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QJEEM25Chfd9d: JEE Main 2025 Chemistry - Class 11 JEEM25Chfd9d Solution: The bond dissociation enthalpy of calculated from the given…

The bond dissociation enthalpy of X2ΔHbond ∘\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ} calculated from the given data is ___________ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} . (Nearest integer)

amp;M+X−(s)→M+(g)+X−(g)ΔHlattice ∘=800 kJ mol−1amp;M( s)→M( g)ΔHsub ∘=100 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}

M( g)→M+(g)+e−(g)ΔHi∘=500 kJ mol−1\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}

X( g)+e−(g)→X−(g)ΔHeg∘=−300 kJ 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}

M( s)+12X2( g)→M+X−(s)ΔHf∘=−400 kJ 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}

[Given : M+X−\mathrm{M}^{+} \mathrm{X}^{-} is a pure ionic compound and X forms a diatomic molecule X2\mathrm{X}_2 in gaseous state]

Step-by-Step Verified Solution

JEE Main 2025 (Online) 23rd January Evening Shift Chemistry - Thermodynamics Question 7 English Explanation

amp;∴ΔHf(MX)amp;=ΔHsub (M)+ I.E. (M)+12[ B.E. (X−X)]amp;+EG(X)+ L.E. (MX)amp;−400=(100)+(500)+12( B.E. )+(−300)+(−800)amp;∴ B.E. =200 kJ mole−1 \begin{aligned}& \begin{aligned}\therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX}) & =\Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] \\& +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX})\end{aligned} \\& -400=(100)+(500)+\frac{1}{2}(\text { B.E. })+(-300)+(-800)\\& \therefore \text { B.E. }=200 \mathrm{~kJ} \mathrm{~mole}^{-1}\end{aligned} ~ \begin{aligned}\end{aligned}

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