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I am learning about lattice enthalpies, and my textbook talks about determining the lattice enthalpy from the standard change of enthalpy of solution and the change of enthalpy of hydration of the ions. It says that the changes in enthalpy of solution and hydration can be determined experimentally but that the lattice enthalpy cannot be.

However, I am struggling to think of how one can measure the enthalpy change when one mole of gaseous ions is dissolved in water under standard conditions. I thought one would encounter the same problem as with the lattice enthalpy, and this involves one mole of an ionic lattice being formed from its gaseous ions under standard condition.

I thought that the problem with this was getting the exact amounts of gaseous ions under standard conditions.

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  • $\begingroup$ You mean the experimental and\or practical way? $\endgroup$
    – M.A.R.
    Feb 17, 2015 at 15:05
  • $\begingroup$ Experimental way $\endgroup$
    – Meep
    Feb 17, 2015 at 15:49

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Usually, one would measure these changes in enthalpy calorimetricly — or at least it is a nice student excersize in physical chemistry lab courses. I was once tasked to determine the solvation enthalpy of copper(II) sulphate pentahydrate.

The calorimeter consisted of an isolated thermos in which we filled a defined amount of water and a concealed inner container in which we filled a sample of the salt; the apparatus contained a stirrer inside the isolated area. We then let the water temperature equilibrate itself, opened in the inner container from the outside (mechanically) and measured the change in temperature. From $\Delta T$ we could calculate $\Delta q$ and from that $\Delta H_\mathrm{solv}$. All corresponded to the equation

$$\ce{CuSO4 . 5 H2O + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

One could then perform a second experiment using anhydrous copper(II) sulphate. According to Hess’ law, the following is true:

$$\ce{CuSO4 + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4})$$ $$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O} ~~~~~~ \Delta H_\mathrm{hyd}(\ce{CuSO4})$$ $$\Delta H_\mathrm{solv}(\ce{CuSO4}) = \Delta H_\mathrm{hyd}(\ce{CuSO4}) + \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

The two $\Delta H_\mathrm{solv}$ can be determined experimentally, thus the $\Delta H_\mathrm{hyd}$ can be calculated from these two values.

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