Question

The amount of energy released when ${10}^{12}$ atoms of $Cl$ (vapour) are converted to $C{l}^{-}$ ions, according to the equation:
$C{l}_{\left(g\right)}+e^{-}\to C{l}_{\left(g\right)}^{-}\text{is}58\times {10}^{-10}\text{J}$
Calculate the ${\Delta }_{eg}{H}^{-}$ of $Cl$ atom in ${\text{eV atom}}^{-1}$.

• A. $-0.036{\text{eV atom}}^{-1}$
• B. $-3.480{\text{eV atom}}^{-1}$
• C. $+0.038{\text{eV atom}}^{-1}$
• D. $-0.361{\text{eV atom}}^{-1}$

Answer 1

The correct option is A $-0.036{\text{eV atom}}^{-1}$

The amount of energy released when 1 mole of $Cl$ is converted to $C{l}^{-}$ ions,
$=\frac{-58\times {10}^{-10}\text{J}\times 6\times {10}^{23}}{{10}^{12}}$
$=-3.48{\text{kJ mol}}^{-1}$
We know that,
$1{\text{eV atom}}^{-1}=96.49{\text{kJ mol}}^{-1}\frac{-3.48}{96.49}=-0.036{\text{eV atom}}^{-1}$