IChO 2008 · Theory problem 7

The dithionate ion

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The dithionate ion (\(\ce{S2O6^2-}\)) is a rather inert inorganic ion. It can be prepared by bubbling sulphur-dioxide continuously into ice-cooled water to which manganese dioxide is added in small increments. Dithionate and sulphate ions are formed under these circumstances.

7.1 Write the balanced chemical equations for the two reactions.

After the reaction is complete, \(\ce{Ba(OH)2}\) is added to the mixture until the sulphate ions are fully precipitated. This is followed by the addition of \(\ce{Na2CO3}\).

7.2 Write the balanced equation for the reaction that takes place upon addition of \(\ce{Na2CO3}\).

Sodium dithionate is then crystallized by evaporating some of the solvent. The prepared crystals dissolve readily in water and do not give a precipitate with \(\ce{BaCl2}\) solution. When the solid is heated and maintained at \(130\,^\circ\mathrm{C}\), 14.88 % weight loss is observed. The resulting white powder dissolves in water and does not give a precipitate with \(\ce{BaCl2}\) solution. When another sample of the original crystals is kept at \(300\,^\circ\mathrm{C}\) for a few hours, 41.34 % weight loss occurs. The resulting white powder dissolves in water and gives a white precipitate with \(\ce{BaCl2}\) solution.

7.3 Give the composition of the prepared crystals and write balanced equations for the two processes that occur during heating.

Although dithionate ion is a fairly good reducing agent thermodynamically, it does not react with oxidants in solution at room temperature. At \(75\,^\circ\mathrm{C}\), however, it can be oxidized in acidic solutions. A series of kinetic experiments were carried out with bromine as an oxidant.

7.4 Write the balanced chemical equation for the reaction between bromine and dithionate ion.

The initial rates (\(v_0\)) of the reaction were determined in a number of experiments at \(75\,^\circ\mathrm{C}\).

\([\ce{Br2}]_0\) \((\mathrm{mmol\,dm^{-3}})\) \([\ce{Na2S2O6}]_0\) \((\mathrm{mol\,dm^{-3}})\) \([\ce{H+}]_0\) \((\mathrm{mol\,dm^{-3}})\) \(v_0\) \((\mathrm{nmol\,dm^{-3}\,s^{-1}})\)
0.500 0.0500 0.500 640
0.500 0.0400 0.500 511
0.500 0.0300 0.500 387
0.500 0.0200 0.500 252
0.500 0.0100 0.500 129
0.400 0.0500 0.500 642
0.300 0.0500 0.500 635
0.200 0.0500 0.500 639
0.100 0.0500 0.500 641
0.500 0.0500 0.400 511
0.500 0.0500 0.300 383
0.500 0.0500 0.200 257
0.500 0.0500 0.100 128

7.5 Determine the order of the reaction with respect to \(\ce{Br2}\), \(\ce{H+}\) and \(\ce{S2O6^2-}\), the experimental rate equation, and the value and unit of the rate constant.

In similar experiments, chlorine, bromate ion, hydrogen peroxide and chromate ion have all been used as oxidizing agents at \(75\,^\circ\mathrm{C}\). The rate equations for these processes are analogous to the one observed with bromine, the units of all rate constants are the same, the values are \(2.53\times10^{-5}\) (\(\ce{Cl2}\)), \(2.60\times10^{-5}\) (\(\ce{BrO3^-}\)), \(2.56\times10^{-5}\) (\(\ce{H2O2}\)), and \(2.54\times10^{-5}\) (\(\ce{Cr2O7^2-}\)).

Experiments were also carried out in acidic sodium dithionate solution without any oxidizing agent. When following the processes by UV spectrophotometry, the slow appearance of a new absorption band around 275 nm was observed. Although hydrogen sulphate ion is a detectable product of the reaction, it does not absorb any light above 200 nm.

7.6 Give the formula of the major species causing the new absorption band and write the balanced equation of the chemical reaction occurring in the absence of oxidants.

