Question

The amount of energy when million atoms of iodine are completely converted into $I^-$ ions in the vapour state according to the equation, $I_(g) + e^- \to I^-_{(g)}$ is $4.9 \times 10^{-13}\, J$. What would be the electron gain enthalpy of iodine in terms of $kJ\, mol^{-1}$ and $eV$ per atom?

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

Answer 1

Answer: (B)

The electron gain enthalpy of iodine is equal to the energy released when 1 mole of iodine atoms in vapour state is converted into $I^-$ ions
$\Delta_{eg}H = \frac{4.9 \times 10^{-13} \times 6.023 \times 10^{23}}{10^{6}}$
$= 29.5 \times 10^{4}\,J\,mol^{-1}$
$= 295\,kJ\,mol^{-1}$
Thus, electron gain enthalpy of iodine $= -295\, kJ\, mol^{-1}$
We know that $1\, eV$ per atom $= 96.49\, kJ \,mol^{-1}$
$\therefore \quad$ Electron gain enthalpy of iodine in $eV$ per atom
$= - \frac{295}{96.49} = -3.06\, eV$ per atom