Find the heat that of sublimation of sodium metal from the following data. Ionisation enegry of sodium `= 502.1`, bond enegry of chlorine `=242.8`, heat of formation of `NaCI = - 411.1`, lattice energy of `NaCI =- 778.0`, and electorn affinity of chlorine is `-365.3`. All units are in `kJ mol^(-1)`.

Explanation

`{:(Na(s) rarr Na(g),, DeltaH = ?),(Na(g) rarr Na^(o+)(g)+e^(-),, DeltaH_(1) = 502.1),(.(1)/(2)[CI(g) rarr 2CI(g)] ,,DeltaH_(2) = 242.8 xx (1)/(2)),( Na(s) +(1)/(2)CI(2)rarr NaCI(s),, DeltaH_(3) = 242.8 xx (1)/(2)),(Na(s)+(1)/(2)CI(g)rarrNaCI(s),,DeltaH_(4)=-411.1),(Na^(o+)(g)+CI^(Theta)rarrCa(g),,DeltaH_(5)=-365.3),("Rewritting":,,),(Na^(o+)(g)+e^(-)rarrNaCI(s),,DeltaH_(1)=-502.1),((1)/(2)[CI_(2)(g)rarr2CI(g)],,DeltaH_(2)=-(242.8)/(2)),(Na(s)+(1)/(2)CI_(2)(g)rarrNaCI(g),,DeltaH_(3)=-411.1),(NaCI(g)rarrna^(o+)(g)+CI^(Theta)(g),,DeltaH_(4)=778.0),(CI^(Theta)(g)rarrCI(g)+e^(-),,DeltaH_(5)=365.3),(ulbar(Na(s)rarrNa(g)),,):}`
or
`DeltaH =- 502.1 - (242.8)/(2) - 411.1 +778 +365.3`
`=- 502.1 -121.4 - 411.1 +1143.3`
`=- 1034.6 +1143.3 = 108.8 kJ mol^(-1)`