# Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants?

I understand Le Chatelier's principle and how every change to an equilibrium system causes an opposing reaction from the system. I also understand how, when pressure is increased, the equilibrium shifts to the side with the lowest number of moles of gas and vice versa.

So, why does this happen? How is it more energetically favourable - or is it something else?

Also, are these explanations of why the ration changes for the other factors right:

• Temperature - equilibrium moves to the endothermic side of the reaction as temperature increases because the activation enthalpy of the exothermic reaction increases as the overall energy of the system is less
• Concentration - ratio of products/reactants changes immediately after one part of the reaction's concentration increases, it then approaches the same ratio (assuming one whole side of the reaction is added)
• Catalysts do nothing more than decrease the time it takes to reach equilibrium as they catalyse both sides of a reversible reaction and reduce the activation enthalpy

With gasses, what you're doing by changing the pressure is you change the partial pressures or the reactants. As long as there's the same moles of gas on either side, the equilibrium is unaffected, but if there's an un unequal number, the reaction quotient is changed. The same would happen if you added water to an aqueous reaction. You can play with the numbers yourself, I'll give you an example to use:

$$\ce{N2(g) + 3H2(g) <->2NH3(g)}$$

We can use the reaction quotient with partial pressures, but it's more clear if we use the one with concentrations:

$$Q_c\ce{ = \frac{[NH3]^2}{[N2][H2]^3}}$$

Using $c=\frac{n}{V}$:

$$Q_c\ce{ = \frac{\frac{n(NH3)^2}{V^2}}{\frac{n(N2)}{V}\times \frac{n(H2)^3}{V^3}}}$$

Take notice of how this fraction depends on volume! So it's really just the system reacting to attempt to reach equilibrium again (making it so that K = Q).

As for temperature. My understanding is that it's not to do with activation energy. It IS related to the enthalpy of the reaction though, and your understanding of what a temperature change means for a particular reaction is correct.

Concentration is the same deal as I've talked about in the beginning of this post, and incidentally it matches your own understanding of it.

You've got catalysts right too. Have a look at the following diagram: • I'm glad you used concentrations over partial pressures because I have actually read about concentrations and equilibrium constants
– Kian
Feb 11, 2013 at 20:55

I'll just explain the effect of changing temperature, as this isn't covered so well as the others in Brian's answer. Gibbs energy and the equilibrium constant of a reaction is given by this identity: $$\ce{\Delta G = -RT * ln(K)}$$

The gibbs energy is also given by: $$\ce{\Delta G = \Delta H - T\Delta S}$$ Equating the two RHS: $$\ce{-RT*ln(K) = \Delta H - T\Delta S}$$ Divide through by $\ce{-RT}$: $$\ce{ln(K) = \Delta H/-RT} + \Delta S/R$$ $$\ln(K) = \frac{-\Delta H}{R}*\frac{1}{T} + \frac{\Delta S}{R}$$ If $\Delta H$ is positive - an endothermic reaction, then $-\Delta H$ is negative. Increasing the temperature makes $\Delta H$ less negative, and therefore makes $ln(K)$ less negative, and thus $K$ bigger- it therefore shifts the equilibrium towards the side of the products: $$K = \frac{Products}{Reagents}$$ Meaning more products. If $\Delta H$ s negative - and exothermic reaction - then $-\Delta H$ is positive. Increasing the temperature makes $-\Delta H$ less positive, and therefore makes $ln(K)$ smaller, and $K$ is therefore smaller, shifting the equilibrium to the side with the reagents on it.

$\Delta H$ and $\Delta S$ also depend on temperature, but over a small temperature range this can be ignored.