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,
$I_{\left(g\right)}+e^{-}\to I_{\left(g\right)}^{-}$ is 4.9 x 10${}^{-13}$ J. What would be the electron gain enthalpy of iodine in terms of $kJmo{l}^{-1}$ and $eV$ per atom?

• A. $295,3.06$
• B. $-295,-3.06$
• C. $439,5.09$
• D. $-356,-7.08$

Answer 1

The correct option is B $-295,-3.06$

The electron gain enthalpy ($E_g$) of an element is defined as the energy released when an atom in gaseous state gains an electron.

$I\left(g\right)+e^{-}\to I^{-}\left(g\right)$

In case of iodine

$E_g$ for 1 million atoms (${10}^{6}atoms$)

$E_g=4.9\times {10}^{-13}J$ for ${10}^{6}atoms$

$E_g=\frac{4.9\times {10}^{-13}}{{10}^6}J/atom$

For one mol of $I$ atoms, amount of energy released in the formation of $I^{-}$ ions by gaining 1 mol of electrons will be:

$E_g=\frac{4.9\times {10}^{-13}}{{10}^6}\times N_A$

where $N_A=6.023\times {10}^{23}atoms/mol$

$E_g=\frac{4.9\times {10}^{-13}}{{10}^6}\times 6.023\times {10}^{23}J/mol$

$=29.51\times {10}^{4}J/atom=295.1kJ/mol$

Since halogens have negative $E_g$

Thus, electron gain enthalpy for one mol iodine $=-295kJmo{l}^{-1}$

In eV units we have the relation:

$1eV=1.6021\times {10}^{-19}J=1.6021\times {10}^{-22}kJ$

$-295kJ/mol=\frac{-295}{1.6021\times {10}^{-22}}eV/mol=184.13\times {10}^{22}eV/mol$

For one atom, $E_g$ in eV is:= $\frac{-184.13\times {10}^{22}}{N_A}eV$

=$\frac{-184.13\times {10}^{22}}{6.023\times {10}^{23}}eV$

$=-3.06eV/atom$