Le Châtelier's Principle and heat

Question

Consider the following reaction at equilibrium. $$\ce{A->B}, \Delta H < 0 $$

Suppose I increase the temperature. Now, quite a few people would invoke Le Châtelier's Principle and say that since "heat" is a product of this reaction, the equilibrium should shift backwards. This is clearly wrong because "heat" is not a species you can have in a reaction. You can't incorporate it into the reaction quotient.

If you take $$\Delta G = \Delta H - T\Delta S$$ for $\Delta S < 0$, you could make a claim of the equilibrium shifting backwards at higher $T$. This doesn't have anything to do with the sign of $\Delta H$ however.

Is there an actual theoretical basis for the following claim that does not invoke a principle that does not apply:

For an exothermic reaction at equilibrium, increasing the temperature will cause the equilibrium to shift towards reactants.

Answer 1

Before to show you that what you said is false (I'm sorry ^^) be sure you understand the constant of a reaction depends on the temperature.

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Let the same reaction you want $$\mathrm{A \rightleftharpoons B}$$

With a constant $\mathrm{K^{\circ}}$. Imagine you heat your system a little with $\mathrm{d}T>0$, then by Van't Hoff's law we get:

$$\frac{\mathrm{d}\ln(K^{\circ})}{\mathrm{d}T}=\frac{\Delta_rH^{\circ}}{RT^2}$$ Where $R$ is the perfect gas constant which is positive. Then $RT^2>0$. So $\mathrm{d}\ln(K^{\circ})$ as the same sign as $\Delta_rH^{\circ}\cdot \mathrm{d}T$.

So if you have an endothermic reaction $\Delta_rH^{\circ}>0$ because $\mathrm{d}T>0$ then, $$\mathrm{d}\ln(K^{\circ})>0$$

Then $K^{\circ}$ will increase with the temperature. If you take $\mathrm{d}T<0$ for an endothermic reaction then the constant of the reaction will decrease with the temperature.

I let you do the same final reasoning for an exothermic reaction. So the "Le Châtelier principle" is still true.

Note: If you know the chemical affinity, you can do the same proof, just a bit longer!

Answer 2

Firstly, we should add standard symbols to the equation, because only $\Delta G^\circ$ is related to the equilibrium constant $K$, whereas $\Delta G$ is related to the instantaneous reaction quotient $Q$. Therefore any change in $K$ has to be related to $\Delta G^\circ$ and not $\Delta G$. So you have

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

and the argument goes, when $T$ increases, $\Delta G^\circ$ increases (given $\Delta S^\circ < 0$), and therefore the equilibrium goes to the left. The issue is that the equilibrium position is not measured by $\Delta G^\circ$ - it is measured by $K$. And the dependence of $K$ on $\Delta G^\circ$ itself has a temperature component:

$$K = \exp{\left(-\frac{\Delta G^\circ}{RT}\right)}$$

so if you increase $T$, not only does the magnitude of the numerator increase, so does the denominator. That's precisely where this argument fails: you can't necessarily relate a larger $\Delta G^\circ$ to a smaller $K$, if the temperature is changing.

(Note that if you have two different reactions, with given values of $\Delta G^\circ$ at the same temperature $T$, then it is fair to say that the reaction with a larger $\Delta G^\circ$ has a smaller $K$. However this has to be thrown out of the window once your $T$ starts changing.)

So, to relate the change in $\Delta G^\circ$ to the change in $K$ you have to effectively "bring the factor of $T$ over to the other side of the equation". You just need a bit of algebraic manipulation:

$$\begin{align} \ln K &= -\frac{\Delta G^\circ}{RT} \\ &= -\frac{\Delta H^\circ - T\Delta S^\circ}{RT} \\ &= -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R} \end{align}$$

From here you can clearly see that if $T$ increases, then the sign of $\Delta H^\circ$ is what controls whether $\ln K$ (and hence $K$) increases or not; $\Delta S^\circ$ has no role to play. This is consistent with the conclusion derived via Le Chatelier's principle.

See also another of my answers on the van 't Hoff equation here.