Answer :
The enthalpy of sublimation of sodium is calculated by adding the enthalpy of fusion (2.6 kJ mol−1) and the enthalpy of vaporization (98.2 kJ mol−1), resulting in a value of 100.8 kJ mol−1.
To calculate the enthalpy of sublimation of sodium, we can use the known values for the enthalpy of fusion and enthalpy of vaporization.
According to given reference sources and understanding of physical chemistry, the enthalpy of sublimation can be approximately determined by summing up the enthalpy of fusion and the enthalpy of vaporization since sublimation is the transition from solid to gas phase, which can be viewed as occurring through a two-step process of melting then vaporizing.
The enthalpy of fusion (2.6 kJ mol−1) and the enthalpy of vaporization (98.2 kJ mol−1) are given. Therefore, the enthalpy of sublimation (ΔHsub) of sodium can be estimated as:
ΔHsub = ΔHfus + ΔHvap
ΔHsub = 2.6 kJ mol−1 + 98.2 kJ mol−1
ΔHsub = 100.8 kJ mol−1
Final answer:
Therefore, the enthalpy of sublimation for sodium is approximately 100.8 kJ mol⁻¹.
Explanation:
To calculate the enthalpy of sublimation of sodium, we can use the information that the sum of the enthalpy of fusion and the enthalpy of vaporization is approximately equal to the enthalpy of sublimation, as stated in Figure 12.6.7. Given that the enthalpy of fusion (ΔfusH) for sodium is 2.6 kJ mol⁻¹ and the enthalpy of vaporization (ΔvapH) is 98.2 kJ mol⁻¹, the calculation is straightforward:
ΔsubH (sodium) = ΔfusH (sodium) + ΔvapH (sodium)
ΔsubH (sodium) = 2.6 kJ mol⁻¹ + 98.2 kJ mol⁻¹
ΔsubH (sodium) = 100.8 kJ mol⁻¹
Therefore, the enthalpy of sublimation for sodium is approximately 100.8 kJ mol⁻¹.
