Question

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 in J/K when 77.2 grams of ethanol at 1 atm condenses to a liquid at the normal boiling point?

825

-184

-263

263

-825

Answer 1

Given:

• Boiling point of ethanol (T): 78.3 °C = 351.45 K (converted to Kelvin)

• Molar enthalpy of vaporization (ΔHvap): 38.56 kJ/mol = 38,560 J/mol

• Mass of ethanol (m): 77.2 g

• Molar mass of ethanol (C₂H₅OH): ~46.07 g/mol

• 
  

Step 1: Calculate moles of ethanol

$$\text{moles}=\frac{\text{mass}}{\text{molar mass}}=\frac{77.2\text{ g}}{46.07\text{ g/mol}}\approx 1.676\text{ mol}$$

Step 2: Determine heat released during condensation

Since condensation is the reverse of vaporization, the heat released (q) is:

$$q=-\text{moles}\times \mathrm{\Delta }{H}_{\text{vap}}=-1.676\text{ mol}\times 38,560\text{ J/mol}\approx -64,630\text{ J}$$

Step 3: Calculate entropy change (ΔS)

$$\mathrm{\Delta }S=\frac{q}{T}=\frac{-64,630\text{ J}}{351.45\text{ K}}\approx -184\text{ J/K}$$

Answer is :

The change in entropy is -184 J/K (negative because the system becomes more ordered during condensation).