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Question

If X is the lattice entahlpy of CaCl2 ,find X. Given that the enthalpy of
(i) sublimation of Ca is 121kJmol1
(ii) dissociation of Cl2 to Cl is 242.8kJmol1;
(iii) ionization of Ca to Ca2+ is 2422kJmol1;
(iv) electron gain enthalpy for Cl to Cl is 355kJmol1;
(iv) ΔfH overall is 795kJmol1.

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Solution

Given:
ΔsubHCa=121kJmol1
ΔHClCl=242.8kJmol1
ΔfHCa2+=2422kJmol1
ΔegHCl=355kJmol1
ΔfHCaCl2=795kJmol1
ΔfHCaCl2=ΔsubHCa+ΔHClCl+ΔfHCa2++2×ΔegHCl+ΔlHCaCl2
or 795=121+242.8+2422(355×2)+ΔlHCaCl2
ΔlHCaCl2=2870.8kJ mol1

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