CHEMISTRY THE CENTRAL SCIENCE

8 BASIC CONCEPTS OF CHEMICAL BONDING

8.8 STRENGTHS OF COVALENT BONDS

The stability of a molecule is related to the strengths of its covalent bonds. The strength of a covalent bond between two atoms is determined by the energy required to break the bond. It is easiest to relate bond strength to the enthalpy change in reactions in which bonds are broken. (Section 5.4) The bond enthalpy is the enthalpy change, ΔH, for the breaking of a particular bond in one mole of a gaseous substance. For example, the bond enthalpy for the bond in Cl2 is the enthalpy change when 1 mol of Cl2(g) dissociates into chlorine atoms:

We use the letter D followed by the bond in question to represent bond enthalpies. Thus, for example, D(Cl—Cl) is the bond enthalpy for the Cl2 bond, and D(H—Br) is the bond enthalpy for the HBr bond.

It is relatively simple to assign bond enthalpies to the bond in a diatomic molecule because in these cases the bond enthalpy is just the energy required to break the molecule into its atoms. However, many important bonds, such as the C—H bond, exist only in polyatomic molecules. For these bonds, we usually use average bond enthalpies. For example, the enthalpy change for the following process in which a methane molecule is decomposed into its five atoms (a process called atomization) can be used to define an average bond enthalpy for the C—H bond, D(C—H):

Because there are four equivalent C—H bonds in methane, the enthalpy of atomization is equal to the sum of the bond enthalpies of the four C—H bonds. Therefore, the average C—H bond enthalpy for CH4 is D(C—H) = (1660/4) kJ/mol = 415 kJ/mol.

The bond enthalpy for a given pair of atoms, say —H, depends on the rest of the molecule containing the atom pair. However, the variation from one molecule to another is generally small, which supports the idea that bonding electron pairs are localized between atoms. If we consider C—H bond enthalpies in many different compounds, we find that the average bond enthalpy is 413 kJ/mol, close to the 415 kJ/mol we just calculated from CH4.

GIVE IT SOME THOUGHT

How can you use the enthalpy of atomization of the hydrocarbon ethane, C2H6(g), along with the value D(C—H) = 413 kJ/mol to estimate the value for D(C—C)?

TABLE 8.4 lists average bond enthalpies for a number of atom pairs. The bond enthalpy is always a positive quantity; energy is always required to break chemical bonds. Conversely, energy is always released when a bond forms between two gaseous atoms or molecular fragments. The greater the bond enthalpy, the stronger the bond. Further, a molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds. For example, N2, which has a very strong N≡N triple bond, is very unreactive, whereas hydrazine, N2H4, which has an N≡N single bond, is highly reactive.

GIVE IT SOME THOUGHT

Based on bond enthalpies, which do you expect to be more reactive, oxygen, O2, or hydrogen peroxide, H2O2?

TABLE 8.4 • Average Bond Enthalpies (kJ/mol)

Bond Enthalpies and the Enthalpies of Reactions

We can use average bond enthalpies to estimate the enthalpies of reactions in which bonds are broken and new bonds are formed. This procedure allows us to estimate quickly whether a given reaction will be endothermic (ΔH > 0) or exothermic (ΔH < 0) even if we do not know  for all the species involved.

Our strategy for estimating reaction enthalpies is a straightforward application of Hess's law. (Section 5.6) We use the fact that breaking bonds is always endothermic and forming bonds is always exothermic. We therefore imagine that the reaction occurs in two steps:

1. We supply enough energy to break those bonds in the reactants that are not present in the products. The enthalpy of the system is increased by the sum of the bond enthalpies of the bonds that are broken.

2. We form the bonds in the products that were not present in the reactants. This step releases energy and therefore lowers the enthalpy of the system by the sum of the bond enthalpies of the bonds that are formed.

The enthalpy of the reaction, ΔHrxn, is estimated as the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed:

Consider, for example, the gas-phase reaction between methane, CH4, and chlorine to produce methyl chloride, CH3Cl, and hydrogen chloride, HCl:

GO FIGURE

Is this reaction exothermic or endothermic?

FIGURE 8.15 Using bond enthalpies to calculate ΔHrxn. Average bond enthalpies are used to estimate ΔHrxn for the reaction in Equation 8.13.

