Question

The figure below is the plot of potential energy versus internuclear distance $(d)$ of $H _{2}$ molecule in the electronic ground state. What is the value of the net potential energy $E_{0}$ (as indicated in figure) in $kJ\, mol ^{-1}$, for $d=d_{0}$ at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of $H$ atom is taken as zero when its electron and the nucleus are infinitely far apart.
Use Avogadro constant as $6.023 \times 10^{23} mol ^{-1}$.

Answer 1

P.E. $=-\frac{ Kq _{1} q _{2}}{ r }$
P.E. of one $H -$ atom in ground state $=-\frac{ K ( e )( e )}{ r }$
P.E. $=\frac{-9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.529 \times 10^{-10}}$
$=-43.5538 \times 10^{-19} J / atom$
So, P.E. of two $H$ - atoms (or one molecules of $H _{2}$ ) $=-2 \times 43.5538 \times 10^{-19}$
$=-87.106 \times 10^{-19} J /$ molecule
So, P.E. of one mole of $H _{2}$ molecules $=-87.106 \times 10^{-19} \times 6.023 \times 10^{23} J$
$=-524.6490 \times 10^{4}$
$=-5246.49 \,kJ / mol ^{-1}$
$\approx-5246.50 \,kJ\,mol ^{-1}$