An experiment was carried out to follow the absorbance at 275 nm with initial concentrations: \([\ce{Na2S2O6}]=0.0022\ \mathrm{mol\,dm^{-3}}\), \([\ce{HClO4}]=0.70\ \mathrm{mol\,dm^{-3}}\), and the temperature was \(75\,^\circ\mathrm{C}\). A pseudo first-order kinetic curve was found with a half-life of 10 hours and 45 minutes.

7.7 Calculate the rate constant of the reaction.

Suggest a balanced chemical equation for the rate determining step of the reactions that used an oxidizing agent.

When periodate ion (which is present as \(\ce{H4IO6^-}\) in aqueous solution) was used as an oxidant for dithionate ion, the two kinetic curves depicted in the graph were detected at \(75\,^\circ\mathrm{C}\) in the same experiment at two different wavelengths. The initial concentrations were \([\ce{H4IO6^-}]=5.3\cdot10^{-4}\ \mathrm{mol\,dm^{-3}}\), \([\ce{Na2S2O6}]=0.0519\ \mathrm{mol\,dm^{-3}}\), \([\ce{HClO4}]=0.728\ \mathrm{mol\,dm^{-3}}\). At 465 nm, only \(\ce{I2}\) absorbs and its molar absorption coefficient is \(715\ \mathrm{dm^3\,mol^{-1}\,cm^{-1}}\). At 350 nm, only \(\ce{I3^-}\) absorbs and its molar absorption coefficient is \(11000\ \mathrm{dm^3\,mol^{-1}\,cm^{-1}}\). The optical path length was 0.874 cm.

Figure 1

7.8 Write balanced chemical equations for the reactions that occur in the region where the absorbance increases at 465 nm, and in the region where the absorbance decreases at 465 nm.

Calculate the expected time for the maximum absorbance of the kinetic curve measured at 465 nm.

Estimate the expected ratio of the slopes of the increasing and decreasing regions in the kinetic curve measured at 465 nm.

Official English text of the problem.

Answers

Species names and SMILES are supplied in the corresponding Solution parts and apply to all equations below.

  • 7.1 \[ \ce{MnO2 + 2SO2 -> Mn^2+ + S2O6^2-} \] \[ \ce{MnO2 + SO2 -> Mn^2+ + SO4^2-} \]

  • 7.2 \[ \ce{MnS2O6 + Na2CO3 -> MnCO3(s) + Na2S2O6} \] Carbonate also precipitates any residual dissolved barium as barium carbonate.

  • 7.3 The crystals are sodium dithionate dihydrate, \(\ce{Na2S2O6 * 2H2O}\).

At \(130\,^\circ\mathrm{C}\): \[ \ce{Na2S2O6 * 2H2O(s) -> Na2S2O6(s) + 2H2O(g)} \] The subsequent decomposition at \(300\,^\circ\mathrm{C}\) is: \[ \ce{Na2S2O6(s) -> Na2SO4(s) + SO2(g)} \]

  • 7.4 \[ \ce{S2O6^2- + Br2 + 2H2O -> 2HSO4^- + 2Br^- + 2H+} \] Equivalently: \[ \ce{S2O6^2- + Br2 + 2H2O -> 2SO4^2- + 2Br^- + 4H+} \]

  • 7.5 Orders in \(\ce{Br2}\), \(\ce{H+}\), and \(\ce{S2O6^2-}\): 0, 1, and 1, respectively. \[ \boxed{v=k[\ce{H+}][\ce{S2O6^2-}]}, \qquad \boxed{k\approx2.56\times10^{-5}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}} \] at \(75\,^\circ\mathrm{C}\).

