Answer:
[tex]\Delta _fS=35.68\frac{J}{K}[/tex]
Explanation:
Hello,
In this case, the entropy of fusion is computed in terms of the enthalpy of fusion considering the fusion temperature in kelvins:
[tex]\Delta _fS=\frac{\Delta _fH}{T}[/tex]
Thus, since the enthalpy of fusion is given in kJ/kg we must compute the grams of benzene in mole of benzene via its molar mass:
[tex]m=1mol*\frac{78.0g}{1mol}=78g[/tex]
Next:
[tex]\Delta _fH=78g*\frac{127.4J}{g}=9937.2J[/tex]
Finally, the entropy:
[tex]\Delta _fS=\frac{9937.2J}{(5.5+273)K}\\\\\Delta _fS=35.68\frac{J}{K}[/tex]
Best regards.
