Unit 5 Worksheet 3 Bond enthalpies
Use the values of bond enthalpies given in Section 11 of the IB Chemistry data booklet.
1. All hydrocarbons burn according to the following equation (the state of the hydrocarbon, CxHy, at
standard conditions varies according to its mass and structure):
CxHy + (x + ¼y)O2(g) → xCO2(g) + ½yH2O(l)
Explain, in terms of bond enthalpies, why the combustion of all hydrocarbons is exothermic.
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2. Nitrogen and hydrogen react to produce ammonia in the Haber process:
N2(g) + 3H2(g) ⇄ 2NH3(g) Calculate the enthalpy of formation of ammonia.
3. Methane reacts with chlorine in the presence of ultraviolet light to produce chloromethane:
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) Calculate the enthalpy change for this reaction.
4. Consider the three compounds cyclohexane, C6H12, cyclohexene, C6H10 and benzene, C6H6.
(a) Calculate the enthalpy change for the hydrogenation reaction of gaseous cyclohexene to form
gaseous cyclohexane. C6H10(g) + H2(g) → C6H12(g)
(b) Both cyclohexene and cyclohexane are liquids at standard conditions. The enthalpies of vaporization
of cyclohexene and cyclohexane are + 33.5 kJ mol-1 and + 32.0 kJ mol-1 respectively. Use these values to
calculate the enthalpy of hydrogenation of cyclohexene, C6H10(l) + H2(g) → C6H12(l), and compare your
value with the literature value of -121.9 kJ mol-1.
(c) Assuming that benzene consists of alternate C=C and C-C bonds, use average bond enthalpies to
calculate the enthalpy of hydrogenation of benzene, C6H6(l) + 3H2(g) → C6H12(l) . (Benzene is a liquid at
standard conditions and the enthalpy of vaporization of benzene is + 33.9 kJ mol-1).
(d) It might be expected that the enthalpy of hydrogenation of benzene would be three times the value
of the enthalpy of hydrogenation of cyclohexene as three double bonds are hydrogenated. In fact the
value is – 210 kJ mol-1. The reason for the difference is that the alternate double and single bonds give
increased stability to the benzene ring through a process known as delocalization. Estimate the value of
the delocalization energy of benzene in kJ mol-1.
Answers
1. The sum of the strong O―H and C=O bonds being formed is always greater than the sum of the
weaker C―H, O=O (and C=C or CΞC where relevant) bonds being broken.
2. Energy in to break NΞN and 3H―H = 945 + (3 x 436) = 2253 kJ
Energy out to form 6N-H = 6 x 391 = 2346 kJ
∆H for reaction (which involves formation of two mol of NH3) = 2253 – 2346 = – 93 kJ
Enthalpy of formation of ammonia = ½ x - 93 = – 46.5 kJ mol-1
3. Energy in to break C―H and Cl―Cl = 414 + 242 = 656 kJ
Energy out to form C―Cl and H―Cl = 324 + 431 = 755 kJ
∆H = 656 – 755 = – 99 kJ mol-1
4. (a) Energy in to break C=C and H―H = 614 + 436 = 1050 kJ
Energy out to form C―C and 2C―H = 346 + (2 x 414) = 1174 kJ
∆H = 1050 – 1174 = – 124 kJ mol-1
(b) Extra energy in to vaporize cyclohexene to gaseous state = 33.5 kJ mol-1
Extra energy out to condense cyclohexane gas to liquid state = 32.0 kJ mol-1
Enthalpy of hydrogenation of cyclohexene = –124 + (33.5 – 32.0) = – 122.5 kJ mol-1
Difference with literature value = [(122.5 – 121.9) / 121.9] x 100 = 0.49%.
This difference is because average bond enthalpies were used for the C-C, C-H and C=C bonds
which are not the exact values for the bonds in cyclohexene and cyclohexane.
(c) Energy in to vaporise C6H6 and to break 3C=C and 3H―H = 33.9 + (3 x 614) + (3 x 436) = 3184 kJ
Energy out to form 3C―C and 6C―H and condense C6H12 = (3 x 346) + (6 x 414) + 32.0 = 3554 kJ
Enthalpy of hydrogenation of benzene = 3184 – 3554) = – 370 kJ mol-1
(d) Benzene is expected to give out either 370 kJ mol-1 (calculated from average bond enthalpies) or 366
kJ mol-1 ( calculated from 3 x literature value) when hydrogenated. In fact it only gives out 210 kJ mol1
as the difference in energy has been used to overcome the delocalization energy. Hence the
delocalization energy of benzene is approximately 160 kJ mol-1.