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The key says the correct answer is B. I am wondering if this is because all exothermic reactions are spontaneous?

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  • $\begingroup$ From the comments in the answers below, it seems that we lack a definition of spontaneous. Does it mean $\Delta_r G < 0$ at the current conditions or $\Delta_r G^\circ < 0$? $\endgroup$ – Karsten Theis Feb 11 at 13:35
  • $\begingroup$ @KarstenTheis Yes, that does seem to be the issue. $\endgroup$ – theorist Feb 11 at 14:57
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    $\begingroup$ While is some technical thermodynamic sense, "spontaneous" might mean something different, in normal language even among chemists it means the reaction happens without an external push. Clearly many exothermic reactions are not spontaneous for the normal deifinition of the word. Burning coal is clearly exothermic, for example, but getting it to burn under normal atmospheric conditions takes significant effort and doesn't just happen. $\endgroup$ – matt_black Feb 11 at 15:30
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    $\begingroup$ @matt_black Note, however, that this question, in giving options involving the entropy of the system and the universe, is specifically concerning itself with the fundamentals of thermodynamics. $\endgroup$ – theorist Feb 11 at 15:57
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The entropy of the system decreases

All the reactants are mono-atomic gases, and the product is a solid. This is a strong indication that the entropy of the system decreases.

The reaction is spontaneous

That statement from the text of the question means that the Gibbs energy of reaction is negative. It is not clear if this refers to standard conditions, or to the current conditions. The question could be more specific in that regard.

Enthalpy of reaction

If the reaction entropy is negative and the reaction Gibbs energy is negative, the reaction enthalpy has to be negative as well because it is the sum of the Gibbs energy and the entropy times the (positive) temperature.

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

$$\Delta_r H = \Delta_r G + T \Delta_r S$$

Are all exothermic reactions spontaneous?

No, but all reactions that are spontaneous at standard state (interpreted to mean that at standard state, equilibrium lies in the forward direction, i.e. K > 1 and $\Delta_r G^\circ < 0$) are either exothermic ($\Delta_r H^\circ < 0$) or show a positive standard entropy of reaction ($\Delta_r S^\circ > 0$), or both. This ensures that the Gibbs energy of reaction is negative and that the entropy of the universe increases (those two are linked in the absence of non-PV work).

[theorist] The direction of spontaneity is merely a function of where the reaction mixture is relative to the equilibrium constant.

The argument I made above does not preclude finding a reaction with positive $\Delta_r H^\circ$ and negative $\Delta_r S^\circ$ that goes forward. In this case, K will be smaller 1, but reaction conditions can be chosen so that Q is smaller than 1 as well, and furthermore smaller than K, which is always possible by increasing reactant concentrations or by removing product.

[Textbook question]Pick the answer that best describes this process.

Answers A) and E) are linked. An endothermic process (A) will decrease the temperature of the system (E). So those two answers are out if there is only one correct answer. Answer C) goes against the second law of thermodynamics, so most teachers would mark it wrong. Above, I made an argument that the reaction has a negative standard entropy of reaction (less freedom for the atoms when they go from mono-atomic gas to a solid compounds); answer D) posits the opposite, so it is out.

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  • $\begingroup$ The fact that the term "exothermic" is used in "B" properly means it is referring to the sign of the standard-state enthalpy change (see IUPAC link in comment to Buck Thorn), which means the entropy change (if used in the same expression) must, for consistency, be standard state as well. And, as I've explained elsewhere, the fact that a rxn has a negative standard-state entropy change, and is spontaneous, places no constraints on whether it is endothermic or exothermic. The direction of spontaneity is merely a function of where the reaction mixture is relative to the equilibrium constant. $\endgroup$ – theorist Feb 11 at 5:41
  • $\begingroup$ And even if standard state wasn't intended, clearly these refer to the enthalpy and entropy changes inherent to the difference between pure reactants and pure products, not to the equilibrium mixture. $\endgroup$ – theorist Feb 11 at 5:46
  • $\begingroup$ @theorist I agree with you that irrespective of the signs of reaction entropy, enthalpy, and Gibbs energy, you can always find conditions where Q < K. I will edit my answer to avoid the term "spontaneous" in my statements. $\endgroup$ – Karsten Theis Feb 11 at 13:16
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No, it is not the case that all exothermic reactions are spontaneous. More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also specifying the concentrations of reactants and products, since for a certain relative concentration of reactants and products, the reaction might be spontaneous in the forward direction, while for another set of relative concentrations it might be spontaneous in the reverse direction. [See illustration below.]

One could, however, make the general statement that chemicals will react spontaneously if, in so doing, the entropy of the universe increases. This, however, is not one of the options you are given.

Why, then, is "B" the correct choice? Answer: It's because this Ga and As are forming a chemical bond, and bond formation is exothermic. In addition, there is a phase change here from a gas to a solid, and that is also exothermic. Furthermore, since Ga and As are both monatomic gases in this reaction, no chemical bonds need to be broken to allow the Ga and As to combine. Thus, since we have chemical bonds being formed but not broken, and a phase change from gas to solid, the reaction is exothermic.

