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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.

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  • $\begingroup$ Then why $K$ the constant of reaction depends on the temperature if "heat" is not a product of the reaction ? $\endgroup$ – ParaH2 Nov 17 '16 at 23:28
  • $\begingroup$ As said already heat does not go into the reaction quotient. The LCP argument (which relates changes in conditions to changes in $Q$) is therefore shaky. $\endgroup$ – orthocresol Nov 18 '16 at 0:16
  • $\begingroup$ I gave this analogy to a duplicate question that asked this: There are two people. One guy likes chocolates and one guy hates them. They decide to transfer the chocolates. If you start throwing chocolates randomly to both of them, in which side will the reaction go? The guy who hates chocolates would keep giving it to the guy who likes them. In a similar way, if you provide heat to a reaction, it would move in the direction of the where the heat is absorbed, i.e the endothermic pathway. $\endgroup$ – Pritt says Reinstate Monica Jun 16 '17 at 15:23
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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!

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  • $\begingroup$ What does K° and ∆H° represent? Is it the equilibrium constant and enthalpy of a standard temperature. If so then why are you calculating for standard conditions $\endgroup$ – Scáthach Sep 28 '18 at 20:52
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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.

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