Calculate the lattice energy of formation of NaClfrom the following data :Na_((s)) +1//2Cl_(2(g)) to NaCl_((s)) Delta H_(f) -411.3 KJ.mol^(-1)Heat of sublimation of Na_((s))= 108 . 7 kJ mol ^(-1)Ionisation energy of Na_((g)) = 49.5.0 kJ mol^(-1)Dissociation energy of Cl_(2(g)) = 244 kJ mol^(-1)Electron affinity of Cl_((g)) = - 349 . 0 kJ mol^(-1)

Calculate the lattice energy of formation of NaClfrom the following da

1.	Calculate the lattice energy of formation of NaClfrom the following data :Na_((s)) +1//2Cl_(2(g)) to NaCl_((s)) Delta H_(f) -411.3 KJ.mol^(-1)Heat of sublimation of Na_((s))= 108 . 7 kJ mol ^(-1)Ionisation energy of Na_((g)) = 49.5.0 kJ mol^(-1)Dissociation energy of Cl_(2(g)) = 244 kJ mol^(-1)Electron affinity of Cl_((g)) = - 349 . 0 kJ mol^(-1)
Answer» Solution :`Delta^(@) H_(1)` = HEAT of sublimation of `Na_((s))` `= 108 kJ mol^(-1)` `Delta^(@) H_(2)`JONISATION energy of `Na_((g))` ` = 495 . 0 kJmol^(-1)` `Delta^(@)H_(3)`dissociation energy of `Cl_(2(g))` ` 244kJ mol^(-1)` `Delta^(@)H_(4)= `Electron affinity of ` Cl_((g))` `= - 349 . 0 kJ mol^(-1)` U = Lattice energy of NaCl `Delta^(@) H_(f) = Delta^(@) H_(1) + Delta^(@)H_(2) + 1//2 Delta^(@)H_(3) + Delta^(@) H_(4) + Delta^(@) H_(5)` `:. Delta^(@)H_(5)= (Delta^(@)H_(f))-(Delta^(@)H_(1)+ Delta^(@)H_(2)+1//2Delta^(@)H_(3)+Delta^(@)H_(4))` `rArr Delta^(@) H_(5) = (-411.3) - (108 .7 + 495.0 + 122-349)` `Delta^(@) H_(5) = (-411.3)-(376.7)` `:. Delta^(@)H_(5) = - 788 kJ mol^(-1)`.