
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
email doc
brown - comments - query?
All my
Group 7/17 halogens advanced chemistry level revision notes
All my advanced A level
inorganic
chemistry revision notes
GCSE Level Group 7
Halogens revision notes
Use your
mobile phone in 'landscape' style?
This is a BIG
website, you need to take time to explore it [SEARCH
BOX]
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 iodineiodide 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 Kc
for the aqueous iodineiodide equilibrium
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 |
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) S2O82(aq) + 2I
(aq) ==> 2SO42(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 ionnegative ion attraction,
reducing repulsion, so this interaction
which has a much lower activation energy.
-
-
Initially,
the 1st step overall for the catalysed reaction is ... (Ea1 in
diagram above)
-
Fe2+
is the 'intermediate', and in the 2nd step overall, it is oxidised
to Fe3+ and the peroxodisulfate ion is reduced to sulphate
ion ... (Ea2 in diagram
above)
-
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.
-
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[S2O82(aq)][I(aq)]
-
so,
what will it be for the catalysed?
-
maybe rate = k[S2O82(aq)][I(aq)][Fe2+(aq)]
?
-
or [rate = k[Fe2+(aq)][I(aq)]
?
-
or rate = k[S2O82(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) + I2(g)
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.
-
- 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:
- The final equilibrium
composition is the same in each case no matter which direction you
started from for the same total moles of gas.
- 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.
-
See also
Le Chatelier's Principle
and
Kc equilibrium expressions
(3) The formation of hydrogen iodide
as an example of 'connecting' kinetics (rates of reaction) and equilibrium
expressions
-
H2(g) + I2(g)
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 250450oC. 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 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
Kc 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) + I2(g)
2HI(g) (ΔH = 10 kJ
mol1, iodine gaseous above 200oC)
-
Temperature and energy change (ΔH)
-
Gas
pressure (ΔV):
-
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
(5) An
example of partition iodine dissolved in
a water/tetrachloromethane system
-
I2(aq)
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 nonpolar
iodine is much more soluble in the nonpolar 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)
-
so the resulting linked
equilibria
-
I2(CCl4)
I2(aq) +
I(aq)
I3(aq)
For
more advanced level chemistry notes on partition see Equilibria Part 4
(6) The equilibrium
constant Kc
for the aqueous iodineiodide equilibrium
-
Iodine is much
more soluble in potassium iodide solution than pure water because of
the equilibrium:
-
I(aq)
+ I2(aq)
I3(aq)
for which Kc = 7.10 x 102 mol1 dm3
at 298K.
-
If the
concentration of the I ion is 0.122 mol dm3, and
that of the I3 ion is 0.153 mol dm3,
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 103
mol dm3
-
For more
see
Advanced level chemistry Equilibria part 2 Kc 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)
[SEARCH
BOX]
Website content © Dr Phil Brown
2000+. All copyrights reserved on revision notes, images, quizzes,
worksheets etc. Copying of website material is NOT permitted. Doc
Brown's Chemistry notes on the Group 17 halogens - physical and
chemical properties of fluorine, chlorine, bromine, iodine and
astatine |
|