Problems

Set up a thermodynamic cycle for determining the enthalpy of hydration of $C{a}^{2+}$ ions using the following data. Some people prefer approaching this type of problem by treating chemical reactions algebraically. Some people prefer approaching this type of problem graphically. Do what works for you. Regardless, figure out the puzzle pieces BEFORE doing any arithmetic.

i$.$ Enthalpy of sublimation of $Ca(s)=178\mathrm{.}2k\frac{J}{m}ol$

ii$.$ First ionization enthalpy of $Ca(g)=589\mathrm{.}7k\frac{J}{m}ol$

iii. Second ionization enthalpy of $Ca(g)=1145k\frac{J}{m}ol$

iv$.$ Enthalpy of vaporization of bromine $=30\mathrm{.}91k\frac{J}{m}ol$

v$.$ Dissociation enthalpy of $B{r}_2(g)=192\mathrm{.}9k\frac{J}{m}ol$

vi$.$ Electron gain enthalpy of $Br(g)=-331\mathrm{.}0k\frac{J}{m}ol$

vii. Enthalpy of solution of $CaB{r}_2(s)=-103\mathrm{.}1k\frac{J}{m}ol$

viii. Enthalpy of hydration of $B{r}^{-}(g)=-289k\frac{J}{m}ol$

ix$.$ Enthalpy of formation of $CaB{r}_2(s)=-675k\frac{J}{m}ol$

Once you have the enthalpy of hydration value for $C{a}^{2+}$ :

a$.$ Comment on the sign of your answer. What does it mean? Would the opposite sign make sense based on what you know about how molecules$/$ions interact?

b$.$ Comment on the magnitude of your answer compared to that of $Br.$ What about these ions makes the values different? Recall that Coulomb's law $)=(k{q}_1\frac{q_2}{r^2}$