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Advanced level chemistry kinetics notes: Acid decomposition of the thiosulfate ion

Doc Brown's Advanced A Level Chemistry Advanced A Level Chemistry - Kinetics-Rates revision notes Part 7

7.7 The acid – thiosulfate reaction precipitating sulfur

Case study 4.7 The acid catalysed decomposition of sodium thiosulfate

  • The reaction between e.g. dilute hydrochloric acid and sodium thiosulfate is a redox reaction catalyzed by hydrogen ions. I do not know of any other catalysts?

  • Its a typical 'rates' reaction at GCSE level to illustrate temperature and concentration factors or used as a coursework investigation. Its followed by the time it takes to form enough sulfur to obscure a black X marked on white paper. The method described in more detail on the How can we measure the rate of a chemical reaction? page.

  • It is possible to follow the reaction with a colorimeter due to the light scattering effect of the colloidal sulfur particles but the absorbance does not follow Beers Law and processing results is apparently difficult!

  • The reaction is ...

  • Na2S2O3(aq) + 2HCl(aq) ==> 2NaCl(aq) + SO2(aq) + S(s) + H2O(l) 

  • which for advanced level is much more appropriately written in the ionic form ...

  • S2O32–(aq) + 2H+(aq) ==>  SO2(aq) + S(s) + H2O(l) 

    • (the correct ionic equation in acid solution)

  • The redox analysis for this reaction is NOT straight forward.

  • However, you can say in the reaction, the following oxidation states do NOT change: Na(+1), H(+1), O(-2), Cl(-1).

  • The thiosulfate ion on face value has an S=S bond, one S=O and two S–O bonds, but all four bonds are 'merged' in the same delocalised pi bonding system in a tetrahedral arrangement about one of the sulfur atoms.

    • (ii) perhaps 'safer' to argue, that on average each sulfur is in the +2 state (oxygen is –2, overall charge on ion 2–).

      • In the products, the oxidation states of sulfur are much clearer to define, +4 in SO2(aq) and 0 in S(s), which, to further complicate matters, is actually S8 molecules!

      • On the basis that the two sulfur atoms start off in the +2 state, you should correctly argue that this is a disproportionation reaction, in which an element in an ion/compound/compound is simultaneously oxidised (+2 to +4 in the sulfite ion/sulfur dioxide) and reduced (+2 to 0 in the sulfur). See Appendix 1

  • You would expect that the rate might be controlled by the interaction of the negative thiosulfate ion and a positive hydrogen ion. You would expect the interaction of oppositely charged ions to have a relatively low activation energy, so in the rate expression:

  • rate = k[S2O32–(aq)]t[H+(aq]h, you might expect the order t and h to be both 1.

    • t=1 is quoted on the web. and found to be so in most reliable experiments (but not all).

  • The reaction has been shown to be a multi-step complex mechanism, so what order h is I don't know?

    • I've come across references that indicates the order h could be 0–1 depending on the relative concentrations of thiosulfate and acid.

    • Whatever, the orders t and h can only be found by experiment and the mechanism is likely to be complex e.g.

    • S2O32–(aq) + H+(aq) ==> '1st' intermediate  (HS2O3, the hydrogensulfite ion is an obvious choice)

    • other intermediates ==> SO32–(aq) or SO2(aq) + S(s)

      • I don't know the details but you would expect the negative thiosulfate ion to combine with a positive proton to form some intermediate that breaks down in one or more steps to give sulfur dioxide, sulfur and water.

  • Found this link, typically it shows the order with respect to the thiosulfate ion is 1, but no mention of order for the hydrogen ion. It also quotes an 'uncited' and 'uncorroborated' complex mechanism at the end.

Appendix 1 The half–cell reactions for the acid–thiosulfate reaction

From I quote, with some editing ...

We can balance the reduction half–reaction as

(i) 6H+(aq) + 4e + S2O32–(aq) → 2S(s) + 3H2O(l)

(here the sulfur is reduced from +2 to zero)

and the oxidation half–reaction as

(ii) 3H2O(l) + S2O32–(aq) → 2SO32–(aq) + 6H+(aq) + 4 e

(here the sulfur is oxidised from +2 to +4)

The overall reaction is

(i) + (ii) = (iii) S2O32–(aq) → S(s) + SO32–(aq)

and noting the regeneration of the hydrogen ion H+(aq) which fits in with the notion of it being an acid catalysed reaction.

This source quotes 1st order of reaction with respect to the thiosulfate, and zero order for the hydrogen ion.

Appendix 2 A major study of the thiosulfate–acid reaction from 1958 by Robert Earl Davis

Is another complex study, rate expression

rate = k[S2O32–(aq)]3/2[H+(aq)]1/2, two fractional orders of reaction, 1.5 and 0.5,

as opposed to rate = k[S2O32–(aq)][H+(aq)], with no mention of 1st order for thiosulfate,

and was explained using a series of nucleophilic displacement reactions at the sulfur atom.

Radiochemical studies (using radioisotopes) have shown the two sulfur atoms retain their identity throughout the reaction..

It is suggested in this paper the reaction involve ions like [HSnO3], where n = 1 to 9, after the reversible formation of the hydrogensulfite ion,

S2O32– + H+  HSO3

the sulfite ion is then formed by reactions such as 

S2O32– + [HSO3]  ==> [HS2O3] + SO32–

this continues to eventually form [HS9O3]

which breaks down to give the sulfur as S8 molecules

 [HS9O3] ==> S8 + HSO3

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