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)

Solution

Chemistry ma’am · 4 steps

Solid to gas is solid to liquid, then liquid to gas, so add the two enthalpies.

1. Two-step path. Sublimation takes the solid straight to vapour. The same change can be done in two steps: melt the solid, then vaporise the liquid.
2. Use Hess's law. Enthalpy is a state function, so the total change depends only on the start (solid) and the end (vapour). So the enthalpies of the two steps add up.

   Δ_(sub)H = Δ_(fus)H + ΔᵥₐₚH

3. Put in numbers. Δ_(sub)H = 2.8 + 98.2 = 101 kJ mol⁻¹.
4. So. The enthalpy of sublimation of X is 101 kJ mol⁻¹.

The common slip: 95. Subtracting instead of adding gives 98.2 - 2.8 = 95.4, about 95. Sublimation needs both steps, so the enthalpies add.

The idea to keep. Sublimation equals fusion plus vaporisation at the same temperature. Whenever you can split a change into steps, Hess's law lets you add their enthalpies.

Answer from the NTA final answer key. Solution by PrepWiser.