Question

The enthalpy and entropy change of a reaction are -3.9 kJ/mole and +56.6 J/mole K respectively at 25°C. What is the free energy change in kJ/mole? Is this reaction always spontaneous, never spontaneous or does it depend on the temperature?

Answer 1

The free energy change (ΔG) of a reaction can be calculated using the equation:

ΔG = ΔH - TΔS

Where: ΔH is the enthalpy change T is the temperature in Kelvin ΔS is the entropy change

Given: ΔH = -3.9 kJ/mol ΔS = +56.6 J/(mol K) Temperature, T = 25°C = 298 K (since the temperature needs to be in Kelvin)

Converting the enthalpy change to kJ/mol: ΔH = -3.9 kJ/mol

Converting the entropy change to kJ/(mol K): ΔS = +56.6 J/(mol K) = 0.0566 kJ/(mol K)

Substituting the values into the equation:

ΔG = ΔH - TΔS ΔG = -3.9 kJ/mol - (298 K)(0.0566 kJ/(mol K))

Calculating ΔG: ΔG = -3.9 kJ/mol - 16.888 kJ/mol ΔG = -20.788 kJ/mol

The free energy change (ΔG) for the reaction is -20.788 kJ/mol.

To determine the spontaneity of the reaction, we can refer to the sign of ΔG:

• If ΔG is negative (as in this case), the reaction is spontaneous under the given conditions (at 25°C).
• If ΔG is positive, the reaction is non-spontaneous under the given conditions.
• If ΔG is zero, the reaction is at equilibrium.

Therefore, based on the given values, the reaction is spontaneous at 25°C. However, it's important to note that the spontaneity of a reaction can depend on temperature. At different temperatures, the sign of ΔG may change, and a reaction that is spontaneous at one temperature might become non-spontaneous at another temperature.