I am studying La-Chatelier's Principle of Chemical Equilibrium. I got a problem of Pressure change in Chemical Equilibrium.

I know that Pressure change If a reversible reaction is in equilibrium state, the effect of Pressure change depends on the number of moles of reactants & products when all are in gaseous state.

But what happens if any of them is liquid or solid? Like a reaction: $$\ce{2A(g) + B(s) <=> C(g) + D(l)}$$

I am pursuing my studies in 10th Grade.In the textbook,I learned that if pressure increases,The Equilibrium state shifts from left to right if the number of moles of gaseous reactants is greater than that of products.And if, Number of moles of Reactants=Number of moles of Products, Then there will be no effect of Pressure change.While Studying,it came to my mind what will be the change if any reactant or Product is in Solid/Liquid state? I didn't got such answer in internet.

What will be the effect of pressure change in chemical equilibrium in the following reaction above?

Eagerly waiting to know the answer.


3 Answers 3


Le Chatelier's principle applies regardless of the phases of the reactants and products.

You just get a bigger effect if gases are involved, since the change in volume is larger.

Specifically, let $\Delta$V be the change in volume in going from reactants to products. If $\Delta$V > $0$, then increasing the pressure will shift the reaction to the left (towards reactants). If $\Delta$V < $0$, then increasing the pressure will shift the reaction to the right.

Also, it's fundamentally about the change in volume, not the change in the number of moles. The reason why you were taught to look at the number of moles is that, if all the reactants and products are in the gas phase, then the side with the greatest number of moles will have the greatest volume.

But this no longer necessarily applies if you have mixed phases. In the latter case, you need to look more closely to determine the sign of $\Delta$V. For your sample reaction,

$$\ce{2A(g) + B(s) <=> C(g) + D(l)},$$

it is clear that $\Delta$V < $0$, because you have two moles of gas on the left, and one mole of gas on the right, and the difference in volume for the gases dominates over the difference in volume for the condensed phases. Hence, for this reaction, as you increase the pressure, it will shift to the right (towards products).

  • $\begingroup$ Ok I understood, But One more question, What happens if there is no gaseous reactants or products in the reaction, like 2A(l)+B(s) => C(l) +D(l) [reversible], What will be the effect of pressure on this following reaction in chemical equilibrium? $\endgroup$
    – Sindid
    Apr 7, 2020 at 7:06
  • $\begingroup$ @Sindid Same thing. You need to sum the molar volumes of the reactants (multiplying the molar volume of A by two), and compare this to the sum of the molar volumes of the products. BTW: If you find an answer helpful, you should consider acknowledging this by upvoting it. $\endgroup$
    – theorist
    Apr 7, 2020 at 7:21

Whenever we talk about equilibrium equations, we deal with active masses of the reactants of the product. All the conclusions regarding the effect on equilibrium caused due to various changes in conditions, including pressure are drawn with respect to these active masses.

What is active mass?

The term active mass means the concentration of the reactants & products expressed in moles per litre (molar concentration). Active mass is usually expressed by enclosing the symbol of the reactant in square bracket [ ]. The active mass of solids and pure liquids is a constant quantity (unity) and solvent (excess) is considered as one, because:

  1. There is no change in activity with the change in quantity or volume of vessel.

  2. Density of pure solids and liquids is constant and molecular mass is also constant.

However, this is not applicable to the substance in aqueous solution or gaseous state because their amount in a given volume can vary, which is exactly the case you are dealing with at your level in general.

Hence, no change in equilibrium is brought out due to change in pressure corresponding to Solid or Liquid reactants only.

  • $\begingroup$ This is incorrect. As a simple example, consider the solid->liquid phase transition in water. Increasing the pressure lowers the temperature at which this occurs (i.e., favors the liquid phase), because the molar volume of the liquid is less than the molar volume of the solid. $\endgroup$
    – theorist
    Apr 5, 2020 at 7:07
  • $\begingroup$ Similarly, increased pressure by adding oxygen to nitrous oxide before compression avoids condensation of nitrous oxide, as chemical potential of liquid increases with pressure. ( used in medicine ). It is rather special case, as the critical T is near room T. $\endgroup$
    – Poutnik
    Apr 5, 2020 at 7:30
  • $\begingroup$ @theorist I get your point. However, it is mentioned that the active mass is taken as unity since "Density of pure solids and liquids is constant and molecular mass is also constant", which implies that the concept is applicable only in cases where during the reaction/process the density remains constant as opposed to the case of 'phase transition' you mentioned above. $\endgroup$ Apr 5, 2020 at 7:32
  • $\begingroup$ @Poutnik There are many details to this concept of active mass and many cases like you mentioned, but since the answer was demanded by a tenth grader, I wrote it the way that would cover most of the problems at that level. $\endgroup$ Apr 5, 2020 at 7:36
  • $\begingroup$ @HappyFeetMe Still no. There is a pressure effect on the relative amount of reactants and products anytime the reactants have a different volume from the products. This is true even if the densities of the products and reactants are assumed to be pressure-independent. $\endgroup$
    – theorist
    Apr 5, 2020 at 16:25

It doesn't matter what aggregate state the reactants and products are in. What matters are the volumes. The rule of thumb about gaseous substances is because only the gaseous substances have significant volume per mole in more or less normal conditions.

Imagine you have a closed container with 1 mole of water and 1 mole of oxygen in 20 liter volume at temp 20C (they don't react, but no worry). Shrink the volume to 10 liter and you will see that only the gas (oxygen) has shrunk, water didn't.

If the conditions are not usual as in the above example and you make the liquids and solids to shrink and expand measurably (say, by applying a GPa range pressure) then all the liquids and solids will also participate in the equilibrium shift.

  • $\begingroup$ LeChatlier's principle is based on the difference in volume between the reactants and products, not on whether their volumes are themselves pressure-dependent. I.e., in a reaction involving condensed phases, it doesn't operate because the condensed phases change density as you change the pressure; rather it operates based on the difference in volume between the reactants and products. I.e., what matters is the value of $\Delta$V, not dV/dp. $\endgroup$
    – theorist
    Apr 6, 2020 at 0:16
  • $\begingroup$ @theorist LeChatlier's principle can be expressed for volume, pressure, temperature or something else (e.g. voltage for some systems). It still holds. $\endgroup$
    – fraxinus
    Apr 6, 2020 at 17:44
  • $\begingroup$ You seem to have misunderstood my comment. I wasn't saying Le Chatelier's principle is restricted to pressure and volume. Rather, I was saying two things: (1) IF you are using it for pressure-dependence, then what matters is $\Delta$V (between products and reactants); and (2) Your last paragraph is incorrect. The liquids and solids don't need to "shrink and expand measurably" to "participate in the equilibrium shift". Indeed, the individual liquids and solids don't need to change density at all! All that matters is that there is a volume difference between products and reactants. $\endgroup$
    – theorist
    Apr 6, 2020 at 21:04

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