Which way Le Chatelier's principle

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?




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.


 2

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.


 1

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.


 0

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.


 0