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Answer :
Final answer:
The enthalpy of sublimation of sodium is calculated by adding the enthalpy of fusion (2.6 kJ/mol) and the enthalpy of vaporization (98.2 kJ/mol), yielding 100.8 kJ/mol.
Explanation:
The question asks about the enthalpy of sublimation of sodium metal. In thermodynamics, the enthalpy of sublimation is the energy required for a substance to change from the solid phase to the gaseous phase without passing through the liquid phase. According to the information provided, and by using a concept similar to Hess's Law, we know that the enthalpy of sublimation is approximately equal to the sum of the enthalpy of fusion and the enthalpy of vaporization for a substance.
For sodium metal, the enthalpy of fusion is given as 2.6 kJ/mol and the enthalpy of vaporization is 98.2 kJ/mol. Therefore, to find the enthalpy of sublimation for sodium, we simply add these two values:
Enthalpy of Sublimation
= Enthalpy of Fusion + Enthalpy of Vaporization
Enthalpy of Sublimation
= 2.6 kJ/mol + 98.2 kJ/mol
= 100.8 kJ/mol
This result represents the total energy required for sodium to change from solid to gas at the molecular level.
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The enthalpy of sublimation for sodium can be calculated by adding the enthalpy of fusion (2.6 kJ/mol) to the enthalpy of vaporization (98.2 kJ/mol), resulting in 100.8 kJ/mol.
The enthalpy of sublimation is understood as the energy required for a substance to transition from the solid phase directly to the gas phase without passing through the liquid phase. This process can be seen as a combination of two steps: first melting (fusion) and then vaporizing.
Therefore, according to Hess's Law, the enthalpy of sublimation can be approximated by summing the enthalpy of fusion and the enthalpy of vaporization.
For sodium metal, the enthalpy of fusion is given as 2.6 kJ/mol and the enthalpy of vaporization is 98.2 kJ/mol. By adding these two values together, we can estimate the enthalpy of sublimation of sodium:
- Enthalpy of fusion: 2.6 kJ/mol
- Enthalpy of vaporization: 98.2 kJ/mol
Therefore, enthalpy of sublimation = enthalpy of fusion + enthalpy of vaporization = 2.6 kJ/mol + 98.2 kJ/mol = 100.8 kJ/mol.
