The enthalpy of fusion for benzene (C₆H₆) is 127.40 kJ/kg, and its melting point is 5.5 $\circ$ C. What is the entropy change when 1 mole of benzene melts at 5.5 $\circ$ C? A)9.95 kJ/K B)35.7 J/K C)1,809 J/K D)1.81 J/K E)127.40 kJ/K

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Quizplus · Accepted Answer

The entropy change ($\Delta S$) can be calculated using the formula $\Delta S = \frac{\Delta H}{T}$, where $\Delta H$ is the enthalpy change and $T$ is the temperature in Kelvin. Given the enthalpy of fusion for benzene is 127.40 kJ/kg and its melting point is 5.5°C (which is 278.65 K when converted), for 1 mole of benzene (78.11 g/mol), the enthalpy change is 9.95 kJ (since 127.40 kJ/kg * 0.07811 kg = 9.95 kJ). Thus, $\Delta S = \frac{9.95 kJ}{278.65 K} = 0.0357 kJ/K = 35.7 J/K$.

The Enthalpy of Fusion for Benzene (C6H6) Is 127 ∘\circ∘ C What Is the Entropy Change When 1 Mole of and Its

The Enthalpy of Fusion for Benzene (C₆H₆) Is 127 $\circ$ C What Is the Entropy Change When 1 Mole of and Its