The first of lithium is 5.41 eV and electron affinity of Cl is `-3.61 eV`. Calculate `Delta H` in kJ `mol^(-1)` for the reaction: `Li_(g) + Cl_(g) rarr Li_(g)^(+) + Cl_(g)^(-)`.

Explanation

Given reaction is `Li_((g)) + Cl_((g)) rarr Li_((g))^(+) + Cl_((g))^(-)`
Formation of `Li_((g))^(+)`
`Li_((g)) rarr Li_((g))^(+) + e^(-) " " Delta H_(1)= 5.4eV`
Formation of `Cl_((g))^(-)`
`Cl_((g)) + e^(-) rarr Cl_((g))^(-) " " Delta H_(1)= -3.61eV`
Overall reaction is `Li_((g)) + Cl_((g)) rarr Li_((g))^(+) + Cl_((g))^(-)`
`Delta H= DeltaH_(1) + Delta H_(2)`
`=5.41 -3.61`
= 1.8eV
`=1.8 xx 9.65 xx 10^(4)` J/mole
`=17.37 xx 10^(4)` J/mole
=173.7 KJ/mole