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Thermodynamics question

2025 · 23 Jan · Shift 2 · Q22
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  5. /2025 · 23 Jan · Shift 2 · Q22

Thermodynamics question

2025 · 23 Jan · Shift 2 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
The bond dissociation enthalpy of X2ΔHbond ∘\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ}X2​ΔHbond ∘​ calculated from the given data is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1. (Nearest integer) M+X−(s)→M+(g)+X−(g)ΔHlattice ∘=800 kJ mol−1M( s)→M( g)ΔHsub ∘=100 kJ mol−1M( g)→M+(g)+e−(g)ΔHi∘=500 kJ mol−1X( g)+e−(g)→X−(g)ΔHeg∘=−300 kJ mol−1M( s)+12X2( g)→M+X−(s)ΔHf∘=−400 kJ mol−1\begin{aligned} & \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-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}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}​M+X−(s)→M+(g)+X−(g)ΔHlattice ∘​=800 kJ mol−1M( s)→M( g)ΔHsub ∘​=100 kJ mol−1​M( g)→M+(g)+e−(g)ΔHi∘​=500 kJ mol−1X( g)+e−(g)→X−(g)ΔHeg∘​=−300 kJ mol−1M( s)+21​X2​( g)→M+X−(s)ΔHf∘​=−400 kJ mol−1[Given : M+X−\mathrm{M}^{+} \mathrm{X}^{-}M+X− is a pure ionic compound and X forms a diatomic molecule X2\mathrm{X}_2X2​ in gaseous state]
Numerical answer
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