Hess's Law
Given: C(s) + O₂(g) → CO₂(g), ΔH = −394 kJ; 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566 kJ. The enthalpy of formation of CO(g) is:
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- hess's law
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
−110.5 kJ
Target reaction for enthalpy of formation of CO(g): C(s)+21O2(g)→CO(g) Given Eq 1: C(s)+O2(g)→CO2(g),ΔH1=−394 kJ Given Eq 2: 2CO(g)+O2(g)→2CO2(g),ΔH2=−566 kJ Divide Eq 2 by 2 and reverse it: CO2(g)→CO(g)+21O2(g),ΔH3=+2566=+283 kJ Add Eq 1 and Eq 3: C(s)+O2(g)+CO2(g)→CO2(g)+CO(g)+21O2(g) Simplifying: C(s)+21O2(g)→CO(g) ΔHf=ΔH1+ΔH3=−394+283=−111 kJ (Exact value is around −110.5 kJ).
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of hess's law. It appeared in the 2025 exam.
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