Question

Enthalpy of fusion of ice is 6.01 kJ mole⁻¹. The enthalpy of vaporization of water is 45.07 kJ mol⁻¹. What is enthalpy of sublimation of ice?
(51.08 kJ mol⁻¹)

Answer 1

The enthalpy of sublimation of ice at 0 degrees Celsius is +51.08 kJ/mol. The enthalpy of vaporization of water at 0 degrees Celsius is +41.07 kJ/mol. Use these values to determine the correct answer for the enthalpy of fusion of ice at 0 degrees Celsius.

Answer 2

Hi Biswajit,

● Answer -

Enthalpy of sublimation = 51.08 kJ.mol⁻¹

● Explaination -

# Given -

∆H(fus) = 6.01 kJ.mol⁻¹

∆H(vap) = 45.07 kJ.mol⁻¹

# Solution -

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 .

Hope this helps you.