Find the electron affinity of chlorine from the following data. Enthalpy of formation of `LiCI` is `-97.5 kcal mol^(-1)`, lattice energy of `LiCI =- 197.7 kcal mol^(-1)`. Dissociation energy of chlirine is `57.6 kcal mol^(-1)`, sublimation enthalpy of lithium `=+ 38.3 kcal mol^(-1)`, ionisation energy of lithium `=123.8 kcal mol^(-1)`.

Explanation

`CI +e^(-) rarr CI^(Theta) DeltaH =?`
Given: `Li(s) +(1)/(2)CI_(2)(g) rarr LiCI(s), DeltaH_(1) =- 97.5 kcal mol^(-1)`
`Li^(o+) +CI^(Theta) (g)rarr LiCI(s), DeltaH_(2) =- 197.7 kcal mol^(-1)`
`CI(g) rarr 2CI(g), DeltaH_(3) = 57.6 kcal mol^(-1)`
`Li(s) rarr Li(g), DeltaH_(4) = 38.3 kcal mol^(-1)`
`Li(g) rarr Li^(o+)(g) +e^(-), DeltaH_(5) = 123.8 kcal mol^(-1)`
Rewritting equations:
`LI(s) +(1)/(2)CI(2)(g) rarr LiCI(s), DeltaH_(1) = - 97.5 kcal mol^(-1)`
`LICI(s) rarr Li^(o+)(g) +CI^(Theta)(g), DeltaH_(2) = 197.7 kcal mol^(-1)`
`(1)/(2)[2CI_(2)(g) rarr CI_(2)(g)], DeltaH_(3) =(1)/(2)xx-57.6 kcal mol^(-1)`
`Li(g) rarr Li(s), DeltaH_(4) =- 38.3 kcal mol^(-1)`
`Li^(o+)(g) +e^(-) rarr Li(g), DeltaH_(5) =- 123.8 kcal mol^(-1)`
`{:ulbar(CI(g)+e^(-)CI^(Theta)(g)):}`
`DeltaH = - 97.5 +197.7 -(1)/(5) xx 57.6 - 38.3 - 123.8`
`=- 90.7 kcal mol^(-1)`
or
Operating:
`DeltaH = DeltaH_(1) - DeltaH_(2) - (1)/(2)DeltaH_(3)-DeltaH_(4)-DeltaH_(5)`
`=97.5 -(-197.7) -(1)/(2)(57.6)-38.3 -123.8`
`=- 90.7 kcal mol^(-1)`