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We take Nacl and dissolve it in water. There is formation of $Na^+$ and $Cl^-$ gaseous ions along with 2H+ and $Cl^{2-}$ aqueous ions.

Then , Na+ is attracted towards $Cl^{2-}$ and 2H+ is attracted towards $Cl^-$ ions.

Energy required by water molecules to separate the NaCl molecules is knows as Lattice enthalpy.

Energy required to form aqueous NaCl molecule is called hydration enthalpy.

My questions are:

  1. Who is the one that converts gaseous ions of NaCl to aqueous ions?

  2. During the dissolution of NaCl, why is there no formation of solid ions first but directly, it states that gaseous ions of NaCl are formed?

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    $\begingroup$ Is this an engineering question? Sounds like physics or chemistry SE. $\endgroup$
    – DKNguyen
    Jul 9, 2021 at 13:40
  • $\begingroup$ @DKNguyen Yes. Engineering thermodynamics. More at a basic level but I am having trouble in understanding it, $\endgroup$
    – S.M.T
    Jul 9, 2021 at 14:37
  • $\begingroup$ I've never heard of ions being described as solid, liquid, or gas. Also, what "states that gaseous ions of NaCl are formed ?" Are you referring to a source that you can provide? $\endgroup$
    – J. Ari
    Jul 9, 2021 at 16:06
  • $\begingroup$ @J.Ari Nor I. OP might have in mind that chlorine is a gas at ATP so maybe should make "gaseous ions" whereas sodium is a solid at ATP so maybe should make "solid ions" and is wondering how they become "liquid ions" when dissolved in water. In which case this question seems based on false premises. $\endgroup$
    – DKNguyen
    Jul 9, 2021 at 16:52
  • $\begingroup$ @DKNguyen I’ll add a photo or link to edit, $\endgroup$
    – S.M.T
    Jul 9, 2021 at 17:02

1 Answer 1

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We define reactions uniquely depending on the reactants and products. Here are examples related to your question.

Formation: Na(s) + (1/2)Cl$_2$(g) $\rightarrow$ NaCl(s)

Lattice Formation: Na$^+$(g) + Cl$^-$(g) $\rightarrow$ NaCl(s)

Solution: NaCl(s) $\rightarrow$ NaCl(aq)

Hydration: Na$^+$(g) $\rightarrow$ Na$^+$(aq)

Atomization: Na(s) $\rightarrow$ Na(g)

A significant observation here is that the STATE of the reactants and products is also required because it is a unique identifier.

As to your questions ...

  1. Who is the one that converts ...

We can convert NaCl(s) to gaseous ions experimentally through processes that may involve multiple steps, including vaporization and ionization.

  1. During the dissolution of NaCl(s), why is there no formation of solid ions first but directly, it states that gaseous ions of NaCl are formed?

The experimental dissolution of NaCl(s) does not involve formation of gaseous ions. It involves formation and solvation of aqueous ions. The two steps (dissolution and solvation) are often not easily separated.

What you refer to is the case where we imagine the dissolution of NaCl(s) happening through a set of steps that include the formation of gaseous ions. The process to combine reaction steps that go from a given set of reactants to a given set of products is formalized in what is called a Born-Haber cycle. This example demonstrates how some of the above reactions fit into a Born-Haber cycle diagram for NaCl(s).

Just because we imagine a reaction path on the Born-Haber cycle does not mean that the reaction occurs experimentally. Indeed, one power to the Born-Haber cycle is to back-calculate an energy for a reaction that does not occur experimentally but that has significant theoretical utility as a building block to obtain other values.

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  • $\begingroup$ Perfect. Thank you very much $\endgroup$
    – S.M.T
    Jul 11, 2021 at 14:30
  • $\begingroup$ So this is talking about a model right? Where in this specific instance, only the inputs and outputs can be verified directly but the way the internal bits and pieces can't be? Instead, the exact workings of the internal bits and pieces are confirmed more directly through other tangential processes and the results of those have proven accurate when applied to this scenario? $\endgroup$
    – DKNguyen
    Jul 11, 2021 at 18:33
  • $\begingroup$ @DKNguyen I believe you are asking whether the reactions in a Born-Haber cycle are required to be elemental steps. The answer is no. $\endgroup$
    – Jeffrey J Weimer
    Jul 11, 2021 at 20:29

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