Here's a graphical illustration of the general concept articulated in the first paragraph:

enter image description here

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  • $\begingroup$ I'll point out that the arrow for the reaction indicates that the reaction is irreversible. Right answer, wrong reason. $\endgroup$ – MaxW Feb 10 at 10:49
  • $\begingroup$ The logic of bond formation being exothermic is that otherwise, constraining the movement of the parts to be bonded would result in a decrease of entropy in the universe. For the hydrophobic effect, you gather non-polar molecules without strong interactions so that the entropy of water increases. $\endgroup$ – Karsten Theis Feb 10 at 22:10
  • $\begingroup$ @MaxW. Nope. All spontaneous reactions are thermodynamically irreversible, thus the type of irreversibility you are speaking about it kinetic irreversibility (KI). So: (1) You can't rely on seeing single vs. dbl. arrows as a signifier of KI, since people often don't bother w/ the dbl. arrows regardless. Further I see no reason this rxn should have KI (i.e., be kinetically trapped). (2) More to the point, the thermodynamics of the rxn are independent of the kinetics. So, adopting your preferred linguistic style: Wrong comment, wrong reason. $\endgroup$ – theorist Feb 11 at 5:12
  • $\begingroup$ @KarstenTheis As I explain in my comment to Buck Thorn, you can have an endothermic reaction in which the entropy of the products is less than that of the reactants, yet it can still proceed spontaneously. Not sure how your reference to the hydrophobic effect applies here. While there continues to be controversy about it, the standard description is that water suffers an entropy penalty (but a favorable decrease in enthalpy) in solvating non-polar solutes, because of the formation of ordered "clathrate shells" in which the water-water bonds are stronger than in bulk. $\endgroup$ – theorist Feb 11 at 5:36
  • $\begingroup$ @theorist Sorry about the non-sequitur in my comment. I was wondering if my statement "any bond formation has to be exothermic" is general, and pondering whether something like aggregation of non-polar species in water would be a counterexample. In the case at hand, there is no solvent, so the entropy balance is much easier to figure out. $\endgroup$ – Karsten Theis Feb 11 at 13:32
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I would add that, given that the entropy change associated with the formation of the solid is almost certainly negative, the only way the reaction can proceed spontaneously at constant T and p is if it is exothermic, invoking the condition that $\Delta G = \Delta H - T \Delta S<0$ or $$\frac{\Delta H}{T} <\Delta S$$ for a spontaneous process.

That the process is exothermic can be justified using the arguments presented by theorist.

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    $\begingroup$ You wrote "the only way the reaction can proceed spontaneously at constant T and p is if it is exothermic (invoking the condition that $\Delta G = \Delta H - T \Delta S<0$ for a spontaneous process)". That's actually not generally true, since one also needs to factor in the entropy of mixing. As a consequence of the entropy of mixing, a reaction could proceed spontaneously even if $\Delta H_{rxn} > 0$ and $\Delta S_{rxn} <0$, if one were to the "left" of the equilibrium position. $\endgroup$ – theorist Feb 10 at 7:46
  • $\begingroup$ @theorist entropy of mixing would contribute to the $\Delta S$ term. But sure, we can split hairs if you like. $\endgroup$ – Buck Thorn Feb 10 at 7:47
  • $\begingroup$ @theorist you split hairs in that you are not reading between the lines. For the reaction as written to proceed in the forward direction (which obviously is implied) then, given that $\Delta S<0$, you must have an exothermic reaction. $\endgroup$ – Buck Thorn Feb 10 at 7:57
  • $\begingroup$ It's not splitting hairs at all. You made an argument that the rxn has to be exothermic based on it being spontaenous + $\Delta S < 0$, where that $\Delta S$ referred only to the inherent entropy difference between the reactants and products ("the entropy change associated with the formation of the the solid is almost certainly negative"). That $\Delta S$ explicitly doesn't take into account the entropy of mixing. As a consequence, your argument that the rxn must be exothermic because of this is simply incorrect...[continued] $\endgroup$ – theorist Feb 11 at 1:13
  • $\begingroup$ As a simple counterexample, consider a reaction, A->B, in which $\Delta H = 10 kJ/mol$, and $\Delta S = - 10 J/(mol K)$, at 298.15K. Then, at that temperature, $\Delta G = 13 kJ/mol => K_{eq} = 0.0053$. Thus, if $\frac{[B]}{[A]} < 0.0053$, the reaction will proceed spontaneouslyly from reactants to products, even though $\Delta S <0$ and it is endothermic. I.e., the fact that it can proceed spontaneously from reactants to products while having a negative $\Delta S$ does not in any way constrain the reaction to be exothermic! $\endgroup$ – theorist Feb 11 at 1:29
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A exothermic reaction has the energy to proceed spontaneously. It may or may not depending on the kinetics. The reaction could be sterically hindered for example, and need a catalyst to happen at all.

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    $\begingroup$ Umm, it could be entropically unfavorable? $\endgroup$ – Buck Thorn Feb 10 at 13:04

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