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)^(-)`.

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)^(-) .

The first ionisation potential of Li is 5.4 e V and the electron affinity of Cl is 3.6 eV . Calculate Delta H in kcal mol^(-1) for the reaction Li (g) + Cl (g) to Li^(+) + Cl^(-) Carried out at such low pressures that resulting ions do not combine with each other.

If the first ionization enthalpy of Li is 5.4 eV and the elec- tron gain enthalpy of chlorine is 3.6 eV then the Delta H in kcal/mole for the reaction will be Li(g)+Cl(g) to Li^(+) (g) Cl^(-) (g) (Presume that the pressure is so low that the ions do not combine with each other)

If the first ionization enthalpy of Li is 5.4 eV and the elec- tron gain enthalpy of chlorine is 3.6 eV then the Delta H in kcal/mole for the reaction will be Li(g)+Cl(g) to Li^(+) (g) Cl^(-) (g) (Presume that the pressure is so low that the ions do not combine with each other)