We know that if a system's equilibrium is exposed to a stress, the system shifts to relieve that stress. According to my teacher, certain kinds of pressures are stresses and others are not.

Normally, if we just say that the total pressure of a container with a reaction occuring inside is increased, the reaction shifts toward the side with a small sum of the mole ratios (i.e. $\rm 2A +3B \longleftrightarrow C + D$, if increasing pressure reaction shifts to the right to decrease pressure). I understood this formerly, but another case erases this understanding.

My teacher also said that if we add a noble gas to the container in which this reaction is occurring, although the pressure increases because of a new gas, the equilibrium does not shift to counteract this pressure increase. She says it has to do with the partial pressures of the reactants and products of only the particles In the reaction, but not in the total balloon. But this makes no sense to me because by increasing the total pressure of the system as described before, how does this cause an equilibrial stress and this case doesn't? How does increasing total pressure even cause an equilibrium shift? By increasing total pressure we have a proportional increase in the collision rates of all particles involved, do we not? If this is true, then why wouldn't the reaction rates both be doubled, and hence cause the equilibrium to not shift?

If I'm missing some fundamental principle, please let me know.



2 Answers 2


Le Chatelier's principle is not about pressure, it is about concentrations and temperature (Please also see the good article on Wikipedia for Le Chatelier's principle).

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the "right" - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.

I'll make a silly comparison. Let's take this reaction: $$\rm Boy+Girl\longleftrightarrow Pair$$ If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...

Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.

  • $P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
  • $P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
  • $P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
  • $P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$

Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$ $$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$

Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:

$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$ where

  • $\upsilon'(A)=2\,mol-2x\,mol$
  • $\upsilon'(B)=3\,mol-3x\,mol$
  • $\upsilon'(C)=1\,mol+x\,mol$
  • $\upsilon'(D)=1\,mol+x\,mol$

Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$

  • $X'(A)=(2-2x)/(7-3x)$
  • $X'(B)=(3-3x)/(7-3x)$
  • $X'(C)=(1+x)/(7-3x)$
  • $X'(D)=(1+x)/(7-3x)$

Solving the eqution gives $x=0.325\,mol$, so now:

  • $\upsilon'(A)=1.35\,mol$
  • $\upsilon'(B)=2.03\,mol$
  • $\upsilon'(C)=1.33\,mol$
  • $\upsilon'(D)=1.33\,mol$

But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right.

  • $P'(A)=0.45\,atm$
  • $P'(B)=0.67\,atm$
  • $P'(C)=0.44\,atm$
  • $P'(D)=0.44\,atm$

Ok, but when we add inert gas, what happens:

$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$

It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction: $$\rm A_g+B_g\longleftrightarrow C_g$$

Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure. $$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$

The partial pressures are:

  • $ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
  • $ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
  • $P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$

Now we will add 1 mol of inert gas: $$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$

Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same):

  • $P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
  • $P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
  • $P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$

As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.

Let's take this reaction: $$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$ You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.

Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:

$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$

will not do practically anything, because concentrations of reagents will not increase under the pressure.

When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.

If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left.

If $\Delta H>0$, you need to think about heat not as "product", but as "reagent".

For better understanding try to "dive" into Equilibrium thermodynamics.

  • 1
    $\begingroup$ I like the astute observation about the Principle's not applying to pressure. Pressure simply influences concentrations, so this is why the shift occurs. Also, though, I need to correct you in the second example about partial pressures. The end idea is still right, but you had an arithmetic error - Ptotal=2/7⋅1.17atm=0.5atm should be = 0.33 atm. $\endgroup$
    – user11629
    Mar 11, 2014 at 16:30
  • $\begingroup$ @user11629 Thank you for correction. And one remark, chemistry is about concentrations always, if something can change concentration it will make effect on reaction. Chemistry is also about activation energies (about temperatures or light). $\endgroup$
    – saldenisov
    Mar 11, 2014 at 16:33
  • $\begingroup$ But one thing that you haven't explained is why decreasing the volume does not affect both reactants and products equally, so why does pressure even cause an equilibrial stress if the likelihood that both reactions will occur is doubled? This is the same effect as a catalyst, which makes both the forward and backward reactions occur faster - and this effect has no net change on the equilibrium position. $\endgroup$
    – user11629
    Mar 11, 2014 at 16:34
  • $\begingroup$ @user11629 I did not catch what you are asking? $\endgroup$
    – saldenisov
    Mar 11, 2014 at 18:10
  • 1
    $\begingroup$ Two rates are equal. More reactants than products exist at equilibrium. We double pressure. Why does this cause a change in the equilibrium position at all? You said it increases the partial pressures of the reactants, which is true, but it also increases the partial pressure of the products, so the ratio between pressures of reactants and products are the same. $\endgroup$
    – user11629
    Mar 12, 2014 at 2:29

I finally figured it out using simple logic for myself as follows. Thank you for the answer though!

Consider an equilibrium as described:

$K_{eq} = \dfrac{[A]^n}{[B]^m}$

If we double the pressure of the system at large, i.e., halving the volume, we double the concentrations of products and reactants. If $n=m$, Le Chatelier's Principle is proven correct because no change occurs, as the mole sum is the same for products and reactants. If $n\neq m$, Le Chatelier's Principle is also correct; doubling and squaring, for instance, will not yield the same resulting as doubling and cubing the concentrations.

In this way we have a short, comprehensive, and intuitive understanding of the reasoning behind Le Chatelier's Principle with regard to concentration changes caused by pressure.

  • $\begingroup$ Well, but one thing is to change the overall pressure by changing the volume (that has an effect on the equilibrium, since partial pressures/concentrations do vary), another is to change the overall pressure by adding an inert gaz without changing the volume (that has no effect on the equilibrium, since partial pressures/concentrations do not vary). So, I do not think your reasoning (which is correct) applies to the arguments you were discussing in your question. $\endgroup$
    – mannaia
    Mar 14, 2014 at 8:36

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