Bond Enthalpy
Given bond enthalpies: C-H = 413 kJ/mol, Cl-Cl = 242 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol. The enthalpy change for CH₄ + Cl₂ → CH₃Cl + HCl is:
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- bond enthalpy
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
−104 kJ
ΔHreaction=∑Bond Enthalpies (Reactants)−∑Bond Enthalpies (Products)
- Bonds to break (Reactants): 1 C−H bond + 1 Cl−Cl bond ∑BEreact=413+242=655 kJ/mol
- Bonds to form (Products): 1 C−Cl bond + 1 H−Cl bond ∑BEprod=328+431=759 kJ/mol ΔHrxn=655−759=−104 kJ/mol
- The reaction is exothermic as stronger bonds are formed than broken.
This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of bond enthalpy. It appeared in the 2025 exam.
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