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 $kJmo{l}^{-1}$ and $eV$ per atom?

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

Answer 1

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
${\Delta }_{eg}H=\frac{4.9\times 10^{-13}\times 6.023\times 10^{23}}{10^6}$
$=29.5\times 10^{4}Jmo{l}^{-1}$
$=295kJmo{l}^{-1}$
Thus, electron gain enthalpy of iodine $=-295kJmo{l}^{-1}$
We know that $1eV$ per atom $=96.49kJmo{l}^{-1}$
$\therefore$ Electron gain enthalpy of iodine in $eV$ per atom
$=-\frac{295}{96.49}=-3.06eV$ per atom