# Help understanding the pressure's effect on equilibrium and Le Chatelier's principle [closed]

So let's say we have a closed system in the following reaction:

$\ce{N2O4 -> 2 NO2}$

The only way to increase the system's pressure without adding any mole of gas and keeping temperature constant is reducing the system's volume. Once we increase the pressure by reducing the total volume, formation of $\ce{N2O4}$ is observed.

In many text books, the explanation for this is that 2 moles of $\ce{NO2}$ occupy a bigger volume than 1 mole of N2O4, thus the equilibrium is restablished when more $\ce{N2O4}$ is formed. What doesn't make sense for me is that both the substances are contained in the same volume, so how 1 mole of $\ce{N2O4}$ would occupy less volume than 2 moles of $\ce{NO2}$? It only makes sense to think about partial pressures, not partial volumes, right?

I'm trying to use entropy and energy microstates to solve this problem, because it seems more fundamental than this mantra in the general chemistry textbooks. From quantum mechanics, I know (can't explain, just know; need further studying on this subject) that translational energy states become closer when a gas expands in a bigger volume, so I'm assuming that in the reaction above, with a bigger pressure (and thus a smaller volume) these energy states will be more spaced between each other. With energy states more far apart, why are there more possible energy microstates when $\ce{N2O4}$ is formed?

I can't get past this! Why does this implies in the formation of the substances with less gaseous coeficients? How are the energy states of reactants and products and their changes on pressure related with that?

Edit: to make myself clearer, I'll use this example from this textbook I'm reading: Chem1 Virtual Textbook, Stephen Lower, link

In the endothermic reaction H–H (g) ⟶ 2 H (g), you only get hydrogen atoms at very high temperatures, because you'll have a higher distribution of populated states on the 2 H side, thus you favor the products at higher temperatures for endothermic reactions. I can completely understand the effects of themperature in Le Chatelier's principle, and I'm trying to understand an analogous thought for changes in pressure in the microscopic point of view.

• Are you familiar with the concept of the Equilibrium Constant for a reaction? – Chet Miller Sep 19 '18 at 2:30
• Yes. I can see from the algebric point of view why it'll increase the concentration or partial pressure on N2O4. But the equilibrium constant comes from gibbs energy which comes from entropy, which is what I can't understand from the microscopic point of view. – Dante Rodella Sep 19 '18 at 2:54
• It's an interesting questions, what happens on the microscopic scale. One point is that at elevated pressure (and identical temperature), the same species spends more time in close contact. Same speed, more collisions per time. I'm not sure if or when that is already enough for a quantitative explanation. – Karl Sep 19 '18 at 6:34
• Which part about the entropy from the microscopic point of view can't you understand (1. entropy change of a pure ideal gas with changes in pressure, 2. entropy of mixing of ideal gases, or 3. entropy change of molecules reacting to produce new substance)? – Chet Miller Sep 19 '18 at 11:13
• I think it's a mix of 1. and 3. as you change the pressure causing the equilibrium to be restablished by a chemical reaction. – Dante Rodella Sep 19 '18 at 14:08

What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.

In order to understand this a little better you can have a look at the ideal gas law:
$$p = \frac{nRT}{V}$$ As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG < 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.

• That makes sense if you, like all of us, have inhaled Le Chatelier as the basis for explaining all chemical experience. I think OP isn't doubting the fact, but is asking "how do the molecules know"? ;-) – Karl Sep 19 '18 at 6:41
• @Karl I tried adding an explanation. Hopefully it is correct (and I didn't miss something) and will help to clarify. – Raven Sep 19 '18 at 7:19
• You misunderstood. I believe OP is looking for a molecular explanation. Molecules haven't heard of Gibbs. ;-) – Karl Sep 19 '18 at 7:30
• Your explanation was great, but it's exactly what @Karl is saying. Also, I can understand ΔrG< 0 since the formation of N2O4 is exothermic, and even a decrease in the system's entropy would cause an increase in the universe's entropy at a certain temperature. But what if it were an endothermic process? – Dante Rodella Sep 19 '18 at 14:05
• @Dante Rodella The whole Gibbs-thing is merely a "visualisation" for what's going on with the energy. This does apply to moles of particles just as well as to individual particles: They'll always try to minimise their energy (which can only be overcome of something else gets an even lower energy state instead). This is the way we experienced our universe to work and I'm afraid I haven't found any explanation as to why that is. As for the endothermic reaction: It doesn't matter whether the reaction is endothermic or exothermic. It only matters that it is exergonic. – Raven Sep 19 '18 at 15:29

A molecule's rotational states do not become closer as the pressure is increased, neither do the vibrational levels, they are a property of the molecules themselves. The population in these levels is determined by the temperature which in your case is constant. More collisions per second as the pressure is increased means more possibilities of reaction. This means that there are more chances that at a given total energy (comprised of translational, rotational & vibrational energy) the molecules will have enough energy to cross the energy barrier (transition state) between reactants and products and vice versa, i.e the number of barrier crossings both ways is increased.

Note that this works both ways, a collision between two $\ce{NO2}$ can form $\ce{N2O4}$ as well as a collision destroying it and forming $2\ce{NO2}$, i.e. a collision between $\ce{N2O4}$ and $\ce{NO2}$ (or between two $\ce{N2O4}$ ) can sometimes convert $\ce{N2O4}$ into $2\ce{NO2}$.

The species are in equilibrium $\ce{N2O4 <=> 2NO2}$ and the equilibrium constant has a fixed value at constant temperature and is unaffected by changes in total pressure, because, as noted above, the population of energy levels depends only on temperature.

[Notes: The equilibrium constant can be expressed as the ratio of partial pressures. At equilibrium $\Delta G^\mathrm{o}=-RT\ln(K_p)$ where for a reaction $\ce{xA + yB<=>rC + sD}$ the equilibrium constant is $\displaystyle K_p= \frac{P_C^rP_D^s}{P_A^xP_B^y}$. The change in $K_p$ with temperature is given by the Van't Hoff isochore, $\displaystyle \frac{d\ln(K_p)}{dT}=\frac{\Delta H^\mathrm{o}}{RT^2}$ ]

• In terms of reaction rates, it makes sense for me but only because the formation of N2O4 is exothermic. The difference in energy between the transition state and N2O4 is higher than the energy difference between the transition state and NO2, so both reaction rates will be higher as the pressure increases, but the formation of N2O4 will be favored only because the formation of N2O4 has a higher reaction rate constant. What would happen in an endothermic reaction? Does this make sense? – Dante Rodella Sep 19 '18 at 14:00
• Also I read this in this textbook that I love: "The larger the volume in which the gas is confined, the more closely-spaced are these [energy] states, resulting in a huge increase in the number of microstates into which the available thermal energy can reside" link. And you're saying "A molecule's rotational states do not become closer as the pressure is increased, neither do the vibrational levels, they are a property of the molecules themselves." So is the textbook wrong or those things just don't relate? – Dante Rodella Sep 19 '18 at 14:02
• Yes; the larger the volume the larger the number of translational energy states, think of this as the number of quantum levels as in a 'particle in a box'; the bigger the box the more closely spaced are the energy levels (density of states increased) and so at a given temperature more levels are populated. The rotational and vibrational states are unaffected by this. – porphyrin Sep 19 '18 at 16:05
• Oh yes, you're right. I corrected exchanging "rotational" by "translational" in my question. – Dante Rodella Sep 19 '18 at 16:44
• The question is about increasing the pressure by decreasing the volume, and that does change partial pressures, and changes the equillibrium, unless the reaction is pressure-invariant! – Karl Sep 19 '18 at 19:45