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Brown's Chemistry Advanced A Level Notes - Theoretical–Physical
Advanced Level
Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 1
Part
1. Examples of dynamic molecular chemical equilibrium and Le Chatelier's principle
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1.0
Introduction
1.1 Reversible reactions
1.2
The equilibrium concept
1.3 Le
Chatelier's Principle – equilibrium rules
1.4
Applying the equilibrium rules
1.0
Introduction
What is a dynamic
equilibrium? How does Le Chatelier's Principle help us predict the shift
in equilibrium position on change of temperature, change in
concentration or change in pressure? What effect does a catalyst have on
the position of an equilibrium?
In contrast with kinetics, which is a
study of how quickly reactions occur, a study of equilibria indicates
how far reactions will go and Le Chatelier’s principle is used to
predict the effects of changes in temperature, pressure and
concentration on the yield of a reversible reaction. Not surprisingly,
this has important consequences for many industrial processes.
1.1
Reversible Reactions
- A reversible reaction is a chemical
change in which the products can be converted back to the original reactants under suitable conditions. A reversible reaction is shown by the sign
,
a half–arrow to the right (forward reaction, L to R), and, a half–arrow to the
left (backward reaction, R to L).
- Reactions which are not reversible (irreversible)
have the usual complete arrow
only pointing to the right.
- -
-
Example 1.1.1 - Thermal decomposition
of ammonium salts
- On heating strongly, the white solid, ammonium chloride,
thermally decomposes into a mixture of two colourless gases,
ammonia and hydrogen chloride. On cooling the reaction is reversed and solid ammonium chloride reforms. This is also
an example of
sublimation and involves both a physical of state as well as a chemical change.
When a substance sublimes it changes directly from a solid into a gas without melting
and on cooling reforms the solid without condensing to form a liquid.
-
Ammonium chloride
+ heat
ammonia + hydrogen chloride
-
NH4Cl(s)
NH3(g) + HCl(g)
-
Note:
Reversing the reaction conditions reverses the direction of chemical change, typical of a reversible reaction.
-
Thermal decomposition means using 'heat' to 'break down' a molecule into smaller ones. The
decomposition of NH4Cl is endothermic, ΔH +ve (heat absorbed, taken in
from the surroundings) and the formation of NH4Cl is exothermic,
ΔH –ve (heat released, given out to the surroundings).
- This means if the direction of chemical change is reversed, the energy change
sign must also be reversed but its numerical value stays the same.
- -
Example 1.1.2 Thermal decomposition of
hydrated copper(II) sulfate crystals
- These are typical examples you
encounter at an earlier study level, but it begs the question, "is it
possible to have a situation, under suitable conditions, in which the reaction
does not completely go in one direction or the other and both reactants and
products co–exist?", and the answer is yes! and the situation is called a
dynamic chemical equilibrium. The word dynamic is used because the
'forward' (L to R) and 'backward' (R to L) reactions do not cease but
match each other in rate so the concentrations of reactants and
products are constant when the equilibrium is established.
There are other examples on my
GCSE chemistry notes page on
reversible reactions
1.2 Reversible reactions and
the concept of a dynamic chemical equilibrium
- Although most reactions you
have encountered at an earlier academic level did go to 100%
completion, it is a fact that many reactions do NOT go to completion
i.e. 100% yield from the forward reaction.
- If ammonium chloride were heated in a
closed system, over a certain temperature range, some of the NH4Cl
will be sublimed into the gases NH3 and HCl and some of the solid
salt remains.
- A closed system means nothing can
enter or leave the system.
- When a reversible reaction occurs in a
closed system, depending on conditions, a chemical equilibrium is formed, in which the
original reactants and products formed coexist. In other words
the reaction (i.e. from left to right as the equation is written)
never goes to completion.
- Eventually the 'system settles down'
and the net concentrations of the reactants and products remain constant
i.e. a state of concentration balance exists.
- BUT the reactions don't stop!
Reactants are continually forming products, and the products are continually
re–forming the original reactants, hence the term dynamic equilibrium.
- In terms of kinetics
('rates of reaction'), it means that the
- rate of formation of
product = rate of re–formation of reactants,
- or the rate of the forward
reaction = rate of the backward reaction
- -
-
Example 1.2.1
The formation/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 Kc expression
and
application of Le Chatelier's Principle
-
Example 1.2.2 The formation
and hydrolysis of the ester ethyl ethanoate.
