Question

For cobalt, Co, the heat of fusion at its normal melting point of 1495 °C is...

For cobalt, Co, the heat of fusion at its normal melting point of 1495 °C is 15.5 kJ/mol.

The entropy change when 1.70 moles of liquid Co freezes at  1495 °C, 1 atm is ___ J/K.

Homework Answers

Answer #1

∆S = - 14.91 J/K mol

Explanation

Free energy change(∆G) is related to enthalphy change(∆H) and entrophy change(∆S) as follows

∆G = ∆H - T∆S

where , T is temperature in Kelvin

For phase change , ∆G = 0

Therefore,

∆S = ∆H/T

enthalphy of fusion , ∆Hfus = 15.5kJ/mol

Melting point , T​​​​​​m = 1495°C = 1768.15K

∆Sfus = ∆Hfus/Tm

∆Sfus = 15.5kJ/mol / 1768.15K

∆Sfus = 0.008766 kJ/K mol

∆Sfus = 8.77 J/ K mol

For freezing

∆Sfreeze = -8.77J/K mol

∆Sfreeze for 1.70 moles of liquid Co

= -8.77J/K mol × 1.70mol

= - 14.91J/K mol

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