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 $kJmo{l}^{-1}$, 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 $6.023\times 10^{23}mo{l}^{-1}$.

Answer 1

Answer: -5246.50

P.E. $=-\frac{K{q}_1{q}_2}{r}$
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 $=-87.106\times 10^{-19}\times 6.023\times 10^{23}J$
$=-524.6490\times 10^4$
$=-5246.49kJ/mo{l}^{-1}$
$\approx -5246.50kJmo{l}^{-1}$