  • 7.6 The absorbing species is dissolved sulfur dioxide, \(\ce{SO2(aq)}\). \[ \ce{S2O6^2- + H+ -> HSO4^- + SO2(aq)} \]

  • 7.7 \[ \boxed{k_{\mathrm{obs}}=1.79\times10^{-5}\ \mathrm{s^{-1}}} \] \[ \boxed{k=\frac{k_{\mathrm{obs}}}{[\ce{H+}]} \approx2.56\times10^{-5}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}} \] A consistent rate-determining reaction is: \[ \ce{S2O6^2- + H+ -> HSO4^- + SO2} \]

  • 7.8 During increasing absorbance, the overall reaction is: \[ \ce{7S2O6^2- + 2H4IO6^- + 2H+ + 2H2O -> 14HSO4^- + I2} \] During decreasing absorbance: \[ \ce{S2O6^2- + I2 + 2H2O -> 2HSO4^- + 2I^- + 2H+} \] The transient triiodide signal results from: \[ \ce{I2 + I^- <=> I3^-} \] The predicted peak time and slope ratio are: \[ \boxed{t_{\max}\approx1.9\times10^3\ \mathrm{s}\approx32\ \mathrm{min}} \] \[ \boxed{|m_{\mathrm{increase}}|:|m_{\mathrm{decrease}}|\approx1:7} \] Thus the signed ratio \(m_{\mathrm{increase}}/m_{\mathrm{decrease}}\approx-1/7\).

Solution

Dot-separated SMILES for salts and hydrates specify their ionic composition, not their crystal structure or hydration geometry.

7.1 — Formation of dithionate and sulphate

The relevant species are:

Species SMILES
Dithionate ion, \(\ce{S2O6^2-}\) [O-]S(=O)(=O)S(=O)(=O)[O-]
Sulfur dioxide, \(\ce{SO2}\) O=S=O
Sulphate ion, \(\ce{SO4^2-}\) [O-]S(=O)(=O)[O-]
Manganese(II) ion, \(\ce{Mn^2+}\) [Mn+2]
Manganese(IV) oxide, \(\ce{MnO2}\) [Mn+4].[O-2].[O-2]

The manganese dioxide SMILES is a formal ionic formula-unit representation of an extended solid. Dithionate contains an S–S bond joining two approximately tetrahedral sulfur centres.

Manganese is reduced from \(+4\) to \(+2\), accepting two electrons. Sulfur starts at \(+4\) in sulfur dioxide.

Formation of dithionate raises each of two sulfur atoms to \(+5\), supplying the required two electrons:

\[ \boxed{\ce{MnO2 + 2SO2 -> Mn^2+ + S2O6^2-}} \]

Alternatively, oxidation of one sulfur atom to \(+6\) supplies two electrons and produces sulphate:

\[ \boxed{\ce{MnO2 + SO2 -> Mn^2+ + SO4^2-}} \]

Both equations balance atoms and charge without additional water or hydrogen ions.

7.2 — Addition of sodium carbonate

The additional species are:

Species SMILES
Manganese(II) dithionate, \(\ce{MnS2O6}\) [Mn+2].[O-]S(=O)(=O)S(=O)(=O)[O-]
Sodium carbonate, \(\ce{Na2CO3}\) [Na+].[Na+].[O-]C(=O)[O-]
Carbonate ion, \(\ce{CO3^2-}\) [O-]C(=O)[O-]
Manganese(II) carbonate, \(\ce{MnCO3}\) [Mn+2].[O-]C(=O)[O-]
Sodium dithionate, \(\ce{Na2S2O6}\) [Na+].[Na+].[O-]S(=O)(=O)S(=O)(=O)[O-]
Barium ion, \(\ce{Ba^2+}\) [Ba+2]
Barium carbonate, \(\ce{BaCO3}\) [Ba+2].[O-]C(=O)[O-]

After removal of sulphate, carbonate precipitates the remaining dissolved manganese(II):

\[ \ce{Mn^2+ + CO3^2- -> MnCO3(s)} \]

The corresponding salt equation shows the formation of soluble sodium dithionate:

\[ \boxed{\ce{MnS2O6 + Na2CO3 -> MnCO3(s) + Na2S2O6}} \]

Any soluble barium remaining from the previous addition is also removed:

\[ \ce{Ba^2+ + CO3^2- -> BaCO3(s)} \]

7.3 — Composition and thermal decomposition

The additional species are:

