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