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, respecti…

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asked Jul 25, 2020 97.9k views

1 Answer

Bachposer · Answer 1 · 2020-07-30T13:07:22+0000

Answer:

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

Step-by-step 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 =
$\Delta S_(fus)$

$\Delta S_(fus)=(\Delta H_(fus))/(83.15 K)=(1.3 kJ/mol)/(83.15 K)=0.015634 kJ/mol K=15.634 J/mol K$

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=
$\Delta S_(vap)$

$\Delta S_(vap)=(\Delta H_(vap))/(87.15K)=(6.3 kJ/mol)/(87.15 K)=0.072289kJ/mol K=72.289 J/mol K$