The entropy of combustion of benzoic acid is zero. This means that...

a)combustion is not spontaneous here.
b)combustion is spontaneous, however, since the reaction is exothermic.
c)combustion is spontaneous, however, since the reaction is endothermic.
d)combustion is nonspontaneous even though the reaction is exothermic.

I think a) combustion is not spontaneous here.
... but I'm really not sure at all

When we burn benzoic acid (for example in a calorimeter to calibrate the calorimeter), the calorimeter temperature rises; therefore, heat must be given off. That means the reaction is exothermic if I'm not on the wrong track here. And that means a isn't right. If delta S = 0, then Tdelta S = 0 and delta G = delta H. If delta H is exothermic, that makes delta H negative and that makes delta G negative. I would go with answer B. One may ask why benzoic is stable at room temperature instead of spontaneously combusting; I would answer that the activation energy had to be overcome to get the reaction started.

To determine the correct answer, we need to understand the concept of entropy and its relationship to combustion.

Entropy is a measure of the disorder or randomness in a system. For a spontaneous reaction to occur, the total entropy of the system and its surroundings should increase. In other words, the products of the reaction should have a higher entropy than the reactants.

Now, let's consider the combustion of benzoic acid. In a combustion reaction, a substance reacts with oxygen to produce carbon dioxide and water. Combustion reactions are typically exothermic, which means they release heat.

Given the information that the entropy of combustion of benzoic acid is zero, this means that the products (carbon dioxide and water) have the same entropy as the reactant (benzoic acid). Consequently, there is no increase in entropy during the combustion process.

Based on this understanding, we can now evaluate the given options:

a) Combustion is not spontaneous here: This is a plausible option since the zero entropy indicates there is no increase in disorder, which suggests that the reaction might not be spontaneous.

b) Combustion is spontaneous, however, since the reaction is exothermic: This option suggests that the reaction is spontaneous due to it being exothermic. While exothermic reactions often favor spontaneity, the given information about entropy being zero contradicts this explanation.

c) Combustion is spontaneous, however, since the reaction is endothermic: This option is less likely because combustion reactions are generally exothermic, not endothermic. It does not align with what we know about combustion.

d) Combustion is nonspontaneous even though the reaction is exothermic: This option is also plausible since the zero entropy indicates no increase in disorder, suggesting the reaction may not be spontaneous despite being exothermic.

Considering the explanations above, option a) "combustion is not spontaneous here" is the most reasonable answer based on the given information.