Question

The amount of energy when a million atoms of iodine are completely converted not ${{{I}}^{{ - }}}$ ions in the vapor state are according to the equation:
${{{I}}_{{{(g)}}}}{{ + }}{{{e}}^{{ - }}} \to {{{I}}^{{ - }}}_{{{(g)}}}$ is ${{4}}{{.9 \times 1}}{{{0}}^{{{ - 13}}}}{{J}}$ . What would be the electron gain enthalpy of iodine in terms of KJ/mol and eV per atom?
(A) ${{295,3}}{{.06}}$
(B) ${{ - 295, - 3}}{{.06}}$
(C) ${{439,5}}{{.09}}$
(D) ${{ - 356, - 7}}{{.08}}$

Answer 1

First we will know about electron gain enthalpy and how it is calculated for a mole and for many moles in a system. Here the enthalpy is given for million atoms so we need to convert it in moles to know the enthalpy in above units.

Complete step by step answer:
When a neutral isolated gaseous atom accepts an extra electron to convert into an anion or negatively charged atom a certain amount of energy is released. This energy is called electron gain enthalpy, ${{{E}}_{{g}}}$.
For the equation above in case of iodine, ${{1}}$ million is equal to ${10^6}$ atoms
${{{E}}_{{g}}}{{ = 4}}{{.9 \times 1}}{{{0}}^{{{ - 13}}}}{{J}}$ for ${10^6}$ atoms
So for one atom, ${{{E}}_{{g}}}{{ = }}\dfrac{{{{4}}{{.9 \times 1}}{{{0}}^{{{ - 13}}}}}}{{{{1}}{{{0}}^{{6}}}}}{{ J ato}}{{{m}}^{{{ - 1}}}}$
When an iodine atom gains electrons, it forms anions. So for one mole of iodine atoms, the amount of energy released is given below. We also know that one mole has Avogadro number of atoms, ${{{N}}_{{A}}}$.
${{{E}}_{{g}}}{{ = }}\dfrac{{{{4}}{{.9 \times 1}}{{{0}}^{{{ - 13}}}}}}{{{{1}}{{{0}}^{{6}}}}}{{ \times }}{{{N}}_{{A}}}$
Substituting the value of Avogadro number, i.e. ${{{N}}_{{A}}} = 6.022 \times {10^{23}}$
${{{E}}_{{g}}}{{ = }}\dfrac{{{{4}}{{.9 \times 1}}{{{0}}^{{{ - 13}}}}}}{{{{1}}{{{0}}^{{6}}}}}{{ \times 6}}{{.023 \times 1}}{{{0}}^{{{23}}}}{{ J mo}}{{{l}}^{{{ - 1}}}}$
${{{E}}_{{g}}}{{ = 29}}{{.51 \times 1}}{{{0}}^{{4}}}{{ J mo}}{{{l}}^{{{ - 1}}}}{{ = 295}}{{.1 kJ mo}}{{{l}}^{{{ - 1}}}}$
Since halogens have negative electron gain enthalpies thus electron gain enthalpy for one mole of iodine ${{ = - 295 kJ mo}}{{{l}}^{{{ - 1}}}}$
For conversion of joules to ${{eV}}$ we have relation between them, i.e. ${{1 eV = 1}}{{.6021 \times 1}}{{{0}}^{{{ - 19}}}}{{ J = 1}}{{.6021 \times 1}}{{{0}}^{{{ - 22}}}}{{ kJ}}$
So ${{ - 295kJmo}}{{{l}}^{ - 1}}{{ = }}\dfrac{{{{ - 295eV}}}}{{{{1}}{{.6021 \times 1}}{{{0}}^{{{ - 22}}}}{{mol}}}}$ ${{ = 184}}{{.13 \times 1}}{{{0}}^{{{22}}}}{{eVmo}}{{{l}}^{{{ - 1}}}}$
${{{E}}_{{g}}}{{ = }}\dfrac{{{{ - 184}}{{.13 \times 1}}{{{0}}^{{{22}}}}}}{{{{{N}}_{{A}}}}}{{ eV = }}\dfrac{{{{ - 184}}{{.13 \times 1}}{{{0}}^{{{22}}}}}}{{{{6}}{{.023 \times 1}}{{{0}}^{{{23}}}}}}{{eV}}$
${{{E}}_{{g}}}{{ = - 3}}{{.06 eV ato}}{{{m}}^{{{ - 1}}}}$

So, the correct answer is Option B.

Additional Information:
The energy released is taken in negative sign. The more the energy released in the reaction more is the electron gain enthalpy for the element. The electron gain enthalpy also predicts the strength by which that extra electron is attached to the element.

Note: The halogens always have negative electron gain enthalpy because after accepting just one electron they can be most stable like noble gases and noble gases have positive electron gain enthalpy because extra electrons will make them highly unstable.