Doc
Brown's Chemistry
TheoreticalPhysical
Advanced Level
Chemistry  Equilibria  Chemical Equilibrium Revision Notes PART 2
2. Chemical equilibrium expressions and calculations
How do we write out chemical equilibrium
expressions. What is the equilibrium constant? How do you do equilibrium
calculations? All is explained with examples including the units for
concentrations (mol dm3) or partial pressures (e.g. atm, Pa or kPa) and
how to solve equilibrium problems using concentration or partial
pressure units.
KS4 Science GCSE/IGCSE reversible reactions
& chemical equilibrium notes
Part 2 subindex: 2.1 K_{c} and K_{p} equilibrium expressions and Le
Chatelier's Principle *
2.2 Exemplar calculations using K_{c}
expressions or K_{p} expressions
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principlerules *
Part 3.
Equilibria and industrial processes * Part 4.
Partition,
solubility product and ionexchange * Part 5.
pH, weakstrong acidbase theory and calculations
* Part 6. Salt hydrolysis,
Acidbase titrationsindicators, pH curves and buffers * Part 7.
Redox equilibria, halfcell electrode potentials,
electrolysis and electrochemical series
*
Part 8. Phase equilibriavapour
pressure, boiling point and intermolecular forces
2.1 Equilibrium expressions and applying Le Chatelier's Principle
IT IS IMPORTANT and ESSENTIAL that
Equilibria Part 1 is studied before working through this page
2.1a
Molar concentration expressions

K_{c} concentration equilibrium
expression INTRODUCTION

It is found
experimentally that the concentrations at the equilibrium point are related by
a simple mathematical equation known as an equilibrium expression
which is governed by an equilibrium constant K, at constant
temperature (K only varies with temperature, and nothing else).

These equilibrium
expressions (many on this page) were originally derived from
experimental analysis of mixtures at equilibrium.

Patterns in these
equilibrium concentrations were 'spotted' and the resulting
mathematical expression of these concentrations were considered to
conform to what was called the 'law of mass action' ( a term
not really used these days!).

A classic study of the
ester formation equilibrium is often described in textbooks.

ALCOHOL + ACID
ESTER +WATER

Knowing the initial
amounts of alcohol and acid, it was easy to titrate the remaining
free carboxylic acid and then logically work out the amounts of the
water, alcohol and ester left in the equilibrium mixture.

They would also have
found out that it took some time to reach equilibrium, but
eventually all the concentrations remained constant i.e. in the
mixture, the point of equilibrium was reached.

See
example 2.2a.1 ester equilibrium calculation
and 2.a.5 too.

The theoretical
justification for K expressions came later in chemical history and
this need not concern us at this level because it involves some
pretty advanced
thermodynamics theory!

From a student's point
of view, here at this level, you are using K in terms of
concentrations or partial pressures only and therefore all other
equilibrium terms, apart from K itself, should be quoted in either ...

For any reaction
in solution or a gaseous mixture:

By convention, the
arithmetical product of the product concentrations^{*} of the forward reaction (RHS) are on the
top line and the arithmetical product of the reactant
concentrations^{*}
from the backward reaction (LHS) are on the bottom line.

^{
*} In all cases the product concentrations are
raised to the appropriate power (a, b, c, .. t, u, w, ..) given by the
stoichiometric mole ratios of the balanced equation.

AND again note that
K_{c} (like K_{p}) is only constant for a
specific constant temperature at which the concentrations of the
equilibrium components might vary from one dynamic equilibrium
situation to another (e.g. in reacting liquid mixtures or in
solutions).

For heterogeneous
equilibria, K expressions do not normally include values for solid phases,
since their chemical potential cannot change since the concentration
of a solid cannot change.

The equilibrium
constant, K_{c} (or K_{p} later), is governed by temperature, which is the only
factor that can alter the internal potential energy of the
reactants or products. The 'rule' for the trend in K value change is
provided by Le Chatelier's Principle.

If the forward
reaction is exothermic, K_{c} (or K_{p} later) will decrease in value
with increase in temperature.

The
equilibrium position will shift more to the left in the endothermic
direction and mathematically, by convention, the top line function
will numerically decrease and the bottom line function must
numerically increase.

If the forward
reaction is endothermic, K_{c} (or K_{p}) will increase in value
with increase in temperature.

