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I have been pondering about the effect of pressure change in regard to Le Chatelier's principle on reactions. For this, I considered the following reaction:

$$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}$$ Le Chatelier's principle states that when the pressure on the whole reaction mixture increases, the equilibrium will shift in the direction where the number of moles of reactant(s)/product(s) is lesser as the partial pressure of a gaseous substance is directly proportional to it's mole fraction. This happens to oppose the effect of pressure being applied that tends to disturb the established equilibrium.

But here is my argument that, if external pressure applier has to be opposed, why can't it just shift backward instead? If it would shift in the backward direction where there are greater number of moles, will it not be more effective to oppose the external pressure?

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    $\begingroup$ What is the meaning of the first sentence in the last paragraph, that I show here : if external pressure applier has to be opposed ? Which applier ? Why oppose ? $\endgroup$
    – Maurice
    Commented May 5, 2023 at 8:35
  • $\begingroup$ Means external pressure applied on the reaction environment is to be opposed in order to revert it's effect on the reaction. That is, I think, what Le Chatelier's principle says. $\endgroup$
    – Proscionexium
    Commented May 5, 2023 at 8:45
  • $\begingroup$ External pressure can be applied anyway, I suppose. Like a piston or so? $\endgroup$
    – Proscionexium
    Commented May 5, 2023 at 8:46
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    $\begingroup$ Rather than “opposes”, say “partially reverses” the change imposed. So you increase the pressure, and the reaction goes (net) in the direction that reduces it. $\endgroup$
    – Karsten ♦
    Commented May 5, 2023 at 10:58
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    $\begingroup$ Expert explanations here is always helpful in clarifying science concepts/problems. Perhaps following question could also shine some light along the way: this and this $\endgroup$
    – bonCodigo
    Commented May 5, 2023 at 12:37

4 Answers 4

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This can get confusing. What you seem to have expressed is you want to reduce the effect of an increased external force by increasing the internal pressure. This can be done in such a situation of a cartridge primer where the firing pin initiates a chemical reaction to give an increase in pressure; that is not a system at equilibrium. A discussion of LeChatliers principle is about the effect on the equilibrium under the changing conditions. Please Simplify your question in your own mind.

At equilibrium the rates of the forward and reverse reactions are equal; the equilibrium constant is the ratio of the rate constants. This reduces to the reaction quotient. A simple increase in pressure will increase each factor in the reaction quotient by the same ratio. There are 4 factors one CO and 3H2 in the forward reaction and two in the reverse. Increasing the pressure accelerates the forward reaction more. In a quick pressure increase it could be possible to follow the reaction by monitoring the pressure.

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So you say, if you kick a ball, the ball should not deform, but you should injure your toes instead. As this would be the principle reversal.

The principle is not limited to chemistry.

The principle means that external change has smaller impact on the system ( in sense of the values of changed parameter, typically temperature or pressure, or change of composition), than it would have if the system did not adapt to it.

If the system response is slow wrt the externally caused change then the system response partially reverses the change. If the system response is fast, it rather weakens the change.

When in doubts, put the qualitative principle aside and just quantitatively evaluate the equilibrium constant equation for partial pressures.

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You can use the equilibrium constant to help. This is $\displaystyle K_e=\frac{P_{CH4}P_{H2O}}{P_{CO}(P_{H2})^3}$. We can express the partial pressures $P_{CH4}$ etc. in terms of the extent of reaction $\alpha$ and the total pressure $P_T$ by using an ICE table but instead of working this out let us write the partial pressures as a function of $\alpha$ as $P_{CH4}=f(\alpha)_{CH4}P_T$ etc. (and we shall not need to know what it is) and so get

$$\displaystyle K_e=\frac{f(\alpha)_{CH4}f(\alpha)_{H2O}}{f(\alpha)_{CO}f(\alpha)_{H2}^3}\frac{1}{P_T^2}$$

The equilibrium constant is constant as pressure changes so if $P_T$ increases, then the denominator, $f(\alpha)_{CO}f(\alpha)_{H2}^3$ must decrease and consequently $f(\alpha)_{CH4}f(\alpha)_{H2O}$ must increase, i.e more product is formed, just to keep $K_e$ constant.

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  • $\begingroup$ Your approach to answer is quite insightful. I am curious to know the source of this kind of approach. Could you please share the source of your approach? $\endgroup$
    – Proscionexium
    Commented May 7, 2023 at 9:21
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    $\begingroup$ No source just me ! $\endgroup$
    – porphyrin
    Commented May 7, 2023 at 13:41
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The shift of the equillibrium of the reaction doesnt fully cancel out the added pressure exterted on the mixture.We can model such a reaction as a system with negative feedback.And in negative feedback the output resists to changes of the input it doesnt completely cancel out any changes of the input.

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