Question

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)

Hint:

Hess's Law works just like addition. If you know the start point (solid) and end point (gas), the path doesn't matter; just sum the energies.

Answer 1

Correct Answer: 101

Step 1: Understanding the Concept:
Sublimation is the transition from solid directly to gas.
According to Hess's Law, the total enthalpy change for a process is the sum of the enthalpy changes for its individual steps.
Sublimation can be envisioned as Solid \(\rightarrow\) Liquid (Fusion) followed by Liquid \(\rightarrow\) Gas (Vaporization).
Step 2: Key Formula or Approach:
\[ \Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} \]
Step 3: Detailed Explanation:
Given:
\(\Delta H_{fus} = 2.8 \text{ kJ mol}^{-1}\)
\(\Delta H_{vap} = 98.2 \text{ kJ mol}^{-1}\)
By Hess's Law:
\[ \Delta H_{sub} = 2.8 + 98.2 = 101.0 \text{ kJ mol}^{-1} \]
Step 4: Final Answer:
The enthalpy of sublimation is 101 kJ \(mol^{-1}\).