Question

10. The enthalpy of fusion for benzene (C6H6, 78.0 g/mol) is 127.40 kJ/kg, and its melting point is 5.5°C. What is the entropy change when 1 mole of benzene melts at 5.5°C?

Answer 1

`\Delta _fS=35.68(J)/(K)`

Step-by-step explanation:

Hello,

In this case, the entropy of fusion is computed in terms of the enthalpy of fusion considering the fusion temperature in kelvins:

`\Delta _fS=(\Delta _fH)/(T)`

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:

`m=1mol*(78.0g)/(1mol)=78g`

Next:

`\Delta _fH=78g*(127.4J)/(g)=9937.2J`

Finally, the entropy:

`\Delta _fS=(9937.2J)/((5.5+273)K)\\\\\Delta _fS=35.68(J)/(K)`

Best regards.