Estimate the boiling point of benzene (C6H6) using its ΔH° and ΔS°.

Question

a. Estimate the boiling point of benzene (C₆H₆) using its ΔH⁰ and ΔS⁰. (C₆H₆ (l) ΔH° = 49.0 kJ/mol, ΔS° = 172.8 J/mol*K, C₆H₆ (g): ΔH° = 82.9 kJ/mol, ΔS° = 269.2 j/mol*K)

b. The actual boiling point of benzene (C₆H₆) is 80.1^oC. How do you explain any differences between your estimated value and the actual value?

Answer 1

1) boiling point is an equilibrium condition : ΔG = 0 = ΔH_vap - T_BPΔ S_vap

We make the assumption that the enthalpy change and the entropy change do not depend on the temperature (This is not correct, but not horrible)

T_BP = ΔH_vap/ΔS_vap in Kelvin Note that ΔH_vap ~ H_f,Benz(g) - H_f,Benz(l) for 25° but expressed in Joules, not kJ

2) The assumption isn't quite right - the delta values need to be at the T_BP, not standard T.

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