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Calculate the lattice energy of formatio...

Calculate the lattice energy of formation of NaCl from 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)`

Text Solution

Verified by Experts

`Delta^(@) H_(1)` = heat of sublimation of `Na_((s))`
`= 108 kJ mol^(-1)`
`Delta^(@) H_(2)` jonisation energy of `Na_((g))`
` = 495 . 0 kJ mol^(-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)`.
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