The
equilibrium position will shift more to the right in the endothermic
direction and mathematically, by convention, the top line function
must numerically increase and the bottom line function must
numerically decrease.

Changes in
pressure or concentration have no effect on a K value for ideal
mixtures of gases/liquids or solutions.

Application of
a catalyst to a reaction also has no effect on a K value.

Why are the values of
equilibrium constants affected by temperature?

As we have noted, it is
very important to quote the specific constant temperature that
applies to any K_{c} or K_{p} value BECAUSE
equilibrium constants vary with temperature AND ONLY temperature.

This is because changes
in concentration, partial pressure (see section
2.1b) or employing a catalyst do NOT affect the energy
content of the molecules themselves (referred to in
thermodynamics as the internal energy of the molecule).

However, if you
change the temperature, you also change the fundamental energy
content of a molecule, and we are now talking about the
thermodynamic property values of the equilibrium components.

Electronic energy is
stored in the chemical bonds of the molecules (chemical energy)
plus their thermal energies of translation (kinetic energy of
movement), rotation (of the whole or parts of a molecule) and
vibration (of bonded atoms).

If you change the
temperature, all the internal energy contents of the molecules are
also changed, but they don't change to the same amount for each
molecule for the same rise in temperature.

If you think of the
internal energy as a sort of chemical potential to effect a chemical
change, the formation of 'reactants' or 'products' may be favoured
one way or the other (l to r or r to l as you write the equilibrium
equation).

Hence the equilibrium
constant not only changes, but may increase or decrease depending on
whether the reaction is endothermic or exothermic.

Therefore the change in
the internal energy (caused by change in temperature), a fundamental
thermodynamic property of a molecule, explains why equilibrium
constants are ONLY affected by temperature and NOT by concentration,
pressure or indeed, the presence of a catalyst.

It is possible to
calculate equilibrium constant from
thermodynamic Gibbs free energy change data for the reaction,
but this may not be part of your course and certainly not
appropriate to study here.

See later specific
examples for the units of K_{c} (or K_{p}
later) and if K has no
units you should state so.

Equilibrium
example 2.1a.1 The formation of hydrogen iodide

Equilibrium
example 2.1a.2 The formation of the ester ethyl ethanoate

Equilibrium
example 2.1a.3 The formation of phosphorus(V) chloride
(gaseous phase)

Equilibrium
example 2.1a.4 The synthesis of ammonia

Equilibrium
example 2.1a.5 The synthesis of phosphorus(V) fluoride

Equilibrium
example 2.1a.7 A cobalt(II) complex ion ligand exchange
reaction

[Co(H_{2}O)_{6}]^{2+}_{(aq)} + 4Cl^{}_{(aq)}
[CoCl_{4}]^{2}_{(aq)} + 6H_{2}O_{(l)}

K_{c} =

[[CoCl_{4}]^{2}_{(aq)}] 

(units mol^{4} dm^{12}) 
[[Co(H_{2}O)_{6}]^{2+}_{(aq)}] [Cl^{}_{(aq)}]^{4} 
 units = (mol dm^{3})/[(mol dm^{3})
x (mol dm^{3})^{4}] = mol^{4} dm^{12}

Note that the
concentration of water is effectively constant in an aqueous solution
and omitted from the equilibrium expression, but is effectively subsumed into the K_{c} value which in complex ion
chemistry is called the stability constant denoted by
K_{stab}.

Equilibrium example 2.1a.8 The
acid behaviour of high oxidation state hexaaqua ion

The
hexaaquiron(III) ion is quite acidic in aqueous solution due to proton
transfer giving the oxonium ion.

[Fe(H_{2}O)_{6}]^{3+}_{(aq)} +
H_{2}O_{(l)}
[Fe(H_{2}O)_{5}OH]^{2+}_{(aq)} +
H_{3}O^{+}_{(aq)}
K_{c} =

[[Fe(H_{2}O)_{5}OH]^{2+}_{(aq)}] [H_{3}O^{+}_{(aq}_{)}] 

(units mol dm^{3}) 
[[Fe(H_{2}O)_{6}]^{3+}_{(aq)}] 
 units = [(mol dm^{3})^{ x
}(mol dm^{3})]/(mol dm^{3}) = mol dm^{3}

Again [H_{2}O_{(l)}]
incorporated into K_{c}.