Our two-step procedure is outlined in FIGURE 8.15. We note that the following bonds are broken and made:

Bonds broken: 1 mol C—H, 1 mol Cl—Cl

Bonds made: 1 mol C—Cl, 1 mol H—Cl

We first supply enough energy to break the C—H and Cl—Cl bonds, which raises the enthalpy of the system (indicated as ΔH1 > 0 in Figure 8.15). We then form the C—Cl and H—Cl bonds, which release energy and lower the enthalpy of the system (ΔH2 < 0). We then use Equation 8.12 and data from Table 8.4 to estimate the enthalpy of the reaction:

The reaction is exothermic because the bonds in the products (especially the H—Cl bond) are stronger than the bonds in the reactants (especially the Cl—Cl bond).

We usually use bond enthalpies to estimate ΔHrxn only if we do not have the needed  values readily available. For the preceding reaction, we cannot calculate ΔHrxn from  values and Hess's law because  for CH3Cl(g) is not given in Appendix C. If we obtain the value of for CH3Cl(g) from another source and use Equation 5.31,

we find that ΔHrxn = —99.8 kJ for the reaction in Equation 8.13. Thus, the use of average bond enthalpies provides a reasonably accurate estimate of the actual reaction enthalpy change.

It is important to remember that bond enthalpies are derived for gaseous molecules and that they are often averaged values. Nonetheless, average bond enthalpies are useful for estimating reaction enthalpies quickly, especially for gas-phase reactions.

TABLE 8.5 • Average Bond Lengths for Some Single, Double, and Triple Bonds

SAMPLE EXERCISE 8.12 Using Average Bond Enthalpies

Using data from Table 8.4, estimate ΔH for the reaction

SOLUTION

Analyze We are asked to estimate the enthalpy change for a chemical reaction by using average bond enthalpies for the bonds broken and formed.

Plan In the reactants, we must break twelve C—H bonds and two C—C bonds in the two molecules of C2H6 and seven O2 bonds in the seven O2 molecules. In the products, we form eight C=O bonds (two in each CO2) and twelve O—H bonds (two in each H2O).

Solve Using Equation 8.12 and data from Table 8.4, we have

Check This estimate can be compared with the value of -2856 kJ calculated from more accurate thermochemical data; the agreement is good.

PRACTICE EXERCISE

Using Table 8.4, estimate ΔH for the reaction

Answer: -86 kJ

Bond Enthalpy and Bond Length

Just as we can define an average bond enthalpy, we can also define an average bond length for a number of common bonds (TABLE 8.5). Of particular interest is the relationship, in any atom pair, among bond enthalpy, bond length, and number of bonds between the atoms. For example, we can use data in Tables 8.4 and 8.5 to compare the bond lengths and bond enthalpies of carbon-carbon single, double, and triple bonds:

CHEMISTRY PUT TO WORK
Explosives and Alfred Nobel

Enormous amounts of energy can be stored in chemical bonds. Perhaps the most graphic illustration of this fact is seen in certain molecular substances used as explosives. Our discussion of bond enthalpies allows us to examine more closely some of the properties of such explosive substances.

A useful explosive substance must (1) decompose very exothermically, (2) have gaseous products so that a tremendous gas pressure accompanies the decomposition, (3) decompose very rapidly, and (4) be stable enough so that it can be detonated predictably. The combination of the first three effects leads to the violent evolution of heat and gases.

To give the most exothermic reaction, an explosive should have weak chemical bonds and should decompose into molecules that have very strong bonds. Table 8.4 tells us that N≡N, C≡O, and C=O bonds are among the strongest. Not surprisingly, explosives are usually designed to produce the gaseous products N2(g), CO(g), I and CO2(g). Water vapor is nearly always produced as well.

Many common explosives are organic molecules that contain nitro (NO2) or nitrate (NO3) groups attached to a carbon skeleton. The Lewis structures of two of the most familiar explosives, nitroglycerin and trinitrotoluene (TNT), are shown here (resonance structures are not shown for clarity). TNT contains the six-membered ring characteristic of benzene.

