For the gaseous reaction
`K+F rarrK^(o+)+F^(ɵ)`
`Delta H = 19 kcal mol^(-1)` under the condition when cations and anions are prevented by electrostatic separation from combining with each other. The `IE_(1)` of `K` is `4.3 eV`. Calculate `Delta_("eg")H^(ɵ)` of `F`.

Explanation

`{:(K_((g))rarrK_((g))^(o+)+e^(-),,,lE_(1)=4.3 eV "atom"^(-1),....(i)),(e^(-)+F_((g))rarrF_((g))^(ɵ),,,"Let" Delta_("eg")H_(1)^(ɵ)= y eV "atom"^(-1),....(ii)):}`
Adding Eqs. (i) and (ii), we get
`K_((g))+ F_((g)) rarr K_((g))^(o+)+F_((g))^(ɵ)`
`:. lE_(1)+Delta_("eg")H_(1)^(ɵ) = 19 kcal mol^(-1) = 0.82 eV "atom"^(-1)` ....(iii)
Note: Use direct relation: `1 eV = 23.06 kcal mol^(-1)`
`:. 19 kcal mol^(-1) = (19.0)/(23.06) = 0.82 eV "atom"^(-1)`
Alternatively:
convert `19 kcal mol^(-1)` in `eV "atom"^(-1)` as follows:
`(1 eV = 1.6 xx 10^(-10)J)`
`19 kcal mol^(-1) = 19 xx 10^(3) cal mol^(-1)`
`= 19 xx 10^(3) xx 4.18 J mol^(-1)`
`= (19 xx 10^(3) xx 4.18)/(1.6 xx 10^(-10))eV mol^(-1)`
`= (19 xx 10^(3) xx 4.18)/(1.6 xx 10^(-10)) xx (1)/(6.023 xx 10^(25))eV "atom"^(-1)`
`= 0.82 eV "atom"^(-1)`
Substituting the value of `lE_(1)` in Eq. (iii), we get
`4.3 eV "atom"^(-1)+Delta_("eg")H_(1)^(ɵ) = 0.82 eV "atom"^(-1)`
`:. Delta_("eg")H_(1)^(ɵ) = (0.82 - 4.3) = -3.48 eV "atom"^(-1)`