Let me free

The change in a system's Gibbs free energy in a reaction is defined as:

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

for system enthalpy $H$ and entropy $S$ changes in the reaction. Note that $\Delta H > 0$ for an endothermic reaction and $\Delta H < 0$ for an exothermic reaction and always temperature $T>0$ . Therefore:

(a) FALSE For an exothermic reaction ( $\Delta H < 0$ ), $G$ can increase ( $\Delta G > 0$ ) if there is an entropy decrease of $\Delta S < -|\Delta H|/T$ .

(b) FALSE For an endothermic reaction ( $\Delta H > 0$ ), $G$ can decrease ( $\Delta G < 0$ ) if there is an entropy increase of $\Delta S > \Delta H/T$ .

(c) TRUE For an exothermic reaction ( $\Delta H < 0$ ), if entropy increases ( $\Delta S > 0$ ), then $\Delta G = -|\Delta H| - T\Delta S < 0$ , so $G$ decreases .

(d) FALSE For an endothermic reaction ( $\Delta H > 0$ ), if entropy increases ( $\Delta S > 0$ ), then $\Delta G = \Delta H - T\Delta S < 0$ only for large enough entropy increase, namely, $\Delta S > \Delta H/T$ , so $G$ may not decrease .

So the answer is (c) only .

The only relevant statement is given in such a way as the equation hence:

ΔG = ΔH - TΔS

The first term, to the right of equality is the term entalpic and the second, entropic. There is a decrease in ΔG, when conventionally, ΔH is a high negative number, characterizing an exothermic reaction. Although, with increase of the term TΔS, with constant T (noting the negativity of this term), the greater the entropy, smaller will be ΔG. The condition of spontaneity occurs with lower ΔG, thus, with a high and negative enthalpic term, indicating an exothermic reaction and a high entropic term (considering the negative sign of this term). The only alternative that contemplates these conditions without opening pretexts to other systems is alternative c.

Logic of guess without making detail reading onto this topic:

If (a) and (b) are correct, then both of them should include in answer together, while both of them seem to be true together or false together. In fact, heat absorbed must become more reactive and useful while heat released must become less useful are not necessarily true. Manufacturing useful substances via fission is valid and via fusion is also valid, which are disregard of energy changes onto chemicals.

But heat released while also become more disordered, the chemical or substance should become less useful.

Finally, heat injected or added in and become highly ordered must not decrease its usefulness!

As a consequence, we should be able to deduce that (a) and (b) are invalid while (c) is the only correct explanation.

Answer: $\boxed{(c)}$

Condition for spontaneity of a process...energy criteria and entropy criteria....should be exothermic and entropy increase...therefore c)