The energy of a hydrogen molecule in its ground state equilibrium configuration is $-31.7\ eV$.Its dissociation energy is_______________$\ eV$. (Up to one decimal place)
Understanding Molecular Dissociation Energy
The dissociation energy is the minimum energy required to break apart a molecule into its constituent atoms starting from its ground state. For a diatomic molecule like hydrogen (H2), it's the energy needed to convert H2 into two separate H atoms.
Given Hydrogen Molecule Energy
The energy of the hydrogen molecule (H2) in its ground state equilibrium configuration is given as $E_{\text{molecule}} = -31.7\ eV$. This energy value represents the molecule's stability relative to two free hydrogen atoms.
The energy of a single hydrogen atom (H) in its ground state is a known fundamental constant: $E_{\text{H}} = -13.6\ eV$.
Calculating H2 Dissociation Energy
First, calculate the total energy of two separated hydrogen atoms:
$ E_{\text{atoms}} = 2 \times E_{\text{H}} = 2 \times (-13.6\ eV) = -27.2\ eV $
The dissociation energy ($D_0$) is the difference between the energy of the separated atoms and the energy of the molecule:
$ D_0 = E_{\text{atoms}} - E_{\text{molecule}} $
Substitute the known values:
$ D_0 = (-27.2\ eV) - (-31.7\ eV) $
$ D_0 = -27.2\ eV + 31.7\ eV $
$ D_0 = 4.5\ eV $
Result Confirmation
The calculated dissociation energy for the hydrogen molecule is $4.5\ eV$. This value falls within the provided range of 4.4 eV to 4.6 eV.
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