Species SMILES
Sodium dithionate dihydrate, \(\ce{Na2S2O6 * 2H2O}\) [Na+].[Na+].[O-]S(=O)(=O)S(=O)(=O)[O-].O.O
Water, \(\ce{H2O}\) O
Sodium sulphate, \(\ce{Na2SO4}\) [Na+].[Na+].[O-]S(=O)(=O)[O-]
Barium sulphate, \(\ce{BaSO4}\) [Ba+2].[O-]S(=O)(=O)[O-]

Let the crystals be \(\ce{Na2S2O6 * nH2O}\). Using standard rounded atomic masses,

\[ M(\ce{Na2S2O6})=206.10\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{H2O})=18.016\ \mathrm{g\,mol^{-1}}. \]

The absence of a barium precipitate after heating to \(130\,^\circ\mathrm{C}\) is consistent with retention of dithionate. Assigning the first mass loss to water gives

\[ \frac{18.016n}{206.10+18.016n}=0.1488. \]

Hence

\[ n=\frac{0.1488(206.10)}{18.016(1-0.1488)} =1.9998\approx2. \]

Thus the crystals are sodium dithionate dihydrate, and dehydration is

\[ \boxed{\ce{Na2S2O6 * 2H2O(s) -> Na2S2O6(s) + 2H2O(g)}}. \]

The hydrate molar mass is approximately \(242.132\ \mathrm{g\,mol^{-1}}\). After the \(300\,^\circ\mathrm{C}\) treatment, the mass remaining per mole of original hydrate is

\[ (1-0.4134)(242.132)=142.03\ \mathrm{g}. \]

This matches one mole of sodium sulphate, \(M\approx142.04\ \mathrm{g\,mol^{-1}}\). The anhydrous dithionate therefore decomposes by disproportionation:

\[ \boxed{\ce{Na2S2O6(s) -> Na2SO4(s) + SO2(g)}}. \]

One sulfur atom changes from \(+5\) to \(+6\), and the other from \(+5\) to \(+4\). Starting from the original crystals, the overall reaction is

\[ \ce{Na2S2O6 * 2H2O(s) -> Na2SO4(s) + SO2(g) + 2H2O(g)}. \]

Its predicted total mass loss is

\[ \frac{64.06+2(18.016)}{242.132}\times100\% =41.34\%. \]

The sulphate residue explains the white precipitate:

\[ \ce{Ba^2+ + SO4^2- -> BaSO4(s)}. \]

7.4 — Oxidation by bromine

The additional species are:

Species SMILES
Bromine, \(\ce{Br2}\) BrBr
Bromide ion, \(\ce{Br^-}\) [Br-]
Hydrogen ion, \(\ce{H+}\) [H+]
Hydrogen sulphate ion, \(\ce{HSO4^-}\) OS(=O)(=O)[O-]

The hydrogen-ion notation represents aqueous acidity; it does not describe proton solvation.

Oxidation of the two sulfur atoms from \(+5\) to \(+6\) releases two electrons. One bromine molecule accepts these electrons to form two bromide ions. Writing hydrogen sulphate as the sulfur product in acidic solution gives

\[ \boxed{\ce{S2O6^2- + Br2 + 2H2O -> 2HSO4^- + 2Br^- + 2H+}}. \]

Using

\[ \ce{HSO4^- <=> H+ + SO4^2-}, \]

the equivalent sulphate-form equation is

\[ \boxed{\ce{S2O6^2- + Br2 + 2H2O -> 2SO4^2- + 2Br^- + 4H+}}. \]

7.5 — Rate law and rate constant

Write the empirical rate law as

\[ v=k[\ce{Br2}]^a[\ce{H+}]^b[\ce{S2O6^2-}]^c. \]

The independently varied concentrations establish the orders:

  • Changing bromine from \(0.100\) to \(0.500\ \mathrm{mmol\,dm^{-3}}\) leaves the rate essentially unchanged, so \(a=0\).
  • Increasing dithionate fivefold, from \(0.0100\) to \(0.0500\ \mathrm{mol\,dm^{-3}}\), changes the rate from 129 to 640, approximately fivefold. Thus \(c=1\).
  • Increasing hydrogen-ion concentration fivefold, from \(0.100\) to \(0.500\ \mathrm{mol\,dm^{-3}}\), changes the rate from 128 to 640, exactly fivefold within the reported values. Thus \(b=1\).

