Bond Enthalpies (Bond Energies)

James Richard Fromm


Bond enthalpies are the enthalpy changes which result when one mole of bonds of a particular type, on average, are formed from isolated gaseous atoms. Since these values are estimates only approximate values can be given and the precision of the values should not be inferred from the precision with which they may be stated. Approximate bond enthalpies for single bonds are given in the Table below. The values of bond enthalpies for double bonds have also been estimated: C=C, 615; C=O, 749; and N=O, 607 kJ/mole. The values of bond enthalpies estimated for triple bonds are carbon-carbon, 812 kJ/mole, and nitrogen-nitrogen, 946 kJ/mole.


Table: Bond Enthalpies of Single Bonds (kJ/mole)
Atom H C N O F Cl S
H 436 413 391 463 563 432 368
C 413 346 305 358 489 328 272
N 391 305 163 222 275 192 ---
S 368 272 --- --- 327 271 251

The formation of methane from the elements carbon and hydrogen can be carried out in many ways. Consider two, with standard conditions being assumed throughout:

Path A: C(s) + 2H2(g) --> CH4(g)

Path B + C + D:

C(s) + 2H2(g) rarrow.gif (63 bytes) C(g) + 2H2(g); vaporization

C(g) + 2H2(g) rarrow.gif (63 bytes) C(g) + 4H(g); dissociation to atoms

C(g) + 4H(g) rarrow.gif (63 bytes) CH4(g); bond formation

Since the same state change occurs either directly or indirectly, the enthalpy change must also be the same, i.e., A = B + C + D, or -74.81 = +716.682 + 4(217.965) - 1663.352 kJ. This value for step D of -1663.352 kJ corresponds to the enthalpy change on formation of methane from the atoms (not from the usual forms of the elements). This is the enthalpy change on the formation of chemical bonds. Methane contains four C-H bonds only, so the heat of formation of one C-H bond is then (-1663.352/4) = -416 kJ/bond. This heat is called the bond enthalpy of the C-H bond in methane.

Bond enthalpies, which are sometimes (erroneously) called bond energies, are found to be very similar from one organic compound to another. Calculations of enthalpies of molecules or parts of molecules can be made on the basis that all bonds of the same type, such as carbon-hydrogen single bonds, have the same value of bond enthalpies. Bond enthalpies are less similar from one inorganic compound to another because the nature of bonding in inorganic compounds varies to a greater extent and inorganic molecules are generally smaller so the bonds at one end of the molecule are closer to, and have a greater effect on, the bonds at the opposite end of the molecule. As a consequence, tables of approximate bond enthalpies are less useful for inorganic compounds.

The heats of combustion of organic compounds can be estimated from bond enthalpies if the bond enthalpy of the diatomic oxygen O=O bond is known; this is quite different from the O-O single bond which is found in organic peroxides. This O=O bond enthalpy is simply the enthalpy change of the reaction

2O(g) rarrow.gif (63 bytes) O2(g),

and under standard conditions would be 0.0 - 2(-249.170) for a bond enthalpy of 498 kJ/mole. Estimations of heats of combustion and other heats made using bond enthalpies are likely to have errors of +/-15% of standard or measured enthalpy values. The significance of bond enthalpies lies more in the very broad extent of their application to organic compounds than in the precision of their estimations.


Example. Let us estimate the molar heat of combustion of ethane using bond enthalpies and compare it with the molar heat of combustion of ethene calculated using standard enthalpies of formation.

In the reaction

H2C=CH2 + 3O2 rarrow.gif (63 bytes) 2CO2 + 2H2O(g)

there are destroyed three O=O bonds in oxygen and four C-H single bonds and one C=C double bond in ethene, while there are formed four C=O double bonds and four H-O single bonds. The original bond enthalpies were 4(413) + 615 + 3(498) = 3761 kJ/mole while the final bond enthalpies were 4(749) + 4(463) = 4848 kJ/mole; the difference, +1087 kJ/mole, is the estimated molar heat of combustion of ethene.

The standard enthalpy change in the combustion reaction of ethene is 2(+52.25) + 0 -2(-393.509) - 2(-241.818) = +1375 kJ/mole, in approximate agreement.


Study Problems

1. Albert Einstein is reported to have said that one match is equivalent to enough energy to melt all the snow in Switzerland. Assume that he is right and that a match weighs exactly 10 grams; calculate the amount of snow in Switzerland on this basis. (One kilogram of snow can be melted by about 333.5 kJ of heat.)

2. Calculate the amount of heat obtained when one mole of (a) gaseous ethane, C2H6, and (b) gaseous ethanol, C2H5OH, are burned. Give a reasonable explanation, in terms of heats associated with specific bonds, for the difference.

3. Calculate the bond enthalpy of the hydrogen-fluorine bond in HF and compare it to the bond enthalpy of the fluorine-fluorine bond in F2.

4. A water calorimeter at 23.51oC contains 1.657 kg of water. Addition of 16.235 g of a compound which dissolves in water without dissociation to ions causes the temperature to decrease to 22.47oC. Calculate the heat change involved. What additional data would be needed to calculate the molar heat of solution of the compound? Given the standard molar heat of solution, what additional data would be needed to calculate the standard molar enthalpy of formation of the dissolved compound?

5. A steel bomb calorimeter has a mass of 6.7952 kg and a heat capacity of 3.9362 kJ/K. A combustion of 1.6257 g of methanol in excess oxygen takes place in the calorimeter. The initial temperature of the calorimeter is 22.961oC. Calculate the final temperature of the calorimeter.

6. Calculate the change in heat when 241.3 g of NO2(g) dimerizes to N2O4(g) at 25oC and one atmosphere pressure.

7. Calculate the change in heat when 316.4 g of chlorine react with excess hydrogen to form HCl at 25oC and one atmosphere pressure.


Copyright 1997 James R. Fromm