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

Correct Answer: -5242.42

\(\text{Potential Energy} = 2 \text{Total Energy}\)

\(E = -13.6 \times \frac{z^2}{n^2} \, \text{eV/atom}\)

= \(-2 \times 13.6 \times \frac{z^2}{n^2} \, \text{eV/atom} + \left(-2 \times 13.6 \times \frac{z^2}{n^2}\right) \, \text{eV/atom}\)

= \(-2 \times 2 \times 13.6 \times 1 \, \text{eV/atom}\)

= \(-4 \times 13.6 \times 1.6 \times 10^{-19} \, \text{J/atom} \times 6.023 \times 10^{23} \, \text{atom/mole}\)

=\(-4 \times 13.6 \times 1.6 \times 6.023 \times 10^4 \, \text{J/mole}\)

= \(-5242.42 \, \text{kJ/mol}\)