Some 'VERY
rough rules of thumb' for an equilibrium K value (K_{c} or
K_{p} in section 2.1b) and the 'position' of
the equilibrium in terms of LHS (e.g. original reactants or
products of backward reaction) and
the RHS (products of the forward reaction):

For: LHS
RHS

(for A + B
C + D the rules below work ok BUT once the ratios of reactants or
products are not 1:1, things are not so simple)

If K_{c}
(or K_{p}) is >> 1 the equilibrium is
mainly on the RHS, maybe virtually 100% completion of the
forward reaction i.e. a very large RHS yield i.e. and likely to be very
thermodynamically feasible.

If K_{c}
(or K_{p}) is approx. 1 the equilibrium
is about 50% RHS and about 50% LHS.

If K_{c}
(or K_{p}) is << 1 the equilibrium is
mainly on the LHS, maybe virtually 0% of products of the forward
reaction i.e. a very low RHS yield i.e. NOT likely to be
thermodynamically feasible).

BUT remember K
changes with temperature considerably changing the position of an
equilibrium, AND, at constant temperature, and therefore constant K, the position of an
equilibrium can change significantly depending on
relative concentrations/pressures of 'reactants' and 'products'.
2.1b
Partial pressure equilibrium law expressions

K_{p} partial equilibrium
expression INTRODUCTION

For any gaseous
reaction: aA_{(g)} + bB_{(g)} + cC_{(g)}
etc.
tT_{(g)} + uU_{(g)} + wW_{(g)} etc.

K_{p} =

p_{T}^{t}
p_{U}^{u} p_{V}^{v} etc. 

(for units see later) 
p_{A}^{a}
p_{B}^{b} p_{C}^{c} etc. 

p_{x}
indicates the partial pressure of x, usually in atm
(atmospheres) or Pa (pascals).

Do NOT put square brackets in K_{p}
expressions!

AND again note that
K_{p} (like K_{c}) is only constant for a
specific constant temperature at which the partial pressures of
the component gases might vary from one equilibrium situation to
another at the same temperature.

The partial
pressure of a gas is defined as the pressure a gaseous component in
a mixture would exert, if it alone occupied the space/volume in
question.

e.g. in air, 21%
by volume is oxygen and 79% is nitrogen etc.

Therefore the
fraction of molecules which are oxygen and nitrogen is 0.21 and 0.79
respectively.

Since air has a total pressure of 1 atm. or
101325 Pa, the partial pressures of oxygen and nitrogen in air are:

p_{O2}
= 0.21 x 1 = 0.21 atm or p_{O2} = 0.21 x
101325 = 21278 Pa,

p_{N2}
= 0.21 x 1 = 0.79 atm or p_{N2} = 0.21 x
101325 = 80047 Pa,

in other words,
the partial pressure of a gas in a mixture p_{gas}
= its % x p_{tot} / 100

In a gaseous
mixture the total pressure equals the sum of all the partial
pressures:

Some other
useful mathematical ideas and expressions for gas mixtures:

Equilibrium
example 2.1b.1 The formation of nitrogen(II) oxide (e.g. in
car engines)

Equilibrium
example 2.1b.2 The synthesis of sulphur(VI) oxide in the
manufacture of sulphuric acid

Equilibrium
example 2.1b.3 The synthesis of ammonia (Haber Process)

Equilibrium
example 2.1b.4 The manufacture of hydrogen (e.g. for
ammonia synthesis)

Equilibrium
example 2.1b.5 The synthesismanufacture of methanol (gas phase)
 CO_{(g)} + 2H_{2(g)}
CH_{3}OH_{(g)}

K_{p} =

p_{CH3OH}


(units atm^{2} or Pa^{2}) 
p_{CO}
p_{H2}^{2} 
 imagine p = partial pressure units
 units = p/(p x p^{2}) = p/p^{3}
= p^{2}
2.2
Exemplar calculations and concept Questions
These exemplar
questions involve both numerical calculations and application of Le
Chatelier's Principle.
2.2a
K_{c} and concentration
calculations

K_{c} Example Q
2.2a.1
Esterification

Given the
esterification reaction: ethanoic acid + ethanol ethyl
ethanoate + water

CH_{3}COOH_{(l)} + CH_{3}CH_{2}OH_{(l)}
CH_{3}COOCH_{2}CH_{3(l)}
+ H_{2}O_{(l)}

A mixture of 1.0
mol of ethanoic acid and 1.0 mol of ethanol was left to reach
equilibrium at 25^{o}C.

