Question

The amount of energy released when a million atoms of iodine are completely converted into $I^{–}$ ions in the vapour state according to the equation, $I\left(g\right)+e^{–}\to I^{–}\left(g\right)$ is $5.0\times {10}^{–13}$ J.
Calculate the electron gain enthalpy of iodine in terms of $kJmo{l}^{–1}$.

• A. $-150.5kJ/mol$
• B. $-301kJ/mol$
• C. $-602kJ/mol$
• D. $-30.1kJ/mol$

Answer 1

The correct option is B $-301kJ/mol$

The electron gain enthalpy of iodine is equal to the energy released when 1 mole of iodine atoms in vapour state are converted into $I^{–}$ions.
= $-\frac{5\times {10}^{-13}\times 6.023\times {10}^{23}}{{10}^6}$
=-$30.1\times {10}^{4}J$
= -$301kJ$