Bond Enthalpy
Mediumchemistry
The bond enthalpies of H–H, Cl–Cl, and H–Cl bonds are 436, 242, and 431 kJ mol⁻¹ respectively. The enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g) is:
Select the correct option:
Solution
Incorrect! Answer:
−184 kJ mol⁻¹
- Bond Enthalpy Method: ΔH=∑(B.E. of bonds broken)−∑(B.E. of bonds formed).
- Bonds Broken (reactant side):
- 1 × H–H = 436 kJ
- 1 × Cl–Cl = 242 kJ
- Total = 678 kJ.
- Bonds Formed (product side):
- 2 × H–Cl = 2 × 431 = 862 kJ.
- Calculate: ΔH=678−862=−184 kJ mol⁻¹.
- Interpretation: The reaction is exothermic because more energy is released during the formation of two strong H–Cl bonds than is consumed in breaking the weaker Cl–Cl and H–H bonds.
- Cross-check: The known experimental value of ΔH for this reaction is approximately −185 kJ mol⁻¹, consistent with our calculation.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- bond enthalpy
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
−184 kJ mol⁻¹
- Bond Enthalpy Method: ΔH=∑(B.E. of bonds broken)−∑(B.E. of bonds formed).
- Bonds Broken (reactant side):
- 1 × H–H = 436 kJ
- 1 × Cl–Cl = 242 kJ
- Total = 678 kJ.
- Bonds Formed (product side):
- 2 × H–Cl = 2 × 431 = 862 kJ.
- Calculate: ΔH=678−862=−184 kJ mol⁻¹.
- Interpretation: The reaction is exothermic because more energy is released during the formation of two strong H–Cl bonds than is consumed in breaking the weaker Cl–Cl and H–H bonds.
- Cross-check: The known experimental value of ΔH for this reaction is approximately −185 kJ mol⁻¹, consistent with our calculation.
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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