Question
The standard molar enthalpy of fusion and the standard molar enthalpy of vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the standard molar entropy changes for the solid ↔ liquid and liquid ↔ vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
A. ΔS°fus = 22.6 J/K·mol; ΔS°vap = 67.1 J/K·mol
B. ΔS°fus = 39.1 J/K·mol; ΔS°vap = 87.8 J/K·mol
C. ΔS°fus = 19.8 J/K·mol; ΔS°vap = 3.87 J/K·mol
D. ΔS°fus = 46.1 J/K·mol; ΔS°vap = 101. J/K·mol
A. ΔS°fus = 22.6 J/K·mol; ΔS°vap = 67.1 J/K·mol
B. ΔS°fus = 39.1 J/K·mol; ΔS°vap = 87.8 J/K·mol
C. ΔS°fus = 19.8 J/K·mol; ΔS°vap = 3.87 J/K·mol
D. ΔS°fus = 46.1 J/K·mol; ΔS°vap = 101. J/K·mol
Answers
I got "B". Can anyone verify?
That's right. Good work.
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