I already know the way to determine whether if the given reaction is exothermic or endothermic by the enthalpy values. But is it possible without enthalpy values, and just by looking at the reaction?

So as far as I know there are two types of chemical reactions, namely one way reactions and reversible reactions.

So how can I determine this factor in one way reactions and reversible reactions?

I guess that for reversible reactions we can use equilibrium constants such as $K_p$, $K_c$, but I am not quite sure whether it is correct. On the other hand, I feel that there is no such way to determine what I want in one way reactions. So how can I decide it?

  • $\begingroup$ Hello everyone , I want to know whether the English that I have used here is good enough or is it horrible?Because I am not a native Englishman.If so please show me that mistakes also, while you helping me to come out of this question. $\endgroup$ – On the way to success Jun 14 '15 at 15:32
  • $\begingroup$ I don't think there is a way. The concept of enthalpy came to be for a reason. $\endgroup$ – Papul Jun 14 '15 at 17:28
  • 5
    $\begingroup$ You can make an educated guess based on the relative strengths of the bonds broken vs bonds formed. $\endgroup$ – orthocresol Jun 14 '15 at 18:20
  • $\begingroup$ @On the way, your English is quite good. It might need some polishing, but then what are the editors for? So no worries. :) Also, regarding your question, if one knows about the bond strengths in both reactants and products (for example they'd say an F-F bond is way weaker than say, an H-H bond) , they could decide the sign of enthalpy, but it would be very very crude and it could lead to wrong results. Unfortunately, I have nothing else related in my mind right now. $\endgroup$ – M.A.R. Jun 14 '15 at 18:21
  • $\begingroup$ @M.A.Ramezani I agree that using bond strengths would lead to a very crude estimate, but if you only want the sign of $\Delta H$ and not the magnitude, for 99% of reactions it should suffice. The 1% would be when your estimated $\Delta H$ is close to 0. Regarding the question itself, I'm just nitpicking here but all reactions are reversible except for reactions where a product escapes from the system. As such, you can theoretically use le Chatelier's principle to determine whether the reaction is exothermic or endothermic based on the change in equilibrium position when temperature changes. $\endgroup$ – orthocresol Jun 14 '15 at 22:57

First of all, the equilibrium constant $K$ in general doesn't tell too much about the enthalpy change of the reaction as it is related to the (Gibbs) free energy change $\Delta G = \Delta H - T\Delta S$. It is true that in some simple cases (such as gas phase reactions) the entropy change can be approximated by looking at the stoichiometry of the reaction but in general it's usually more complicated.

Regarding your distinction between one-way and reversible reactions I think you may as well assume all reactions are reversible, but some have incredibly huge $K$ so the equilibrium is shifted totally to one side of the reaction - this is the result of a very negative $\Delta G$.

In case of a "reversible" reaction you can always look at the change of the equilibrium constant with temperature. The constant is

$K = \exp\left(-\frac{\Delta G}{RT}\right)=\exp\left(-\frac{\Delta H}{RT}\right)\exp\left(\frac{\Delta S}{R}\right)$

so from the temperature dependence you can deduce the sign of $\Delta H$. Naturally, this is assuming that $\Delta H$ and $\Delta S$ doesn't depend on the temperature in the temperature range you're looking at.

equilibrium constant vs temperature

The figure shows the change of equilibrium constant with temperature (arbitrary units) for an exothermic (yellow line) and an endothermic (blue line) reaction.

Just by looking at the reaction I don't think you can tell the enthalpy change. Normally you have bonds breaking on the reactant side and others forming on the product side and normally there is a delicate balance between the two. People have come up with bond strength values (which could be determined from known $\Delta H$ values) and you can use those in your reaction.

  • $\begingroup$ How about comparing equilibrium constants at two different temperatures ? $\endgroup$ – On the way to success Jun 17 '15 at 10:30
  • $\begingroup$ But can't we compare Kp ,Kc for different temperatures , and then come to a conclusion? $\endgroup$ – On the way to success Jun 17 '15 at 10:34
  • $\begingroup$ Yes, you are right. I have modified the post so it makes much more sense. $\endgroup$ – albapa Jun 17 '15 at 10:40
  • $\begingroup$ If you would give few examples that would be great. $\endgroup$ – On the way to success Jun 17 '15 at 10:48
  • $\begingroup$ Then how about comparing two Kp values at two different temperatures ? $\endgroup$ – On the way to success Jun 20 '15 at 1:42

If the enthalpies of formation for all reactants and products are known then you can compute the standard enthalpy change ($\Delta H^\circ$).

If they are not known you can estimate $\Delta H^\circ$ using bond energies as described in this video.

Finally if you happen to know that the reaction is spontaneous ($\Delta G^\circ < 0$) and if the standard entropy change is (likely) negative (e.g. if the number of particles decrease) then it must be an exothermic reaction.

  • $\begingroup$ not sure what you mean by "one way" and "reversible" reactions $\endgroup$ – Jan Jensen Jun 17 '15 at 8:36
  • $\begingroup$ @ Jan Jensen:I used one way to indicate the reactions which have one solid forward arrow, while reversible for ones which are in equilibrium. $\endgroup$ – On the way to success Jun 17 '15 at 10:32
  • $\begingroup$ $\ce{A -> B}$ means that at equilibrium there is more $\ce{A}$ than $\ce{B}$, while $\ce{A} \rightleftarrows \ce{B}$ suggests that there is roughly an equal amount of $\ce{A}$ and $\ce{B}$. However, both usually imply equilibrium $\ce{A <=> B}$. Either way, thus does not affect how you would estimate endo/exothermicity. $\endgroup$ – Jan Jensen Jun 17 '15 at 12:25

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.