Question

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.

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 , ∆H_{fus} = 15.5kJ/mol

Melting point , T_{m} = 1495°C = 1768.15K

∆S_{fus} = ∆H_{fus}/T_{m}

∆S_{fus} = 15.5kJ/mol / 1768.15K

∆S_{fus} = 0.008766 kJ/K mol

∆S_{fus} = 8.77 J/ K mol

For freezing

∆S_{freeze} = -8.77J/K mol

∆S_{freeze} for 1.70 moles of liquid Co

= -8.77J/K mol × 1.70mol

= - 14.91J/K mol

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