3
$\begingroup$

Consider a reaction:$$\ce{$aA + bB$ <=> $cC + dD$}$$

The value of reaction quotient at a certain time $t$, $${Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}}$$ where the concentrations $[A], [B], [C]$ and $[D]$ are at time $t$.

Let the reaction start initially at $t = 0$, with only reactants $i.e.$ $[A]$ and $[B]$ equal to say $1$ mol and $[C]$ and $[D]$ equal to $0$ mol. Hence, $$Q_c = 0$$

We know that the value of equilibrium constant $K_c$ must be such that, $$K_c > 0$$ Thus, $$Q_c < K_c$$ Which is the only condition for the advancement of reaction in forward direction. This condition does not consider the value of change in Gibbs energy $\Delta G$.

Now, considering the relation $$\Delta G = \Delta G^o + RT~\mathrm{ln}~Q_c$$ When $Q_c = 0+$ then $\mathrm{ln}~Q_c \to -\infty$, which means $\Delta G << 0$ and reaction is spontaneous in forward direction.

Hence, can it be concluded that every reaction is spontaneous in forward direction if it starts with only reactants?

If so, how can we define a non-spontaneous reaction?

$\endgroup$
  • 3
    $\begingroup$ No reaction is at equilibrium when one of the species (reactant or product) is zero. So you can make any reaction that goes to equilibrium go forward (start with no products) or in reverse (start with no reactants). $\endgroup$ – Karsten Theis Mar 26 at 15:54
  • $\begingroup$ This spontaneous / non-spontaneous thing seems to be a big bee in some teachers´ bonnets. I never found it useful after I had understood what chemical equillibrium means. $\endgroup$ – Karl Mar 26 at 19:30
  • 1
    $\begingroup$ @KarstenTheis It seems very logical that "if a reaction is going to be in equilibrium in future than that has to be spontaneous in forward direction at start with only reactants." This means that any reversible reaction has to be spontaneous in forward direction at start if we start with only reactants. Thanks for your approach! $\endgroup$ – Apurvium Mar 27 at 10:45
7
$\begingroup$

It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not.

The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. When we make the general statement that a reaction is spontaneous or not, we are usually referring to whether this $\Delta G^\circ$ is greater or less than 0.

If $\Delta G^\circ < 0$, then the reaction is described as spontaneous, and it will proceed in the forward direction when starting from a mix of 1 M reactants and 1 M products until equilibrium is reached.

In contrast, $\Delta G$ is the free energy change for the reaction in whatever state is of interest. In your case, you have chosen a state with no product present at all. When we say that a reaction is spontaneous based on $\Delta G$, we mean whether it will go in the forward or reverse direction from this specific state of interest, not the standard state.

Obviously, if there is no product present, the reaction cannot go in the reverse direction (forming reactant from product), so all reactions will go in the forward direction if started in a state of $0$ product. That is what you have shown quantitatively using $Q=0$ so that $\Delta G <<0$.

So you are absolutely correct that every reaction in a state in which no product is present is spontaneous. That does not mean, however, that the reaction will necessarily happen anytime soon. Very often, the activation energy barrier is high enough that the reaction rate is extremely slow. Or it is possible that the equilibrium constant may favor reactants so much that even one molecule of product is sufficient to reach equilibrium and the reaction stops very quickly.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Or the equilibrium constant may favor reactants so much that one molecule of product is already way past equilibrium, and you'll be back to zero in short order even if any product is ever produced. Or some other reaction might consume everything and leave you with no product or reactant for the reaction in question. $\endgroup$ – user2357112 supports Monica Mar 27 at 8:13
  • $\begingroup$ (1) That's an insight that for comparing $\Delta G^o$ starting point should be a mix of 1 M reactants and 1 M products (standard states). (2) Both $\Delta G$ and $\Delta G^o$ can be compared for spontaneity, depending on the conditions. (3) $\Delta G^o = -RT~\mathrm{ln}~K$ only at equilibrium when $\Delta G = 0$. Can we equate $\Delta G^o = \Delta H^o - T\Delta S^o = -RT~\mathrm{ln}~K$ at any (constant) temperature? (4) Only an irreversible reaction can be non-spontaneous at start because a reversible reaction (at start) will go in forward direction to attain equilibrium. $\endgroup$ – Apurvium Mar 27 at 10:35
  • $\begingroup$ @Andrew Can you please verify the above four statements? $\endgroup$ – Apurvium Mar 29 at 12:31
  • $\begingroup$ (1) yes (2) yes (3) No. $\Delta G^\circ = -RT \ln K$ is always true. It does not matter what $\Delta G$ is, since that isn't in the equation. For the second part, true technically only if $\Delta H^\circ$ and $\Delta S^\circ$ have been determined for your T of interest. They don't vary a lot with T, though, so we treat them as invariant if the temp change isn't too great from standard. (4) Not sure what you're saying here. Irreversible usually means forward is so favored that reverse is not observed, so it would definitely be spontaneous. $\endgroup$ – Andrew Mar 29 at 21:00
  • $\begingroup$ @Andrew So the question remains "What is a non-spontaneous reaction?" Can we say that any reaction can be spontaneous or non-spontaneous depending upon temperature and pressure conditions? $\endgroup$ – Apurvium yesterday
4
$\begingroup$

In general it is necessary to consider any entropy changes in determining whether a system is at equilibrium or if a spontaneous change will occur.

