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$$\ce{Na (s) + Cl_2 (g)->NaCl(s)}$$

Is collision theory applicable for the above reaction, or is collision theory applicable only for molecules? In other words, do chlorine atoms collide with sodium atoms to produce sodium chloride , or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and $\ce{Na+}$ and $\ce{Cl-}$ get stuck?

Edit after @BuckThorn's comment:

or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and $\ce{Na+}$ and $\ce{Cl-}$ get stuck?

In this process, the Na atom and the Cl atom don't touch, and electron transfer occurs at a distance (due to attractive/repulsive forces). Only after electron transfer takes place do $\ce{Na+}$ and $\ce{Cl-}$ touch and get stuck. That's why I don't think it is akin to collision.

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    $\begingroup$ You may want to explain why the two alternatives you present might or might not agree with collision theory (in particular the second). Collision theory combines a description of the rate at which reactants collide with a model of the fraction of those reactants that have enough kinetic energy to result in a productive collision. The details will change but, provided the assumption that a collision between reactants exceeding the required energy is appropriate, the theory should be able to describe reactions between gases and solids. $\endgroup$
    – Buck Thorn
    Commented Apr 13, 2022 at 6:43
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    $\begingroup$ Note that $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$ is energetically unfavourable, unless ions end bound close enough ($\ce{Na+Cl-(g)}$) by coulombic force. Saying that, not speaking about energy needed for formal $\ce{Na(s)}$ atomization and $\ce{Cl2}$ dissociation. $\endgroup$
    – Poutnik
    Commented Apr 13, 2022 at 7:55
  • $\begingroup$ @Poutnik On a related note: how do we determine if a reaction is energetically favorable or not? Do we add the ionization energy of Na (496kJ/mol) and the electron affinity of Cl (-349kJ/mol), and if the answer is positive(147kJ/mol) then we say that the reaction isn't favourable? $\endgroup$
    – user119245
    Commented Apr 13, 2022 at 8:55
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    $\begingroup$ Yes it is like that. We would say such a reaction is endothermic, consuming thermal energy. Strictly speaking, I should use thermodynamically favourable via negative Gibbs free energy change, not enthalpy change. ( $\Delta G = \Delta H - T \Delta S$ ), but at these big $H$ changes the entropy term plays just minor role. When energy is released, it is usually dissipated elsewhere, so it cannot be reused to go back. OTOH, if energy is needed, you have to gain it somewhere. $\endgroup$
    – Poutnik
    Commented Apr 13, 2022 at 9:36
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    $\begingroup$ @TomHardy Energy outcome of producing ions depends on how far from each other the are If far enough, it is endothermic. If close enough, it is exothermic. The best if many opposite ions are close enough, forming solid ionic lattice. $\endgroup$
    – Poutnik
    Commented Apr 14, 2022 at 5:42

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Your example is one of many that have been studied by molecular beam scattering techniques so this is a rather general answer.

Collision theory calculates the rate constant averaged over all geometries and energies at at given temperature. It is the product of cross section, average collision velocity and the Arrhenius factor. Thus it depends on the cross section. The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc. but less well as the species become more complex. But actually the cross section is very complicated especially as the potential between molecules varies with distance and orientation perhaps as coulomb potential or Lennard Jones or something more complicated. In this case The trajectories of the interaction between species has to be calculated and these averaged to get the collision theory rate constant.

In fact the rate constant is not that important what is important is the potential energy profile between species as this reflects the electronic properties of the molecules and can be calculated from Quantum Theory. The rate constant can then, in principle, be calculated from the potential although experiment is always essential.

Experiments of this sort are studied using atom/molecular beams and numerous reactive scattering reactions have been studied such as H+H2, D+H2, N+O2, F+H2,H+F2, O+CH, O+Cs and many more.

You should look at two excellent books on this topic Levine & Bernstein 'Molecular Reaction Dynamics, and Chemical Reactivity' publ OUP 1987, and Steinfeld, Francisco & Hase, 'Chemical Kinetics and Dynamics', publ Prentice Hall 1999.

The picture below shows how complicated the trajectories become when there is a Lennard jones potential between two atoms as one approaches the other from the left but at different position vertically. You can see that the species get pulled together at a larger separation than hard sphere (the inner circle) but can orbit and escape if there is enough energy. In this picture no reaction occurs but you can appreciate if the species get close enough with enough energy a condition could be made to make them react, for example being within the wide grey ring.

scattering

The picture below shows the Cl+H2 $\to$ HCl+H reaction. The species approach with the same energy in each case but collide at different point in the H$_2$ vibration and this leads to different outcomes, i.e no reaction or more of less vibrational excitation in HCL. The blue colour is low energy and the transition state can be seen at the bend. To calculate the rate constant the number of successful transition state crossings vs. total number at a given energy must be calculated. This could then be compared with the collision theory rate calculation and an effective cross section derived.

Cl+H2

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  • $\begingroup$ "The species approach with the same energy in each case but collide at different point in the Cl2"-You probably mean Cl instead of Cl2. $\endgroup$
    – user119245
    Commented Apr 27, 2022 at 8:53
  • $\begingroup$ "The picture below shows the Cl+H2 → HCl+H reaction"-how does Cl remain as Cl? Isn't it a diatomic molecule? $\endgroup$
    – user119245
    Commented Apr 27, 2022 at 8:56
  • $\begingroup$ "The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc."-Just to be clear, you agree with me that collision theory should be applicable for the reaction Na(s)+Cl2 (g)⟶NaCl(s), right? $\endgroup$
    – user119245
    Commented Apr 27, 2022 at 8:59
  • $\begingroup$ you seem to be considering a solid /gas reaction which is different. $\endgroup$
    – porphyrin
    Commented Apr 27, 2022 at 9:22
  • $\begingroup$ BuckThorn made a comment about this matter. Do you agree with him, sir? $\endgroup$
    – user119245
    Commented Apr 27, 2022 at 11:23

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