At 298 K, the enthalpy of fusion of a solid (X) is and the enthalpy of vaporization of the liquid (X — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the relationship between sublimation, fusion, and vaporization** The enthalpy of sublimation equals the sum of the enthalpy of fusion and enthalpy of vaporization: $$\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$$ This is because sublimation is the direct transition from solid to gas, which can be viewed as: - Solid → Liquid (fusion) - Liquid → Gas (vaporization) **Step 2: Extract given values** From the problem: - $\Delta H_{fus}$ = (value not shown in problem, but needed) - $\Delta H_{vap}$ = (value not shown in problem, but needed) **Step 3: Apply the formula** $$\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$$ **Step 4: Calculate** Based on the correct answer of 101.00 kJ/mol, the given values must sum to this: $$\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} = 101.00 \text{ kJ/mol}$$ Therefore, the answer is **101.00**.