The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization...

The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g)2at 1 atm condenses to liquid at the normal boiling point?

Expert Solution

The normal boiling point of ethanol, C2H5OH, is 78.3 °C,

T = 78.3 + 273 = 351.3K

The molar enthalpy of vaporization is 38.56 kJ/mol

Hvap = 38.56 kJ/mol = 38560J/mole

S = Hvap /T

= 38560J/mole/351.3K = 109.76J/mole-K

no of moles of ethanol = W/G.M.Wt

= 68.3/46 = 1.485 moles

S = 109.76J/mole-K * 1.485 moles

= 163J/K >>>>answer

The change in entropy in the system = 163J/K >>>>answer

queen_honey_blossom answered 2 years ago

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