1
$\begingroup$

I understand that the lattice energy is the energy released when a crystal forms and that's the reason it takes the negative sign. But do we need to add this energy again when we are trying to vaporize an ionic compound ?

For example, let's assume we want to completely vaporize a mole of NaCl starting at room temperature, how much energy do we need ?

I don't know neither the specific heat nor the latent heat of vaporization of NaCl, but I understand we would need:

1- Energy to raise the temperature of mole of NaCl from 298 °K to its boiling point.

2- Energy equal to the latent heat of vaporization of a mole of NaCl.

I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol. So my question here is: do we need to add this lattice energy to the calculated steps above to account for all the energy required to vaporize a mole of NaCl ?

$\endgroup$
  • $\begingroup$ Well you left out steps...(1) Energy to raise NaCl to melting point (2) Energy to melt solid NaCl to liquid NaCl at its melting point (3) Energy to raise liquid NaCl from melting point to boiling point (4) Energy to vaporize NaCl at its boiling point. // So the lattice energy is (2) $\endgroup$ – MaxW Nov 18 '15 at 23:49
  • $\begingroup$ @MaxW, I wasn't sure if NaCl would turn to liquid first. But anyway, is't (2) the latent heat of fusion ? Also, once an ionic compound is melted, this means that there are no more lattices to break, is that correct ? $\endgroup$ – Abanob Ebrahim Nov 18 '15 at 23:54
  • $\begingroup$ Sorry to have mislead you. The lattice energy is defined as the energy liberated when gaseous ions are united to form the solid lattice. This would be at a constant temperature. So the whole heating it to some other temperature doesn't apply. $\endgroup$ – MaxW Nov 19 '15 at 0:12
  • $\begingroup$ I'm sure that measuring lattice energy from ions is impractical so that the Born-Haber cycle is used. en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle // I upvoted you question in hope that someone who has studied thermodynamics more recently than I can answer your question. I'm a bit confused about the temperature dependence of the lattice energy. $\endgroup$ – MaxW Nov 19 '15 at 0:42
  • $\begingroup$ @MaxW, its also important to find out what we will end up with once we vaporize the NaCl mole. Will we get Na+ and Cl- or just molecules of NaCl in gas phase ? $\endgroup$ – Abanob Ebrahim Nov 19 '15 at 2:47
-1
$\begingroup$

The energy to vapourize one mole of NaCl is the negative of its lattice energy, assuming you are starting at 25 degrees C. Can be calculated via born habor cycle- The Born-Haber Cycle can be reduced to a single equation:

Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy

*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.

Rearrangement to solve for lattice energy gives the equation:

Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities

$\endgroup$
  • $\begingroup$ But the lattice energy is a huge number compared to the latent heat of vaporization. For example, Al2O3 has a lattice energy of -15600 kJ/mole, so if you are correct, then it would take 15.6 MJ of energy to vaporize 102 grams of Al2O3. When comparing this to iron, you will notice that it takes about 450 kJ/mole to or about 820 kJ to vaporize 102 grams of iron. Can you see the difference ? $\endgroup$ – Abanob Ebrahim Nov 19 '15 at 12:35
  • $\begingroup$ You are also assuming that once NaCl is vaporized, the gas will be made of its gaseous ions and not NaCl molecules. Could you please explain why this will happen ? $\endgroup$ – Abanob Ebrahim Nov 19 '15 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.