1. Determine the average bond energy of a bond using given enthalpies of reaction.
2. Use a table of bond energies to estimate H for a reaction.

Average Bond Energies. Occasionally the enthalpy change H is needed for a reaction for which enthalpy of formation data does not exist. One approach that allows an estimate of H° uses the concept of bond energy. Consider the gas phase atomization process:

CH₄(g) C(g) + 4H(g)     DH^o_rxn = 1664 kJ

In this reaction, four C—H bonds are broken; therefore we can define the average C—H bond energy as one-fourth of DH^o_rxn for the reaction. Hence the average C—H bond energy in CH₄ is 416 kJ/mol.

The actual bond energies of the individual C—H bonds in CH₄ are not the same as the average value. Even so, the use of average bond energies makes it possible to estimate the enthalpy changes of certain reactions. Some of the bond energies in Table 9.4 of the textbook are listed here in Table 9.4.

Table 9.4 Bond Energies of Some Common Bonds

Bond	Bond Energy (kJ/mol)
H—H	436
C—H	414
N—H	392
O O	499
C O	802
O—H	460
I—Cl	210
N N	941
N O	630

We can estimate H for the combustion of methane as follows:

CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O(g)

CH₄ + 2O₂ C + 4H + 4O CO₂ + 2H₂O

First we break 4 C—H bonds and 2 O O bonds, where SBE (reactants) is the sum of the bond energies of bonds broken. Then let the atoms recombine to form the 2 C O and 4 O—H bonds of the products. SBE (products) is the sum of the bond energies of all bonds formed from the atoms. Keep in mind that breaking bonds is an endothermic process, and making new bonds is an exothermic process. Therefore

H = SBE (reactants) – SBE (products)

H = [4 BE (C—H) + 2 BE (O O)] – [2 BE (C O) + 4 BE (O—H)]

Inserting values from Table 9.5 gives:

DH	= [4(414) + 2(499)] – [2(802) + 4(460)]
	= 2654 kJ – 3444 kJ

DH = –790 kJ

If we compare this value with the standard enthalpy of combustion of methane, we get

CH₄(g) + 2O₂(g) CO₂(g) + 2H₂O(l)    DH^o_rxn = –890 kJ

At first it seems there is a large discrepancy of 100 kJ. However, in our calculation, H₂O is present as a gas! But H° refers to the standard state of H₂O as a liquid. Therefore, the 100 kJ difference is largely due to the H of vaporization of two moles of H₂O, which is 81.4 kJ. Two important conclusions can be drawn here; (1) Bond energies are used to estimate enthalpies of reaction of gas phase reactions, and (2) enthalpies of reaction calculated from average bond energies are only approximate values.

EXAMPLE Use of Bond Energies to Estimate Hrxn

Estimate Hrxn for the reaction:

Cl₂(g) + I₂(g) 2ICl(g)

using bond energies given in Table 9.4 (text), and given that BE (I—Cl) = 210 kJ/mol.

kJ

          Correct!

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Complete the following questions to check your understanding of the material. Select the check button to see if you answered correctly.

1. Given the N N and H—H bond energies in Table 9.5 and standard enthalpy of reaction:

   N₂(g) + H₂(g) NH₃(g)    DH^o_rxn = –46.3 kJ/mol

   calculate the average N—H bond energy in ammonia.

2. Given the bond energies in Table 9.5, calculate the enthalpy of formation for ammonia, NH₃.

   N₂(g) + H₂(g) NH₃(g)    DH^o_f(NH₃) = ?