The standard molar entropy of vaporization ( S°vap ) for benzene is 88 J/mol•K a

Question

The standard molar entropy of vaporization (S°vap) for benzene is 88 J/mol•K at 25°C. Consider this information, along with the following standard molar enthalpies of formation (H°f) for benzene at 25°C, and estimate the normal boiling point of benzene. (H°f [C6H6(l)] = 49 kJ/mol; H°f[C6H6(g)] = 80 kJ/mol)

The standard molar entropy of vaporization (S°vap) for benzene is 88 J/mol•K at 25°C. Consider this information, along with the following standard molar enthalpies of formation (H°f) for benzene at 25°C, and estimate the normal boiling point of benzene. (H°f [C6H6(l)] = 49 kJ/mol; H°f[C6H6(g)] = 80 kJ/mol)

Explanation / Answer

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dG liquid = dG vapor

then

dG = 0 in equilibrium

dH = Hvap - Hliq = 80 - 49 = 31 kJ = 31000 J

dS = 88 J/K

then

if dG = 0

dG = dH - T*dS

dH = T*dS

T = dH/dS = 31000 J / mol /88 J/ molK= 352.272 K

T = 352.272-273.15

T = 79.122 °C

best asnwer is 79

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