# Effect of Pressure on Equilibrium

This might be a little bit of a basic question, but I've been trying to get my head around pressure and Le Chatelier's principle and I can't quite understand it.

Currently, my understanding of Le Chatelier's principle is that: if a change to the position of the equilibrium's position is made, the system will shift to restore the equilibrium position to be consistent with the equilibrium constant.

My current understanding of equilibrium position is essentially the ratio between concentrations of reactants and products.

So here's my problem: I understand that adding, say, more of a reactant will change the equilibrium position directly by increasing the concentration of reactants, resulting in more products being produced to restore the equilibrium position back to consistency with equilibrium constant. However, I do not understand how altering the pressure of a system has any impact on the equilibrium position? It doesn't appear to increase the concentration of either the reactants or the products? If the volume of a reaction vessel is decreased, wouldn't the concentration of the reactants and products increase proportionately?

For example, if I have 4 reactants and 2 products in a reaction vessel of 10 units cubed. Let's say the concentration of reactants is 4/10 which is 0.4, while the concentration of products is 2/10 which is 0.2. The ratio of reactants to products is 0.4:0.2 which is 2:1.

If I increased the pressure by decreasing the volume of the reaction vessel to 5 units cubed, the concentration of reactants would be 4/5 which 0.8 while the concentration of products would be 2/5 which is 0.4. The ratio of reactants to products is 0.8:0.4 which is still 2:1.

I hope I have phrased my confusion clearly. Please, if you can help explain this to me or point out a conceptual misunderstanding, I would highly appreciate it.

Thank you!

• Speaking about pressure impact on equilibrium, it is much more common to use rather partial pressures, as there is usually gaseous context. For concentrations in condensed phases without involved gases, pressure has minimal impact. Commented May 5, 2022 at 10:11
• @Poutnik so, does this partial pressure have an effect on the concentration of the gases and if so how? Commented May 5, 2022 at 10:22
• Considering it as your rhetorical question which you answer yourself. Relation of gas partial pressure and and its formal concentration is determined for for ideal gas approximation by the state equation pV=nRT and concentration definition c=n/V // Consider $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$ and how the change of total pressure affects the reaction quotient. Commented May 5, 2022 at 10:24
• Have a look at this answer chemistry.stackexchange.com/questions/153545/… which gives an example showing that the extent of reaction changes as pressure changes. Commented May 6, 2022 at 10:48

Equilibrium problems involve chemical species in (g) and (aq) states (gas and aqueous solution, respectively).

For an ideal gas:

PV = nRT

Concentration is defined as:

C = n/V

If we algebraically manipulate the first equation, we get:

n/V = P/RT

Therefore:

C = P/RT

In other words, the concentration of an ideal gas is a function of both temperature (T) and Pressure (P):

C = f(T,P)

However, at isothermal conditions (constant temperature), the concentration of an ideal gas is only dependent on Pressure:

C = f(P)

For a reversible reaction of the form:

aA(g) + bB (g) ↽−−⇀ cC(g) + dD(g)

The ratio of products to reactants you're describing is a constant (K) which can be expressed in terms of concentration (Kc), partial pressures (Kp), or even molar fractions (Kx):

Kc = (CC^c)(CD^d)/(CA^a)(CB^b)

Kp = (PC^c)(PD^d)/(PA^a)(PB^b)

Kx = (XC^c)(XD^d)/(XA^a)(XB^b)

And the three of them are related in the following way:

Kp = Kx(P)^Δn = Kc(RT)^Δn

(where Δn = c+d-a-b)

From here, we can observe that at constant temperature (T), all K's are a function of pressure (P), unless Δn = 0.

When Δn = 0, all K's become independent with respect to "P", and they all become equal to each other:

Kp = Kx = Kc

In conclusion, we have two cases:

1. If Δn = 0, then the equilibrium constant will not depend on pressure:

K ≠ f(P)

1. If Δn ≠ 0, then the equilibrium constant will change if pressure changes:

K = f(P)

First examine the equilibrium constant expression. The numerator is the forward rate, the denominator the reverse rate. If all are the same phase and the number of product and reactant moles in the equation the same the effect of pressure is minimal.

Example: gas phase H2 + I2 = 2HI Same moles no major effect of pressure there will be a minor effect from nonideality.

below the iodine triple point: I2(s) + H2 = 2HI Now the [I2] in the vapor is almost constant and pressure change affects [H2] and [HI] and will change the equilibrium position in this case a pressure increase removes HI.

Reactions involving phase changes must consider the volume changes in the phase change and the latent energy change. There is also a small change in vapor pressure with inert gas pressure.

In a gas phase reaction when the equation moles of products and reactants differ increased pressure favors the side with the fewer moles.

Solids and liquids have the same considerations fewer moles are favored by higher pressures altho the behaviors of some high-pressure metallic phases are hard to explain. The pressures involved are high for condensed phases. In a phase change for a pure substance increased pressure favors the denser phase. Increased pressure by an inert gas raises the vapor pressure of solids and liquids.