# Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide.?

CaCl2 lattice energy is -2247 kj/mol delta H of solution is -46 kj/mol

CaI2 lattice energy is -2059 kj/mol delta H of solution is -104 kj/mol

The enthalpy of energy should just be in kJ.

which ion Cl ‾ or I ‾, is more strongly attracted to water?

Please explain how you figured it out because I’m lost. Thank you

• basically add up the two numbers

hydration is 2247 + 46

as hydration is able in this case to overcome the lattice energy AND then release some energy to spare.

Cl- is more attrracted as the total number above is greater

• 34. (a) Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium

iodide. Lattice Energy ∆Hsoln

CaCl2(s) –2247 kJ/mol –46 kJ/mol

CaI2(s) –2059 kJ/mol –104 kJ/mol

Using the equations for lattice enthalpy and the enthalpy of solution for each ionic solid, we can calculate the

enthalpy of hydration of each compound.

CaCl2(s) → Ca2+(g) + 2Cl–(g) ∆H = –LE = –(–2247) kJ/mol = 2247 kJ/mol

Ca2+(g) + 2Cl–(g) → Ca2+(aq) + 2Cl–(aq) ∆Hhyd = ? kJ/mol

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

CaCl2(s) → Ca2+(aq) + 2Cl–(aq) ∆Hsoln = –46 kJ/mol

∆Hhyd (CaCl2) = –2293 kJ/mol

CaI2(s) → Ca2+(g)I–(g) ∆H = –LE = –(–2059) kJ/mol = 2059 kJ/mol

Ca2+(g) + 2I–(g) → Ca2+(aq) + 2I–(aq) ∆Hhyd = ? kJ/mol

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

CaI2(s) → Ca2+(aq) + 2I–(aq) ∆Hsoln = –104 kJ/mol

∆Hhyd (CaI2) = –2163 kJ/mol

(b) Based on your answers to part a, which ion, Cl– or I–, is more strongly attracted to water?

The enthalpy of hydration of calcium chloride is more exothermic than that of calcium iodide. Since these

compounds contain a common cation Ca2+, this difference arises from the halide ion. Since more exothermic

energy is associated with the hydration of the chloride ion, it is more strongly attracted to water.

Source(s): http://personal.stevens.edu/~mtoledo/Ch11-HW.pdf