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Question
The enthalpy change of the following reaction:
\[\ce{CH_{4(g)} + Cl_{2(g)} -> CH3Cl_{(g)} + HCl_{(g)}ΔH^0 = –104 kJ}\]
Calculate C – Cl bond enthalpy. The bond enthalpies are:
| Bond | C − H | Cl − Cl | H − Cl |
| ∆H°/kJ mol−1 | 414 | 243 | 431 |
Solution
Given: ΔrH0 = –104 kJ
ΔH0 (C – H) = 414 kJ mol–1
ΔH0 (Cl – Cl) = 243 kJ mol–1
ΔH0 (H – Cl) = 431 kJ mol–1
To find: C – Cl bond enthalpy, ΔH0 (C – Cl)
Formula: ΔrH0 = ∑ΔH°(reactant bonds) − ∑ΔH0 (product bonds)
Calculation: ΔrH0 = Σ ΔH0 (reactant bonds) – Σ ΔH0 (product bonds)
= [4 ΔH0(C – H) + ΔH°(Cl – Cl)] – [3 ΔH0(C – H) + ΔH0(C – Cl) + ΔH0(H – Cl)]
= ΔH0(C – H) + ΔH0(Cl – Cl) – ΔH0(C – Cl) – ΔH0(H – Cl) –104 kJ
∴ = 1 mol × (414 kJ mol–1) + 1 mol × (243 kJ mol–1) – ΔH0(C – Cl) – 1 mol × (431 kJ mol–1)
∴ –104 kJ = 226 kJ – ΔH°(C – Cl)
∴ ΔH0(C – Cl) = 226 kJ + 104 kJ
∴ ΔH0(C – Cl) = 330 kJ mol–1
C – Cl bond enthalpy is 330 kJ mol–1.
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