Question

The value of enthapy of vapourization of a liquid is 32kj/mol.The standard enthalpy of formation of that liquid is 20kj/mol .Calculate the standard enthalpy of formation of vapourization state of that liquid?

Answer 1

Explanation:

△H= enthalpy of vapourisation =30 kJ/mol =30000 J/mol

△S= entropy of vaporisation =75 J/mol/K

At boiling point,

The reversible process liquid ⇋ Vapour is in equilibrium at one atmospheric pressure.

∴△G=0

△G=△H−T△S

∴T=

△S

△H

=

75 Jmol

−1

K

−1

30000 Jmol

−1

=400K