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The enthalpy change of the following reaction: CHA4(g)+ClA2(g)⟶CHA3ClA(g)+HClA(g)ΔHA0=−104kJ Calculate C – Cl bond enthalpy. The bond enthalpies are: Bond C − H Cl − Cl H − Cl ∆H°/kJ mol−1 414 243 431 - Chemistry

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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
Numerical
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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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Chapter 4: Chemical Thermodynamics - Short answer questions (Type- I)

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