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JEE Advanced 2018 Paper 2 · Chemistry · Chemical Thermodynamics · hard

Use the following data (all in kJ mol^-1): enthalpy of sublimation of Mg =147.7; first ionisation enthalpy of Mg =737.7; second ionisation enthalpy…

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Use the following data (all in kJ mol−1\mathrm{kJ\,mol^{-1}}): enthalpy of sublimation of Mg =147.7=147.7; first ionisation enthalpy of Mg =737.7=737.7; second ionisation enthalpy of Mg =1450.7=1450.7; enthalpy of atomisation of Cl\mathrm{Cl} (per mole of Cl atoms) =121.5=121.5; electron gain enthalpy of Cl =−349=-349; ΔfH∘(MgCl2, s)=−641.3\Delta_fH^\circ(\mathrm{MgCl_2,\,s})=-641.3. The hypothetical compound MgCl\mathrm{MgCl} (rock-salt type) is estimated to have a lattice enthalpy of −753-753. The magnitude of the enthalpy change of the reaction 2MgCl(s)→Mg(s)+MgCl2(s)\mathrm{2MgCl(s)\rightarrow Mg(s)+MgCl_2(s)} is ______ kJ\mathrm{kJ} per mole of MgCl2\mathrm{MgCl_2} formed. Give the answer correct to two decimal places.
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Answer: 451.10

Build the Born-Haber cycle for hypothetical MgCl\mathrm{MgCl}: ΔfH∘=ΔsubH(Mg)+IE1+ΔatH(Cl)+ΔegH(Cl)+ΔlatticeH\Delta_fH^\circ=\Delta_{sub}H(\mathrm{Mg})+\mathrm{IE_1}+\Delta_{at}H(\mathrm{Cl})+\Delta_{eg}H(\mathrm{Cl})+\Delta_{lattice}H. Substituting, ΔfH∘(MgCl)=147.7+737.7+121.5−349−753=−95.1 kJ mol−1\Delta_fH^\circ(\mathrm{MgCl})=147.7+737.7+121.5-349-753=-95.1\ \mathrm{kJ\,mol^{-1}}, so MgCl is stable with respect to its elements. For the disproportionation, ΔH=ΔfH∘(MgCl2)+ΔfH∘(Mg)−2ΔfH∘(MgCl)=−641.3+0−2(−95.1)=−451.1 kJ\Delta H=\Delta_fH^\circ(\mathrm{MgCl_2})+\Delta_fH^\circ(\mathrm{Mg})-2\Delta_fH^\circ(\mathrm{MgCl})=-641.3+0-2(-95.1)=-451.1\ \mathrm{kJ}. The second ionisation of Mg, although it costs 1450.7 kJ, is more than repaid by the much larger lattice energy of MgCl2\mathrm{MgCl_2} (more highly charged Mg2+\mathrm{Mg^{2+}} ion), so disproportionation is strongly exothermic. The magnitude is 451.10 kJ451.10\ \mathrm{kJ}.
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