Question 8.10: USING BOND DISSOCIATION ENERGIES TO CALCULATE ΔH° Use the da...
USING BOND DISSOCIATION ENERGIES TO CALCULATE ΔH°
Use the data in Table 7.1 on page 219 to find an approximate ΔH° in kilojoules for the industrial synthesis of chloroform by reaction of methane with Cl_{2}.
CH_{4}(g) + 3 Cl_{2}(g) → CHCl_{3}(g) + 3 HCl(g)TABLE 7.1 Average Bond Dissociation Energies, D (kJ/mol)
| H—H 436^{a} C—H 410 N—H 390 O—F 180 I—I 151^{a}
H—C 410 C—C 350 N—C 300 O—Cl 200 S—F 310 H—F 570^{a} C—F 450 N—F 270 O—Br 210 S—Cl 250 H—Cl 432^{a} C—Cl 330 N—Cl 200 O—I 220 S—Br 210 H—Br 366^{a} C—Br 270 N—Br 240 O—N 200 S—S 225 H—I 398^{a} C—I 240 N—N 240 O—O 180 H—N 390 C—N 300 N—O 200 F—F 159a H—O 460 C—O 350 O—H 460 Cl—Cl 243^{a} H—S 340 C—S 260 O—C 350 Br—Br 193^{a} Multiple covalent bonds^{b} C=C 728 C≡C 965 C=O 732 O=O 798^{a} N≡N 945^{a} |
^{a}Exact value
^{b} We’ll discuss multiple covalent bonds in Section 7.5.
STRATEGY
Identify all the bonds in the reactants and products, and look up the appropriate bond dissociation energies in Table 7.1.Then subtract the sum of the bond dissociation energies in the products from the sum of the bond dissociation energies in the reactants to find the enthalpy change for the reaction.
The reactants have four C—H bonds and three Cl—Cl bonds; the products have one C—H bond, three C—Cl bonds, and three H-Cl bonds. The bond dissociation energies from Table 7.1 are:
C—H D = 410 kJ/mol Cl—Cl D = 243 kJ/mol
C—Cl D = 330 kJ/mol H—Cl D = 432 kJ/mol
Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the enthalpy change for the reaction:
ΔH° = [3 D_{Cl—Cl} + 4 D_{C—H}] – [D_{C¬H} + 3 D_{H—Cl} + 3 D_{C—Cl}]
= [(3 mol)(243 kJ/mol) + (4 mol)(410 kJ/mol)] – [(1 mol)(410 kJ/mol) + (3 mol)(432 kJ/mol) + (3 mol)(330 kJ/mol)]
= -327 kJ
The reaction is exothermic by approximately 330 kJ.