Therefore,

\[ \boxed{v=k[\ce{H+}][\ce{S2O6^2-}]}. \]

Using the first experiment,

\[ k= \frac{640\times10^{-9}\ \mathrm{mol\,dm^{-3}\,s^{-1}}} {(0.500\ \mathrm{mol\,dm^{-3}}) (0.0500\ \mathrm{mol\,dm^{-3}})} = \boxed{2.56\times10^{-5}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}}. \]

The overall empirical order is two.

7.6 — Reaction without an oxidant

The absorbing species is sulfur dioxide, \(\ce{SO2(aq)}\), SMILES O=S=O.

Dithionate can disproportionate: one sulfur atom is oxidized from \(+5\) to \(+6\), producing hydrogen sulphate, while the other is reduced to \(+4\). In the strongly acidic medium, the latter product is predominantly dissolved sulfur dioxide.

The balanced equation is

\[ \boxed{\ce{S2O6^2- + H+ -> HSO4^- + SO2(aq)}}. \]

Hydrogen sulphate does not absorb at 275 nm according to the statement, so the growing band is assigned to the sulfur dioxide product.

7.7 — Half-life and the rate-determining reaction

Hydrogen ion is in large excess over dithionate, so its concentration remains approximately constant. Consequently,

\[ -\frac{d[\ce{S2O6^2-}]}{dt} =k[\ce{H+}][\ce{S2O6^2-}] =k_{\mathrm{obs}}[\ce{S2O6^2-}], \]

where

\[ k_{\mathrm{obs}}=k[\ce{H+}]. \]

The half-life is

\[ t_{1/2}=10(3600)+45(60)=38700\ \mathrm{s}. \]

For first-order decay,

\[ k_{\mathrm{obs}} =\frac{\ln2}{t_{1/2}} =\frac{0.693147}{38700} =\boxed{1.79\times10^{-5}\ \mathrm{s^{-1}}}. \]

The corresponding second-order constant is

\[ k=\frac{1.79\times10^{-5}}{0.70} =\boxed{2.56\times10^{-5}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}}. \]

This agrees with the bromine result and with the constants for the other oxidants. Together with the zero order in bromine, this supports a common slow reaction that does not involve the oxidant:

\[ \boxed{\ce{S2O6^2- + H+ -> HSO4^- + SO2}} \qquad\text{(slow)}. \]

The released sulfur dioxide is then rapidly oxidized. For example, with bromine,

\[ \ce{SO2 + Br2 + 2H2O -> HSO4^- + 2Br^- + 3H+} \qquad\text{(fast)}. \]

Adding these two reactions gives the overall equation in 7.4. This is a consistent kinetic mechanism; the rate data alone do not uniquely establish its microscopic elementary steps.

7.8 — Periodate, iodine formation, and iodine consumption

The additional species are:

Species SMILES
Tetrahydrogen orthoperiodate ion, \(\ce{H4IO6^-}\) [O-][I+](O)(O)(O)(O)[O-]
Iodine, \(\ce{I2}\) II
Iodide ion, \(\ce{I^-}\) [I-]
Triiodide ion, \(\ce{I3^-}\) I[I-]I

The periodate SMILES is a charge-separated representation; iodine has six oxygen neighbours in an approximately octahedral arrangement. Triiodide is linear and has three-centre, four-electron bonding, which the single-bond SMILES does not fully describe.

Increasing absorbance at 465 nm.

The slow reaction in 7.7 supplies sulfur dioxide. Periodate rapidly oxidizes this sulfur dioxide while being reduced to iodine:

\[ \boxed{\ce{2H4IO6^- + 7SO2 + 2H2O -> I2 + 7HSO4^- + 5H+}}. \]

The stoichiometry follows from electron balance: two iodine atoms each fall from oxidation state \(+7\) to \(0\), accepting fourteen electrons; seven sulfur dioxide molecules each supply two electrons.

