Verified Solution • 2025–26 Session moderate QJEEM23Ch16d6: JEE Main 2023 Chemistry - Class 11 JEEM23Ch16d6 Solution: The enthalpy change for the conversion of to (aq) is ( )… The enthalpy change for the conversion of 12Cl2( g)\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})21Cl2( g) to Cl−\mathrm{Cl}^{-}Cl− (aq) is ( −-− ) ___________ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1 (Nearest integer) Given : ΔdisHCl2( g)θ⊖=240 kJ mol−1,ΔegHCl(g)⊖=−350 kJ mol−1\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}ΔdisHCl2( g)θ⊖=240 kJ mol−1,ΔegHCl(g)⊖=−350 kJ mol−1 , ΔhydHCl(g)−Θ=−380{\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}ΔhydHCl(g)−Θ=−380 kJ mol−1\mathrm{kJ~mol^{-1}}kJ mol−1 Step-by-Step Verified Solution amp;12Cl2( g)⟶Cl(aq) −ΔH=?amp;ΔHamp;=12ΔdissHCl2∘+ΔegΔHCl(g)∘+ΔhydHCl(g)−∘amp;=12×240+(−350)+(−380)amp;=−610 kJ mol−1\begin{aligned}& \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}_{\text {(aq) }}^{-} \quad \Delta \mathrm{H}=? \\\\& \begin{aligned}\Delta \mathrm{H} & =\frac{1}{2} \Delta_{\mathrm{diss}} \mathrm{H}_{\mathrm{Cl}_2}^{\circ}+\Delta_{\mathrm{eg}} \Delta \mathrm{H}_{\mathrm{Cl}(\mathrm{g})}^{\circ}+\Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}^{-}}^{\circ} \\\\& =\frac{1}{2} \times 240+(-350)+(-380) \\\\& =-610 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}\end{aligned}amp;21Cl2( g)⟶Cl(aq) −ΔH=?amp;ΔHamp;=21ΔdissHCl2∘+ΔegΔHCl(g)∘+ΔhydHCl(g)−∘amp;=21×240+(−350)+(−380)amp;=−610 kJ mol−1 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring