Calculate the values of the molar entropy of fusion and vaporization for water at these two phase transitions at a constant pressure = 1 atm.

Data: ΔH_fus [H₂O] = 6.01 kJ.mol^-1

ΔH_vap [H₂O] = 40.65 kJ.mol^-1

Question

Calculate the values of the molar entropy of fusion and vaporization for water at these two phase transitions at a constant pressure = 1 atm.

Data: ΔH_fus [H₂O] = 6.01 kJ.mol^-1

ΔH_vap [H₂O] = 40.65 kJ.mol^-1

| General Chemistry 3

Accepted Answer

Entropy change on fusion:

ΔS_fus = $\frac{∆ H_{fus}}{T_{m}}$

ΔS_fus = molar entropy of fusion (in J.K^-1.mol^-1)

ΔH_fus = molar enthalpy of fusion (in J.mol^-1)

T_m = melting point (in K) = 273 K for water

ΔS_fus =

\frac{6.01 \times 10^{3}}{273}

= 22.0 J.mol^-1.K^-1

Entropy change on vaporization:

ΔS_vap =

\frac{∆ H_{vap}}{T_{b}}

ΔS_vap = molar entropy of vaporization (in J.K^-1.mol^-1)

ΔH_vap = molar enthalpy of vaporization (in J.mol^-1)

T_b = boiling point (in K) = 373 K for water

ΔS_vap =

\frac{40.65 \times 10^{3}}{373}

= 109 J.mol^-1.K^-1

Exercise 5 | Entropy, Free Energy, and Equilibrium