Question

Enthalpy of fusion of ice at 0°C and 1 atmosphere is 6.01 kJ mol ^{- 1} and enthalpy of evaporation of water at 0°C and 1 atmosphere is 45.07 kJ mo1 ^{- 1}. Calculate the enthalpy of sublimation of one mole of ice at 0°C and 1 atmosphere. Write the equation for the same.

Answer 1

1. Enthalpy of fusion of ice: H₂O(S) → H₂O(l), ΔH = 6.01 kJ mo1 ^{- 1}
2. Enthalpy of vaporization of water: H₂O → H₂O(g), ΔH = 45.07 kJ mo1 ^{- 1}
3. Enthalpy of sublimation of ice: H₂O(S) →H₂O(g), ΔH = ?
   Addition of reactions (i) and (ii) gives
   \[\ce{H2O(S) + \cancel{H_2O(l)} -> \cancel{H_2O(l)} + H2O(g), ΔH = 6.01 + 45.07 kJ mol^{-1}}\]
   H₂O(S) → H₂O(g), → ΔH = 51.08kJ mol ^{- 1}
   This is the same as reaction (c).
   ∴ enthalpy of sublimation of ice  = 51.08 kJ mol ^{- 1}.