Δg And Equilibrium
For a reaction at 298 K, ΔG° = -9.9 kJ mol⁻¹. The equilibrium constant K is approximately (R = 8.314 J mol⁻¹ K⁻¹):
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- δg and equilibrium
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
50
- Formula: ΔG∘=−RTlnK.
- Rearrangement: lnK=−RTΔG∘.
- Calculation:
- ΔG∘=−9,900 J/mol.
- RT=8.314×298≈2,477.5 J/mol.
- lnK=−(−9,900)/2,477.5≈3.996≈4.0.
- Solve for K: K=e4.
- Estimation: Since e≈2.718, e4≈(2.7)4≈50–60.
- Result: The closest option is 50.
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of δg and equilibrium. It appeared in the 2025 exam.
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