The figure below is the plot of potential energy versus internuclear distance (𝑑) of H₂ molecule in the electronic ground state. What is the value of the net potential energy 𝐸0 (as indicated in the figure) in kJ mol⁻¹, for 𝑑=𝑑₀ 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 × 10²³ mol⁻¹.

Moderate

Solution

At d = d0, nucleus-nucleus & electron-electron repulsion is absent.
Hence potential energy will be calculated for 2 H atoms. (P.E. due to attraction of proton & electron)
$P.E. = \underset{(Bohar radius)}{\frac{- {Kq}_{1} q_{2}}{r}} = \frac{(9 \times 10^{9}) {(1.6 \times 10^{- 19})}^{2}}{0.529 \times 10^{- 10}} = - 4.355 \times 10^{- 21} kJ$
For 1 mol = –4.355 × 10^–21 × 6.023 × 10²³ = –2623.249 kJ/mol
For 2 H atoms = –5246.49 kJ/mol

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