Enthalpy And Hess's Law
Given: C(s) + O₂(g) → CO₂(g); ΔH = −393 kJ mol⁻¹ CO(g) + ½O₂(g) → CO₂(g); ΔH = −283 kJ mol⁻¹ The enthalpy of formation of CO(g) from its elements is:
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- enthalpy and hess's law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
−110 kJ mol⁻¹
- Target Reaction: C(s) + ½O₂(g) → CO(g); ΔH_f = ?
- Apply Hess's Law: Subtract the second equation from the first.
- Equation (i): C(s) + O₂(g) → CO₂(g); ΔH₁ = −393 kJ.
- Reverse Equation (ii): CO₂(g) → CO(g) + ½O₂(g); ΔH₂' = +283 kJ.
- Add: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂(g) from both sides and combine O₂ terms: C(s) + ½O₂(g) → CO(g).
- ΔH_f = −393 + 283 = −110 kJ mol⁻¹.
- Verification: The value is exothermic (negative), consistent with the known standard enthalpy of formation of CO.
This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of enthalpy and hess's law. It appeared in the 2025 exam.
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