# Mathematical justification for Le Chatelier's principle

At equilibrium

\begin{align} K &= \exp\left(\cfrac{TΔS^\circ - ΔH^\circ}{RT}\right)\\ ⇒ \frac{\mathrm d \ln K}{\mathrm dT} &= \frac{ΔH^\circ}{RT^2} \end{align}

If enthalpy change is positive, change in $$\ln K$$ w.r.t. $$T$$ is positive. Hence, $$\ln K$$ and therefore $$K$$ increase (position of equilibrium shifts to the right) as $$T$$ increases. And, $$\ln K$$ and therefore $$K$$ decrease (position of equilibrium shifts to the left) as $$T$$ decreases.

If enthalpy change is negative, change in $$\ln K$$ w.r.t. $$T$$ is negative. Hence, $$\ln K$$ and therefore $$K$$ increase (position of equilibrium shifts to the right) as $$T$$ decreases. And, $$\ln K$$ and therefore $$K$$ decrease (position of equilibrium shifts to the left) as $$T$$ increases.

Can you give a similarly mathematical reasoning for why a reaction

$$\ce{A(g) <=> 2 B(g)}$$

has its equilibrium position shifted to the left as the pressure of the surroundings increases?

• You cannot use that sort of relation to rationalise changes in equilibrium position. The reason is that the van ‘t Hoff relation (the one with d ln K/dT) tells you how the equilibrium constant changes with temperature. When you change the pressure, the equilibrium constant K does not change. – orthocresol Sep 26 '19 at 11:39
• The question is unclear. You have to carefully specify what is in the gas phase (just A and B, or inert gasses as well). The connection between temperature change and pressure change is that for both, the Gibbs energy of reaction is zero before and after, and equilibrium concentrations/partial pressures change. Of course, @orthocresol is right in saying that in one case, K changes and Q will change to match it again, and in the other case K does not change and Q will return to its starting value (but again, the concentrations change in either case). – Karsten Theis Sep 26 '19 at 15:06

## 1 Answer

I think that I might have an answer to my own question:

$$K=exp \left( \frac{TΔS^o-ΔH^o}{RT} \right)$$

implies that K is dependent on T, but not on total pressure of system, p (nor concentrations).

For $$A_{(g)}⇌2B_{(g)}$$ $$K= \frac{{p_{B}}^2}{p_{A}}$$ Partial pressure of a gas = mole fraction (mf) x total pressure of system

If p is increased then K initially increases: $$K=\frac{(mf_B.p)^2}{mf_A.p}$$ $$⇒ K=\frac{{mf_B}^2.p}{mf_A}$$

But p doesn't affect K, yet K has increased. So the mole fraction of A must be increased by shifting the position of equilibrium to the left (thus decreasing the mole fraction of B), in order to re-establish the value of K to before p was increased.