Question

For the particular reaction, ΔH°rxn = -238.4 kJ/mol and ΔS° = -80.0 J/mol·K, calculate ΔG° for this reaction at 298 K. What can be said about the spontaneity of the reaction at 298 K?

a) ΔG° = -238.4 kJ/mol, the reaction is spontaneous at 298 K.
b) ΔG° = -238.4 kJ/mol, the reaction is not spontaneous at 298 K.
c) ΔG° = -238.4 kJ/mol, the spontaneity of the reaction at 298 K cannot be determined from the given information.
d) ΔG° cannot be calculated from the given information.

Answer 1

To calculate ΔG° at 298 K for the reaction, the entropy change must be converted to the same units as the enthalpy change. The formula ΔG° = ΔH° - TΔS° is used to determine that the reaction is spontaneous at 298 K, with a ΔG° of -214.56 kJ/mol.

Step-by-step explanation:

To calculate the standard Gibbs free energy change (ΔG°) at 298 K for the reaction with given ΔH° = -238.4 kJ/mol and ΔS° = -80.0 J/mol·K, we use the formula:

ΔG° = ΔH° - TΔS°

First, it's important to ensure the units for enthalpy and entropy are consistent. Since enthalpy is given in kJ/mol and entropy in J/mol·K, we need to convert entropy to kJ by dividing by 1000:

ΔS° = -80.0 J/mol·K / 1000 = -0.0800 kJ/mol·K

Now we use the formula:

ΔG° = (-238.4 kJ/mol) - (298 K * -0.0800 kJ/mol·K)

ΔG° = -238.4 kJ/mol - (-23.84 kJ/mol)

ΔG° = -214.56 kJ/mol

The reaction has a negative ΔG°, indicating that it is spontaneous at 298 K.