Does the enthalpy of solution formula (LE + Hyd) change depending on the question? Hey guys. I have been working on enthalpies and literally every video and website I’ve seen,
the formula for calculating enthalpy of solution = lattice enthalpy of dissociation + enthalpy of hydration, using which, I got a few questions correct. However, this particular question (picture added), I have used the same formula and gotten the answer -5273 $\pu{kJ mol-1}$.
The mark scene for this question states that the answer is -51 $\pu{kJ mol-1}$.
I know what they have done, I know that they have used        -(-2662) because the answer sheet has the working. I don’t understand why they gave done this Why have they done so?????
Thank you!!
Apparently your problem is related to the improper use of negative signs. Let's solve it differently.
Suppose you start from the separated ions $\ce{Ca^{2+}}$ and $\ce{2 F^-}$ in the vacuum. This state has the energy zero by definition.
Now suppose you join the ions to produce $1$ mole of solid $\ce{CaF2}$.  The process is exothermic, and the final state $\ce{CaF2(s)}$ is at the energy level $-2611 kJ/mol.$
Now suppose you are doing another transformation with the same ions. You put them directly into water, so that the ions become hydrated in solution. You obtain a solution of $\ce{CaF2}$ in water, or a solution of $\ce{Ca^{2+}(aq)}$ and $\ce{2 F^-(aq)}$. This transformation is also exothermic, and at the end $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ are at the energy level $- 2662 kJ/mol$.
The final levels after these two transformations are $-2611 kJ/mol$ for the final solid, and $-2662 kJ/mol$ for the solution. These energy levels are not very different from one another.
So the passage from the solid state $\ce{CaF2}$ to the solution $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ is slightly exothermic, with $\Delta H = -51 kJ/mol$