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Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies $\ldots$ - My comment

This makes me wonder. I remember seeing $\Delta H_{\mathrm{lattice}}$ of "common" ionic compounds like $\mathrm{NaCl}$ and considered them fairly positivea: \begin{array}{|c|c|} \hline \mathbf{Ionic~compound} & \mathbf{\Delta H^\circ (kJ~mol^{-1})} \\\hline \ce{LiF} & 1022^\mathrm{\,b} \\\hline \ce{NaCl} & 771 \\\hline \ce{KBr} & 670 \\\hline \ce{CsI} & 585 \\\hline \end{array}

Yeah$\ldots$ so judging by the usual trends $\ce{CsI}$ should have the least positive lattice enthalpy in ionic compounds derived only from monoatomic ions. But that's still some lattice enthalpy!

So, could there be negative lattice enthalpies?


Notes:
a: Note that there are two major definitions of lattice enthalpy. Here, we're using the definition relevant to $\Delta H_r$ of the following endothermic reaction: $$\ce{MX(s) → M+(g) + X^- (g)}$$ While the other definition uses the enthalpy of the reverse process as the reference, only the sign might differ. Note that I had the other defintion in my mind when posting that comment.
b: All data obtained at $298~\mathrm{K}$. source

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  • $\begingroup$ I really want to say no because you have to break bonds in order to ionize the lattice and you aren't forming any (as far as I know) but I don't have any hard theory to back this up. $\endgroup$ – bon Jun 16 '15 at 21:15
  • $\begingroup$ Yes @Bon $\ldots$ The "no" coming at first to my mind fascinated me even more towards asking this question. $\endgroup$ – M.A.R. Jun 16 '15 at 21:17
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Short answer: Not for "normal" compounds.

First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.

As you see from your table, the energies decrease with:

  • Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)
  • Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).

The Born-Lande and Kaputinskii equations give some numeric estimates, e.g. (Born-Landé)

$E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$

where:

  • $N_A$ = Avogadro constant;
  • M = Madelung constant
  • $z^+$ = charge number of cation
  • $z^−$ = charge number of anion
  • e = charge on an electron (elementary charge)
  • $\epsilon_0$ = permittivity of free space
  • $r_0$ = distance to closest ion
  • n = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.

(Note that I've inverted the formula from Wikipedia's version to match your reaction.)

None of these can give you a negative lattice enthalpy.

How could it happen? Well, you'd need something where the solid is less stable than the gas-phase ions. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.

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    $\begingroup$ I've been trying to think if an unusual compound (e.g., $\ce{Xe+ F-}$ might create this, but it doesn't line up. $\endgroup$ – Geoff Hutchison Jun 17 '15 at 19:54
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    $\begingroup$ Even without assuming lattice energy formulas, just going by regular thermodynamics, the only way for lattice formation from an ion gas to be exergonic ($\Delta G<0$) while being endothermic ($\Delta H>0$) is if entropy increased going from the gas to the solid ($\Delta S>0$). That is probably impossible, so either thermodynamics under constant pressure isn't enough, or an endothermic lattice can only exist in non-equilibrium conditions or with some form of work being continuously performed on the system. $\endgroup$ – Nicolau Saker Neto Jun 17 '15 at 20:02
  • $\begingroup$ @NicolauSakerNeto Agreed. I wanted to go from the lattice description, but your comments are easily worth a separate answer. I'd vote for it. :-) $\endgroup$ – Geoff Hutchison Jun 17 '15 at 20:48
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Geoff's answer is enlightening because it shows several of the factors influencing lattice formation energetics and how they cannot be easily made to output endothermic ionic lattices. However, an even stronger statement can be made by forgoing any reference to lattice energy formulas.

Considering thermodynamics as we chemists are most used to, we have a system which is free to evolve under constant pressure and in the absence of external potentials such as gravitational, electric or magnetic fields, and which is given infinite time to reach an equilibrium condition. As such, we can use our good old friend, the Gibbs free energy equation, in its simplest form:

$$\Delta G = \Delta H -T\Delta S$$

For a process to be spontaneous, we say it is exergonic ($\Delta G<0$). This requires one of three conditions to be met:

  1. $\Delta H<0$, $\Delta S>0$ : The process is exothermic and increases system entropy; spontaneous process at any temperature.
  2. $\Delta H<0$, $\Delta S<0$ : The process is exothermic and decreases system entropy; spontaneous process only at sufficiently low temperatures.
  3. $\Delta H>0$, $\Delta S>0$ : The process is endothermic and increases disorder; spontaneous process only at sufficiently high temperatures

Now consider the process where gaseous ions combine to form an ionic solid:

$$\ce{y\ A^{+x}(g) + x\ B^{-y}(g) -> A_{y}B_{x}(s)}$$

The question essentially asks whether this process can be exergonic while being endothermic. For this to be the case, the only condition which has any chance of being met is the third. Therein lies the problem. It is hard to imagine a situation where a crystalline solid with long-range ordering, as is the case with most common ionic compounds, could possibly have a higher entropy than a gas of ions. This is a strong, general argument against the stability and existence of ionic solids with endothermic lattice energies.

This thermodynamic argument, however, does not constitute ultimate proof that endothermic lattices are impossible, just unlikely. For a stable endothermic lattice to be made, one of the constraints mentioned at the start must be relaxed. For example, perhaps thermodynamics under constant pressure is not adequate, but under constant volume (with the associated Helmholtz free energy considerations) it could work.

Another possibility is that additional potentials can be present in the system, in which case other terms must be added to the Gibbs free energy expression. As a loosely related example, water can liquefy/freeze at unusual temperatures under the action of a very strong electric field, even at atmospheric pressure.

Yet another way is if, even though formation of an endothermic lattice is not spontaneous, once formed it could somehow be kinetically stabilized against dissociation. This delicate interplay of kinetics and thermodynamics in Chemistry is certainly thoroughly exploited, but whether it can applied to this particular situation, I don't know.

One more opportunity for the existence of a negative energy lattice can be based on non-equilibrium thermodynamics, either in systems close to (but not quite at) equilibrium, or in systems very far from equilibrium. Thermodynamics in these systems is much harder to describe mathematically, however.

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