Question

The normal boiling point of benzene is 80.1 c and the heat of evaporation h is 30.7 kj/mol calculate s for the reaction

Answer 1

Answer is: entropy of evaporization of benzene is 86,7 J/mol·K.
T(benzene - C₆H₆) = 80,1 °C = 80,1 + 273,15 = 353,25 K.
ΔHvap(C₆H₆) = 30,7 kJ/mol.
ΔSvap(C₆H₆) = ?
ΔS(C₆H₆) = ΔH ÷ T
ΔS(C₆H₆) = 30,7 kJ/mol ÷ 353,25 K
ΔS(C₆H₆) = 0,087 kJ/mol·K = 86,7 J/mol·K.
ΔS - entropy of evaporization.