QJEEM21Chdea1: JEE Main 2021 Chemistry - Class 11 JEEM21Chdea1 Solution: At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol − …

The enthalpy of sublimation is the total amount of energy required to convert a solid directly into a gas. This can be calculated by summing the enthalpy of fusion (solid to liquid) and the enthalpy of vaporization (liquid to gas). Mathematically, this relationship is represented as:

$$\mathrm{\Delta }{H}_{\text{sublimation}}=\mathrm{\Delta }{H}_{\text{fusion}}+\mathrm{\Delta }{H}_{\text{vaporization}}$$

Given:

Enthalpy of fusion, $$\mathrm{\Delta }{H}_{\text{fusion}}=2.8{\text{kJ mol}}^{-1}$$

Enthalpy of vaporization, $$\mathrm{\Delta }{H}_{\text{vaporization}}=98.2{\text{kJ mol}}^{-1}$$

Substitute these values into the equation:

$$\mathrm{\Delta }{H}_{\text{sublimation}}=2.8{\text{kJ mol}}^{-1}+98.2{\text{kJ mol}}^{-1}=101.0{\text{kJ mol}}^{-1}$$

Therefore, the enthalpy of sublimation of the substance (X) is approximately 101 kJ mol^$$-1$$.