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Advanced Level Inorganic Chemistry Periodic Table Revision Notes

Part 9. Group 7/17 The Halogens

9.12 Miscellaneous aspects of halogen chemistry

Some kinetics and equilibrium examples e.g. iron(II)/iron(III) ions catalyse the oxidation of iodide ions by peroxodisulphate, the formation/decomposition of hydrogen iodide, he formation of hydrogen iodide as an example of 'connecting' rates of reaction and equilibria, applying Le Chatelier's Principle to the HI/H2/I2 equilibrium, an example of partition – iodine dissolved in a water/tetrachloromethane system and the iodine–iodide equilibrium.

PLEASE NOTE KS4 Science GCSE/IGCSE/O Level GROUP 7 HALOGENS NOTES are on a separate webpage

INORGANIC Part 9 Group 7/17 Halogens sub–index: 9.1 Introduction, trends & Group 7/17 data * 9.2 Halogen displacement reaction and reactivity trend  * 9.3 Reactions of halogens with other elements * 9.4 Reaction between halide salts and conc. sulfuric acid * 9.5 Tests for halogens and halide ions * 9.6 Extraction of halogens from natural sources * 9.7 Uses of halogens & compounds * 9.8 Oxidation & Reduction – more on redox reactions of halogens & halide ions * 9.9 Volumetric analysis – titrations involving halogens or halide ions * 9.10 Ozone, CFC's and halogen organic chemistry links * 9.11 Chemical bonding in halogen compounds * 9.12 Miscellaneous aspects of halogen chemistry

 

Advanced Level Inorganic Chemistry Periodic Table Index * Part 1 Periodic Table history * Part 2 Electron configurations, spectroscopy, hydrogen spectrum, ionisation energies * Part 3 Period 1 survey H to He * Part 4 Period 2 survey Li to Ne * Part 5 Period 3 survey Na to Ar * Part 6 Period 4 survey K to Kr and important trends down a group * Part 7 s–block Groups 1/2 Alkali Metals/Alkaline Earth Metals * Part 8  p–block Groups 3/13 to 0/18 * Part 9 Group 7/17 The Halogens * Part 10 3d block elements & Transition Metal Series * Part 11 Group & Series data & periodicity plots * All 11 Parts have their own sub–indexes near the top of the pages

 
Pd s block d blocks and f blocks of metallic elements p block elements
Gp1 Gp2 Gp3/13 Gp4/14 Gp5/15 Gp6/16 Group7/17 Gp0/18
1

1H

2He
2 3Li 4Be ZSymbol, z = atomic or proton number

highlighting position of Group 7/17 Halogens

outer electrons ns2np5

5B 6C 7N 8O 9F

fluorine

10Ne
3 11Na 12Mg 13Al 14Si 15P 16S 17Cl

chlorine

18Ar
4 19K 20Ca 21Sc 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu 30Zn 31Ga 32Ge 33As 34Se 35Br

bromine

36Kr
5 37Rb 38Sr 39Y 40Zr 41Nb 42Mo 43Tc 44Ru 45Rh 46Pd 47Ag 48Cd 49In 50Sn 51Sb 52Te 53I

iodine

54Xe
6 55Cs 56Ba 57-71 72Hf 73Ta 74W 75Re 76Os 77Ir 78Pt 79Au 80Hg 81Tl 82Pb 83Bi 84Po 85At

astatine

86Rn
7 87Fr 88Ra 89-103 104Rf 105Db 106Sg 107Bh 108Hs 109Mt 110Ds 111Rg 112Cn 113Uut 114Fl 115Uup 116Lv 117Uus

ununseptium

118Uuo

9.12 Miscellaneous aspects of halogen chemistry

Some kinetics and equilibrium examples


Iron(II)/iron(III) ions catalyse the oxidation of iodide ions by peroxodisulphate

  • The uncatalysed reaction is overall is ...

  • (a) S2O8(aq) + 2I– (aq) ==> 2SO4(aq) + I2(aq) 

  • However, this 'direct' uncatalysed reaction involves the collision of two highly repelling negative ions and so has a very high activation energy (Ea3 in the diagram below). 

  • BUT, the collision of an Fe3+ ion and an I– ion involves a positive ion–negative ion attraction, reducing repulsion, so this interaction which has a much lower activation energy.

  • doc b

  • Initially, the 1st step overall for the catalysed reaction is ... (Ea1 in diagram above)

    • (b) 2Fe3+(aq) + 2I–(aq) ==> 2Fe2+(aq) + I2(aq) 

  • Fe2+ is the 'intermediate', and in the 2nd step overall, it is oxidised to Fe3+ and the peroxodisulphate ion is reduced to sulphate ion ... (Ea2 in diagram above)

    • (c) 2Fe2+(aq) + S2O8(aq) ==> 2SO4(aq) + 2Fe3+(aq) 

  • So, the iron(III) ion is regenerated in the catalytic cycle, showing the iron(II/III) ions act in a genuine catalytic cycle but remember it cannot be simply two steps, the above must represent the summations of at least four steps.

