The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system in J/K when 97.2 grams of ethanol at 1 atm condenses to a liquid at the normal boiling point?

The  change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate  change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The  change in entropy in the system is -231.5 J/K

answer: h+ its less than oh

explanation: