CALCULATE the enthalpy of vapourisation per mole for ethanol given delta s =109.8J/KM AND THE BOILING POINT OF ETHANOL 78.5 CELSIUS

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

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