Question

Enthalpyof a fusionof ice is 6.01 KJ÷mol.The enthalpy of vapourization of water is 45.07KJ÷mol.what is enthalpy of sublimation of ice

Answer 1

here is the answer.................

Answer 2

Answer:

Enthalpy of sublimation = 51.08 kJ.mol⁻¹

Explanation:

∆H(fus) = 6.01 kJ.mol⁻¹

∆H(vap) = 45.07 kJ.mol⁻¹

Enthalpy of sublimation of substance is calculated as the sum of its vaporization enthalpy & fusion enthalpy.

∆H(sub) = ∆H(fus) + ∆H(vap)

∆H(sub) = 6.01 + 45.07

∆H(sub) = 51.08 kJ.mol⁻¹

Hence, enthalpy of sublimation of ice is 51.08 kJ.mol⁻¹ .