Exemplar Problems
Problem 14 : A chemical reaction has an equilibrium constant Kc = 0.025 at 25°C (298 K). Calculate ∆G° for this reaction at the given temperature. (R = 8.314 J/(mol•K))
Solution :
The relationship between ∆G° and Kc is given by the equation: ∆G° = -RT ln(Kc)
∆G° = -8.314 J/(mol•K) * 298 K * ln(0.025)
∆G° ≈ -8.314 J/(mol•K) * 298 K * (-3.689)
∆G° ≈ 8,710.58 J/mol ≈ 8.71 kJ/mol
So, the standard Gibbs free energy change (∆G°) for the reaction at 298 K is approximately 8.71 kJ/mol.
BETA
