Spontaneity And Gibbs Energy
For a reaction, (\Delta G^\circ = -30) kJ/mol at 298 K. Which of the following is the correct equilibrium constant (K) for this reaction? (R = 8.314 J/mol·K)
Select the correct option:
Solution
1.7×105
The relationship between standard Gibbs free energy and the equilibrium constant is: (\Delta G^\circ = -RT\ln K), or equivalently (K = e^{-\Delta G^\circ / RT}). This arises from the condition of equilibrium in chemical thermodynamics. Given (\Delta G^\circ = -30{,}000) J/mol, (R = 8.314) J/mol·K, (T = 298) K: (\frac{\Delta G^\circ}{RT} = \frac{-30000}{8.314 \times 298} = \frac{-30000}{2477.6} = -12.11). Therefore (K = e^{12.11}). Converting: (e^{12.11} = e^{12} \times e^{0.11} \approx 162755 \times 1.116 \approx 1.7 \times 10^5). Option (3.3 \times 10^{-5}) is (1/K) — it arises from using the wrong sign in the exponent. Option (6.0 \times 10^4) results from using (\log_{10}) without converting to natural logarithm. Option (2.1 \times 10^3) uses an incorrect value of (RT). This is a JEE standard application of the (\Delta G^\circ = -RT\ln K) relationship from NCERT Chapter 6. Plausibility check: a large negative (\Delta G^\circ) implies a strongly product-favored equilibrium, consistent with (K \gg 1).
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- spontaneity and gibbs energy
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.7×105
The relationship between standard Gibbs free energy and the equilibrium constant is: (\Delta G^\circ = -RT\ln K), or equivalently (K = e^{-\Delta G^\circ / RT}). This arises from the condition of equilibrium in chemical thermodynamics. Given (\Delta G^\circ = -30{,}000) J/mol, (R = 8.314) J/mol·K, (T = 298) K: (\frac{\Delta G^\circ}{RT} = \frac{-30000}{8.314 \times 298} = \frac{-30000}{2477.6} = -12.11). Therefore (K = e^{12.11}). Converting: (e^{12.11} = e^{12} \times e^{0.11} \approx 162755 \times 1.116 \approx 1.7 \times 10^5). Option (3.3 \times 10^{-5}) is (1/K) — it arises from using the wrong sign in the exponent. Option (6.0 \times 10^4) results from using (\log_{10}) without converting to natural logarithm. Option (2.1 \times 10^3) uses an incorrect value of (RT). This is a JEE standard application of the (\Delta G^\circ = -RT\ln K) relationship from NCERT Chapter 6. Plausibility check: a large negative (\Delta G^\circ) implies a strongly product-favored equilibrium, consistent with (K \gg 1).
This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of spontaneity and gibbs energy. It appeared in the 2025 exam.
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