Question

The energy released during conversion of million atoms of iodine in gaseous state to iodide ion in gaseous state is $4.9\times {10}^{13}J$ . What is the electron gain enthalpy (nearest integer value) in $eV/atom$.

Answer 1

${10}^6$ atoms require $4.9\times {10}^{13}J$.

$6.023\times {10}^{23}$ atoms will require $I_{\left(g\right)}=\frac{4.9\times {10}^{13}\times 6.023\times {10}^{23}}{{10}^6}J$.

$\therefore$ $I_{\left(g\right)}=\frac{4.9\times {10}^{13}\times 6.023\times {10}^{23}}{{10}^6}J$
$E_{g{a}^{}}=295.127kJ/mol$
Also,
$E_{g{a}^{}}=\frac{295.127\times {10}^3}{1.60\times {10}^{-19}\times 6.023\times {10}^{23}}=3.06eV/atom$

The correct answer is $3$.