Question

Ethanol boils at 78.4° centigrade and ∆Hvap of ethanol is 42.4k/mole .Calculate entropy of vaporisation?

Answer 1

We know that,

Entropy of vaporization, 

ΔvapS = ΔvapH/ Tb where, ΔvapH = enthalpy of vaporization and Tb= boiling point (temperature) in Kelvin 

Here, ΔvapS = 109.8J/Kmol, and boiling point Tb = 273 + 78.5 =351.5 K

So, enthalpy of  vaporization of ethanol per mole is ,  ΔvapH = ΔvapS x Tb = 109.8 x 351.5 = 38594.7  J/mol

Answer 2

ANSWER

IT INCREASES

43.4J/K-MOL