Chemistry-Gibbs Free Energy
NEET Chemistry Gibbs Free Energy MCQ Question
Type: MCQ-numerical-Medium-Class 11
Calculate the standard Gibbs free energy change (∆rG°) for the following reaction at 298 K, given that the equilibrium constant (Kp) for the reaction is 3.14 × 10⁻²: 2NO(g) + O₂(g) ⇌ 2NO₂(g).
A
-7.16 kJ mol⁻¹
B
-5.89 kJ mol⁻¹
C
-8.63 kJ mol⁻¹
D
-6.45 kJ mol⁻¹
Correct Answer
Option A
Detailed Explanation
Using the formula ∆rG° = -RT ln Kp, where R = 8.314 J K⁻¹ mol⁻¹ and T = 298 K, the calculation gives ∆rG° = -7.16 kJ mol⁻¹.
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