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Question

The amount of energy when a million atoms of iodine are completely converted into I ions in the vapor state according to the equation,
I(g)+eI(g) is 4.9 x 1013 J. What would be the electron gain enthalpy of iodine in terms of kJ mol1 and eV per atom?

A
295, 3.06
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B
295, 3.06
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C
439, 5.09
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D
356, 7.08
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Solution

The correct option is B 295, 3.06
The electron gain enthalpy (Eg) of an element is defined as the energy released when an atom in gaseous state gains an electron.
I(g)+eI(g)
In case of iodine
Eg for 1 million atoms (106atoms)
Eg=4.9×1013J for 106atoms
Eg=4.9×1013106J/atom
For one mol of I atoms, amount of energy released in the formation of I ions by gaining 1 mol of electrons will be:
Eg=4.9×1013106×NA
where NA=6.023×1023atoms/mol
Eg=4.9×1013106×6.023×1023J/mol
=29.51×104J/atom=295.1kJ/mol
Since halogens have negative Eg
Thus, electron gain enthalpy for one mol iodine =295kJmol1
In eV units we have the relation:
1eV=1.6021×1019J=1.6021×1022kJ
295kJ/mol=2951.6021×1022eV/mol=184.13×1022eV/mol
For one atom, Eg in eV is:= 184.13×1022NAeV
=184.13×10226.023×1023eV
=3.06eV/atom

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