Bond Enthalpy Approximation
Hardchemistry
Estimate ΔH (kJ mol⁻¹) for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) using average bond enthalpies. (Data approximate: C–H: 413, O=O: 498, C=O in CO2: 799, O–H: 463).
Select the correct option:
Solution
Incorrect! Answer:
-800 kJ mol⁻¹
- Formula: ΔH=∑BHbroken−∑BHformed.
- Bonds Broken (Reactants):
- 4×C−H = 4×413=1652 kJ.
- 2×O=O = 2×498=996 kJ.
- Total reactant energy = 2648 kJ.
- Bonds Formed (Products):
- 2×C=O (in CO2) = 2×799=1598 kJ.
- 2×2(O−H) (in 2H2O) = 4×463=1852 kJ.
- Total product energy = 3450 kJ.
- Calculation: ΔH=2648−3450=−802 kJ.
- Result: Roughly −800 kJ/mol.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- bond enthalpy approximation
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
-800 kJ mol⁻¹
- Formula: ΔH=∑BHbroken−∑BHformed.
- Bonds Broken (Reactants):
- 4×C−H = 4×413=1652 kJ.
- 2×O=O = 2×498=996 kJ.
- Total reactant energy = 2648 kJ.
- Bonds Formed (Products):
- 2×C=O (in CO2) = 2×799=1598 kJ.
- 2×2(O−H) (in 2H2O) = 4×463=1852 kJ.
- Total product energy = 3450 kJ.
- Calculation: ΔH=2648−3450=−802 kJ.
- Result: Roughly −800 kJ/mol.
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of bond enthalpy approximation. It appeared in the 2025 exam.
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