- ethanoic acid + ethanol
ethyl ethanoate + water (hydrolysis <==> esterification)
- CH3COOH(l) +
CH3CH2OH(l)
CH3COOCH2CH3(l)
+ H2O(l)
- L to R forward reaction: If equimolar
amounts of pure ethanoic acid and pure ethanol are refluxed with a few drops of conc. H2SO4(l)
catalyst, about 2/3rds of the initial reactants are converted into the ester.
- The forward or backward
reactions are slow at room temperature without heating and employing
a catalyst, but equilibrium would eventually be reached at room
temperature even in the absence of a catalyst.
- R to L backward reaction: If equimolar
amounts of pure ethyl ethanoate and pure water are mixed, eventually about
1/3rd of the ester reverts back to ethanoic acid and ethanol.
- Note: If the ester is refluxed with
lots of acidified water, the ester
is 100% hydrolysed back to the original acid and alcohol.
-
See also Equilibria Part 2 Kc expressions
-
Example 1.2.3 The
formation/decomposition of ammonia.
Some important outcomes from
experimentally studying dynamic equilibrium reactions
- It doesn't matter whether you start
with the 'reactants' or the 'products', either way, if the
conditions are suitable, both are present when a state of equilibrium
exists.
- Eventually the net concentrations of the reactants and products remain the
same BUT the forward and backward reactions don't stop.
- When a dynamic equilibrium
is achieved
there is a state of balance between the constant concentrations of the reactants and
products because the rate at which the reactants change into products is exactly equal to the rate at which the products change back to the original reactants.
- However, the actual relative amounts of the
original reactants left, and products formed, at equilibrium, depend on the
particular reaction and reaction conditions
e.g. the initial concentrations, temperature and pressure (if gaseous reactants or
products are involved) and the value of the equilibrium constant (see on
this page
Le Chatelier's Principle
below and
Equilibria Part 2).
- A catalyst does not affect the
position of the equilibrium, i.e. the final constant concentrations are
the same with or without a catalyst, you
simply get to the equilibrium point faster with a catalyst!
- In some cases you
can adjust reaction conditions sufficiently to make the reaction go virtually 100% in one direction e.g.
example
hydrolysis of an ester.
- At a given constant
temperature, all the final equilibrium concentrations are
mathematically governed
by the equilibrium expression and the equilibrium constant and
these are dealt with in detail in
Equilibria Part 2.
1.3 Le Chatelier's Principle
1.4 Applying Le
Chatelier's Principle and the equilibrium rules
-
These are
initially described without reference to equilibrium constants and
equilibrium expressions which are fully explained and described in
Equilibria Part 2 (where some
examples are repeated with their Kc or Kp
equilibrium expressions)
-
Example 1.4.1 - limestone
decomposition
-
The thermal
decomposition of calcium carbonate (limestone) to make calcium oxide
(quicklime):
-
CaCO3(s)
CaO(s) + CO2(g) (ΔH = +178 kJ
mol–1)
-
Note: By
convention, the ΔH value quoted corresponds to the forwards
reaction (L to R), reversing the sign gives the ΔH for the
backward reaction (R to L).
-
For more details
of the industrial process see
Equilibria Part 3.1
-
Rule 1
– temperature and energy change (ΔH)
-
Rule 2 – gas
pressure (ΔV)
-
Rule 3 –
concentration
-
Rule 4 – catalyst:
Not applicable.
-
Example
1.4.2 - Synthesis of ammonia
-
The synthesis
of ammonia: nitrogen + hydrogen
ammonia
-
N2(g) + 3H2(g)
2NH3(g) (ΔH = –92 kJ mol–1)
- Rule 1 – temperature and energy change (ΔH)
- The forward, and desirable
reaction, to form ammonia, is exothermic, so lowering the
temperature favours its formation.
- Rule 2 – gas pressure (ΔV)
- Increase in pressure
favours ammonia formation since 4 mol of gaseous reactants ==> 2
mol gaseous products.
- Rule 3 – concentration:
In terms of enforced change => system response
- If the nitrogen or hydrogen concentration
was increased, some of this 'extra' gases would change to
ammonia.
- If the nitrogen or hydrogen concentration
was decreased, some of ammonia would change to nitrogen and hydrogen.
- A 1:3 ratio N2:H2
mixture is used in industry, but for academic reasoning
practice, specifically ...