    • The oxidation state of iron alternates between +2 and +3 in the catalytic cycle.

    • Iodine's oxidation state changes from –1 to 0.

  • Note 1: It doesn't matter whether you start with the iron(II) or iron(III) ion, catalysis will occur because the peroxodisulphate would oxidise some Fe2+ to Fe3+ (reaction b) and the Fe3+ then oxidises the iodide!

  • Note 2: If you added up the two equations (b + c) of the cycle you get equation (a) showing the overall reaction change.

  • Note 3: The full catalysis mechanism must be quite complex e.g. at least 4 steps because the chances of three particles colliding in the right way (a termolecular collision) and with sufficient frequency is unlikely. Most mechanisms proceed by bimolecular collisions, whatever the overall order of the reaction!

  • The rate expression for the uncatalysed reaction is:

    •   rate = k[S2O8(aq)][I–(aq)]

    • so, what will it be for the catalysed?

    • maybe rate = k[S2O8(aq)][I–(aq)][Fe2+(aq)] ?  

    • or [rate = k[Fe2+(aq)][I–(aq)] ?  

    • or rate = k[S2O8(aq)][Fe2+(aq)]?

    • I don't know!

  • Advanced Level Chemistry Notes – Rates of Reaction – Kinetics Notes

  • A Level Notes on Transition Metals – Iron


 

The formation/decomposition of hydrogen iodide

  • hydrogen + iodine (c) doc b hydrogen iodide (2 mol gas ==> 2 mol gas)
  • H2(g) + I2(g) (c) doc b 2HI(g) (all gases above 200oC)
  • L to R forward reaction: If you start with pure hydrogen and pure iodine, so much of them combines to form hydrogen iodide.
  • R to L backward reaction: If you start with pure hydrogen iodide, some, but not all of it,  will decompose into hydrogen and iodine.
  • Starting with the same total number of moles of either H2 + I2 or  HI, the final equilibrium concentrations will be the same at the same temperature, volume and pressure. This is illustrated in the diagram below showing the fate of 2 mol of reacting gases.
  • (c) doc b
  • Graph lines (1) and (2) show what happens if you start with 2.0 mol of pure hydrogen iodide which decomposes 50%, for the sake of argument and mathematical simplicity, into hydrogen and iodine.
    • Graph line (1) shows the gradual 50% reduction of HI from 2 mol to 1 mol.
    • Graph line (2) shows the gradual formation, from 0 mol of each, of 0.5 mol H2 and 0.5 mol I2.
  • Graph lines (3) and (4) show what happens if you start with 1.0 mol of hydrogen plus 1.0 mol of iodine and no hydrogen iodide.
    • Graph line (3) shows the 50% reduction of 1.0 mol of H2 or I2 to 0.5 mol of each.
    • Graph line (4) shows the formation of 1.0 mol of HI from the net reaction of 0.5 mol H2 and 0.5 mol I2.
  • Note:
    1. The final equilibrium composition is the same in each case no matter which direction you started from for the same total moles of gas.
    2. Where the graph lines first become horizontal, meaning no further net change in concentration, the equilibrium point was first reached i.e. here, after about 32 minutes.
    3. See also Le Chatelier's Principle and Kc equilibrium expressions


 

The formation of hydrogen iodide as an example of 'connecting' rates of reaction and equilibria

  • H2(g) + I2(g) (c) doc b 2HI(g)

  • Kc =

      [HI(g)]2
    ––––––––––––––––––– (no units)
    [H2(g)] [I2(g)]
  • Kc has no units as all the concentration units cancel out.
  • An example of the quantitative connection between kinetics (rates of reaction) and equilibrium expressions.
    • This, historically, has been one of the most studied reactions in terms of kinetics and equilibrium and is a good example to study for comparing and amalgamating two important conceptual frameworks in chemistry. (If you haven't studied kinetics – rate expressions etc. then just miss out this paragraph.)
    • The concentrations of reactants and products have been followed quantitatively by starting with either hydrogen iodide or a hydrogen iodine gas mixture at temperatures of 250–450oC. The graphs below show in principle what happens.
    • Both the forward (f) and backward (b) reactions occur via a simple one step mechanism i.e. via a single bimolecular collision and this simple reaction mechanism leads to simple and verifiable second order kinetics rate expressions.
    • ratef = kf [H2(g)] [I2(g)] and rateb = kb [HI(g)]2
    • Now at the point of dynamic equilibrium, with no net change in concentrations, the rate of the forward reaction = rate of the backward reaction, so
    • ratef = kf [H2(g)] [I2(g)] = rateb = kb [HI(g)]2
    • therefore [H2(g)] [I2(g)] = ratef / kf  and  [HI(g)]2 = rateb / kb
    • and substituting into the equilibrium expression, with the 'rates' cancelling out, gives
    • Kc =

        rateb x kf        kf
      –––––––––––––––––– = –––––
       kb x ratef         kb
    • So the equilibrium constant is equal to the ratio of the two rate constants of the forward and backward reaction.
  • See also Le Chatelier's Principle and Kc equilibrium expressions


 

Applying Le Chatelier's Principle to the HI/H2/I2 equilibrium

  • The formation of hydrogen iodide from hydrogen and iodine:

  • H2(g) + I2(g) (c) doc b 2HI(g) (ΔH = –10 kJ mol–1, iodine gaseous above 200oC)

  • Temperature and energy change (ΔH)

    •  Increasing temperature favours the LHS, i.e. increases the endothermic decomposition of hydrogen iodide.