On analysis of the
equilibrium mixture it was found that by titration with standard
sodium hydroxide solution 0.333 mol of the ethanoic acid was left
unreacted.

Calculate the
value of the equilibrium constant K_{c} and give its units.
 The reactantproduct mole
ratios are 1:1 ==> 1:1
 If 0.333 mol ethanoic acid was
left unreacted, then 0.667 mol of the acid had reacted.
 Therefore 0.667 mol of ethanol
must also have reacted, leaving 0.333 unreacted.
 If 0.667 of the acid/ethanol
reacted, 0.667 mol of ethyl ethanoate and 0.667 mol of water
must be formed.
 The concentrations = mol /
volume, but the volume terms cancel each other out, so substitution in
the equilibrium expression can be done in terms of final numbers of
moles reactants and products

K_{c} =

[CH_{3}COOCH_{2}CH_{3(l)}]
[H_{2}O_{(l)}] 
 
[CH_{3}COOH_{(l)}]
[CH_{3}CH_{2}OH_{(l)}] 

K_{c} =

0.667 x 0.667 

= 4.01 (no units) 
0.333 x 0.333 

K_{c} Example Q
2.2a.2
Formation of hydrogen iodide or the decomposition of hydrogen iodide

In these examples, don't
forget that all the V's cancel, so you can work your logic in moles
when solving these equilibrium problems. In the first example,
concentrations are given, BUT after that its all moles, logic and
maybe algebra!

Example
(a) For the reaction:
H_{2(g)} + I_{2(g)
}
2HI_{(g)}

An equimolar
mixture of hydrogen and iodine was heated in a sealed flask at 491^{o}C
and the concentration of iodine was found to be 2.5 x 10^{2} mol dm^{3}
and that of hydrogen iodide 1.71 x 10^{1} mol dm^{3}.

(a) Calculate
the value of the equilibrium constant K_{c}.

The concentration
of the remaining hydrogen and iodine must be the same since they
started at a 1:1 ratio and react in a 1:1 ratio.

Therefore
substituting in the equilibrium expression:

K_{c} =

[HI_{(g)}]^{2} 
 
[H_{2(g)}]
[I_{2(g)}] 

K_{c} =

(1.71 x 10^{1})^{2} 

= 46.8
(no units) 
(2.5 x 10^{2}) x
(2.5 x 10^{2}) 



Examples
(b) More logicmathematical solutions to the hydrogen, iodine
and hydrogen iodide equilibrium

HI/H2/I2 moles logic
to the fore!

(i) For the equilibrium
reaction: 2HI_{(g)}
H_{2(g)} + I_{2(g)
}

Suppose we start
with 1 mole of hydrogen iodide and fraction x of it decomposes
into hydrogen and iodine (x x 100 = % decomposition).

For every mole of HI
that decomposes, you will form 0.5 moles of hydrogen and 0.5
moles of iodine.

This
arises from the stoichiometry of the equation i.e. the 2 : 1 : 1
mole ratio, so we can tabulate this logic as follows ...


HI 
H_{2} 
I_{2} 
initial moles 
1.0 
0.0 
0.0 
moles
at equilibrium 
1x 
0.5x 
0.5x 
 Noting that all the V's cancel,
so you can write the equilibrium expression in terms of the
relative moles of hydrogen iodide, hydrogen and iodine, giving
.....

K_{c} =

[H_{2(g)}]
[I_{2(g)}] 
 
[HI_{(g)}]^{2} 

K_{c} =

(0.5x)(0.5x)
0.25x^{2} 
 =
 = ..... (no units) 
(1x)^{2} (1x)^{2}

 So, if for example, the hydrogen
iodide was 20% decomposed, then obviously the proportion
decomposed x = 0.20, and simple substitution into the equation
enables you to calculate the equilibrium constant K without any
difficulty.
 Incidentally, since K is
dimensionless, Kc = Kp even if you were working in partial
pressures.

However,
if you were given K, to solve the equation for x, the proportion
decomposed, this requires some carefully worked our algebra and
the formula for solving quadratic equations (shown on the
right).
 Rearranging the equilibrium
expression gives
 K(1x)^{2} = 0.25x^{2}
 K(x^{2}  2x + 1) =
0.25x^{2}
 Kx^{2}  2Kx + K =
0.25x^{2}
 (K  0.25)x^{2} 
2Kx + K = 0
 So, in the quadratic
equation formula
 a = K  0.25, b = 2K
and c = K, to solve for x.
 bon voyage and watch the
sign!

