# Is +6 really the most stable form of sulfur?

I think everyone might be aware that +6 is considered the most stable form of sulfur. But I think why this reaction happens very slowly and at such a high temperature and even the help of a catalyst is required, but in almost all redox reactions sulfur goes to +6:

(controversial where +6 is forms very forcefully): $$\ce{SO2 +O2->[\ce{Pt/V2O5}][\text{high temp.}]SO3}$$ And also does it reduces easily (controversial where +6 is reduced easily): $$\ce{P +SO3->P4O10 +SO2}$$ The reactions in my favour are(where stable +6 is formed): $$\ce{SO2 +Cr2O7^2- ->SO4^2- +Cr^3+}$$ $$\ce{BaSO3- ->[\text{warming/\ce{Br2-}water}][\ce{/HNO3/H2O2/\text{dil.}HCl}]BaSO4 }$$ $$\ce{[AgSO3]- ->[\text{boil}]Ag + SO4^2- +SO2}$$ $$\ce{Ag2SO3+H2O ->Ag + SO4^2- +H+}$$ $$\ce{SO3^2- +MnO4- ->Mn^2+ +SO4^2- }$$ $$\ce{PbSO3->[\text{boil}]PbSO4}$$ $$\ce{S +HNO3->[\text{boil}]SO4^2-}$$ etc.

• IDK what is happening with the silver compounds, but all the other reactions oxidizing sulfur use strong oxidation reagents. The energy released from the reduction may be used to created less stable oxidized products. – LDC3 Nov 28 '14 at 17:50
• I would clarify the title. :) – entropid Nov 29 '14 at 16:38

I think it's a matter of activation energy and reaction kinetics. Although the product is much more thermodynamically stable than the reagent(s), the reaction path could feature a high activation energy which requires those harsh conditions and even a catalyst to make it happen. As shown in the graph below, despite having a favourable difference of free energy ($\Delta G < 0$), the activation energy $E_a$ required to go from reactants to products could be not negligible and hard to overcome.

The catalysts allows the reaction to happen more easily by lowering the free energy of the transition state (by stabilising it). The activation energy is also reduced by the high temperature, according to the Arrhenius equation:

$$k = A e^{{-E_a}/{RT}}$$