Xg+ →Xg → Xg−;ΔH for the entire process -1600 kJ/mole.If IP of ‘X' is 1260 kJ/mole then Electron gain enthalpy of ‘X’ would be
+2860 kJ/mole
+340 kJ/mole
–340kJ/mole
–2860 kJ/mole
Given Xg → Xg+, ΔH = +1260kJ
∴Xg+ → Xg ΔH = −1260kJ/moleXg+ →−1260Xg →ΔH2 Xg−–1260+∆H2=–1600 ΔH2 = −340kJ/mole