This question is a typical application of thermodynamic cycles to estimate a bond dissociation entha — Physical Chemistry — Thermodynamics Chemistry Question
The bonds in dibenzyl
This question is a typical application of thermodynamic cycles to estimate a bond dissociation enthalpy.
The first step in the pyrolysis of toluene (methylbenzene) is the breaking of the C6H5CH2–H bond. The activation enthalpy for this process, which is essentially the bond dissociation enthalpy, is found to be 378.4 kJ mol–1.
Write a balanced equation for the complete combustion of toluene.
Model Answer
C7H8 + 9 O2 → 7 CO2 + 4 H2O.
Standard enthalpies are given below, using the recommended IUPAC notation (i.e. f = formation, c = combustion, vap = vaporisation, at = atomisation)
∆fH°(CO2, g, 298K) = –393.5 kJ mol–1
∆fH°(H2O, l, 298K) = –285.8 kJ mol–1
∆cH°(C7H8, l, 298K) = –3910.2 kJ mol–1
∆vapH°(C7H8, l, 298K) = +38.0 kJ mol–1
∆atH°(H2, g, 298K) = +436.0 kJ mol–1.
i) Calculate ∆fH°(C7H8, l, 298K)
ii) Estimate ∆fH° for the benzyl radical C6H5CH2·(g) at 298 K.
Model Answer
(all at 298 K)
i) ∆cH°(C7H8, l) = 7 ∆cH°(CO2, g) + 4 ∆fH°(H2O, l) − ∆fH°(C7H8, l)
⇒ ∆fH°(C7H8, l) = +12.2 kJ mol–1
ii) ∆fH°(Bz, g) = ∆fH°(C7H8, l) + ∆vapH°(C7H8) + ∆bondH°(Bz−H) − ½ ∆atH°(H2, g) = 210.6 kJ mol–1.
The standard entropy of vaporisation of toluene is 99.0 J K–1 mol–1.
iii) Calculate ∆vapG° for toluene at 298 K.
iv) What is the reference state of toluene at 298 K?
v) Calculate the normal boiling point of toluene.
Model Answer
iii) ∆vapG° = ∆vapH° – T∆vapS° = 8.50 kJ mol−1
iv) liquid (∆vapG° > 0)
v) TB = ∆vapH° / ∆vapS° = 384 K
The standard enthalpy of formation of dibenzyl (1,2–diphenylethane) is 143.9 kJ mol–1.
Calculate the bond dissociation enthalpy for the central C–C bond in dibenzyl, C6H5CH2–CH2C6H5.
Model Answer
∆bondH°(Bz–Bz) = 2 ∆fH°(Bz, g) – ∆fH°(Bz–Bz, g) = 277.3 kJ mol−1