Chemistry-Gibbs Free Energy
NEET Chemistry Gibbs Free Energy MCQ Question
Type: MCQ-numerical-Hard-Class 11
If the standard Gibbs energy change (ΔrG°) for a reaction is -13.6 kJ mol⁻¹, what is the equilibrium constant (K) at 298 K? (Consider R = 8.314 J mol⁻¹ K⁻¹)
A
K = 1.89 × 10²
B
K = 3.26 × 10⁻¹
C
K = 1.00
D
K = 0.53
Correct Answer
Option A
Detailed Explanation
The relationship between Gibbs free energy and the equilibrium constant is given by ΔrG° = -RTlnK. Solving for K gives K = exp(-ΔrG° / RT). Substituting ΔrG° = -13.6 kJ mol⁻¹ = -13600 J mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹, and T = 298 K, we find K ≈ 1.89 × 10².
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