(ii) For the equilibrium reaction:
H_{2(g)} + I_{2(g)
}
2HI_{(g)}

Suppose we start
with 1 mole of hydrogen and 1 mole of iodine and no hydrogen
iodide.

For every x moles of
hydrogen or iodine that react, 2x moles of hydrogen iodide will
be formed, so we can tabulate this argument as follows ...

H_{2} 
I_{2} 
HI 
initial moles 
1.0 
1.0 
0.0 
moles
at equilibrium 
1x 
1x 
2x 
K_{c} =

[HI_{(g)}]^{2} 
 
[H_{2(g)}]
[I_{2(g)}] 
K_{c} =

(2x)(2x)
4x^{2} 

=  = .... (no units) 
(1x)^{2}
(1x)^{2} 

Again,
if you know the amount of HI formed OR the amount of iodine or
hydrogen reacted, you can readily calculate K via the molar
logic.
 BUT if you are given K, to solve
for x, this requires solving a quadratic equation.
 Rearranging the equilibrium
expression
 K(1x)^{2} = 4x^{2}
 K(x^{2} 2x + 1) =
4x^{2}
 Kx^{2}  2Kx + K =
4x^{2}
 (K  4)x^{2}  2Kx +
K = 0
 So, in the quadratic
equation formula
 a = K4, b = 2K and
c = K, to solve for
x.
 watch the signs and enjoy!
(iii) For the
equilibrium reaction: H_{2(g)} + I_{2(g)
}
2HI_{(g)}

A general solution
for ANY combination of hydrogen and iodine forming hydrogen
iodide.

Suppose we start
with A moles of hydrogen and B moles of iodine and no hydrogen
iodide.

For every x moles of
hydrogen OR iodine that react, 2x moles of hydrogen iodide will
be formed, so we can tabulate this argument as follows ...

H_{2} 
I_{2} 
HI 
initial moles 
A 
B 
0.0 
moles
at equilibrium 
Ax 
Bx 
2x 
K_{c} =

[HI_{(g)}]^{2} 
 
[H_{2(g)}]
[I_{2(g)}] 
K_{c} =

(2x)^{2}


= .... (no units) 
(Ax)(Bx) 

Again,
if you know the amount of HI formed (2x) OR the amount of
iodine/hydrogen reacted (x), you can readily calculate K.
 Note that x is NOT a
fraction here!
 Its the actual moles of
hydrogen or iodine that have reacted to form hydrogen
iodide.
 BUT if you are given K, to solve
for x, this requires solving a quadratic equation.
 Rearranging the equilibrium
expression
 K(Ax)(Bx) = 4x^{2}
 K(x^{2} Ax  Bx +
AB) = 4x^{2}
 Kx^{2} AKx BKx +
ABK = 4x^{2}
 (K4)x^{2} (A +
B)Kx + ABK = 0
 So, in the quadratic
equation formula
 a = K4, b = (A+B)K
and c = ABK, to
solve for x.
 This methodology is not
designed for late night working! and watch the signs!

(iv) For the equilibrium
reaction: 2HI_{(g)}
H_{2(g)} + I_{2(g)
}

A general solution
for the decomposition of hydrogen iodide.

Suppose we start
with A moles of hydrogen iodide and x moles of it decomposes
into 0.5x moles of hydrogen and 0.5x moles of iodine.

This
arises from the stoichiometry of the equation i.e. the 2 : 1 : 1
mole ratio, so we can tabulate this logic as follows ...


HI 
H_{2} 
I_{2} 
initial moles 
A 
0.0 
0.0 
moles
at equilibrium 
Ax 
0.5x 
0.5x 
 Noting that all the V's cancel,
so you can write the equilibrium expression in terms of the
relative moles of hydrogen iodide, hydrogen and iodine, giving
.....

