Based on the following thermodynamic data, calculate the boiling point of ethanol in degrees Celsius. Substance ΔH∘f (kJ/mol) S∘f [J/(K⋅mol)] C2H5OH(l) −277.7 160.6 C2H5OH(g) −235.1 282.6 Express the boiling point numerically in degrees Celsius.

76.03 °C.

Explanation:

Equation:

C2H5OH(l) --> C2H5OH(g)

ΔHvaporization = ΔH(products) - ΔH (reactants)

= (-235.1 kJ/mol) - (-277.7 kK/mol)

= 42.6 kJ/mol.

ΔSvaporization = ΔS(products) - ΔS(reactants)

= 282.6 J/K.mol - 160.6 J/K.mol

= 122 J/K.mol

= 0.122 kJ/K.mol

Using gibbs free energy equation,

ΔG = ΔH - TΔS

ΔG = 0,

T = ΔH/ΔS

T = 42.6/0.122

= 349.18 K.

Coverting Kelvin to °C,

= 349.18 - 273.15

= 76.03 °C.