4
$\begingroup$

If you have an endothermic reaction with a negative entropy change, is it still possible to induce the reaction in some way despite the fact that the Gibbs free energy change is positive or all temperatures?

$\endgroup$

1 Answer 1

6
$\begingroup$

Yes, $\Delta G = -RT\ln K$.

If a reaction is endothermic with a negative entropy change, $\Delta G$ is positive.

That $\Delta G$ is positive only means that $K<1$.

$\Delta G$ would need to approach infinity for $K$ to approach zero.

Therefore, at equilibrium, there will always be some products in principle no matter how unfavorable the reaction.

As a practical matter, the reaction would need to be only slightly endothermic and having only a slight negative entropy change to get a significant amount of product.

However, if the small amount of product is continuously removed from the system, as for example by precipitation from a liquid phase, or gas bubbling out of a liquid phase, or by liquid-liquid extraction, one can keep the reaction going forward.

$\endgroup$
4
  • $\begingroup$ Thank you for your reply! Is K the equilibrium constant in this equation? I have not come across this equation before... $\endgroup$
    – Meep
    Feb 17, 2015 at 17:17
  • $\begingroup$ The equation is actually the definition of standard equilibrium constant, so it is a very important equation: goldbook.iupac.org/S05915.html $\endgroup$
    – DavePhD
    Feb 17, 2015 at 17:31
  • $\begingroup$ Why does the ozone layer exist although the Gibbs energy for its formation is positive? $\endgroup$
    – Shub
    Mar 20, 2022 at 16:39
  • $\begingroup$ @Shub diatomic oxygen in dissociated in to monoatomic oxygen by ultraviolet light, then the monoatomic oxygen reacts with diatomic oxygen to form ozone: csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q2.pdf $\endgroup$
    – DavePhD
    Mar 21, 2022 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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