- Increasing nitrogen ==>
decreases hydrogen and increases ammonia.
- Increasing hydrogen ==>
decreases nitrogen and increases ammonia.
- Decreasing ammonia ==>
decreases nitrogen and hydrogen.
- Decreasing nitrogen ==>
increases hydrogen and decreases ammonia.
- Decreasing hydrogen ==>
increases nitrogen and decreases ammonia.
-
Rule 4 – catalyst:
An iron oxide catalyst is used, time = money for industrial
chemical production!
-
More details of
the Haber process are given in
Equilibria Part 3.
-
Example 1.4.3
- A historic classic equilibrium study by Max Bodenstein, first
published in 1899
-
He made some of the first
accurate calculations of equilibrium constants.
-
The formation of hydrogen
iodide from hydrogen and iodine:
-
H2(g) + I2(g)
2HI(g) (ΔH = –10 kJ
mol–1, iodine gaseous above 200oC)
-
Rule 1 –
temperature and energy change (ΔH)
-
Rule 2 – gas
pressure (ΔV):
-
Rule 3 –
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 concentration at the new equilibrium
point.
-
Rule 4 – catalyst:
Not applicable.
-
HISTORIC NOTE:
-
The equilibrium
2HI (g) H2(g)
+ I2(g) was the first gas phase equilibrium to be
thoroughly studied. It was investigated by the chemist Max Bodenstein (1871-1942) in the late 1890's in Germany. He found that
the reaction could be studied between about 380 and 500 °C, but that
at higher temperatures the reaction was shifted too far to the
product side - endothermic decomposition from left to right.
-
To study the reaction, Bodenstein sealed various amounts of HI in
glass containers. These were then heated to the desired temperature
for various intervals of time in an accurately thermostated oven.
The glass containers were then removed and cooled rapidly, broken
open and the contents analyzed for all three components. Bodenstein
was able to determine the equilibrium constant for the reaction from
the quantities of hydrogen gas, iodine vapor that had formed, and
hydrogen iodide that remained.
-
I don't think he actually analyse directly the hydrogen and iodine
directly, but you can titrate the hydrogen iodide with sodium
hydroxide. From one molar measurement and the starting quantities,
you can work out the quantities of the other components in the final
equilibrium mixture. See
calculations in Part 2
-
Example 1.4.4
- Ester formation and hydrolysis
- Esterification: e.g. ethanoic acid + ethanol
ethyl ethanoate + water
- CH3COOH(l) +
CH3CH2OH(l)
CH3COOCH2CH3(l)
+ H2O(l) (ΔH =
–2 kJ
mol–1)
-
Rule 1 –
temperature and energy change (ΔH)
-
Rule 2 – gas
pressure (ΔV):
Not applicable, no gases involved.
-
Rule 3 –
concentration
-
Rule 4 – catalyst
-
The forward
esterification reaction is catalysed by acids e.g. a few drops of
conc. sulphuric acid.
-
The reverse
reaction i.e. hydrolysis of the ester back to the acid and alcohol
is catalysed by dilute acids.
-
Example 1.4.5
- Equilibria involving the oxides of nitrogen (teacher
demonstrations in fume cupboard)
-
The equilibria
involving oxygen O2, nitrogen (II) oxide NO, nitrogen(IV) oxide
NO2 and its
dimer N2O4.
-
Nitrogen dioxide, NO2
can be made from the irreversible thermal decomposition
of lead(II) nitrate in a pyrex boiling tube connected to a 100 cm3
gas syringe in a fume cupboard.
-
Preparation of nitrogen(IV) oxide
-
You can conveniently prepare a sample of
nitrogen(IV) oxide by heating lead(II) nitrate - which crackles!
-
lead(II)
nitrate ==> lead(II) oxide + nitrogen(IV) oxide + oxygen
-
2Pb(NO3)2(s)
==> 2PbO(s) + 4NO2(g) + O2(g)
-
Both lead nitrate and nitrogen dioxide are poisonous, for
teachers only!
-
(a) The thermal decomposition
of nitrogen(IV) oxide
-
2NO2(g,
brown)
2NO(g,
colourless) + O2(g, colourless) (ΔH
= +113 kJ
mol–1)
-
Transfer some NO2 to
a pyrex boiling tube and keep the gas in with a wad of mineral
wool.