As there must be an increase in entropy in actual processes then $dS_{system}+dS_{surr}=dS_{irrev} \ge 0$. By using the first law with the last expression and after several steps, we find that in an isothermal reaction (surrounding temperature $T$) in which no work is done other than $pV$ (gas expansion) work the condition becomes

$$dG + TdS_{irrev}=0$$

thus a process can take place spontaneously with entropy production only if the free energy decreases. If there is no change in free energy, no change can take place and the system is at equilibrium.

This is what you have determined in your question but with the implicit assumption that reaction will occur.

A non-spontaneous process is one that is already at equilibrium or that will require input of some work to make it change.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Thanks! I mentioned another form of this relation in this question. For spontaneity, $\Delta S_{total} > 0$ but $\Delta S_{sys}$ can be positive or negative. $\endgroup$ – Apurvium Mar 27 at 11:28
  • $\begingroup$ I think that my $S_{irrev}$ is your $S_{total}$ so we agree. For a reversible process $dS_{irrev}=0$. $\endgroup$ – porphyrin Mar 27 at 15:25
2
$\begingroup$

I agree with Andrew that it depends on whether you define spontaneity based on $\Delta G^\circ <0$ or $\Delta G < 0$. Typically, freshman chemistry books use the former.

However, I've never liked equating spontaneity with the sign of $\Delta G^\circ$, prefering to instead use the sign of $\Delta G^\circ$ as an indicator of whether reactants or products are "favored".

The reason I don't like using the sign of $\Delta G^\circ$ (as opposed to that of $\Delta G$) as an indicator of spontaneity is it causes the very confusion you are experiencing. It also obscures the concept of chemical equilibrium.

I thus favor equating spontaneity with the sign of $\Delta G$ (which in turn depends not only on $K_{eq}$, but also upon where the reaction mixture is relative to it), which is the view you are articulating. For clarifying this latter view, I think the following diagram, which I've borrowed and modified from one of my earlier answers (Are all exothermic reactions spontaneous?), is helpful: enter image description here

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ I deleted my comment. Your approach makes sense, and I think everyone learning about Gibbs energy of reaction should be exposed to the graph you posted. I'd be happy if the term "spontaneous" disappears from textbooks because it always sounds like it is talking about kinetics, which is not the intent. For relating the direction of the reaction to the Gibbs energy, I think textbooks should include the following relationship: $$\Delta_r G = RT \ln{\frac{Q}{K}}$$ which relates the Gibbs energy to the ratio of the reaction quotient and the equilibrium constant. $\endgroup$ – Karsten Theis Mar 28 at 16:09
  • 1
    $\begingroup$ ... If you start with that relationship, you can plug in $Q=1$ (i.e. standard state) to get the relationship between standard Gibbs energy of reaction $\Delta_r G^\circ$ and equilibrium constant. The relationship as is shows the sign of Gibbs energy of reaction when $Q < K$ and $Q > K$, something students are already familiar with at that point, so it is a quantitative extension of that concept. For biochemistry, you can use the relationship directly to get the Gibbs energy available from a proton gradient or any other gradient (by plugging in K = 1). $\endgroup$ – Karsten Theis Mar 28 at 16:14
1
$\begingroup$

This question shows that you have probably not really understood what the free enthalpy (or Gibbs energy, or free energy) is. I will try to explain it qualitatively without too much thermodynamics. Let's go !

The origin of the Gibbs energy is coming from Gibbs' reflexions on the spontaneity of chemical reactions. He was trying to find a potential energy similar to the mechanical energy for predicting a spontaneous reaction. In mechanics, if an object is situated at a high level, let's say on a table, and that it has the possibility of going to a lower level, on the ground, it will spontaneously do it and fall, without any effort greater than a flip from the observer. Energy is produced in the fall. The contrary is never spontaneous. You may heat it, it will never jump back up higher onto the table. You always have to pick up the object, give it some amount of mechanical work to put it back on the table.

So Gibbs was trying to find some sort of potential energy that atoms and molecules may have that will always decrease in spontaneous reactions. Enthalpy, or heat content, is not this energy, because endothermic reactions may be spontaneous. For example, the mixture of NaHCO3 and HCl solution reacts spontaneously, but the reaction is endothermic. The temperature of the system decreases. To go back to the original temperature, you have to heat the final system.

