The change in enthalpy $\left[\Delta \mathrm{H}\right]$ in ${\mathrm{kJmol}}^{-1}$ for the reaction is $\mathrm{Mg}+2 \mathrm{F}\to {\mathrm{MgF}}_2$

Given: $\mathrm{EA}$ of $\mathrm{F}=328 \mathrm{kJ} {\mathrm{mol}}^{-1},{\mathrm{IE}}_1$ of $\mathrm{Mg}=737 \mathrm{kJ} {\mathrm{mol}}^{-1},{\mathrm{IE}}_2$ of $\mathrm{Mg}=1451 \mathrm{kJ} {\mathrm{mol}}^{-1}$

The enthalpy change accompanying a reaction is called the reaction enthalpy.

Enthalpy of a reaction depends on the conditions under which a reaction is carried out.

The change in enthalpy of the reaction$\mathrm{Mg}+2 \mathrm{F}\to {\mathrm{MgF}}_2$ is calculated as given below

$$\begin{gathered} {\mathrm{Mg}}_{\left(\mathrm{g}\right)}\to {\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{+2}+2{\mathrm{e}}^{-}, ∆\mathrm{H}=∆{\mathrm{H}}_{\mathrm{I}.{\mathrm{E}}_1}+∆{\mathrm{H}}_{\mathrm{I}.{\mathrm{E}}_2} \\ 2\mathrm{F}+2{\mathrm{e}}^{-}\to 2{\mathrm{F}}^{-}, ∆\mathrm{H}=2∆{\mathrm{H}}_{\mathrm{E}.\mathrm{A}} \\ ------------------- \\ {\mathrm{Mg}}_{\left(\mathrm{g}\right)}+2\mathrm{F}\to {\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{+2}+2{\mathrm{F}}^{-}, ∆\mathrm{H}=∆{\mathrm{H}}_{\mathrm{I}.{\mathrm{E}}_1}+∆{\mathrm{H}}_{\mathrm{I}.{\mathrm{E}}_2}+2∆{\mathrm{H}}_{\mathrm{E}.\mathrm{A}} \\ ∆\mathrm{H}=737+1451+2(-328) \\ =1532\mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$