-
The
temperature effect can be observed by strongly heating the gases
in the pyrex tube above 400oC.
-
The brown colour should fade at
the lowest hottest part of the boiling tube.
-
The equilibrium will shift from
left to right, in the endothermic direction - absorbing the heat
to counteract the increase in temperature.
-
This experiment can be done
after you have done the lower temperature boiling tube
experiment below.
-
(b) Playing around with
nitrogen(IV) oxide and its dimer ('dinitrogen tetroxide')
-
2NO2(g,
brown)
N2O4(g,
colourless) (ΔH = –58 kJ
mol–1)
-
(b)(i) Again set up a boiling
tube of NO2 plus wad of mineral wool to contain it.
-
Cool the tube in ice and the
brown colour is considerably decreased - equilibrium shifts
in the exothermic direction and more colourless N2O4
formed - heat released to counteract the decrease in
temperature.
-
If you then warm up the boiling
tube of NO2 in hot water, the brown colour
intensity increases as some the N2O4
dimer is decomposed to NO2 - the equilibrium shifts
in the 'right to left' endothermic direction
-
Rule 1 –
temperature and energy change (ΔH)
-
Increases
in temperature favours the endothermic decomposition of NO2
to NO and O2, so at high temperatures the brown colour fades.
-
Decrease
in temperature favours the exothermic formation of the dimer N2O4
from NO2, so the brown colour fades on cooling the gas
mixture.
-
Rule 2 – gas
pressure (ΔV)
-
(b)(ii) The pressure effect can
be observed by sealing the gases at room temperature in the
gas syringe and compressing and decompressing it
-
I just doubled up the rubber
tubing and clamped it together, BUT you need to hold the gas
syringe very firmly when compressing the gaseous contents.
-
You can do this on an OHP,
take care, its a bit awkward for all of an A level class to see
it at close quarters in the confines of a fume cupboard.
-
2NO2(g,
brown)
N2O4(g,
colourless)
-
Increase
in pressure favours the RHS, more N2O4, because 2 mol
gas ==> 1 mol gas, so theoretically the mixture would get
lighter in colour - reduction in the number of gas molecules
to counteract the increase in pressure.
-
Decrease
in pressure favours LHS NO2 formation,
1 mol gas ==> 2 mol gas, so theoretically the mixture would get
darker in
colour - increase in the number of gas molecules to counteract
the decrease in pressure.
-
(b)(ii) can
be demonstrated by compressing/decompressing the gas mixture in
the syringe to see the brown colour intensity increase/decrease
- but the observations are quite subtle!
-
Compressing the NO2/N2O4
mixture:
-
When you press the gas
syringe plunger in, you increase the pressure of the gases.
-
The orange-brown colour
becomes more darker because ALL concentrations are
increased.
-
BUT, within a few seconds
the orange-brown colour fades just a little bit as the
equilibrium moves from left to right - a little more N2O4
formed - as predicted from the pressure rule.
-
Decompressing the NO2/N2O4
mixture:
-
When you pull the gas syringe
plunger out, you decrease the pressure of the gases.
-
The orange-brown colour
becomes lighter because ALL concentrations are decreased.
-
BUT, within a few seconds,
the colour gets a little bit darker as the equilibrium moves
from right to left - a little more NO2 formed -
as predicted from the pressure rule.
-
In fact
you can even see the dynamic equilibrium 'kinetics' in
directly in action
here. There is a time lag
of about 1–2 seconds before the new equilibrium position is
established as the 'imposed' colour intensity change becomes
constant.
-
Example 1.4.6
- Ligand displacement reaction involving two cobalt(II) complex ions
-
Pink hexa–aqua
cobalt(II) ions form a complex with chloride ions, the
blue
tetrachlorocobaltate(II) ion.
-
[Co(H2O)6]2+(aq,
pink) + 4Cl–(aq)
[CoCl4]2–(aq, blue) + 6H2O(l)
(ΔH +ve, don't know value)
-
Rule 1 –
temperature and energy change (ΔH)
-
Increase
in temperature favours the endothermic blue complex formation,
cooling favours the exothermically form pink ion.
-
If prepared at
higher temperature, with just enough chloride added to turn the
solution blue, on cooling it becomes pink.
-
Rule 2 – gas
pressure (ΔV): Not
applicable, no gases involved.
