Question

For a specific reaction with ΔH=−111.4kJ/mol and ΔS=−25.0J/(mol\cdotpK), perform the following tasks:

Calculate ΔG for this reaction at 298 K (in kJ).

Answer 1

The ΔG for this reaction at 298 K is -12.9 kJ/mol.

Step-by-step explanation:

To calculate the standard free-energy change (ΔG°) for a reaction at a given temperature (T), you can use the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. In this case, the values are given as ΔH° = -111.4 kJ/mol and ΔS° = -25.0 J/(mol·K).

Substituting these values into the equation, we get ΔG° = -111.4 kJ/mol - (298 K)(-25.0 J/(mol·K)). Solving this equation gives us ΔG° = -111.4 kJ/mol - (-7450 J/mol) = -1.29 × 10^4 J/mol, or -12.9 kJ/mol.

Therefore, the ΔG for this reaction at 298 K is -12.9 kJ/mol.