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In case of the decomposition of hydrogen molecule into two hydrogen atoms, the enthalpy of atomization is same as the _________
  • a)
    bond dissociation enthalpy
  • b)
    enthalpy of formation
  • c)
    enthalpy of combustion
  • d)
    enthalpy of sublimation
Correct answer is option 'A'. Can you explain this answer?
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In case of the decomposition of hydrogen molecule into two hydrogen at...
Hydrogen Molecule Decomposition and Enthalpy of Atomization

Introduction:
The decomposition of a hydrogen molecule (H2) into two hydrogen atoms (2H) is a chemical process that involves breaking the bond between the two hydrogen atoms. This process can be represented by the equation: H2(g) → 2H(g).

Enthalpy of Atomization:
The enthalpy of atomization is the enthalpy change that occurs when one mole of a substance in the gaseous state is completely separated into its constituent atoms in the gaseous state. It is denoted by ΔHatom.

Bond Dissociation Enthalpy:
The bond dissociation enthalpy, also known as bond energy, is the energy required to break a particular bond in one mole of a gaseous substance. It is denoted by ΔHdissociation.

Comparison:
The enthalpy of atomization and bond dissociation enthalpy are related but not exactly the same. However, in the case of the decomposition of a hydrogen molecule into hydrogen atoms, the enthalpy of atomization is the same as the bond dissociation enthalpy.

Explanation:
When a hydrogen molecule (H2) is decomposed into two hydrogen atoms (2H), the bond between the two hydrogen atoms is broken. This process requires energy, which is supplied as heat. The enthalpy change associated with this process is the bond dissociation enthalpy, ΔHdissociation.

On the other hand, the enthalpy of atomization is the enthalpy change when one mole of a substance in the gaseous state is completely separated into its constituent atoms in the gaseous state. In the case of the decomposition of a hydrogen molecule, one mole of H2 is converted into two moles of H atoms.

Since the process of breaking the bond between the hydrogen atoms is the same as separating the hydrogen atoms in the gaseous state, the enthalpy change for both processes is the same. Therefore, the enthalpy of atomization is equal to the bond dissociation enthalpy for the decomposition of a hydrogen molecule into hydrogen atoms.

Conclusion:
In summary, the enthalpy of atomization is the same as the bond dissociation enthalpy for the decomposition of a hydrogen molecule into two hydrogen atoms. This is because the process of breaking the bond between the hydrogen atoms is equivalent to separating the hydrogen atoms in the gaseous state.
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In case of the decomposition of hydrogen molecule into two hydrogen at...
In case of the decomposition of hydrogen molecule the enthalpy is the same as enthalpy of atomization as well as bond dissociation enthalpy because the bond association enthalpy refers to the breakage of H-H Bond. So that two hydrogen atoms are formed whereas enthalpy of atomization is the breakage of hydrogen molecules in order to form two atoms.
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In case of the decomposition of hydrogen molecule into two hydrogen atoms, the enthalpy of atomization is same as the _________a)bond dissociation enthalpyb)enthalpy of formationc)enthalpy of combustiond)enthalpy of sublimationCorrect answer is option 'A'. Can you explain this answer?
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