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.

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

With a constant $\mathrm{K^{\circ}}$. Imagine you heat a little your system 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 reasonning for an exothermic reaction. So the "Le Châtelier principle" is still true.

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

I hope it can help you if you want I can add the proof with affinity. Have a good day I'm going to sleep ! :-)

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.

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!

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.

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

With a constant $\mathrm{K^{\circ}}$. Imagine you heat a little your system 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 reasonning for an exothermic reaction. So the "Le Châtelier principle" is still true.

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

I hope it can help you if you want I can add the proof with affinity. Have a good day I'm going to sleep ! :-)

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.

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

With a constant $\mathrm{K^{\circ}}$. Imagine you heat a little your system 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 reasonning for an exothermic reaction. So the "Le Châtelier principle" is still true.

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

I hope it can help you if you want I can add the proof with affinity. Have a good day I'm going to sleep ! :-)