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Question

If X is the lattice enthalpy of CaCl2, find X. Given that the enthalpy of
(i) Sublimation of Ca(s) is 121 kJmol1
(ii) Dissociation of Cl2(g) to 2Cl(g) is 242.8 kJmol1
(iii) Ionization of Ca(g) to Ca2+(g) is 2422 kJmol1
(iv) Electron gain enthalpy for Cl(g) to Cl1(g) is 355 kJmol1
(v) fH overall is 795 kJmol1

A
2870.8 kJ/mol
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B
-2372.6 kJ/mol
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C
-2870.8 kJ/mol
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D
2372.6 kJ/mol
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Solution

The correct option is C -2870.8 kJ/mol
Given :
subHCa=121 kJmol1
dissHCl2=242.8 kJmol1
IEHCa=2422 kJmol1
EGHCl=355 kJmol1
fHCaCl2=795 kJmol1
lHCaCl2= Lattice energy

using Hess's law :

fHCaCl2=subHCa+dissHClCl+IEHCa+2×EGHCl+lHCaCl2
795=121+242.8+2422(355×2)+lHCaCl2
Thus,
lHCaCl2=2870.8 kJ/mol

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