Chemistry · Structure of Atom

JEE Advanced 2020 — Paper 2 — Question 30

The figure below is the plot of potential energy versus internuclear distance (d) of $\mathrm{H}_{2}$ molecule in the electronic ground state. What is the value of the net potential energy $E_{0}$ (as indicated in figure) in $\mathrm{kJ} \mathrm{mol}^{-1}$ , 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 $6.023 \times 10^{23} \mathrm{~mol}^{-1}$ .

Answer: -5246.50

Numerical answer — enter this value.

Step-by-step solution

P.E. $=-\frac{K q_{1} q_{2}}{r}$

P.E. of one H - atom in ground state $=-\frac{\mathrm{K}(\mathrm{e})(\mathrm{e})}{r}$

P.E. $=\frac{-9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.529 \times 10^{-10}}=-43.5538 \times 10^{-19} \mathrm{~J} /$ atom

So, P.E. of two H - atoms (or one molecules of $\mathrm{H}_{2}$ ) $=-2 \times 43.5538 \times 10^{-19}=-87.106 \times 10^{-19} \mathrm{~J} / \mathrm{molecule}$

So, P.E. of one mole of $\mathrm{H}_{2}$ molecules $=-87.106 \times 10^{-19} \times 6.023 \times 10^{23} \mathrm{~J}$

\begin{aligned} & =-524.6490 \times 10^{4} \\& =-5246.49{\mathrm{~kJ} / \mathrm{mol}^{-1}} \\& {\approx-5246.50 \mathrm{kJmol}^{-1}} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Chemistry
Chapter: Structure of Atom
Topic: Bohr's Model of Atom