born haber cycle for enthalpies of solutionWatch this thread
You have to be able to construct and calculate cycles based on what you refer to as BH cycles and solution cycles.
Qs from the old F325 paper and the new paper 1 (and maybe 3) are relevant.
The way you draw them is probably more simple than how you draw them for producing an ionic lattice. I've quickly thrown together the layouts for them:
This is a Born-Haber cycle for if the enthalpy change of solution is endothermic.
As you can see, the ionic compound (MX) can be dissolved which causes it to undergo an endothermic enthalpy change of solution (ΔHSOLN, arrows point upwards as it's endothermic); this process results in two aqueous ions (M+(aq) and X-(aq)). However, MX also has a lattice enthalpy associated with it (ΔHLE), which going against results in two gaseous ions (M+(g) and X-(g)). These two ions can undergo their respective enthalpy changes of hydration (ΔHHYD, arrows point downwards as these processes are exothermic), which result in the aqueous ions.
And this is the Born-Haber cycle for if the enthalpy change of solution is exothermic.
The only difference with this is that the ΔHSOLN arrow points downwards as it's exothermic this time, and so the structure changes slightly.
One arrow being longer than last time does not mean that the enthalpy change for this process is now larger, I just had to draw it that way to make everything fit together.
(P.S. Apologies if the pictures are a little low in quality, I'll reupload them if you want me to).
To visualise this mathematically: ΔHLE > (ΔHHYD (M+) + ΔHHYD (X-)).
This is because the energy available to be released when hydrating the gaseous ions must be greater than the energy keeping the lattice together, so that the lattice can be 'torn apart', so to speak.
On the flip side, if not enough energy would be released when the gaseous ions are hydrated, the lattice won't be broken up, meaning that the substance will stay as a solid and so won't dissolve.
Take for instance NaCl and KCl: KCl is less soluble than NaCl at the same temperature.
This is because the sum of the hydration enthalpies of K+ and Cl- is relatively much less negative than its lattice enthalpy, compared to if you were to do this for NaCl.
What you can observe is that the difference between the sum of the hydration enthalpies and the lattice enthalpy is actually the enthalpy change of solution. So, as this gets more positive, solubility will decrease at a given temperature.
This is due to the Gibbs free energy change, :
must be negative for a reaction to be spontaneous at a temperature T, and in these reactions, is positive too.
So, becomes more positive as becomes more positive, meaning that the reaction becomes less and less spontaneous.