Determination of the standard enthalpy of formation of liquid benzene Bond O=O H—H D° 498.3 436.0 St — Physical Chemistry — Thermodynamics Chemistry Question
Study of liquid benzene hydrogenation
Determination of the standard enthalpy of formation of liquid benzene
Bond O=O H—H
D° 498.3 436.0
Standard latent heat at 298 K in kJ mol‒1
∆subH°(C(graphite)) = 716.70 kJ mol‒1
∆vapH°(C6H6) = 33.90 kJ mol‒1
The different steps of benzene hydrogenation into cyclohexane are given in the scheme 1.
Scheme 1: benzene hydrogenation
Write down the balanced chemical equation for the formation of liquid benzene from its constituent elements in their standard states.
Model Answer
6 C(graphite) + 3 H2(g) = C6H6(l)
Calculate the standard enthalpy of formation of liquid benzene ∆fH°(C6H6(l)) using liquid cyclohexane.
Model Answer
ΔfH°(C6H6(l))
6 C(graphite) + 3 H2(g) C6H6(l)
Calculate the standard enthalpy of formation of liquid benzene ∆fH°(C6H6(l)) using Hess law.
Model Answer
∆fH°(C6H6(l)) = 6 ∆combH°(C(graphite)) – 3 ∆combH°(H2(g)) – ∆combH°(C6H6(l))
∆fH°(C6H6(l)) = –6 × 393.5 – 3 × 285.6 + 3268.0
∆fH°(C6H6(l)) = 50.2 kJ mol–1
Calculate the difference between the ∆fH°(C6H6(l)) values obtained in the two previous questions. Choose the correct explanation for this difference.
- The difference is due to experimental errors on the values of standard enthalpies of combustion reactions.
- The method used at question 2 does not take into account the nature of bonds in benzene.
- The Hess law is only rigorously applicable with standard enthalpies of formation.
- The method used in question 3 does not take into account the electronic delocalization.
Model Answer
Eresonance = 50.2 – 201.3 = –151.1 kJ mol–1
Correct statement: The method used at question 2 does not take into account the nature of bonds in benzene.
Calculate the enthalpy of reaction for the full hydrogenation of liquid benzene into liquid cyclohexane.
Model Answer
C6H6(l) + 3H2(g) = C6H12(l)
∆rH°(hydrogenation) = – ∆fH°(C6H6(l)) – 3∆fH°(H2(g)) + ∆fH°(C6H12(l))
∆rH° = –50.2 – 3 × 0.0 – 156.4 = ‒206.6 kJ mol–1
Complete this scheme by calculating the standard enthalpy of hydrogenation of
Model Answer
∆rH°(hydrogenation) = x – 112.1 – 119.7 = ‒ 206.6
x = 25.2 kJ mol–1
Correct statement:
The breaking of benzene aromaticity.
Model Answer
2 H2
2 H2
206.6 kJ mol-1
Eres = 7.6 kJ mol-1