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Calculate the lattice energy of NaCl using Born-Haber cycle. |
Answer» Solution : `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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