As heat content is not the energy Gibbs was looking for, he tried to discover the energy which is always exported in a chemical reaction, namely both in endothermic and exothermic reaction. Is there such an energy somewhere ? Yes, there is such an energy, and this is electricity. A cell works out by a chemical reaction. This reaction may be exo- or endothermic. But the galvanic cell will always deliver electric energy. A cell cannot absorb electric energy. It would be a non-sense. So Gibbs decided that in all spontaneous chemical reaction, "his energy", called G, must decrease, whatever the thermic effect. And $\ce\Delta$${G}$, the variation of $\ce{G}$, is related to the emf $\ce{E}$ of the cell by the relation : $\ce{\Delta G = - zEF}$

As a consequence, he established the free energy of formation $\ce{\Delta G_{form}}$ for all substances consumed and produced in galvanic cells. And $\ce\Delta$${G}$ of any chemical reaction can be worked out by substracting the $\ce{\Delta G_{form}}$ of the products minus those of the reactants. A difficulty happened when Gibbs realized that the emf $\ce{E}$ of a cell depends upon the concentrations of the substances, via the Nernst equation. So $\ce{\Delta G_{form}}$ must also be changing with the concentrations. As a consequence, the tabulated values of $\ce{\Delta G_{form}}$ are always reported for pure substances. In this case, $\ce{\Delta G_{form}}$ is called $\ce{\Delta G°_{form}}$, and it is this $\ce{\Delta G°_{form}}$ which is equal to $\ce{ -zEF}$.

Later on, but it is too long to explain it here, Gibbs was also able to calculate $\ce{\Delta G°_{r}}$ for reactions happening out of galvanic cell. He found that $\ce{\Delta G°_{r}}$ is related to the equilibrium constant of the reaction, by the expression $\ce{\Delta G°_{r}}$ = - RT ln K.

Surprisingly, $\ce{\Delta G°_{form}}$ and $\ce{\Delta H°_{form}}$ are not very different from one another. More surpringly, the emf and of course $\ce{\Delta G°_{r}}$ changes with the temperature T, although $\ce{\Delta H°_{r}}$ is nearly independant on T. And if $\ce{\Delta G°_r}$ and $\ce{\Delta H°_{r}}$ are reported on the same graph versus T, you will see that both values vary linearity with T, $\ce{\Delta H°_r}$ being nearly horizontal. Most surprisingly, both lines have a crossing point exactly at O K.

Apparently the difference $\ce{\Delta G°_{form}}$ - $\ce{\Delta H°_{form}}$ is proportional to T. And the ratio ($\ce{\Delta G°_{form}}$ - $\ce{\Delta H°_{form}}$)/T is constant. This ratio is called $\ce\Delta S$ and it is the entropy difference of the reaction. But this is another story.

Have you followed my explanation ?

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Yes, most of it, except "$\Delta G^o_{form}$ and $\Delta H^o_{form}$ are not very different from one another." Nice way to relate $\Delta G^o_r$ with mechanics. You have mentioned that $\Delta H^o_r$ is nearly independent on temperature. But what about Kirchhoff's law related to it? $\endgroup$ – Apurvium Mar 27 at 9:26
  • $\begingroup$ One more thing. The standard state of a substance is its pure form at 1 bar pressure at a specified temperature. Is temperature not defined for standard state? $\endgroup$ – Apurvium Mar 27 at 9:32
  • $\begingroup$ Can you please check the above statements? $\endgroup$ – Apurvium yesterday
0
$\begingroup$

I think it is best to start with a real-world example of a spontaneous reaction as cited in electrochemistry, namely, as occurring in an electrochemical (or battery) cell. In particular, comments from Wikipedia:

A spontaneous electrochemical reaction (change in Gibbs free energy less than zero) can be used to generate an electric current in electrochemical cells. This is the basis of all batteries and fuel cells.

However, for those with experience with the startup of a so-called electrochemical cell, there can be an obvious inception (or waiting) period for these 'spontaneous electrochemical reactions' to proceed!

To explain the lag effect, I further note this comment from an educational source on entropy:

Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system in so called 'spontaneous' reactions.

Apparently, an increase in entropy corresponds to an increase in disorder. And, in the case of the battery cell with a salt bridge, the removal of electrons via a wire from the cell also results in a corresponding movement of cations and anions (to maintain charge balance in the respective half-cells). The latter occurs more slowly, but apparently is required for the continued efficient flow of electrons.

So, as illustrated by the so called 'spontaneous' electrochemical cells, there is also an increase in entropy component to balance the decrease in Gibbs free energy.

Interestingly, my cited example/argument seems to suggest that the word 'spontaneous' as coupled to 'spontaneous reactions' may implied, in particular cases, a more relative measure in terms of actual passage of time.

|improve this answer|||||
$\endgroup$

Your Answer

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.