Verified Solution • 2025–26 Session moderate

QJEEM25Che26d: JEE Main 2025 Chemistry - Class 11 JEEM25Che26d Solution: The formation enthalpies, for and are 220.0 and , respectively, …

The formation enthalpies, ΔHf⊖\Delta \mathrm{H}_{\mathrm{f}}^{\ominus} for H(g)\mathrm{H}_{(\mathrm{g})} and O(g)\mathrm{O}_{(\mathrm{g})} are 220.0 and 250.0 kJ mol−1250.0 \mathrm{~kJ} \mathrm{~mol}^{-1} , respectively, at 298.15 K , and ΔHf⊖\Delta \mathrm{H}_{\mathrm{f}}^{\ominus} for H2O(g)\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} is −242.0 kJ mol−1-242.0 \mathrm{~kJ} \mathrm{~mol}^{-1} at the same temperature. The average bond enthalpy of the O−H\mathrm{O}-\mathrm{H} bond in water at 298.15 K is _______ kJ mol−1\mathrm{kJ} \mathrm{~mol}^{-1} (nearest integer).

Step-by-Step Verified Solution

12H2( g)→H( g);ΔfH(H(g))=220KJ/mol12O2( g)→O( g);ΔfH(O(g))=250KJ/mol\begin{array}{ll}\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{H}(\mathrm{~g}) \quad ;\Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{H}_{(\mathrm{g})}\right)=220 \mathrm{KJ} / \mathrm{mol} \\\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{O}(\mathrm{~g}) \quad ; \Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{O}_{(\mathrm{g})}\right)=250 \mathrm{KJ} / \mathrm{mol}\end{array}

JEE Main 2025 (Online) 28th January Morning Shift Chemistry - Thermodynamics Question 2 English Explanation

amp;ΔHf(H2O(l))=−242=440+250−2( B.E.(O−H))amp;BE(O−H)=466KJ/mol\begin{aligned}& \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\right)=-242=440+250-2(\mathrm{~B} . \mathrm{E} .(\mathrm{O}-\mathrm{H})) \\& \mathrm{BE}(\mathrm{O}-\mathrm{H})=466 \mathrm{KJ} / \mathrm{mol}\end{aligned}

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