-
Rule 3 –
concentration
-
Increase in
chloride concentration decreases the pink ion concentration and
increases the blue ion concentration i.e. shifts the equilibrium
position from left to right.
-
Diluting with
water shifts the equilibrium to the left so solution is less blue
and more pink.
-
Rule 4 – catalyst:
Not applicable here, despite involving transition metal complexes!
-
Similarly if concentrated sodium chloride
solution or hydrochloric acid is added to copper(II) sulphate solution the
pale yellow–brown tetrachlorocuprate(II) complex ion is
formed (seen as green at first due to the mixture of blue from the original Cu2+
ion).
-
Example 1.4.7
- The high temperature formation of nitrogen(II)
oxide.
- N2(g) + O2(g)
2NO(g) (ΔH = +181 kJ
mol–1)
- Rule 1 –
temperature and energy change (ΔH)
- Increase in temperature
favours the endothermic formation of NO.
- This reaction does not happen
at room temperature but is formed at the high temperatures in car engines.
- Unfortunately when released
through the car exhaust, it cools to normal temperatures when NO
irreversibly reacts with oxygen in air to form nitrogen(IV) oxide, NO2,
which is acidic, a lung irritant and a reactive free radical molecule
involved in the chemistry of photochemical smog not good!
- Rule 2 – gas
pressure (ΔV)
- Since 2 mol gaseous reactants
gives 2 mol gaseous products, pressure does not affect the position of
the equilibrium.
- Rule 3 –
concentration
- The concentration of nitrogen
is high from air, but although the concentration of oxygen is low in
the exhaust gases, there is sufficient present in the combustion
process to ensure a small % of NO is formed.
- Rule 4 – catalyst: Not
applicable.
-
Example 1.4.8
-
The oxidation of sulfur
dioxide to sulfur trioxide
- The oxidation of sulfur
dioxide to sulfur trioxide
- e.g. in the Contact Process
for manufacturing sulphuric acid, see
Equilibria Part 3.3.
- 2SO2(g) + O2(g)
2SO3(g) (ΔH = –196 kJ
mol–1)
- Rule 1 –
temperature and energy change (ΔH)
- The exothermic formation of
sulphur trioxide is favoured by low temperature.
- Rule 2 – gas
pressure (ΔV)
- Higher pressure favours a
higher yield of sulphur trioxide as 3 gas moles ==> 2 gas moles,
though 1–2 only atm is used in practice because the equilibrium is
already so far to the right (about 99%).
- Rule 3 –
concentration
- Air is used as the source of
oxygen and despite its dilution with nitrogen the concentration,
the oxygen concentration is high enough to move the equilibrium very much
to the RHS.
- Rule 4 – catalyst
- A vanadium(V) oxide, V2O5,
catalyst ensures the high yield of 99% SO3 is attained
fast, but no more!
-
Example 1.4.9
- Reforming reaction to manufacture hydrogen
- One way to produce
hydrogen
for the Haber synthesis of ammonia is to react methane gas with steam.
- CH4(g) + H2O(g)
3H2(g)
+ CO(g) (ΔH = +206 kJ
mol–1)
- Rule 1 –
temperature and energy change (ΔH)
- Increase in temperature
favours the endothermic formation of hydrogen (and carbon monoxide).
- Rule 2 – gas
pressure (ΔV)
- For the desired forward
reaction, 2 mol of reactant gases ==> 4 mol of product gases, so the
increase in product volume is favoured by lower pressure.
- Rule 3 –
concentration
- Theoretically increase in
methane and steam concentrations will increase the hydrogen
concentration, but this essentially means increasing pressure
favouring the LHS, so you might not gain as much hydrogen as you like
to!
- Rule 4 – catalyst
- A nickel catalyst is used, but
cannot affect the yield, only the rate of reaction – speeds it up for
more efficient production.
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WHAT NEXT?
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principle–rules
* Part 2. Kc and Kp equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product Ksp, common ion effect,
ion–exchange systems *
Part 5. pH, weak–strong acid–base theory and
calculations * Part 6. Salt hydrolysis,
acid–base titrations–indicators, pH curves and buffers * Part 7.
Redox equilibria, half–cell electrode potentials,
electrolysis and electrochemical series
*
Part 8.
Phase equilibria–vapour
pressure, boiling point and intermolecular forces watch out for sub-indexes
to multiple sections or pages
TOP OF PAGE
Chemical
Equilibrium Notes Index
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