For a given chemical reaction, the enthalpy change is exactly and the entropy change is . Assuming t — Thermodynamics and Thermochemistry Chemistry Question
Question
For a given chemical reaction, the enthalpy change is exactly $\Delta H = +35.5 \text{ kJ mol}^{-1}$ and the entropy change is $\Delta S = +83.6 \text{ J K}^{-1}\text{mol}^{-1}$. Assuming that both $\Delta H$ and $\Delta S$ do not vary with temperature, under what temperature condition will the reaction proceed spontaneously?
💡 Solution & Explanation
The criterion for spontaneity at constant pressure and temperature is $\Delta G = \Delta H - T\Delta S < 0$. When both $\Delta H$ and $\Delta S$ are positive, the reaction is driven by entropy and becomes spontaneous only at high temperatures. Setting $\Delta G < 0$ gives $T > \frac{\Delta H}{\Delta S}$. Converting enthalpy to Joules: $T > \frac{35500 \text{ J mol}^{-1}}{83.6 \text{ J K}^{-1}\text{mol}^{-1}} \approx 424.64 \text{ K}$. Thus, the reaction is spontaneous exclusively when $T > 425 \text{ K}$.