# Entropy change of evaporating NaCl [closed]

What is $\Delta S$ for this process?

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

(A reference that I could quote would be ideal.)

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• The heat of sublimation at 900 K is 211.3 J/mole – Chet Miller May 23 '18 at 17:52

## 1 Answer

Under what conditions? And do you want a numerical answer or an equation for its calculation? If the former, a starting point is to google "standard molar entropy NaCl" and you'll find references to the standard entropies of both NaCl(s) and NaCl(g). I would recommend the NIST Chemistry WebBook, which should be in the first few hits, as an authoritative source. That will get you the answer under standard conditions.

If you want an equation, you can google "Third Law entropy calculation" and you'll find equations for integrating the heat capacity of the various phases up from 0 K to find the entropy at some other temperature. In your case, since you only want dS, you just have to integrate between your two state points (where your NaCl is solid, and where it's a gas).

You'll also need those equations if you want a numerical answer at some other temperature than the standard entropy, because you'll need to integrate the heat capacities of NaCl(s) and NaCl(g) from 298K to the temperature of interest. I would guess you can just use Cp of NaCl(s) at 298K, but for NaCl(g) Cp may change a fair amount. Maybe the ideal gas approximation is good enough. Anyway, fortunately the bulk of dS is going to be in the phase change anyway, and the standard molar entropy difference will get you that.

• Thank you Christopher!*<br> I was looking for a number.<br> Background: I'm teaching a class (~undergraduate college level) that includes some thermodynamics. We discussed the lattice energy of NaCl (Madelung etc.) and how it describes the reaction NaCl(s) --> Na+(g) + Cl-(g); and how dissolving NH4NO3 in water results in a cold solution. So I meant to discuss the reason why NH4NO3 dissolves, but NaCl does not evaporate, even though both processes are "bad" energy-wise.<br> So I'd say standard conditions.<br> __________<br> *I'm told not to say that but I can't help myself. – Martin May 24 '18 at 12:23
• Given all that, would you say it's reasonable to take standard molar entropies for gas (229.79 J/(mol·K)) and solid (72 J/(mol·K)) from the Wikipedia data page for NaCl, and say that ΔS = 229 - 72 = 157 J/mol/K? – Martin May 24 '18 at 12:27
• Now I'm reading that NaCl(g) is actually mostly NaCl molecules, with some Na2Cl2, and no ions. Maybe I'm better off using a different example... – Martin May 24 '18 at 12:30