  • Gas pressure (ΔV):

    • No effect on position of equilibrium, 2 mol gas ==> 2 mol gas, no net change in gas moles.

  • Concentration

    •  e.g. if more iodine was added to a constant volume container, the hydrogen concentration or partial pressure would decrease as some reacts with added iodine to give more hydrogen iodide as the system tries to minimise the iodine increase.

    • Please note that there would still be an overall increase in iodine at the new equilibrium point.

  • See also Le Chatelier's Principle and Kc equilibrium expressions


 

An example of partition – iodine dissolved in a water/tetrachloromethane system

  • I2(aq) (c) doc b I2(CCl4)

  • Kpartition = [I2(aq)] / [I2(CCl4)] = 0.0116 at 298K

  • The ratio or Kpartition, is ~constant whatever the total amount of iodine dissolved, as long as the temperature is constant.

  • The non–polar iodine is much more soluble in the non–polar organic solvent than in the highly polar water solvent.

  • The iodine cannot disrupt few of the strong intermolecular forces of hydrogen bonding between water molecules.

  • This system can be analysed by titrating extracted aliquots with standardised sodium thiosulphate and starch indicator.

  • If the aqueous solution is replaced by aqueous potassium iodide, the simple Kpartition expression does not hold because of a 2nd homogeneous equilibrium and both equilibria expressions must be satisfied (see also below)

    • I–(aq) + I2(aq) (c) doc b I3–(aq) as well as I2(aq) (c) doc b I2(CCl4)

    • so the resulting linked equilibria

    • I2(CCl4) (c) doc b I2(aq) + I–(aq) (c) doc b I3–(aq)

  • For more advanced level chemistry notes on partition see Equilibria Part 4


 

Kc Iodine–iodide equilibrium

  • Iodine is much more soluble in potassium iodide solution than pure water because of the equilibrium:

  • I–(aq) + I2(aq) (c) doc b I3–(aq) for which Kc = 7.10 x 102 mol–1 dm3 at 298K.

  • If the concentration of the I– ion is 0.122 mol dm–3, and that of the I3– ion is 0.153 mol dm–3, calculate the concentration of free iodine.

  • Kc =

      [I3–(aq)]
    ––––––––––––––––––––––––
    [I–(aq)] [I2(aq)]
  • Rearranging gives ...

  • [I2(aq)] =

      [I3–(aq)]
    ––––––––––––––––––––
    Kc x [I–(aq)]
  • [I2(aq)] = 0.153 / (7.10 x 102 x 0.122) = 1.77 x 10–3 mol dm–3

  • For more see Advanced level chemistry Equilibria part 2 Kc expressions

 

PLEASE NOTE KS4 Science GCSE/IGCSE/O Level GROUP 7 HALOGENS NOTES are on a separate webpage

WHAT NEXT?

INORGANIC Part 9 Group 7/17 Halogens sub–index: 9.1 Introduction, trends & Group 7/17 data * 9.2 Halogen displacement reaction and reactivity trend  * 9.3 Reactions of halogens with other elements * 9.4 Reaction between halide salts and conc. sulfuric acid * 9.5 Tests for halogens and halide ions * 9.6 Extraction of halogens from natural sources * 9.7 Uses of halogens & compounds * 9.8 Oxidation & Reduction – more on redox reactions of halogens & halide ions * 9.9 Volumetric analysis – titrations involving halogens or halide ions * 9.10 Ozone, CFC's and halogen organic chemistry links * 9.11 Chemical bonding in halogen compounds * 9.12 Miscellaneous aspects of halogen chemistry


keywords phrases formula: S2O82–(aq) + 2I– (aq) ==> 2SO42–(aq) + I2(aq) 2Fe3+(aq) + 2I–(aq) ==> 2Fe2+(aq) + I2(aq) 2Fe2+(aq) + S2O82–(aq) ==> 2SO42–(aq) + 2Fe3+(aq) rate = k[S2O82–(aq)][I–(aq)] H2(g) + I2(g) 2HI(g) ratef = kf [H2(g)] [I2(g)] and rateb = kb [HI(g)]2 ratef = kf [H2(g)] [I2(g)] = rateb = kb [HI(g)]2 [H2(g)] [I2(g)] = ratef / kf and [HI(g)]2 = rateb / kb I2(aq) I2(CCl4) Kpartition = [I2(aq)] / [I2(CCl4)] = 0.0116 at 298K I–(aq) + I2(aq) I3–(aq) as well as I2(aq) I2(CCl4) I2(CCl4) I2(aq) + I–(aq) I3–(aq)

 

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