Lattice and hydration energies

Question

Consider the dissolving of calcium chloride crystals in water:

CaCl2(s) → Ca(aq) 2+ + 2Cl(aq) −

∆H= –75.6 kJ

(i) Compare the lattice energy and hydration energy for calcium chloride.

(ii) If 50.0 g of solid calcium chloride is dissolved in 250 mL of water in a well insulated calorimeter, what temperature change would theoretically occur?

Answer 1

i) Google the energies and compare them. Note that lattice energy is crystal to ionized gases whereas the hydration energy is solid to ions in water. (Clearly, the lattice energy is much greater (Think melting and sublimating vs dissolving a salt))

ii) You can model the heating of the water mixed with ions as the energy released from 50 g of CaCl₂ hydrating and going into the mixture (assume c_w applies):

-nΔH_hyd = m_mixc_wΔT

Now n = m_CaCl2/M_CaCl2 = 50g/110.98 g/mole

Solve for ΔT = -nΔH/(m_mixc_w) where m_mix = 50g + 250g of water and c_w = 4.184 J/g°C

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