Question

The molar heats of fusion and vaporization of argon are 1.3 kJ/mol and 6.3 kJ/mol respectively, and argon's melting point and boiling point are -190°C and -186 °C, respectively. Calculate the entropy changes for the fusion and vaporization of argon.

Answer 1

Answer:

The entropy changes for the fusion and vaporization of argon is 15.634 J/mol K and 72.289 J/mol K respectively.

Explanation:

The molar heats of fusion =$\Delta H_{fus}= 1.3 kJ/mol$

Melting point of argon = -190°C = 83.15 K

Entropy changes for the fusion = $\Delta S_{fus}$

$\Delta S_{fus}=\frac{\Delta H_{fus}}{83.15 K}=\frac{1.3 kJ/mol}{83.15 K}=0.015634 kJ/mol K=15.634 J/mol K$

The molar heats of vaporization of argon = $\Delta H_{vap}=6.3 kJ/mol$

Boiling point of argon = -186°C = 87.15 K

Entropy changes for the vaporization= $\Delta S_{vap}$

$\Delta S_{vap}=\frac{\Delta H_{vap}}{87.15K}=\frac{6.3 kJ/mol}{87.15 K}=0.072289kJ/mol K=72.289 J/mol K$