# Will equilibrium ever establish if one of the products starts to disassociate?

When $$\ce{N_2O_5}$$ is heated at a certain temperature, it disassociates as $$\ce{N_2O_5 (g)<=>N_2O_3 (g) + O_2 (g); K_c = 2.5}$$ At the same time $$\ce{N_2O_3}$$ also decomposes as $$\ce{N_2O_3 (g)<=> N_2O (g) + O_2 (g)}$$If initially 4 moles of $$\ce{N_2O_5}$$ were taken in a $$\pu{1 L}$$ flask and allowed to disassociate, and at equilibrium $$\ce{[O_2]=2.5 M}$$ then find equilibrium concentration of $$\ce{N_2O_5}$$

Lets say the first reaction reached equilibrium and then $$\ce{N_2O_3}$$ started decomposing. This means more amount of $$\ce{O_2}$$ will be produced and the equilibrium for the first reaction will get disturbed.

The first reaction will then proceed backwards as predicted by Le Chatelier's principle (since the concentration of $$\ce{O_2}$$ is more than the equilibrim conc.).

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $$\ce{N_2O_3}$$ will decompose further to form oxygen.

But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

• Where did you see that $\ce{N2O3}$ gets decomposed into $\ce{N2O}$ and $\ce{O2}$ ? This is improbable. As far as I know it gets decomposed into $\ce{NO + NO2}$. A good proof is the fact that $\ce{N2O3}$ reacts with $\ce{NaOH}$ to produce a mixture of nitrite and nitrate through $$\ce{N2O3 + 2 NaOH -> NaNO3 + NaNO2 + H2O}$$ See Mary Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, NewYork 1994, p. 698. Oct 17, 2020 at 9:34
• This is a question in my textbook. But let's just suppose hypothetically that it got decomposed into $\ce{N_2O and O_2}$. Would equilibrium ever be established in this case? Oct 17, 2020 at 10:37
• @Eyyboss this might have some useful information. (ChemLibreTexts on multistep reactions).
– user98623
Oct 17, 2020 at 11:09
• Because N2O3 is at equilibrium both reactions come to equilibrium, i.e. the N2O5 is not completely used up. You can add both equations together, eliminate N2O3 and carry on. Oct 17, 2020 at 11:20
• In fact, N2O3 is the adduct $\ce{NO + NO2 <=> ON-NO2}$ with N-N bond, formed at very low temperature, similar to N2O4.. Oct 17, 2020 at 18:42

Let's make this an abstract problem with species A, B, C, and F:

$$\ce{A <=> B + F}\tag{1}$$

$$\ce{B <=> C + F}\tag{2}$$

We can also write down the sum of the two reactions, which will also be at equilibrium:

$$\ce{A <=> C + 2 F}\tag{3 = 1+2}$$

And let's use a for the concentration of A and so on.

### Finding the solution using reactions 1 and 3

The concentration of F is equal to the equilibrium constant for reaction (1), so the concentrations of A and B have to be equal:

$$a = b$$

For every mole of B made, we also get a mole of F. For every mole of C made, we get two moles of F. We know the concentration of F is 2.5 M, so we get:

$$b + 2c = 2.5 M$$

Finally, we started with 4 M of A, which makes equimolar B and makes equimolar C, so the sum of all three has to be 4 M:

$$a + b + c = 4 M$$

Now we have three equations and three unknowns. We can combine the first and third equation to eliminate $$a$$:

$$2b + c = 4M$$

We can multiply the second equation in preparation for solving for $$c$$:

$$2b + 4c = 5 M$$

Subtracting the latter equation gives the solution for $$c$$:

$$c = \frac{1}{3} M$$

$$a$$ and $$b$$ come out as 1.833 M.

### Le Chatelier

Lets say the first reaction reached equilibrium and then $$\ce{N_2O_3}$$ started decomposing. This means more amount of $$\ce{O_2}$$ will be produced and the equilibrium for the first reaction will get disturbed

Without the second reaction, the first will attain equilibrium when $$a$$ is about 1.85 M. The second reaction makes F and uses up B, so the first reaction almost stays at equilibrium, and the concentration of A does not change much.

### Will it ever stop?

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $$\ce{N_2O_3}$$ will decompose further to form oxygen.

But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

If you look at this step-wise (figure out how much one reaction would go, disregarding the other), you are never done. This is similar to Zeno's paradox. However, the corrections get smaller and smaller, and your approximation gets better and better.

• My doubt is not regarding the solution to the problem, but the mechanism of equilibrium establishment. Please read my question again. Oct 17, 2020 at 14:37
• @Eyy boss: There is no solvent here, all are gases. Volume is $\pu{1.0 L}$ and it's okay to consider concentrations as given moles. Oct 17, 2020 at 15:56
• @Eyyboss You now have my completed answer.
– Karsten
Oct 17, 2020 at 20:29

The equation $$\ce{N_2O_5 <=> N_2O_3 + O_2}$$ is really two equations $$\ce{N_2O_5 \to N_2O_3 + O_2}$$ and the reverse $$\ce{ N_2O_3 + O_2 \to N_2O_5}$$. Suppose we start with pure $$\ce{N_2O5}$$ frozen at 77K so no reaction can occur then suddenly heat it up to room temperature so that reaction starts. As soon as $$\ce{N_2O3}$$ is formed there is the possibility that it will react with oxygen and reform $$\ce{N2O5}$$ and so on back and forth until equilibrium is reached where the rates of forwards and backwards reaction are equal and all three species co-exist in proportions depending on the equilibrium constant.

In you case there is a second equilibrium $$\ce{N2O3 <=>N2O + O2}$$ so as soon as any $$\ce{N2O3}$$ is formed there is a chance that it will form $$\ce{N2O}$$ and a chance that it will form $$\ce{ N2O5}$$ depending on the rate constants. However, the $$\ce{N2O3}$$ can reform by the second equilibrium as well as by the first so some of it is present at equilibrium as are all the other species. See first figure.

Now consider the case that the second reaction is not at equilibrium $$\ce{N2O3 \to N2O + O2}$$ then once $$\ce{N2O3}$$ is formed it reacts irreversibly and is lost. This will also deplete all the $$\ce{N2O5}$$. Eventually only $$\ce{N2O}$$ and oxygen will remain.

It is possible to solve the differential rate equations for these reactions and technically equilibrium is never reached, but this is an artefact of using this approach and not real. Instead, using what is called a Monte-Carlo approach uses the fact that at any given time a molecule may react or may not and by counting the number of molecules present at any time the kinetics can be determined. As we use only a few molecules this shows that there is some random fluctuation in the numbers of reactants and products, this fluctuation represents what happens with small numbers of molecules and means that the population properly reached equilibrium but fluctuates about this. This may seem silly when there are $$10^{20}$$ molecules reacting but if a small region is examined, say a sphere of some nm radius then this fluctuation will be apparent. This is, of course, the same throughout the sample.

The first figure shows the equilibrium calculation with both reactions, I have arbitrarily chosen rate constants to show the populations clearly. The second figure (with a longer time scale) shows how the species are depleted when the $$\ce{N2O3}$$ equilibrium is made 'one-way' by making the reverse rate constant zero. Above, the log populations vs time under equilibrium conditions. Log populations when the second equilibrium is made irreversible, i.e. $$\ce{N2O3 \to N_2O + O2}$$. Only $$\ce{N2O}$$ and $$\ce{O2}$$ remain as there is no further reaction possible for them. Notice that the time scales are different in the two plots.