Adding seven occurrences of the slow dithionate decomposition gives

\[ \boxed{\ce{7S2O6^2- + 2H4IO6^- + 2H+ + 2H2O -> 14HSO4^- + I2}}. \]

Thus seven dithionate ions produce one iodine molecule, and the 465 nm absorbance rises.

Decreasing absorbance at 465 nm.

Once periodate is exhausted, newly generated sulfur dioxide reduces iodine:

\[ \ce{SO2 + I2 + 2H2O -> HSO4^- + 2I^- + 3H+}. \]

Including the slow dithionate decomposition gives

\[ \boxed{\ce{S2O6^2- + I2 + 2H2O -> 2HSO4^- + 2I^- + 2H+}}. \]

Now one dithionate ion consumes one iodine molecule.

As iodide accumulates, some remaining iodine forms triiodide:

\[ \boxed{\ce{I2 + I^- <=> I3^-}}. \]

This accounts for the transient 350 nm signal. Triiodide is ultimately reduced as well:

\[ \ce{I3^- + SO2 + 2H2O -> 3I^- + HSO4^- + 3H+}. \]

Time of maximum iodine absorbance.

The maximum occurs approximately when all periodate has been converted to iodine. The required amount of decomposed dithionate per unit volume is

\[ \Delta[\ce{S2O6^2-}] =\frac{7}{2}[\ce{H4IO6^-}]_0 =\frac{7}{2}(5.3\times10^{-4}) =1.855\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

This is only

\[ \frac{1.855\times10^{-3}}{0.0519}\times100\%=3.57\% \]

of the initial dithionate, so an approximately constant rate is appropriate:

\[ v\simeq (2.56\times10^{-5})(0.728)(0.0519) =9.67\times10^{-7}\ \mathrm{mol\,dm^{-3}\,s^{-1}}. \]

Therefore,

\[ t_{\max}\simeq \frac{1.855\times10^{-3}}{9.67\times10^{-7}} = \boxed{1.9\times10^3\ \mathrm{s}\approx32\ \mathrm{min}}. \]

Allowing for dithionate depletion while holding acidity constant gives the modest refinement

\[ t_{\max} = \frac{-\ln\!\left(1-\dfrac{(7/2)[\ce{H4IO6^-}]_0} {[\ce{S2O6^2-}]_0}\right)} {k[\ce{H+}]} \approx1.95\times10^3\ \mathrm{s}. \]

Both estimates are consistent with the graph.

The expected peak height provides an independent stoichiometric check. Neglecting the small amount of triiodide near the iodine maximum,

\[ [\ce{I2}]_{\max}\simeq\frac{5.3\times10^{-4}}{2} =2.65\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

Beer–Lambert’s law then gives

\[ A_{465,\max} =\varepsilon l[\ce{I2}]_{\max} =(715)(0.874)(2.65\times10^{-4}) \approx0.166, \]

matching the plotted maximum.

The apparently large triiodide absorbance does not imply a similarly large concentration: its peak \(A_{350}\approx0.19\) corresponds to only

\[ [\ce{I3^-}]_{\max}\approx \frac{0.19}{11000(0.874)} \approx2.0\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \]

Ratio of slopes.

Let \(v\) denote the approximately constant rate of dithionate decomposition. Ignoring the transient redistribution into triiodide,

\[ \left(\frac{d[\ce{I2}]}{dt}\right)_{\mathrm{increase}} =\frac{v}{7}, \qquad \left(\frac{d[\ce{I2}]}{dt}\right)_{\mathrm{decrease}} =-v. \]

Because \(A_{465}=\varepsilon l[\ce{I2}]\), the same ratio applies to the absorbance slopes:

\[ \boxed{ \frac{m_{\mathrm{increase}}}{m_{\mathrm{decrease}}} \approx-\frac17 }. \]

Thus the decreasing region is expected to be approximately seven times steeper in magnitude. Triiodide formation and the small decline in dithionate concentration cause departures from this simple estimate.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.