Exemplar Problems
Problem 3 : A reaction has an equilibrium constant (Kc) of 0.25 at 25°C. Calculate the standard Gibbs free energy change (∆G°) for the reaction at this temperature.
Solution :
The relationship between ∆G° and Kc is given by the equation: ∆G° = -RT ln(Kc)
Given that Kc = 0.25 and T = 25°C (which is 298 K), we can calculate ∆G°:
∆G° = -8.314 J/(mol•K) * 298 K * ln(0.25) ∆G° ≈ -8.314 J/(mol•K) * 298 K * (-1.386)
∆G° ≈ 4,156.24 J/mol ≈ 4.16 kJ/mol
So, the standard Gibbs free energy change (∆G°) for the reaction at 25°C is approximately 4.16 kJ/mol.
BETA
