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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?

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    $\begingroup$ 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. $\endgroup$ – orthocresol Sep 26 at 11:39
  • $\begingroup$ 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). $\endgroup$ – Karsten Theis Sep 26 at 15:06
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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.

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