A process has $$\Delta H = 200$$ J mol$$^{-1}$$ and $$\Delta S = 40$$ J K$$^{-1}$$ mol$$^{-1}$$. Out of the values given below choose the minimum temperature above which the process will be spontaneous:

The spontaneity of a process at a constant temperature and pressure is determined by the change in Gibbs Free Energy ($$\Delta G$$), defined by the Gibbs-Helmholtz equation:

$$\Delta G = \Delta H - T\Delta S$$

For any process to be spontaneous, the change in Gibbs Free Energy must be strictly negative:

$$\Delta G < 0 \implies \Delta H - T\Delta S < 0$$

Step-by-Step Calculation:

Rearranging the criteria for spontaneity to solve for the absolute temperature ($$T$$):

$$\Delta H < T\Delta S$$

$$T > \frac{\Delta H}{\Delta S}$$

Substitute the given thermodynamic values ($$\Delta H = 200\text{ J mol}^{-1}$$ and $$\Delta S = 40\text{ J K}^{-1}\text{ mol}^{-1}$$) into the inequality:

$$T > \frac{200\text{ J mol}^{-1}}{40\text{ J K}^{-1}\text{ mol}^{-1}}$$

$$T > 5\text{ K}$$

Conclusion:

The equilibrium threshold is reached at exactly $$5\text{ K}$$. For the reaction to cross this threshold and become spontaneous, the operating temperature must be maintained above this value.

Answer: Option B — 5 K