Question

17.3 #2

For ethanol, C₂H₅OH, the heat of fusion at its normal melting point of -115 °C is 5.0 kJ/mol.

The entropy change when 1.51 moles of liquid C₂H₅OH freezes at -115 °C, 1 atm is ______ J/K.

Answer 1

entropy change at melting point = dHfusion/Tm

given ,Tm = -115 C= 273 K -115 = 158 K

dHfusion = 5 KJ/mol = 5000 J/mol , tho=is is for 1 mole ethanol

entropy change for 1.51 mol ethanol , dS= 1.51* 5000 j/mol/158 K = 47.7848 j/K

since liquid changes to solid , entropy reduced. Hence sign entropy will be negative

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