The amount of energy released when one million of atoms of iodine in vapour state are converted to l'ions is 4.9x10-13 J according to the reaction: I(g) + F(g) [1] Express the electron gain enthalpy of iodine in terms of kJ mol-1 and eV per atom.
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Answers
Explanation:
The electron gain enthalpy (E g ) of an element is defined as the energy released when an atom in gaseous state gains an electron.
I(g)+e
−
→I
−
(g)
In case of iodine
E
g
for 1 million atoms (10
6
atoms)
E
g
=4.9×10
−13
J for 10
6
atoms
E
g
=
10
6
4.9×10
−13
J/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:
E
g
=
10
6
4.9×10
−13
×N
A
where N
A
=6.023×10
23
atoms/mol
E
g
=
10
6
4.9×10
−13
×6.023×10
23
J/mol
=29.51×10
4
J/atom=295.1kJ/mol
Since halogens have negative E
g
Thus, electron gain enthalpy for one mol iodine =−295kJmol
−1
In eV units we have the relation:
1eV=1.6021×10
−19
J=1.6021×10
−22
kJ
−295kJ/mol=
1.6021×10
−22
−295
eV/mol=184.13×10
22
eV/mol
For one atom, E
g
in eV is:=
N
A
−184.13×10
22
eV
=
6.023×10
23
−184.13×10
22
eV
=−3.06eV/atom
