JEE Challenger
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ChemHardAdv 2020 • Paper 2

Potential Energy Net Value for Hydrogen Molecule Ground State

The graph below illustrates potential energy against internuclear distance (dd) for an H2\text{H}_2 molecule in its electronic ground state. Find the net potential energy E0E_0 (in kJ mol−1\text{kJ mol}^{-1}) at d=d0d = d_0 where electron-electron repulsion and nucleus-nucleus repulsion energies are absent. Take the potential energy of an isolated H\text{H} atom as zero when its electron and nucleus are separated infinitely. Use Avogadro's constant as 6.023×1023 mol−16.023 \times 10^{23}\text{ mol}^{-1}.

Question Diagram 1
Official Numerical Answer-5248.97

Topics & Concepts

Chemical Bonding and Molecular StructureMolecular Orbital TheoryBohr Model

Step-by-Step Solution

To find the net potential energy E0E_0 (in kJ mol−1\text{kJ mol}^{-1}) of an H2\text{H}_2 molecule at internuclear distance d=d0d = d_0 where interatomic repulsions (electron-electron and nucleus-nucleus) are absent, we consider two isolated, non-interacting hydrogen atoms in their electronic ground state (n=1n = 1).

Step 1: Potential Energy of a Single Ground-State Hydrogen Atom

According to the Bohr model of the hydrogen atom, the total energy E1E_1 of an electron in the ground state (n=1n = 1) is: E1=−13.6 eVE_1 = -13.6 \text{ eV}

The total energy EE is the sum of the kinetic energy KK and the potential energy VatomV_{\text{atom}}: E=K+VatomE = K + V_{\text{atom}}

For a inverse-square force law (Coulomb potential), by the Virial Theorem: ⟨K⟩=−12⟨Vatom⟩\langle K \rangle = -\frac{1}{2} \langle V_{\text{atom}} \rangle

Thus, the potential energy VatomV_{\text{atom}} of a single hydrogen atom is twice its total ground-state energy: Vatom=2E1=2×(−13.6 eV)=−27.2 eVV_{\text{atom}} = 2 E_1 = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}

Step 2: Potential Energy of the Two Non-Interacting H Atoms

At d=d0d = d_0, the two hydrogen atoms are sufficiently far apart such that interatomic repulsion and attraction terms are absent. Therefore, the system consists of two independent hydrogen atoms.

The potential energy per pair of H atoms (H2\text{H}_2 molecule at d0d_0) is: Vmolecule=2×Vatom=2×(−27.2 eV)=−54.4 eVV_{\text{molecule}} = 2 \times V_{\text{atom}} = 2 \times (-27.2 \text{ eV}) = -54.4 \text{ eV}

Step 3: Conversion to kJ mol−1\text{kJ mol}^{-1}

We convert the potential energy from eV\text{eV} per molecule to kJ mol−1\text{kJ mol}^{-1} using the fundamental constants:

  • 1 eV=1.602×10−19 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}
  • NA=6.023×1023 mol−1N_A = 6.023 \times 10^{23} \text{ mol}^{-1}

E0=Vmolecule×NA×1.602×10−19 J eV−1×10−3 kJ J−1E_0 = V_{\text{molecule}} \times N_A \times 1.602 \times 10^{-19} \text{ J eV}^{-1} \times 10^{-3} \text{ kJ J}^{-1}

Substituting the values: E0=−54.4 eV×1.602×10−19 J eV−1×6.023×1023 mol−1×10−3 kJ J−1E_0 = -54.4 \text{ eV} \times 1.602 \times 10^{-19} \text{ J eV}^{-1} \times 6.023 \times 10^{23} \text{ mol}^{-1} \times 10^{-3} \text{ kJ J}^{-1}

E0=−54.4×1.602×6.023×10 kJ mol−1E_0 = -54.4 \times 1.602 \times 6.023 \times 10 \text{ kJ mol}^{-1}

E0=−5248.97 kJ mol−1E_0 = -5248.97 \text{ kJ mol}^{-1}

Final Answer:

The net potential energy E0E_0 at d=d0d = d_0 is −5248.97 kJ mol−1-5248.97 \text{ kJ mol}^{-1} (or approximately −5249 kJ mol−1-5249 \text{ kJ mol}^{-1}).

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