Question

At 298 K, ∆H = -314 kJ/mol and ∆S = -0.372 kJ/(K•mol). What is the Gibbs free energy of the reaction?

Answer 1

-203 kJ/mol

Step-by-step explanation:

Step 1: Given data

• Standard enthalpy of the reaction (ΔH°): -314 kJ/mol

• Standard entropy of the reaction (ΔS°): -0.372 kJ/K.mol

• Absolute temperature (T): 298 K

Step 2: Calculate the standard Gibbs free energy of the reaction (ΔG°)

We will use the following expression.

ΔG° = ΔH° - T × ΔS°

ΔG° = (-314 kJ/mol) - 298 K × (-0.372 kJ/K.mol) = -203 kJ/mol

By convention, when ΔG° < 0, the reaction is spontaneous.