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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 8.3 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)

STRATEGY

Identify all the bonds in the reactants and products, and look up the appropriate bond dissociation energies in Table 8.3. 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.

TABLE 8.3 Average Bond Dissociation Energies, D (kJ/mol)
H–H 436a C–H 410 N–H 390 O–F 180 I–I 151a
H–C 410 C–C 350 N–C 300 O–Cl 200 S–F 310
H–F 570a C–F 450 N–F 270 O–Br 210 S–Cl 250
H–Cl 432a C–Cl 330 N–Cl 200 O–I 220 S–Br 210
H–Br 366a C–Br 270 N–Br 240 O–N 200 S–S 225
H–I 298a 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 243a
H–S 340 C–S 260 O–C 350 Br–Br 193a
Multiple covalent bonds
C=C 728 C≡C 965 C=O 732 O=O 498a N≡N 945a
a Exact value
Step-by-Step
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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 approximate bond dissociation energies from Table 8.3 are:

\begin{array}{llll}\mathrm{C}-\mathrm{H} & D=410 \mathrm{~kJ} / \mathrm{mol} & \mathrm{Cl}-\mathrm{Cl} & D=243 \mathrm{~kJ} / \mathrm{mol} \\\mathrm{C}-\mathrm{Cl} & D=330 \mathrm{~kJ} / \mathrm{mol} & \mathrm{H}-\mathrm{Cl} & D=432 \mathrm{~kJ} / \mathrm{mol}\end{array}

Subtracting the product bond dissociation energies from the reactant bond dissociation energies gives the approximate enthalpy change for the reaction:

\begin{aligned}\Delta H^{\circ}= & {\left[3 D_{\mathrm{Cl}-\mathrm{Cl}}+4 D_{\mathrm{C}-\mathrm{H}}\right]-\left[D_{\mathrm{C}-\mathrm{H}}+3 D_{\mathrm{H}-\mathrm{Cl}}+3 D_{\mathrm{C}-\mathrm{Cl}}\right] } \\= & {[(3 \mathrm{~mol})(243 \mathrm{~kJ} / \mathrm{mol})+(4 \mathrm{~mol})(410 \mathrm{~kJ} / \mathrm{mol})]-[(1 \mathrm{~mol})(410 \mathrm{~kJ} / \mathrm{mol})} \\& +(3 \mathrm{~mol})(432 \mathrm{~kJ} / \mathrm{mol})+(3 \mathrm{~mol})(330 \mathrm{~kJ} / \mathrm{mol})]\end{aligned} \\ = -327 \ kJ

The reaction is exothermic by approximately 330 kJ.

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