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for the entire process -1600 kJ/mole.If IP of ‘X' is 1260 kJ/mole then Electron gain enthalpy of ‘X’ would be
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a
+2860 kJ/mole
b
+340 kJ/mole
c
–340kJ/mole
d
–2860 kJ/mole
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Correct option is C
Given Xg → Xg+, ΔH = +1260kJ∴Xg+ → Xg ΔH = −1260kJ/moleXg+ →−1260Xg →ΔH2 Xg−–1260+∆H2=–1600 ΔH2 = −340kJ/moleTalk to our academic expert!
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