Question

The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.

Answer 1

Answer:

The change in entropy is -1.66 kJ/K.

Explanation:

Heat of vaporization of acetic acid = 41.0 kJ/mol

Heat of condensation of acetic acid = -41.0 kJ/mol

Mass of acetic acid = 954 g

Condensing temperature of acetic acid =118.1 °C = 391.25 K

Moles of acetic acid = $\frac{954 g}{60.05 g/mol}=15.8867 mol$

Heat evolved during condensation of 15.8867 moles of acetic acid:

$-41.0 kJ/mol\times 15.8867 mol=-651.35 kJ$

Entropy change = ΔS = $\frac{Heat}{Temperature}$

$\Delta S=\frac{-651.35 kJ}{391.25 K}=-1.664 kJ/K\approx -1.66 kJ/K$

The change in entropy is -1.66 kJ/K.