Chemistry - Thermochemistry Question with Solution | TestHub

The bond dissociation enthalpy of ${\mathrm{X}}_2\Delta {\mathrm{H}}_{\text{bond }}$ calculated from the given data is $\mathrm{kJ}{\mathrm{mol}}^{-1}$. (Nearest integer) $\begin{array}{rl}& {\mathrm{M}}^{+}{\mathrm{X}}^{-}(\mathrm{s})\to {\mathrm{M}}^{+}(\mathrm{g})+{\mathrm{X}}^{-}(\mathrm{g})\Delta {\mathrm{H}}_{\text{lattice }}^{\ast }=800\mathrm{kJ}{\mathrm{mol}}^{-1}\\ & \mathrm{M}(\mathrm{s})\to \mathrm{M}(\mathrm{g})\Delta {\mathrm{H}}_{\text{sub }}^{\circ }=100\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$ $\mathrm{M}(\mathrm{g})\to {\mathrm{M}}^{+}(\mathrm{g})+{\mathrm{e}}^{-}(\mathrm{g})\Delta {\mathrm{H}}_{\mathrm{i}}=500\mathrm{kJ}{\mathrm{mol}}^{-1}$ $\mathrm{X}(\mathrm{g})+{\mathrm{e}}^{-}(\mathrm{g})\to {\mathrm{X}}^{-}(\mathrm{g})\Delta {\mathrm{H}}_{\mathrm{eg}}^{\ast }=-300\mathrm{kJ}{\mathrm{mol}}^{-1}$ $\mathrm{M}(\mathrm{s})+\frac{1}{2}{\mathrm{X}}_2(\mathrm{g})\to {\mathrm{M}}^{+}{\mathrm{X}}^{-}(\mathrm{s})\Delta {\mathrm{H}}_f^{\circ }=-400\mathrm{kJ}{\mathrm{mol}}^{-1}$ [Given : ${\mathrm{M}}^{+}{\mathrm{X}}^{-}$is a pure ionic compound and X forms a diatomic molecule ${\mathrm{X}}_2$ in gaseous state]