A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following:

For checking spontaneity we always examine the Gibbs free-energy change, whose expression is first stated as

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

where $$\Delta G$$ is the Gibbs free energy change, $$\Delta H$$ the enthalpy change, $$\Delta S$$ the entropy change and $$T$$ the absolute temperature in kelvin.

A process is said to be spontaneous when $$\Delta G<0$$, non-spontaneous when $$\Delta G>0$$ and at equilibrium when $$\Delta G=0$$.

According to the question the reaction is non-spontaneous at low temperature, so for small $$T$$ we must have

$$\Delta G_{\text{low }T} > 0$$

At high temperature the same reaction becomes spontaneous, therefore for large $$T$$ we must have

$$\Delta G_{\text{high }T} < 0$$

We now analyse the general expression $$\Delta G = \Delta H - T\Delta S$$ term by term to see which signs of $$\Delta H$$ and $$\Delta S$$ can satisfy both of the above requirements.

Step 1: put $$T=0$$ conceptually (or “very low” $$T$$). When $$T$$ is very small, the product $$T\Delta S$$ will also be very small regardless of the sign of $$\Delta S$$. The value of $$\Delta G$$ will then be controlled mainly by $$\Delta H$$ itself:

$$\Delta G_{\text{low }T} \approx \Delta H - (0)\cdot\Delta S = \Delta H$$

Because the reaction is non-spontaneous at low temperature, $$\Delta G_{\text{low }T}$$ must be positive, so

$$\Delta H > 0\quad\text{(enthalpy must be positive)}$$

Step 2: examine $$T\rightarrow\infty$$ conceptually (or “very high” $$T$$). For a very large $$T$$ the term $$T\Delta S$$ dominates the expression. We have

$$\Delta G_{\text{high }T} = \Delta H - T\Delta S$$

Substituting the fact that $$\Delta H>0$$ found above gives

$$\Delta G_{\text{high }T} = (\text{positive}) - T\Delta S$$

To make $$\Delta G_{\text{high }T}<0$$ (spontaneous) the second term must outweigh the positive $$\Delta H$$, and because the minus sign stands in front, $$\Delta S$$ itself must be positive so that $$T\Delta S$$ will be a positive quantity and its subtraction will push $$\Delta G$$ into the negative region:

$$\text{If}\;\Delta S > 0,\quad \Delta G_{\text{high }T} = (\text{positive}) - (\text{large positive}) < 0$$

If, instead, $$\Delta S<0$$, then $$-T\Delta S$$ becomes positive, making $$\Delta G$$ even more positive; that would never create spontaneity at high temperature and hence is ruled out.

Therefore the only consistent assignment of signs is

$$\boxed{\Delta H > 0,\quad \Delta S > 0}$$

Let us match this conclusion with the given options:

A. $$\Delta H$$ negative, $$\Delta S$$ positive — rejected.

B. Both negative — rejected (gives opposite temperature dependence).

C. $$\Delta H$$ positive, $$\Delta S$$ negative — always non-spontaneous, rejected.

D. Both positive — exactly the condition derived above.

Hence, the correct answer is Option D.