Bond Enthalpies and Exothermic or Endothermic Reactions
We are in a position to appreciate the general principle which determines whether a gaseous reaction will be exothermicDescribes a process in which energy is transferred to the surroundings as a result of a temperature difference. or endothermicIn chemical thermodynamics, describes a process in which energy is transferred from the surroundings to the system as a result of a temperature difference.. If less energyA system's capacity to do work. is needed to break up the reactantA substance consumed by a chemical reaction. molecules into their constituent atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. than is released when these atoms are reconstituted into productA substance produced by a chemical reaction. molecules, then the reaction will be exothermic. Usually an exothermic reaction corresponds to the breaking of weak bonds (with small bond enthalpies) and the making of strong bonds (with large bond enthalpies). In the hydrogen-fluorine reaction (Fig. 1 from Bond Enthalpies) one quite strong bond (DH–H = 436 kJ mol–1) and one very weak bond (DF–F = 159 kJ mol–1) are broken, while two very strong H―F bonds are made. (Note that with a bond enthalpyThe change in enthalpy when a mole of chemical bonds of a particular type are broken; the molecules whose bonds are broken must be in the gas phase. Closely related to bond energy. of 566 kJ mol–1, H―F is the strongest of all single bonds.) The combustionVigorous combination of a material with oxygen gas, usually resulting in a flame. of methane discussed in Example 1 from Bond Enthalpies is another example of the formation of stronger bonds at the expense of weaker ones. The bond enthalpy of the O―H bond is not much different in magnitude from those of the C―H and bonds which it replaces: All lie between 400 and 500 kJ mol–1. The determining factor making this reaction exothermic is the exceedingly large bond enthalpy of the bond which at 803 kJ mol–1 is almost twice as great as for the other bonds involved in the reaction. Not only this reaction but virtually all reactions in which CO2 with its two very strong bonds is produced are exothermic.
In other cases the number of bonds broken or formed can be important. A nice example of this is the highly exothermic (ΔH°(298 K) = –483.7 kJ mol–1) reaction between hydrogen and oxygen to form water:
- 2H2(g) + O2(g) → 2H2O(g)
All three types of bonds involved have comparable bond enthalpies:
- DH―H = 436 kJ mol–1 DO=O = 498 kJ mol–1 DO―H = 467 kJ mol–1
but the reason the reaction is exothermic becomes obvious if we rewrite it to make the bonds visible:
While three bonds must be broken (two H―H and one bond), a total of four bonds are made (four O―H bonds). Since all the bonds are similar in strength, making more bonds than are broken means the release of energy. In mathematical terms
- ΔH° = 2DH―H + DO=O + 4DO―H
- = 2 × 436 kJ mol–1 + 1 × 498 kJ mol–1 – 4 × 467 kJ mol–1
- = – 498 kJ mol–1
- ΔH° = 2DH―H + DO=O + 4DO―H
In summary, there are two factors which determine whether a gaseous reaction will be exothermic or not: (1) the relative strengths of the bonds as measured by the bond enthalpies, and (2) the relative number of bonds broken and formed. An exothermic reaction corresponds to the formation of more bonds, stronger bonds, or both.
Since the strength of chemical bonds is a factor in determining whether a reaction will release energy or not, it is obviously important to know which kinds of bonds will be strong and which weak, and we can make some empirical generalizations about the magnitudes of bond enthalpies. The first and most obvious of these is that triple bonds are stronger than double bonds which in turn are stronger than single bonds. As can be seen from Table 1 in Bond Enthalpies, triple bonds have bond enthalpies in the range of 800 to 1000 kJ mol–1. Double bonds range between 400 and 800 kJ mol–1 and single bonds are in the range of 150 to 500 kJ mol–1.
A second generalization is that the strengths of bonds usually increase with polarity. The bond enthalpies of the hydrogen halides, for instance, increase in the order HI < HBr < HCl < HF, and a similar order can be noted for bonds between carbon and halogens. There are exceptions to this rule, though. One would expect the C―N bond to be intermediateIn chemical kinetics, a species that is formed in an early step in a reaction mechanism and then consumed in a later step; evidence of existence of an intermediate may be important for the interpretation of a rate law. in strength between the C―C bond and the C―O bond. As the table shows, it is actually weaker than either.
A third factor affecting the strength of bonds is the size of the atoms. For the most part smaller atoms form stronger bonds. The smallest atom, hydrogen, forms four of the five strongest single bonds in the table. This is not entirely a matterAnything that occupies space and has mass; contrasted with energy. of size since hydrogen is also the most electropositive elementA substance containing only one kind of atom and that therefore cannot be broken down into component substances by chemical means. featured. Difference in electronegativity, however, does not explain why the H―H bond enthalpy is so large. If we look at the halogens (VIIA elements), we find that the bond enthalpies of I―I, Br―Br, and Cl―Cl increase as expected with decreasing size. The F―F bond is an exception, though: DF―F (158 kJ mol–1) is significantly smaller than DCl―Cl (242 kJ mol–1). Other notable exceptions to this rule are the N―N and O―O bonds. The occurrence of these weak single bonds is of considerable importance to the chemistry of compounds which contain them.The value of a particular bond enthalpy in a given molecule can sometimes be very informative about the nature of the bonding in the molecule. This is especially true of molecules in which resonanceA situation that arises when two or more Lewis structures can be drawn to represent covalent bonding in an ion or molecule. In each Lewis structure the atomic nuclei remain in the same positions; only the electrons move. When resonance applies, bonding is considered to be an average superposition of the individual Lewis structures. is a possibility, as the following example shows.
EXAMPLE 1 When benzene is burned in oxygen according to the equation
C6H6(g) + 15/2 O2(g) → 6CO2(g) + 3H2O(g)
calorimetric measurements give the value of [ΔH°(298 K) as –3169 kJ mol–1 benzene. Use this information together with Table 1 from the Bond Enthalpies section to calculate the mean bond enthalpy for the carbon-carbon bond in benzene.
SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Indicating the required bond enthalpy by the symbol DCC and carefully counting the bonds broken and bonds formed, we have
ΔH° = 6DC―H + 6DCC + 15/2DO=O – DC=O + 6DO―H
or –3169 kJ mol–1 = 6 (416 kJ mol–1) + 6DCC + 15/2 (498 kJ mol–1) – 12 (803 mol–1) – 6 (467 kJ mol–1) = –6207 kJ mol–1 + 6DCC
Thus 6DCC = (–3169 + 6207) kJ mol–1 = 3038 kJ mol–1
or DCC = 506 kJ mol–1
As expected for a resonance structureAny of two or more Lewis structures that represent covalent bonding in an ion or molecule where the positions of the atomic nuclei remain the same. No single resonance structure represents the actualy bonding; rather the bonding is an average superposition of the individual Lewis structures. the bond-enthalpy value is intermediate between that given in the table for a C―C single bond (348 kJ mol–1) and that given for a double bond (614 kJ mol–1).