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Worked Examples · Example 4.13
Q.

The enthalpy change of the following reaction CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g)\mathrm{CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)}, ΔrH0\Delta_r H^0 = -104 kJ. Calculate C-Cl bond enthalpy. The bond enthalpies are

BondC-HCl-ClH-Cl
ΔH0\Delta H^0/kJ mol−1^{-1}414243431
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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−104=[414+243]−[ΔH0(C−Cl)+431]-104 = [414 + 243] - [\Delta H^0(\mathrm{C{-}Cl}) + 431], so ΔH0(C−Cl)=330\Delta H^0(\mathrm{C{-}Cl}) = 330 kJ mol−1^{-1}.

Step 1. In CH4+Cl2→CH3Cl+HCl\mathrm{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl} the net change is: 1 C-H bond and 1 Cl-Cl bond break; 1 C-Cl bond and 1 H-Cl bond form (the other three C-H bonds survive unchanged in CH3_3Cl and cancel).

Step 2. ΔrH0=[ΔH0(C−H)+ΔH0(Cl−Cl)]−[ΔH0(C−Cl)+ΔH0(H−Cl)]\Delta_r H^0 = [\Delta H^0(\mathrm{C{-}H}) + \Delta H^0(\mathrm{Cl{-}Cl})] - [\Delta H^0(\mathrm{C{-}Cl}) + \Delta H^0(\mathrm{H{-}Cl})]. …

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