The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by $$\Delta G° = A - BT$$ where A and B are non-zero constants. Which of the following is true about this reaction?

We recall the fundamental thermodynamic relation that connects the standard Gibbs energy change, the standard enthalpy change and the standard entropy change of a reaction:

$$\Delta G^\circ \;=\; \Delta H^\circ \;-\; T\,\Delta S^\circ.$$

In the question the same quantity is expressed empirically as

$$\Delta G^\circ \;=\; A \;-\; B\,T,$$

where $$A$$ and $$B$$ are given to be non-zero constants. We now identify the two expressions term by term.

Comparing

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

with

$$\Delta G^\circ \;=\; A \;-\; B\,T,$$

we equate the temperature-independent parts and the coefficients of the temperature term separately:

$$\Delta H^\circ \;=\; A,$$

$$-\,\Delta S^\circ \;=\; -\,B \;\;\Longrightarrow\;\; \Delta S^\circ \;=\; B.$$

Thus, the constant $$A$$ represents the standard enthalpy change $$\Delta H^\circ$$, while the constant $$B$$ represents the standard entropy change $$\Delta S^\circ$$.

Next, we recall the criterion for the heat character of a reaction:

• If $$\Delta H^\circ < 0,$$ the reaction is exothermic (heat is released).
• If $$\Delta H^\circ > 0,$$ the reaction is endothermic (heat is absorbed).

Since $$\Delta H^\circ = A,$$ the sign of $$A$$ alone decides whether the reaction is endothermic or exothermic. In particular,

$$A > 0 \;\;\Longrightarrow\;\; \Delta H^\circ > 0 \;\;\Longrightarrow\;\; \text{reaction is endothermic}.$$

Looking at the options, Option B states exactly this condition:

“Endothermic if $$A > 0$$.”

No other option relates the sign of $$A$$ (which equals $$\Delta H^\circ$$) correctly to the heat character of the process.

Hence, the correct answer is Option B.