1.

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

Answer» `Delta H//`molecule of `Li^+` and `Cl^- =IF_1 +` Electron gain enthaply .
` = 5.41 - 3. 61 =1. 80 eV`
` :. Li rarr Li^+ +e , IP_1 +ve`
and ` Cl + e rarr Cl^(-1) ,` Electron gain enthalpy `= -ve`
` :. Delta H//mol=1.8 xx 6. 023xx 10^(23) eV`
`=1.8xx 6. 023 xx 10^(233) xx 1.602 xx 10^(19) J`
` =1. 8 xx 6.023 xx 10^(23) xx 1. 602 xx 10^(19) xx 10^(-3) kJ`
` = 173. 7 kJ`.


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