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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.

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  • $\begingroup$ Do this reaction requires heat/ occurs in presence of sunlight? $\endgroup$
    – ashu
    Aug 9, 2013 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. $\endgroup$ Aug 9, 2013 at 19:49

3 Answers 3

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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.

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  • $\begingroup$ +1 plus the production of NaOCl is carried out at -40 degree Celsius as standard procedure. $\endgroup$
    – blackSmith
    Aug 9, 2013 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, 2013 at 20:31
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    $\begingroup$ 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, 2013 at 20:43
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    $\begingroup$ @user2514631, I'm not sure what answer you're looking for. Thermodynamics is essentially the overriding factor in all chemical reactions at the aggregate scale. An immense number of molecules are colliding and interacting in the course of the reaction. Some interactions result in the breakage of chemical bonds, and some in their formation. Once an energetically favorable state is achieved (i.e., that minimizes free energy), a reversal of that state is unlikely, because the products are at the bottom of an energy curve. Given enough time, the preponderance of molecules will reach this state. $\endgroup$
    – Greg E.
    Aug 10, 2013 at 2:12
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    $\begingroup$ @user2514631, that's a simplified description of the process from a statistical standpoint and on a macroscopic scale, and there's really no deeper answer, at least not that I can provide. If you want mathematical rigor, you must study thermodynamics and physical chemistry. If you're interested in the mechanism of the reaction on the scale of individual molecules, I could plausibly imagine that it would begin with nucleophilic attack by the hydroxide on the chlorine, with chloride as the leaving group. This forms hypochlorous acid, which is deprotonated by the second hydroxide, making water. $\endgroup$
    – Greg E.
    Aug 10, 2013 at 2:23
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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

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

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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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