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Calculate the lattice energy of NaCl usi...

Calculate the lattice energy of `NaCl` using Born-Haber cycle.

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`DeltaH_f` =heat of formation of sodium chloride =`-411.3 "kJ mol"^(-1)`
`DeltaH_1`=heat of sublimation of Na(g) =`108.7 "kJ mol"^(-1)`
`DeltaH_2` = ionisation energy of Na(g)=`495.0 "kJ mol"^(-1)`
`DeltaH_3` =dissociation energy of `Cl_(2(s))=244 "kJ mol"^(-1)`
`DeltaH_4`=Electron affinity of Cl(s) =`-349.0 "kJ mol"^(-1)`
U=lattice energy of NaCl
`DeltaH_f = DeltaH_1+DeltaH_2 +1//2DeltaH_3 +DeltaH_4 +U`
`therefore U=(DeltaH_f)-(DeltaH_1+DeltaH_2+1//2DeltaH_3 +DeltaH_4)`
`rArr` U=(-411.3 )-(108.7+495+122-349)
U=(-411.3)-(376.7)
`therefore U=-788 "kJ mol"^(-1)`
This negative sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gases ions `Na^+` and `Cl^-`
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