Doc Brown's Chemistry - 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 peroxodisulfate, 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.

Sub-index for this page

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

(2) The formation and thermal decomposition of hydrogen iodide

(3) The formation of hydrogen iodide as an example of 'connecting' kinetics (rates of reaction) and equilibrium expressions

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

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

(6) The equilibrium constant K_c for the aqueous iodine–iodide equilibrium

9.12 Miscellaneous aspects of halogen chemistry

Some kinetics and equilibrium examples

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

• The uncatalysed reaction is overall is ...

• (a) S₂O₈^2–(aq) + 2I^– (aq) ==> 2SO₄^2–(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 (E_a3 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.

• 

• Initially, the 1st step overall for the catalysed reaction is ... (E_a1 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 peroxodisulfate ion is reduced to sulphate ion ... (E_a2 in diagram above)

  • (c) 2Fe²⁺(aq) + S₂O₈^2–(aq) ==> 2SO₄^2–(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 peroxodisulfate 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[S₂O₈^2–(aq)][I^–(aq)]

  • so, what will it be for the catalysed?

  • maybe rate = k[S₂O₈^2–(aq)][I^–(aq)][Fe²⁺(aq)] ?  

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

  • or rate = k[S₂O₈^2–(aq)][Fe²⁺(aq)]?

  • I don't know!

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

• A Level Notes on Transition Metals – Iron

(2) The formation and thermal decomposition of hydrogen iodide

• hydrogen + iodine hydrogen iodide (2 mol gas ==> 2 mol gas)
• H₂(g) + I_2(g) 2HI(g) (all gases above 200^oC)
• 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.
• 
• 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 K_c equilibrium expressions

(3) The formation of hydrogen iodide as an example of 'connecting' kinetics (rates of reaction) and equilibrium expressions

• H_2(g) + I_2(g) 2HI_(g)

• Kc =	[HI(g)]2
  	–––––––––––– (no units)
  	[H2(g)] [I2(g)]

• K_c 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 250–450^oC. 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.
  • rate_f = k_f [H_2(g)] [I_2(g)] and rate_b = k_b [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
  • rate_f = k_f [H_2(g)] [I_2(g)] = rate_b = k_b [HI_(g)]²
  • therefore [H_2(g)] [I_2(g)] = rate_f / k_f  and  [HI_(g)]² = rate_b / k_b
  • and substituting into the equilibrium expression, with the 'rates' cancelling out, gives the equilibrium constant as the product of dividing the forward rate constant by the backward rate constant.

  • 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 K_c equilibrium expressions

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

• The formation of hydrogen iodide from hydrogen and iodine:

• H₂(g) + I_2(g) 2HI(g) (ΔH = –10 kJ mol^–1, iodine gaseous above 200^oC)

• 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 K_c equilibrium expressions

• I_2(aq) I_2(CCl4)

• K_partition = [I_2(aq)] / [I_2(CCl4)] = 0.0116 at 298K

• The ratio or K_partition, 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 K_partition expression does not hold because of a 2nd homogeneous equilibrium and both equilibria expressions must be satisfied (see also below)

  • I^–_(aq) + I_2(aq) I₃^–_(aq) as well as I_2(aq) I_2(CCl4)

  • so the resulting linked equilibria

  • I_2(CCl4) I_2(aq) + I^–_(aq) I₃^–_(aq)

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

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

• I^–(aq) + I₂(aq) I₃^–(aq) for which K_c = 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 K_c expressions

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)