K_{c} =

[H_{2(g)}]
[I_{2(g)}] 
 
[HI_{(g)}]^{2} 

K_{c} =

(0.5x)(0.5x)
0.25x^{2} 
 =  = ..... (no units) 
(Ax)^{2} (Ax)^{2}

 So, if you know how much HI
reacteddecomposed, OR, the amount of iodine formed, then simple
substitution into the equilibrium equation enables you to
calculate the equilibrium constant K without any difficulty.
 Note that x is NOT a
fraction here!
 Its the actual moles of
hydrogen iodide that have decomposed to form hydrogen and
iodine.

However,
if you were given K, to solve the equation for x, the amount
decomposed, this requires some carefully worked our algebra and
the formula for solving quadratic equations (shown on the
right).
 Rearranging the equilibrium
expression
 K(Ax)^{2} = 0.25x^{2}
 K(x^{2} 2Ax + A^{2})
= 0.25x^{2}
 Kx^{2} 2KAx + KA^{2}
= 0.25x^{2}
 (K  0.25)x^{2}
2KAx + KA^{2} = 0
 So, in the quadratic
equation formula
 a = K0.25, b = 2KA
and c = KA^{2},
to solve for x.
 keep a clear head and watch
the signs!

K_{c} Example Q
2.2a.3
Formation of complex ion

Aqueous iron(II)
ions can complex with chloride ions to form the
tetrachloroferrate(III) ion.

For the
equilibrium: Fe^{3+}_{(aq)} + 4Cl^{}_{(aq)}
FeCl_{4}^{}_{(aq)}

K_{c} =
8.0 x 10^{2} mol^{1} dm^{3} at 298K.

(a) If the
concentration of the free chloride ion is 0.80 mol dm^{3} and that
of the free iron(III) ion 0.20 mol dm^{3}, calculate the
concentration of the tetrachloroferrate(III) complex ion.

K_{c} =

[FeCl_{4}^{}_{(aq)}] 
 
[Fe^{3+}_{(aq)}] [Cl^{}_{(aq)}]^{4} 
 rearranging the equilibrium
expression gives ...
 [FeCl_{4}^{}_{(aq)}] =
K_{c} x [Fe^{3+}_{(aq)}] x [Cl^{}_{(aq)}]^{4}
 [FeCl_{4}^{}_{(aq)}] = 8.0 x 10^{2} x 0.20 x (0.80)^{4} = 6.55 x 10^{3}
mol dm^{3}
 (b) Equal volumes of 5 molar
sodium chloride solution and 0.02 molar iron(III) nitrate Fe(NO_{3})_{3}
solution were mixed together.
 (i) Assuming the chloride ion
concentration changes very little and c represents the
concentration of the tetrachlorateferrate(III) ion, show how the
concentration of the complex ion can be approximately calculated.
 The nitrate ion is a spectator ion and
can be ignored.
 Let [FeCl_{4}^{}_{(aq)}]_{equilib}
= c, and
 since 1 mole of Fe^{3+}
forms 1 mole of FeCl_{4}^{} complex
 then [Fe^{3+}_{(aq)}]_{equilib}
= [Fe^{3+}_{(aq)}]_{init}  c, and assuming
 [Cl^{}_{(aq)}]_{equilib}
~ [Cl^{}_{(aq)}]_{init} since [Cl^{}_{(aq)}]
>> [Fe^{3+}_{(aq)}] + [FeCl_{4}^{}_{(aq)}]
 (ii) Calculate the
concentration of the [FeCl_{4}^{}_{(aq)}]
ion.
 The dilution factor on mixing
equal volumes is 2, therefore
 [Fe^{3+}_{(aq)}]_{init}
= 0.02/2 = 0.01 mol dm^{3} and
 Cl^{}_{(aq)}]_{equilib}
~ [Cl^{}_{(aq)}]_{init} = 5.0/2 = 2.5 mol
dm^{3}
 from (a) [FeCl_{4}^{}_{(aq)}] =
K_{c} x [Fe^{3+}_{(aq)}] x [Cl^{}_{(aq)}]^{4}
 c = 8.0 x 10^{2} x (0.01 
c) x (2.5)^{4}
 c = 3.125 x (0.01  c) = 0.03125 
3.125c
 4.125c = 0.03125
 [FeCl_{4}^{}_{(aq)}]
= c = 0.03125/4.125 = 0.7575 = 7.58 x 10^{3}
mol dm^{3}
 (iii) What percentage of the
original Fe^{3+} ion is converted into the complex?
 If all of the Fe^{3+} ion had
been converted to the chloro complex, the concentration of the
complex would be 0.01 mol dm^{3}
 Therefore the % conversion =
7.58 x 10^{3} x 100/0.01 = 75.8%

K_{c} Example Q
2.2a.4
Iodineiodide equilibrium

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

I^{}_{(aq)}
+ I_{2(aq)}
I_{3}^{}_{(aq)}
for which K_{c} = 7.10 x 10^{2} mol^{1} dm^{3}
at 298K.

