Question:
The equilibrium constant for a reaction is 10. \(\Delta G^\circ\) will be (\(R = 8 \, \text{J K}^{-1} \text{mol}^{-1}, T = 300 \, \text{K}\))
The equilibrium constant for a reaction is 10. \(\Delta G^\circ\) will be (\(R = 8 \, \text{J K}^{-1} \text{mol}^{-1}, T = 300 \, \text{K}\))
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Remember: When \(K>1\), \(\Delta G^\circ\) is negative, indicating the reaction is spontaneous in the forward direction under standard conditions.
Updated On: May 2, 2026
- \(-5527 \, \text{kJ mol}^{-1}\)
- \(-5.527 \, \text{kJ mol}^{-1}\)
- \(-55.27 \, \text{kJ mol}^{-1}\)
- \(+5.527 \, \text{kJ mol}^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The relationship between standard Gibbs free energy change and equilibrium constant is \(\Delta G^\circ = -RT \ln K\).
Step 2: Detailed Explanation:
Given: \(K = 10\), \(R = 8 \, \text{J K}^{-1} \text{mol}^{-1}\), \(T = 300 \, \text{K}\). \(\Delta G^\circ = -RT \ln K = -RT \times 2.303 \log K\) \(\Delta G^\circ = -8 \times 300 \times 2.303 \times \log 10\) \(\Delta G^\circ = -8 \times 300 \times 2.303 \times 1\) \(\Delta G^\circ = -2400 \times 2.303 = -5527.2 \, \text{J mol}^{-1}\) \(\Delta G^\circ = -5.527 \, \text{kJ mol}^{-1}\).
Step 3: Final Answer:
\(\Delta G^\circ = -5.527 \, \text{kJ mol}^{-1}\), option (B).
The relationship between standard Gibbs free energy change and equilibrium constant is \(\Delta G^\circ = -RT \ln K\).
Step 2: Detailed Explanation:
Given: \(K = 10\), \(R = 8 \, \text{J K}^{-1} \text{mol}^{-1}\), \(T = 300 \, \text{K}\). \(\Delta G^\circ = -RT \ln K = -RT \times 2.303 \log K\) \(\Delta G^\circ = -8 \times 300 \times 2.303 \times \log 10\) \(\Delta G^\circ = -8 \times 300 \times 2.303 \times 1\) \(\Delta G^\circ = -2400 \times 2.303 = -5527.2 \, \text{J mol}^{-1}\) \(\Delta G^\circ = -5.527 \, \text{kJ mol}^{-1}\).
Step 3: Final Answer:
\(\Delta G^\circ = -5.527 \, \text{kJ mol}^{-1}\), option (B).
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