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Calcium chloride on icy roads

Ion charge and radius, dissolving energetics, Hess's law with hydrates, entropy

Road crews in Central Otago spread anhydrous calcium chloride, CaClX2\ce{CaCl2}, on icy roads; ordinary NaCl is less effective in very cold conditions. ΔsolH\Delta_{sol}H (kJ mol−1^{-1}): CaClX2(s)\ce{CaCl2(s)} −82.8; NaCl(s) +3.9; CaClX2 ⋅ 6 HX2O(s)\ce{CaCl2.6H2O(s)} +15.0. M(CaClX2\ce{CaCl2}) = 111.1 g mol−1^{-1}. c=4.18c = 4.18 J g−1^{-1} °C−1^{-1}. Ionic radii (pm): CaX2+\ce{Ca^2+} 100, NaX+\ce{Na+} 102, KX+\ce{K+} 138, ClX−\ce{Cl-} 181.
Facts you may need
  • electron configurations of ions
  • ion–dipole attractions
  • q = mcΔT
  • Hess's law
  • entropy and spontaneity

(a)

Ion **X**2+^{2+} and ion **Y**−^- both have the electron configuration 1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6. Identify **X** and **Y**, justify your choice (including why no other 2+ and 1− ions fit), and explain the difference in their radii (100 pm and 181 pm).

(b)

Discuss, in terms of the attractions broken and formed, why dissolving CaClX2\ce{CaCl2} is strongly exothermic while dissolving NaCl is slightly endothermic.

(c)

50.0 g of anhydrous CaClX2\ce{CaCl2} is dissolved in 200 g of water. Calculate the temperature rise, assuming the solution's mass is 250 g, its specific heat capacity is that of water, and no heat is lost.

(d)

Explain why NaCl still dissolves, even though dissolving it is endothermic.

(e)

Calculate ΔH\Delta H for CaClX2(s)+6 HX2O(l)→CaClX2 ⋅ 6 HX2O(s)\ce{CaCl2(s) + 6H2O(l) -> CaCl2.6H2O(s)}. Use this to explain why the hydrate dissolves endothermically, and why crews store CaClX2\ce{CaCl2} in sealed bags.