Answer 1

Relation of enthalpy of sublimation with enthalpy of fusion and enthalpy of vaporisation:

The enthalpy of sublimation of ice at 0⁰C and 1 atm pressure is 51.08 kJ mol⁻¹.H₂O(s) → H₂O(g), ΔH = 51.08 kJ mol⁻¹ at 0⁰C.

When solid is converted to vapour, either in one step or two steps, the solid first gets converted to the liquid state and then to the vapour state. The enthalpy change remains the same. This is because enthalpy is a state function.

For example,
H₂O(s) → H₂O(l), Δ_fus H = +6.01 kJ mol⁻¹ at 0°C
H₂O(l) → H₂O(g), Δ_vap H = +45.07 kJ mol⁻¹ at 0°C

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H₂O(s) → H₂O(g), Δ H = 51.08 kJ mol⁻¹ at 0°C

Therefore, it follows that
Δ_sub H = Δ_fus H + Δ_vap H