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:
Select the correct option:
Solution
−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).
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More hess's law Practice Questions
About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- hess's law
- Difficulty
- Hard
- Year
- 2025
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of hess's law. It appeared in the 2025 exam. Practice this and similar questions to strengthen your understanding of chemical thermodynamics concepts.
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