}$ is very close to zero, which implies an equilibrium constant close to 1 in ambient conditions. Not only that, but nitrogen dioxide is intensely brown, whereas dinitrogen tetroxide is colourless, so the reversible shifting between the different species can be directly visualised.

Eframgoldberg, Nitrogen dioxide at different temperatures.jpg, commons.wikimedia.org

Image source: Eframgoldberg, Nitrogen dioxide at different temperatures.jpg, commons.wikimedia.org. From left to right, the vial temperatures are -196 °C, 0 °C, 23 °C, 35 °C and 50 °C.

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There exist much simpler examples, like the synthesis of water according to

$$\ce{2 H2(g) + O2(g) -> 2 H2O(l)}$$

This reaction is exothermic. As the entropies of $\ce{H2},$ $\ce{O2}$ and $\ce{H2O(l)}$, in $\pu{J mol^{-1} K^{-1}}$, are respectively $131,$ $205$ and $70$, the reaction has a negative change of entropy.

A contrario, exothermic chemical reactions which have an increase of entropy are not so common. Apart from explosion of TNT or other explosives...


The nitridation/oxidation of transition elements is exothermic and decreases the entropy, since that one of the reactants ($\ce{O2}$, $\ce{N2}$) is gaseous, but at the end is dissolved in the crystal structure of the resulting product.

For example the formation of titanium nitride:

$$\ce{Ti(s) + 1/2 N2(g) -> TiN(s)}$$

It is higly exothermic and it evidently decreases the overall entropy.


It is not exactly a reaction, but consider crystallization of $\ce{KClO3}$ solution. If you put together solid $\ce{KClO3}$ with its solution of particular concentration, the crystalization becomes spontaneous below particular temperature, where solution is the saturated solution for given temperature.

It is exothermic ($\Delta H \lt 0$), as the dissolution is endothermic, with $\Delta S \lt 0$.

$$\Delta G = \Delta H - T \cdot \Delta S$$

$$T \lt \frac{\Delta H}{\Delta S} \implies \text{spontaneous}$$

$$T \gt \frac{\Delta H}{\Delta S} \implies \text{non spontaneous}$$


A more general rule is any polymerization reaction. Since you chain molecules together you reduce the number of freely moving particles. Thus such reaction always leads to a decrease of entropy. This also means that any polymerization reaction needs to exothermic or it would not work. In some special cases like polymethylmethacrylate (PMMA, also known as plexiglas) the polymer can even be depolymerized into its monomers by heating it above the so-called ceiling temperature, causing the entropy penalty to outweigh the enthalpic benefit (most polymers decompose before reaching this temperature).


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