The molar heats of fusion and vaporization of a certain substance are 2.26 kJ/mol and 8.16 kJ/mol, and its melting point and boiling point are −184°C and −108°C, respectively. Calculate the entropy changes for fusion (solid to liquid) and vaporization (liquid to gas). ΔSfus = J/K·mol ΔSvap = J/K·mol

ΔSfus = 25 J/K·mol

ΔSvap = 49.5 J/K·mol

Explanation:

We can calculate the entropy change for fusion (ΔSfus) using the following expression.

ΔSfus = ΔHfus/Tm

where,

ΔHfus: molar heat of fusion

Tm: melting point (-184°C + 273.15 = 89 K)

ΔSfus = ΔHfus/Tm

ΔSfus = (2.26 × 10³ J/mol)/89 K

ΔSfus = 25 J/K·mol

We can calculate the entropy change for vaporization (ΔSvap) using the following expression.

ΔSvap = ΔHvap/Tb

where,

ΔHvap: molar heat of vaporization

Tb: boiling point (-108°C + 273.15 = 165 K)

ΔSvap = ΔHvap/Tb

ΔSvap = (8.16 × 10³ J/mol)/165 K

ΔSvap = 49.5 J/K·mol

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.