3
$\begingroup$

Suppose I have a reversible reaction \begin{align} \ce{A(g) + B(g) <=> C(g)}\\ \end{align} with equilibrium constant K.

Its ΔΗ is positive throughout.

Now, suppose I start with 1 mole each of gases A and B, and zero (negligible) moles of C. The temperature is constant.

According to $\Delta G = RT\ln(Q/K)$ , $\Delta G$ will be highly negative at my initial conditions and will increase (become less negative) as reaction proceeds forward.

Also, using the same expression, $$\Delta S = \frac{\Delta H}{T} - R \ln\left(\frac{Q}{K}\right)$$

Hence, $\Delta S$ is highly positive initially and decreases as the reaction proceeds forward. When $\Delta S = \Delta H/T$ (at $Q=K$) , no net reaction takes place further and equilibrium is maintained.

I understand it.

But how is $\Delta S$ positive? How can it be explained theoretically?

Theoretically, shouldn't $\Delta S$ be negative as number of gaseous moles decrease on going forward?

Or is this type of reaction not possible?

Source:https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.10%3A_Entropy_Changes_in_Gaseous_Reactions

New contributor
Half Infinity is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
3
  • 1
    $\begingroup$ I understand the desire to get an answer, but please (next time) don't repost. Instead edit your earlier Q. Since you got an answer to this post please remove your earlier post. Also, your use of notation is still flawed (perhaps my fault - my comments seem to have confused you). $\endgroup$
    – Buck Thorn
    Aug 5 at 14:31
  • $\begingroup$ I didn't know the rules... Will take care next time $\endgroup$ Aug 5 at 15:14
  • $\begingroup$ If A +B = C the reaction Must be exothermic even if the only energy loss is a temporary change of kinetic to potential in an entirely elastic collision So delta H is negative and delta S is also negative. This has always bothered me if a reaction is to reach equilibrium meaning delta G = 0, delta H and delta S must have the same sign and the common penalty box with 4 options is bogus. There is really only one situation if standard conditions are ignored and a reversible reaction at equilibrium is perturbed in either direction delta G becomes negative and the reaction returns to equilibrium. $\endgroup$
    – jimchmst
    Aug 5 at 19:00

1 Answer 1

4
$\begingroup$

But how is dS positive? How can it be explained theoretically?

The reaction entropy rules usually taught first are about the physical state of the pure substances. If you go from a condensed state (solid or liquid) to a gas, the molecules are less constrained. Similarly, if you open up a ring, the molecule itself has more conformational freedom, and this is also reflected in an increase in entropy.

There is a second, concentration-dependent contribution to the entropy of reaction. This is the entropy of mixing. A mixture has more microstates than the separate two pure substances. This is behind the experience that ideal mixtures (when the intra- and intermolecular interactions have the same strength) don't unmix. Entropy increases whem you mix something. This strangely is the case when you go from pure reactant to a mixture of reactant and product as the reaction proceed. Similarly, there is some "unmixing" when all the reactant turns into product.

This entropy of mixing is where the $R \ln(Q)$ (or $R T \ln(Q)$) comes in. This is what makes the reaction entropy and the reaction Gibbs energy concentration-dependent.

Theoretically, shouldn't dS be negative as number of gaseous moles decrease on going forward?

Yes, going from pure reactants to pure products, the entropy change is negative. However, if reactants and products are in the same phase, you have a very high positive change in entropy as the first couple of reactant molecules turn into product. If you want, you can explain this with kinetics as well, with no back reaction (no product yet) and a robust forward reaction.

[OP in comments: What if they are in different phases? Like gases combining to form a liquid?]

One example where all "reactants" and "products" are in separate, pure phases is ice melting. If you are above the melting point, all ice will melt and the system will not reach a dynamic equilibrium. This is different for water evaporating. In a water bottle that is half-full and capped, after a while you reach equilibrium, and the same amount of water evaporates and condenses, so there is no longer a net change. The entropy of water in the gas phase is dependent on the partial pressure (the "concentration"). If you write the equilibrium expression for these processes, you will see the difference ("leaving out", or setting to one, pure solids and liquids, but not gases).

Conversely, if the last few reactant molecules turn into product and there is no more reactant left, the entropy change is negative (this is hypothetical because it is past the equilibrium, i.e. in a direction away from equilibrium). Again, using kinetics, the forward reaction rate has almost "dried up", the the reverse reaction rate is robust.

Or is this type of reaction not possible?

This type of reaction is possible. The standard Gibbs reaction energy will have a high positive value, though, so the equilibrium constant will be very low. Equilibrium will be reached quickly if you start out with pure reactants.

[OP in comments] How do you say that the Standard Gibbs Energy will be highly positive for these type of reactions?

The standard enthalpy of reaction is positive (given) and the standard entropy is negative (less particles on product side). Adding them up according to $$\Delta G = \Delta H - T \Delta S$$ gives a positive value at all temperatures, at least as high as $\Delta H$

Nomenclature

You have to distinguish $$\Delta S^\circ$$ and $$\Delta_r S = \frac{\partial \Delta S}{\partial \xi}$$ The former is what you can calculate from tabulated data. The latter is a more formal way of writing your dS, signifying the infinitesimal small change in entropy as the reaction moves forward an infinitesimal amount, i.e. when some reactant turns into some product according to the balanced chemical equation.

Textbooks, especially introductory once, are inconsistent in their nomenclature (there is an entire article on this topic for Gibbs energy of reaction). For the nomenclature and symbols for Gibbs energy, see the first slide in this lecture.

[OP in comments] In your formal notation of dS , what is the physical quantity you took partial differential with?

It is the extent of reaction $\xi$ (greek lowercase xi). It measures the amount that has reacted ($\xi$ = 1 mol means one mole has reacted if the coefficient is one, two when the coefficient is two and so on, with reactants having negative values and products positive values).

$\endgroup$
13
  • 1
    $\begingroup$ @BuckThorn I commented on the other post that editing and reopening is the way it is supposed to work. I think I understand the question though: Why does the rule (less gas particles on product side) say entropy should be negative, yet the change in (system) entropy as the reaction reaches equilibrium is positive? $\endgroup$ Aug 5 at 15:02
  • $\begingroup$ I have started to understand things now.... What do you mean by "last few reactants" in fifth paragraph? The reactants are about to finish? Or equilibrium is about to be attained? $\endgroup$ Aug 5 at 15:24
  • $\begingroup$ Also, in fourth paragraph, you wrote there is a high change in entropy if reactants and products are in same phase.... What if they are in different phases? Like gases combining to form a liquid? $\endgroup$ Aug 5 at 15:29
  • $\begingroup$ Can you also suggest me some standard book for understanding the concepts of Entropy(and Entropy of Mixing) , Enthaply and Gibbs Energy correctly and deeply....Thanks a lot $\endgroup$ Aug 5 at 16:08
  • $\begingroup$ @KarstenTheis One question more plz... How do you say that the Standard Gibbs Energy will be highly positive for these type of reactions? $\endgroup$ Aug 5 at 16:59

Your Answer

Half Infinity is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.