One reaction used to form sodium hypochlorite (A type of bleach) is as follows:

$\ce{2NaOH(aq) +Cl2(g)->NaOCl(aq) + NaCl(aq) + H2O(l)}$

Provided the sodium hydroxide is dissolved in some substance.

I know that the sodium hydroxide dissociates to form the $\ce{Na+}$ and $\ce{OH-}$ ions, but I do not know why the $\ce{Cl2}$ breaks down into 2 separate atoms instead of just staying the way it is. My questions are: what are the driving forces behind this reaction and why does the $\ce{Cl2}$ break down? I see how it happens, I just don't know why.

  • $\begingroup$ Do this reaction requires heat/ occurs in presence of sunlight? $\endgroup$ – ashu Aug 9 '13 at 18:51
  • $\begingroup$ I don't think so. Here is a link to how it is made: link. Alternatively, there is the Wikipedia page. $\endgroup$ – user2514631 Aug 9 '13 at 19:49

At constant pressure and temperature, the spontaneity of all chemical reactions is controlled by the difference in Gibbs free energy ($\Delta G$), as described by the following equation:

$\Delta G = \Delta H - T\Delta S$

Where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change, and $T$ is the absolute temperature. For a reaction to take place spontaneously, the free energy change must be negative (i.e., $\Delta G < 0$). In the specific reaction you're asking about, it's very probable that the change in entropy is negative, since a gas is being consumed and an equal total number of moles products are being formed as moles of reactants, so that the end result is an increase in order with respect to the reaction mixture. However, the formation of products results in a net release of heat (i.e., $\Delta H$ is negative), which increases the entropy of the surroundings sufficiently, such that, overall, entropy is increased in a global sense.

Ultimately, what drives this reactions forward is its exothermic nature (under standard conditions). In this reaction, the formation of stable chemical bonds releases energy as heat. If one were to attempt this reaction at very high temperatures, it would eventually fail, as the increase of the $T\Delta S$ term would at some point dwarf the enthalpy change.

There are other factors at work as well, including chemical equilibrium. By Le Chatelier's principle, the process can be made favorable by, e.g., maintaining a continual excess of reactants, or removing products formed from the reaction mixture as it proceeds.

Edit: Here's one source which indicates 348cal/g energy released, to put a specific number on the process. You could also calculate the enthalpy change using tables of standard thermodynamic data.

  • $\begingroup$ +1 plus the production of NaOCl is carried out at -40 degree Celsius as standard procedure. $\endgroup$ – blackSmith Aug 9 '13 at 20:28
  • $\begingroup$ @blackSmith, thanks. To your knowledge, is the reaction violently exothermic, or is that temperature merely to prevent the degradation of the hypochlorite/formation of other chlorates? $\endgroup$ – Greg E. Aug 9 '13 at 20:31
  • 1
    $\begingroup$ Formation of chlorate is the main issue for production, but it also favors the dissolution equilibrium in the forward direction. You've already mentioned Le Chatelier. $\endgroup$ – blackSmith Aug 9 '13 at 20:43
  • $\begingroup$ @blackSmith, thanks, I appreciate the response. $\endgroup$ – Greg E. Aug 9 '13 at 20:44
  • $\begingroup$ Thank you, but why does the chlorine break down from Cl2 in order to be used in NaCL and NaOCl? $\endgroup$ – user2514631 Aug 10 '13 at 0:29

Also, why does it react with the OH− ion after being broken down?

Does it? Chlorine is not stable in water, it undergoes disproportionation :

$\ce{Cl2 + 3 H2O <=> OCl- + Cl- + 2 H3O+}$

Performing the reaction in alkaline media simply shifts the equilibrium to the right (Le Chatelier principle) and prevents a possible decomposition of hypochlorous acid ($\ce{HOCl}, pK_a = 7.49$) according to

$\ce{2 HOCl <=> Cl2O + H2O}$


According to the Frost Diagram of Chlorine in basic medium, it can be seen that Cl is at a convex side. So it will undergo disproportionation.


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