Question

How many $Cl$ atoms can you ionise in the process: $Cl⟶{Cl}^{+}+e$ by the energy liberated for the process $Cl+e⟶{Cl}^{-}$ for 1 mole atoms? Given; $IE=12.967eV$ and $EA=3.61eV$. If answer in multiple of $X\times {10}^{23}$, find the value of X to the nearest integer value.

Answer 1

Let $n$ atoms of chlorine be ionised in a process $Cl⟶{Cl}^{+}+e$ by the energy liberated during gain of electron by $N$ atoms following,

$Cl+e⟶{Cl}^{-}$

$\therefore$ Energy liberated during addition of electron for ${}^{\prime }{N}^{\prime }$ $Cl$ atoms$=$Energy used during removal of electron from ${}^{\prime }{n}^{\prime }$ $Cl$ atoms.

or, $N\times EA=n\times IE$

or, $n=\frac{(6.023\times {10}^{23}\times 3.61)}{12.967}=1.678\times {10}^{23}\cong 2\times {10}^{23}$. Hence, the value of $X$ is 2.