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Exercises · 3.15

Q.Energy of an electron in the ground state of the hydrogen atom is −2.18×10−18-2.18 \times 10^{-18} J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol−1^{-1}. (Hint: Apply the idea of mole concept to derive the answer.)

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Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in their ground state. Multiply the single-atom ionization energy by Avogadro's number to get 1.312×106 J mol−1\boxed{1.312 \times 10^{6} \text{ J mol}^{-1}}.

Why the mole concept matters here

The ground-state energy you're given, −2.18×10−18-2.18 \times 10^{-18} J, describes one hydrogen atom. The negative sign tells us the electron is bound to the nucleus—it sits in a potential well 2.18×10−182.18 \times 10^{-18} J below the reference state (a free electron infinitely far from the proton, defined as zero energy).

Ionization means pulling that electron completely out of the atom's grip, taking it from E=−2.18×10−18E = -2.18 \times 10^{-18} J to E=0E = 0 J. The energy input required for this process is the ionization energy. For a single atom, that's simply +2.18×10−18+2.18 \times 10^{-18} J.

But chemistry deals with macroscopic quantities. When we talk about ionization enthalpy in thermodynamics or compare it across the periodic table, we always quote it per mole of atoms. That's where Avogadro's number bridges the atomic and molar scales.

Step-by-step calculation

  1. Identify the ionization energy for one atom. The electron in the ground state has energy E1=−2.18×10−18E_1 = -2.18 \times 10^{-18} J. To ionize it (move it to E=0E = 0), we must supply:

ΔEatom=0−(−2.18×10−18)=+2.18×10−18 J\Delta E_{\text{atom}} = 0 - (-2.18 \times 10^{-18}) = +2.18 \times 10^{-18} \text{ J}

  1. Scale up to one mole of hydrogen atoms. One mole contains NA=6.022×1023N_A = 6.022 \times 10^{23} atoms (Avogadro's number). If each atom requires 2.18×10−182.18 \times 10^{-18} J, then one mole requires:

ΔHionization=(2.18×10−18 J/atom)×(6.022×1023 atoms/mol)\Delta H_{\text{ionization}} = (2.18 \times 10^{-18} \text{ J/atom}) \times (6.022 \times 10^{23} \text{ atoms/mol})

  1. Perform the multiplication.

ΔHionization=2.18×6.022×10−18+23=13.12796×105 J mol−1\Delta H_{\text{ionization}} = 2.18 \times 6.022 \times 10^{-18 + 23} = 13.12796 \times 10^{5} \text{ J mol}^{-1}

Rounding to three significant figures (matching the precision of the given energy):

ΔHionization=1.31×106 J mol−1=1312 kJ mol−1\Delta H_{\text{ionization}} = 1.31 \times 10^{6} \text{ J mol}^{-1} = 1312 \text{ kJ mol}^{-1} …

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