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For ethyl alcohol, C2H5OH, the enthalpy of fusion is 108.9 J/g, and the entropy of fusion is 31.6 J/mol •K. The enthalpy of vaporization at the boiling point is 837 J/g, and the molar entropy of vaporization is 109.9 J/mol •K.

Answer :

sebassandin

Answer:

[tex]T_m=157.22=-115.8^oC\\T_b=350.34K=77.34^oC[/tex]

Explanation:

Hello,

In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:

[tex]T_m=\frac{\Delta _fH}{\Delta _fS} =\frac{108.9\frac{J}{g}*\frac{46g}{1mol} }{31.6\frac{J}{mol*K} } \\\\T_m=157.22=-115.8^oC\\\\T_b=\frac{\Delta _bH}{\Delta _bS} =\frac{837\frac{J}{g}*\frac{46g}{1mol} }{109.9\frac{J}{mol*K} } \\\\T_b=350.34K=77.34^oC[/tex]

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