For the reaction `K(g)+F(g) +K^(o+)+F^(Theta)` (separated ions `DeltaH = 19 kcal mol^(-1)`), if the ionisation potential of `K` and the electron affinity of `F^(Theta)` have a geometric means of `3.88eV` and `IP gt EA`, calculate the values fo ionisation potential and electron affinity.

Explanation

`K(g) rarr K^(o+) (g) +e^(-) …(i)IE =x`
`F(g) +e^(-) rarr F^(Theta)(g) …(ii) EA =- y`
Add equations (i) and (ii),
`K(g) +F(g) rarr F^(o+)(g) +F^(Theta)(g)`
`DeltaH = 19 kcal mol^(-1) =0.82 eV atom^(-1)`
`[{:(19kcal =19xx10^(3)cal =19 xx10^(3)xx4.18J),(=(19xx10^(3)xx4.18)/(1.6xx10^(-19))eVmol^(-1)),(=(19xx10xx4.18)/(1.6xx10^(-19))xx(1)/(6.23xx10^(23)atm)eVat om^(-1)),(=0.82eV):}]`
`:. x-y = 19 kcal mol^(-1) = 0.82 eV` ...(ii)
`sqrt(xy) = (3.88) eV` (given)
`xy = (3.88)^(2)`
`xy = 15.0544`
`x = (15.0544)/(y)`
Substituting the value of `x` in equation (iii), we get
`(15.0544)/(y) -y = 19`
Solving, we get `y = 3.48 eV`
`x+y = 7.48 eV`
`x - y = 0.82 eV`
`IE = x = 4.313`
`EA = y = 3.48 eV`