Imagine a typical thermodynamic system, a gas container, with bouncing particles or molecules, all identical. The particularity of such molecules is that after interacting at high speeds (which would not be rare), they might join forming bigger molecules. For example, three molecules A, B and C would become A and BC+vibration. Clearly, BC+vibration can be perceived as a concentration of energy and as organization.

What is the behavior of entropy in this case?

Considering the Boltzmann approach, it all depends on probabilities, so, if the process is a result of a spontaneous evolution, entropy will grow naturally even if the system gets organized.

Considering Clausius' approach, the number of parts of the system reduces, so perhaps entropy could decrease (not an expert here).

Considering common interpretations as "energy dissipation" or "disorder", entropy would decrease: energy is getting "concentrated" or order is produced, which would be opposite to the perception of disorder.

What is the right approach here?

In addition, we can imagine the opposite behavior: a lot of compound molecules, like AB+energy and CD+energy, which after bouncing at high speed become A, B and CD+energy, with a net increase of energy due to the liberated energy of AB. Does entropy also grow in this case?


[Edited following advice in comment]

You are basically considering a dimerization reaction in the gas phase $$ 2\text{A} \rightleftharpoons \text{A}_2 $$ and in the simplest case, A is a monatomic gas, and A$_2$ a simple diatomic molecule. Chemists think of this as a "dynamical equilibrium", and I believe that this is the best approach here. It allows one to discuss the entropies and energies of the reactants (on the left) and products (on the right), without needing to consider the details of the collisions.

The largest contribution to the entropy change is the loss of three degrees of translational freedom when two atoms are converted into a single molecule. It is true that the molecule will have some entropy associated with rotational motion, and some more associated with vibration of the bond between the atoms. But these will be small contributions compared to the translational term.

The thermodynamic driving force is a free energy, which will include an energetic contribution as well as entropic terms. The biggest energy term is usually the energy gain associated with forming the bond: this will favour the A$_2$ molecule.

The extent to which the equilibrium favours one side or another will be a balance between these terms, and this will depend on the temperature. At high $T$, the entropy terms will win, and the gas will be mostly atoms; at low temperature, the bond energy will win, and you should expect to see mostly molecules. Interestingly, you will never see 100% atoms or 100% molecules; there is always an extra entropy term corresponding to the mixing of the reactants with the products, and this always causes some nonzero concentration of both species, even if one of the concentrations is very low.

This can all be quantified. It is one of the great successes of statistical mechanics that it is possible to calculate the partition functions of the atoms and molecules in gas reactions of this kind, and hence the Gibbs free energy, with both energy and entropy terms. Some of the details are given below, in case they are of interest.

This problem is tackled in most standard statistical mechanical texts, and indeed in most general Physical Chemistry textbooks at undergraduate level: for example PW Atkins and J de Paula (and now Keeler), Atkins' Physical Chemistry (Oxford University Press).

The monatomic species A has a translational partition function (giving its ideal gas behaviour) and there may be a degeneracy associated with its electronic ground state. The diatomic molecule A$_2$ has a translational partition function, possible electronic ground state degeneracy, and additionally, partition functions associated with molecular rotation (typically with a symmetry number) and vibration of the bond, as well as the (ground state) energy associated with the formation of the chemical bond. All of these can be calculated fairly accurately: the vibration can be represented as a quantum harmonic oscillator, and the rotation as a quantum linear rotor. With this information, you can calculate the free energy of a mixture of the two species, for any temperature, pressure, and composition.

What is more, the equilibrium constant for the above reaction can also be calculated from these quantities, so you can actually work out the equilibrium composition of a reacting mixture (for a given temperature, pressure etc). This is done as a worked example for sodium in Atkins and de Paula 7th edition, p678 (which is the edition I have; I presume something similar is done in later editions).

The key equation is (if we neglect electronic degeneracies for simplicity) $$ \text{equilibrium constant} = K = \frac{p^\circ q_{\text{rot}} q_{\text{vib}} \Lambda_A^6}{k_BT\Lambda_{A_2}^3} \exp(+D_0/k_BT) $$ where $D_0$ is the dissociation energy (per molecule) of A$_2$, $k_B$ is Boltzmann's constant, $T$ the temperature, and $p^\circ$ is the standard pressure (1 bar). The $\Lambda$ factors are thermal de Broglie wavelengths for the two species, $\Lambda=h/\sqrt{2\pi mk_BT}$ where $m$ is the mass of A or A$_2$ as appropriate; these take care of the translational contributions. $q_{\text{rot}}$ and $q_{\text{vib}}$ are the partition functions mentioned above, which also depend on $T$. The link with thermodynamics is provided by $$ \Delta_r G^{\circ} = -k_BT \ln K $$ which is the standard Gibbs reaction free energy per molecule (chemists usually write it per mole, replacing $k_B$ by $R$) and this represents the difference in standard Gibbs energies of formation between A$_2$ and $2A$. The entropy and enthalpy contributions may be calculated separately from standard thermodynamic relations such as $\Delta_r S^{\circ}=-d(\Delta_r G^{\circ})/dT$. The dissociation energy $D_0$ only contributes to the energetic (enthalpic) part of $\Delta_r G^{\circ}$, not the entropy.

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    $\begingroup$ I do not think the OP is at that level, you should adapt your answer to his level, more intuitive $\endgroup$ – Wolphram jonny Oct 26 '18 at 18:36
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    $\begingroup$ @Wolphramjonny thanks for your advice. You are right, of course, my answer was too formal and mathematical. I have changed it, trying to be more descriptive, but I have left the details in place for anyone who is interested in following them up. $\endgroup$ – user197851 Oct 26 '18 at 21:25

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