Nitroglycerin is a pale yellow, oily liquid. It is highly shock-sensitive: Merely shaking the liquid can cause its explosive decomposition into nitrogen, carbon dioxide, water, and oxygen gases:

The large bond enthalpies of N2 (941 kJ/mol), CO2 (2 × 799 kJ/mol), and H2O (2 × 463 kJ/mol) make this reaction enormously exothermic. Nitroglycerin is an exceptionally unstable explosive because it is in nearly perfect explosive balance: With the exception of a small amount of O2(g) produced, the only products are N2, CO2, and H2O. Note also that, unlike combustion reactions  (Section 3.2), explosions are entirely self-contained. No other reagent, such as O2(g), is needed for the explosive decomposition.

Because nitroglycerin is so unstable, it is difficult to use as a controllable explosive. The Swedish inventor Alfred Nobel (FIGURE 8.16) found that mixing nitroglycerin with an absorbent solid material such as diatomaceous earth or cellulose gives a solid explosive (dynamite) that is much safer than liquid nitroglycerin.

RELATED EXERCISES: 8.98 and 8.99

FIGURE 8.16 Alfred Nobel (1833-1896), Swedish inventor of dynamite. By many accounts Nobel's discovery that nitroglycerin could be made more stable by absorbing it onto cellulose was an accident. This discovery made Nobel a wealthy man. He was also a complex and lonely man, however, who never married, was frequently ill, and suffered from chronic depression. He had invented the most powerful military explosive to date, but he strongly supported international peace movements. His will stated that his fortune be used to establish prizes awarding those who “have conferred the greatest benefit on mankind,” including the promotion of peace and “fraternity between nations.” The Nobel Prize is probably the most coveted award that a scientist, writer, or peace advocate can receive.

As the number of bonds between the carbon atoms increases, the bond length decreases and the bond enthalpy increases. That is, the carbon atoms are held more closely and more tightly together. In general, as the number of bonds between two atoms increases, the bond grows shorter and stronger. This trend is illustrated in FIGURE 8.17 for N—N single, double, and triple bonds.

GO FIGURE

The line segments in the graph both have negative slopes. Why does this make sense?

FIGURE 8.17 Bond strength versus bond length for N—N bonds.

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

Phosgene, a substance used in poisonous gas warfare during World War I, is so named because it was first prepared by the action of sunlight on a mixture of carbon monoxide and chlorine gases. Its name comes from the Greek words phos (light) and genes (born of). Phosgene has the following elemental composition:12.14% C, 16.17% O, and 71.69% Cl by mass. Its molar mass is 98.9 g/mol. (a) Determine the molecular formula of this compound. (b) Draw three Lewis structures for the molecule that satisfy the octet rule for each atom. (The Cl and O atoms bond to C.)(c) Using formal charges, determine which Lewis structure is the dominant one. (d) Using average bond enthalpies, estimate ΔH for the formation of gaseous phosgene from CO(g) and Cl2(g).

SOLUTION

(a) The empirical formula of phosgene can be determined from its elemental composition. (Section 3.5) Assuming 100 g of the compound and calculating the number of moles of C, O, and Cl in this sample, we have

The ratio of the number of moles of each element, obtained by dividing each number of moles by the smallest quantity, indicates that there is one C and one O for each two Cl in the empirical formula, COCl2.

The molar mass of the empirical formula is 12.01 + 16.00 + 2(35.45) = 98.91 g/mol, the same as the molar mass of the molecule. Thus, COCl2 is the molecular formula.

(b) Carbon has four valence electrons, oxygen has six, and chlorine has seven, giving 4 + 6 + 2(7) = 24 electrons for the Lewis structures. Drawing a Lewis structure with all single bonds does not give the central carbon atom an octet. Using multiple bonds, three structures satisfy the octet rule:

(c) Calculating the formal charges on each atom gives

The first structure is expected to be the dominant one because it has the lowest formal charges on each atom. Indeed, the molecule is usually represented by this single Lewis structure.

(d) Writing the chemical equation in terms of the Lewis structures of the molecules, we have

Thus, the reaction involves breaking a C≡O bond and a Cl—Cl bond and forming a C=O bond and two C—Cl bonds. Using bond enthalpies from Table 8.4, we have

Notice that the reaction is exothermic. Nevertheless, energy is needed from sunlight or another source for the reaction to begin, as is the case for the combustion of H2(g) and O2(g) to form H2O(g) (Figure 5.14).