As T approaches infinity, ln K approaches

Question

Quizplus · Accepted Answer

The correct answer is based on the thermodynamic equation $ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $. As $ T $ approaches infinity, the $ T\Delta S^{\circ} $ term dominates, and $ \Delta G^{\circ} $ approaches $ -T\Delta S^{\circ} $. Since $ \ln K = -\frac{\Delta G^{\circ}}{RT} $, substituting $ \Delta G^{\circ} $ with $ -T\Delta S^{\circ} $ gives $ \ln K = \frac{\Delta S^{\circ}}{R} $, indicating that as $ T $ approaches infinity, $ \ln K $ approaches $ \frac{\Delta S^{\circ}}{R} $.

As T Approaches Infinity, Ln K Approaches A)infinity ΔS∘R \frac{\Delta S^{\circ}}{R} RΔS∘​

As T Approaches Infinity, Ln K Approaches
A)infinity $\frac{\Delta S^{\circ}}{R}$