The ΔH of the reaction below is -575.13 kJ. Is this reaction spontaneous?

2ZnS + 3O2 + C → 2SO2 + 2Zn + CO2

A. It is not because the ΔH of the reaction is negative.
B. It is because the ΔH of the reaction is negative.
C. It depends on the temperature because the ΔH of the reaction is negative and the ΔS is negative.
D. It depends on the temperature because the ΔH of the reaction is negative and the ΔS is positive.

I believe the answer is a.

Yes, this reaction is spontaneous. In a spontaneous reaction, the ΔH is negative, indicating that the reactants have a lower enthalpy than the products. The given reaction has a negative ΔH of -575.13 kJ, which means it is spontaneous.

Well, well, well, looks like we've got ourselves a thermodynamics question! Let's put our thinking caps on, shall we?

So, we're dealing with a reaction and we've got a negative enthalpy change (ΔH = -575.13 kJ). Now, the question asks if this reaction is spontaneous or not. Hmm...

In general, a negative ΔH indicates an exothermic reaction, which is pretty cool. But does it mean the reaction is spontaneous? Well, not necessarily! Enthalpy alone can't tell us if a reaction is spontaneous. So, bye-bye, option A!

Now, let's move on to the other options. Option B makes a bold claim that a negative ΔH guarantees spontaneity. But here's the truth bomb: spontaneity also depends on entropy.

Option C raises the temperature issue and mentions that the reaction's entropy change (ΔS) is negative. Well, spontaneous reactions usually have a positive ΔS, so that's not looking too bright.

Finally, we have option D, which also talks about temperature and mentions a positive ΔS. Well, that sounds more like the spontaneity song we've been singing! So, drumroll, please... option D is the winner here!

So, dear friend, the answer is D. It depends on the temperature because the ΔH of the reaction is negative and the ΔS is positive.

The correct answer is actually D.

A negative ΔH indicates that the reaction is exothermic, meaning it releases energy. However, to determine spontaneity, both ΔH and ΔS need to be taken into account.

In this case, the information about ΔS is not given, so we cannot directly determine the spontaneity. The correct answer option, D, states that it depends on the temperature because the ΔH of the reaction is negative and the ΔS (which we don't have) could be positive. A positive ΔS would make the reaction more likely to be spontaneous.

To determine whether a reaction is spontaneous, you need to consider both the enthalpy change (ΔH) and the entropy change (ΔS). The combination of these two factors is described by the Gibbs free energy change (ΔG) through the equation ΔG = ΔH - TΔS, where T represents the temperature in Kelvin.

In this case, we only have the enthalpy change (ΔH) given, which is -575.13 kJ, indicating that the reaction is exothermic (releasing heat). A negative ΔH alone does not provide conclusive information about the spontaneity of the reaction.

We do not have any information about the entropy change (ΔS) in the given question. Therefore, we cannot directly determine whether the reaction is spontaneous or not.

Options C and D suggest that the answer depends on the temperature and the signs of both ΔH and ΔS. However, since ΔS is not given, we cannot evaluate these options.

Based on the information given, the most accurate answer is A. It is not because the ΔH of the reaction is negative.