What is the change in entropy when 0.210 mol of potassium freezes at 63.7°C? (ΔHfus = 2.39 kJ/mol) kJ/K

1.75 J/K.

Explanation:

Using Gibbs free energy equation,

ΔG = ΔH - TΔS

Where,

ΔG = free energy

ΔH = enthalpy change

T = absolute temperature

ΔS = entropy change

Note: ΔG of fusion = 0

Therefore,

ΔH = TΔS,

ΔS = ΔH/T

Given:

T = °C + 273.15

= 273.15 + 69

= 342.15 K

n = 0.25 mol.

ΔH = 2.39 kJ/mol

= 2.39 / 342.15

= 6.99 J/Kmol

ΔS = ΔSmolar * n

= 6.99 * 0.25

= 1.75 J/K.