Question

You are given Avogadro's no. atoms of $X$. If half of the atoms of $X$ transfer one electron to the other half of $X$ atoms, $409kJ$ must be added. If these $X^{-}$, an additional $733kJ$ energy must be added. If $IE$ and $EA$ are $a$ and $b$ of $X$. Find $a+b$ (nearest integer value) in $eV$. Use ($1eV=1.602\times {10}^{-19}J$ and $N=6.023\times {10}^{23}$)

Answer 1

$X⟶X^{+}+e;\Delta H=I{E}_1=aeV$
$X+e⟶X^{-};\Delta H=E_{g{a}_1}=-beV$

If $N/2$ atoms of $X$ lose electrons which are taken up by remaining $N/2$ of $X$ to give $X^{-}$, then

$a\times \frac{N}{2}-b\times \frac{N}{2}=\frac{409\times {10}^3}{1.602\times {10}^{-19}}eV$

or $a-b=\frac{409\times {10}^3}{1.602\times {10}^{-19}\times 6.023\times {10}^{23}}$

$\therefore a-b=8.477$

Now $N/2$ of $X^{-}$ ;lose two electrons to give $X^{+}$
$X^{-}⟶X+e;\Delta H=+b$

$X⟶X^{+}+e;\Delta H=+{IE}_2=+a$

$a\times \frac{N}{2}+b\times \frac{N}{2}=\frac{733\times {10}^3}{1.602\times {10}^{-19}}$

$a+b=\frac{733\times {10}^3\times 2}{1.602\times {10}^{-19}\times 6.023\times {10}^{23}}$

or $a+b=15.194$

$a=11.835eV;b=3.358eV$

$a+b=15.19$
Nearest integer value of $a+b=15$
Ans -$15$