Gibbs Free Energy

Question

At 298 K, a reaction has ΔH = +25 kJ mol⁻¹ and ΔS = +120 J mol⁻¹ K⁻¹. Is the reaction spontaneous at 298 K?

RankGuru · Accepted Answer

1. **Formula**: $\Delta G = \Delta H - T \Delta S$.
2. **Values**:
   - $\Delta H = +25,000$ J/mol.
   - $T = 298$ K.
   - $\Delta S = +120$ J/mol·K.
3. **Calculation**:
   - $T \Delta S = 298 	imes 120 = 35,760$ J/mol.
   - $\Delta G = 25,000 - 35,760 = -10,760$ J/mol.
4. **Criterion for Spontaneity**: If $\Delta G < 0$, the reaction is spontaneous.
5. **Conclusion**: Since $\Delta G$ is negative, the reaction is **spontaneous** even though it is endothermic.

Gibbs Free Energy

Select the correct option:

🔒 Solution Hidden from View

More chemical thermodynamics Practice Questions

The relationship between Gibbs free energy change and cell potential is:

For a reaction to be spontaneous at all temperatures, which condition must be satisfied?

About This Question

Solution