Many chemistry reactions are reversible in nature. In such reactions initially reactants react to form products but after some time products also react to give reactants back again. And an equilibrium is established where forward and backward reactions occur with equal rates.

If products are more stable, why will they react again to give reactants?

  • 3
    $\begingroup$ If the Pacific ocean is deeper (which it is), why wouldn't all water from Atlantic pour into it? $\endgroup$ Commented Feb 17, 2020 at 8:29
  • 2
    $\begingroup$ " but after some time ..." No. Directly, always. $\endgroup$
    – Karl
    Commented Feb 17, 2020 at 8:35
  • 2
    $\begingroup$ This is one version of the central question of thermodynamics. The solution was to introduce the concept of entropy. $\endgroup$
    – Buck Thorn
    Commented Feb 17, 2020 at 11:35
  • 2
    $\begingroup$ Does this answer your question? Why does equilibrium exist? $\endgroup$
    – Mithoron
    Commented Feb 17, 2020 at 17:45

2 Answers 2


It is best to think of reversible reactions statistically

A good understanding about equilibrium can be had by thinking about the detailed molecular level processes involved at the molecule level.

Take, for example, the simple equilibrium between liquid water and water vapour in a closed vessel. At room temperature, the equilibrium clearly favours liquid water. But there is always some water vapour present (that's what we call humidity in the wider atmosphere). Why is there any vapour if the equilibrium favours liquid? The reason is that the water molecules don't all have the same amount of energy. some are travelling fast and some some are slow. A small percentage of the fast ones can escape the liquid and become vapour (ie gas-phase water molecules not liquid phase water molecules). In the gas above the liquid, the same is true. Not all the gas molecules have the same speed. Some are travelling slowly and, if they hit the liquid surface, stay in the liquid. Both processes are dynamic and are happening fairly fast. So molecules are constantly exchanging between the vapour and the liquid.

Equilibrium is not reached when everything stops: at room temperature there is a lot of kinetic energy shared out among the molecules of the system. Equilibrium happens when the rates of evaporation and liquefaction are the same. So the number of molecules per second going into the vapour from the liquid equals the number going the other way.

The same is true for other, more complex, reversible reactions. Equilibrium is not stasis but an average state resulting from dynamic reactions in both directions.

This point of view makes it clear why, even when one product is far more stable, the reverse reaction can still happen.


It is no use comparing the stability of the reactants and products.

Maybe you would better understand with an experimental example. Suppose you prepare $\pu{10 mL}$ of an alcoholic solution of cobalt chloride hydrate $\ce{CoCl2.6H2O}$. This solution is blue due to the formation of $\ce{CoCl4^2-}$ according to the following equation: $$ \ce{2 CoCl2.6H_2O -> [Co(H2O)6]^2+ + [CoCl4]^2-}$$ The ion $\ce{[Co(H2O)6]^2+}$ is pale pinkish, and the ion $\ce{CoCl4^2-}$ is dark blue. So the color of the whole solution is blue. Now you add two or three drops of water. You observe that the solution turns pinkish, because of the following equation, which I will show that it is an equilibrium: $$\ce{[CoCl4]^2- + 6H2O <=> [Co(H2O)6]^2+ + 4 Cl^-}$$

This equation is written in the exothermic sense.

Now if you heat this pink solution to $\pu{60 ^\circ C}$ or $\pu{70 ^\circ C}$, the solution turns blue. $\ce{[Co(H2O)6]^2+}$ is transformed into $\ce{CoCl4^2-}$. The reaction goes to the left hand side. This is a consequence of the Le Chatelier principle: Heating a system favors the endothermic side, which is also the side where the entropy is maximum.

Cooling the hot blue solution will change its color to pink again, because $\ce{[Co(H2O)6]^2+}$ is favored since cooling favors the exothermic transformation. This can be repeated by successive heating and cooling.

  • 1
    $\begingroup$ It is no use comparing the stability of the reactants and products. And in the course of your answer you are comparing the stabilities of those reactions... $\endgroup$ Commented Feb 18, 2020 at 12:59

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.