Question

The bond dissociation enthalpy of $\mathrm{X}_{2} \quad \Delta \mathrm{H}_{\text {bond }}^{\mathrm{o}}$ calculated from the given data is \_\_\_\_ $\mathrm{kJ} \mathrm{~mol}^{-1}$.
(Nearest integer)

$\mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}_{ \text {lattice }}=800 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{s}) \rightarrow \mathrm{M}(\mathrm{g}) \Delta \mathrm{H}^{\circ}_{ \text {sub }}=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})^{-}+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}_{ \mathrm{i}}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{X}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}^{\circ}_{ \text {eg }}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{M}(\mathrm{s})+\frac{1}{2} \mathrm{X}_{2}(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}^{\circ}_{ \mathrm{f}}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$

[Given : $\mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\mathrm{X}_{2}$ is gaseous state]

Accepted Answer

200

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Chemical Thermodynamics

Solution

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