8
$\begingroup$

How is the spontaneity of an exothermic reaction changed when temperature is decreased? Assuming the entropy is negative.

Using Gibbs free energy equation, $\Delta G=\Delta H-T\Delta S$, $\Delta G$ should become less negative as temperature is decreased meaning the reaction is less spontaneous. However, using Le Chatelier's principle, as temperature decreases the reaction should shift towards the right towards the products, becoming more spontaneous.

What's wrong here?

$\endgroup$
  • 3
    $\begingroup$ They are not directly correlated. The Le Chatelier's principle counts from one equilibrium state, considers the changing due to different condition. The Gibbs free energy counts from reactant (haven't start reaction), considers the amount of free energy to reach equilibrium. They could be related if one starts from Gibbs free energy-> set up van't Hoff equation at spefic condition-> derive Le Chatelier's principle. (I don't remember some details, though) $\endgroup$ – user26143 Dec 10 '13 at 13:13
  • 1
    $\begingroup$ Here, $H$ and $S$ are constants for particular reactions, so $T$ is the only variable, which is always positive (measured in K). As $S$ is said to be negative, $-TS$ will be positive. Now, $H$ is negative (exothermic reaction). Therefore, we can write $G=H+(-TS)$ as $G=(-)+(+)$, as $T$ increases, negative value decreases, so reaction will tend to be less spontaneous, if $T$ value decreases, negative value increases, so reaction will tend to be spontaneous. So, Le Chatelier's principle is in agreement with gibbs energy. Anyway, you have good thinking in linking concepts. $\endgroup$ – Immortal Player Dec 10 '13 at 18:58
4
$\begingroup$

The key is $\Delta G = -RT \ln K$.

The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$.

$-RT \ln K = \Delta H-T\Delta S$

$-R \ln K = \frac{\Delta H}{T}-\Delta S$

In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent.

For an exothermic reaction, $\Delta H$ is negative. If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.