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What is the change in entropy when 0.170 mol of potassium freezes at 62.4

What is the change in entropy when 0.170 mol of potassium freezes at 62.4

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Expert Solution

Given freezing point of potasium is , Tf = 62.4 oC = 62.4 + 273 = 335.4 K

Enthalpy change for fusion , Hfus = 2.39 kJ/mol

This is for one mole.For 0.170 mol of Potassium the Enthalpy change for fusion ,

Hfus = 0.170 x 2.39 kJ

         = 0.4063 KJ

         = 406.3 J

We know that change in entropy of fusion , Sfus =Hfus / Tf

                                                                        = 406.3 J / 335.4K

                                                                        = 1.211 J/K

queen_honey_blossom answered 2 years ago
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