At 298 K, the enthalpy of fusion of a solid X is 2.8 kJ mol$$^{-1}$$ and the enthalpy of vaporisation of the liquid X is 98.2 kJ mol$$^{-1}$$. The enthalpy of sublimation of the substance X in kJ mol$$^{-1}$$ is ___. (in nearest integer)

We are given that at $$298\ \text{K}$$ the enthalpy of fusion of the solid X is $$\Delta H_{\text{fus}} = 2.8\ \text{kJ mol}^{-1}$$ and the enthalpy of vaporisation of the liquid X is $$\Delta H_{\text{vap}} = 98.2\ \text{kJ mol}^{-1}$$.

First, let us recall Hess’s Law, which states that the total enthalpy change for a process depends only on the initial and final states; it is the same whether the transformation occurs in a single step or through several intermediate steps.

To sublime a solid directly into its vapour, we can imagine the process occurring in two consecutive steps:

1. The solid melts to form the liquid.

2. The liquid vaporises to form the gas.

Using Hess’s Law, the overall enthalpy change for sublimation, denoted $$\Delta H_{\text{sub}}$$, is the algebraic sum of the enthalpy changes for these two steps:

$$\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$$

Substituting the given numerical values, we have

$$\Delta H_{\text{sub}} = 2.8\ \text{kJ mol}^{-1} + 98.2\ \text{kJ mol}^{-1}$$

Adding the two enthalpy terms carefully,

$$\Delta H_{\text{sub}} = 101.0\ \text{kJ mol}^{-1}$$

When expressed to the nearest integer, this value remains

$$\Delta H_{\text{sub}} \approx 101\ \text{kJ mol}^{-1}$$

So, the answer is $$101$$.