thermodynamics - Le Châtelier's Principle and heat

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

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!