Answer the following question (Based on `EA, Delta_(eg)H^(ɵ)` and `IE`).
`IE_(1)` of `Li` is `5.4 eV "atom"^(-1)` and the `EA` of `Cl` is `3.6 eV "atom"^(-1)`. Calculate `Delta_( r)H^(ɵ)` in `"kcal mol"^(-1)` and `kJ mol^(-1)` for the reaction
`Li_((g))+Cl_((g)) rarr Li_((g))^(o+)+Cl_((g))^(ɵ)`
formed at such a low pressure that resulting ions do not combine with each other.

Explanation

To solve the problem, we need to calculate the standard enthalpy change (ΔH) for the reaction:

\[ \text{Li}_{(g)} + \text{Cl}_{(g)} \rightarrow \text{Li}^{+}_{(g)} + \text{Cl}^{-}_{(g)} \]

We are given the following data:
- Ionization Energy (IE₁) of Lithium (Li) = 5.4 eV
- Electron Affinity (EA) of Chlorine (Cl) = 3.6 eV

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