If the
concentration of the I^{} ion is 0.122 mol dm^{3}, and
that of the I_{3}^{} ion is 0.153 mol dm^{3},
calculate the concentration of free iodine.

K_{c} =

[I_{3}^{}_{(aq)}] 
 
[I^{}_{(aq)}]
[I_{2(aq)}] 

Rearranging gives
...

[I_{2(aq)}] =

[I_{3}^{}_{(aq)}] 
 
K_{c}
x [I^{}_{(aq)}] 

[I_{2(aq)}] =
0.153 / (7.10 x 10^{2} x 0.122) = 1.77 x 10^{3}
mol dm^{3}

K_{c}
Example Q
2.2a.5 Ester equilibrium  titrimetric
analysis

A titration
method for determining the equilibrium constant for an
esterification reaction.

12.0g of pure ethanoic acid was mixed with 11.5g of pure ethanol and left to stand
for a week at room temperature (298K/25^{o}C). The mixture was then mixed with deionised water and made
up to 250 cm^{3} in a calibrated volumetric flask. When a
25.00 cm^{3} aliquot of the mixture was titrated with 0.50
mol dm^{3} sodium hydroxide solution using phenolphthalein
indicator (colourless ==> 1st permanent pink endpoint), 10.60 cm^{3}
of the alkali was needed for complete neutralisation.

(i) CH_{3}COOH_{(l)} + CH_{3}CH_{2}OH_{(l)}
CH_{3}COOCH_{2}CH_{3(l)}
+ H_{2}O_{(l)} (esterification reaction)

(ii) CH_{3}COOH_{(l)}
+ NaOH_{(aq)}
==> CH_{3}COO^{}Na^{+}_{(aq)}
+ H_{2}O_{(l)} (neutralisation titration reaction)

(a) Calculate
the moles of ethanoic acid unreacted in the original mixture.

1 mole CH_{3}COOH
: 1 mole NaOH from equation (i) above.

moles = molarity
x vol(dm^{3}), so moles acid = 0.50 x 10.6/1000 = 0.0053 mol

since the 25.00
cm^{3} aliquot titrated is equal to 1/10th of the total
mixture,

the total moles
of unreacted acid = 10 x 0.0053 = 0.053 mol CH_{3}COOH
left

(b) Calculate
the moles of ethanoic acid and ethanol in the starting mixture.

M_{r}(CH_{3}COOH)
= 60, mol ethanoic acid = 12/60 = 0.20

M_{r}(CH_{3}CH_{2}OH)
= 46, mol ethanol = 11.5/46 = 0.25

(c) Calculate
the moles of ethanol left unreacted and the moles of ethyl ethanoate
ester and water formed.

equation 
CH_{3}COOH_{(l)} + CH_{3}CH_{2}OH_{(l)}
CH_{3}COOCH_{2}CH_{3(l)}
+ H_{2}O_{(l)} 
moles at start 
0.20 0.25 0 0 
mol
at equilibrium 
0.20  x 0.25  x x x 
mol at equilibrium 
0.053
0.103
0.147
0.147 

x = moles of
ester or water formed,

so x mol
of acid or alcohol must be used to reach equilibrium,

since mol ratios
are 1:1 ==> 1:1 in the reaction equation.

Therefore 0.20 
x = 0.053 from calculation (a), therefore x = 0.20  0.053 =
0.147

so all the
molar quantities can be logically deduced and are shown in the final
line of the table.

TIP In an
exam, doing the working of this part of the Q under the equation is
not a bad idea.

(d) Calculate
the equilibrium constant K_{c} for this esterification.

K_{c} =

[CH_{3}COOCH_{2}CH_{3(l)}]
[H_{2}O_{(l)}] 
 
[CH_{3}COOH_{(l)}]
[CH_{3}CH_{2}OH_{(l)}] 

K_{c} =

(0.147/V) x (0.147/V) 

= 3.96 (no units) 
(0.053/V) x (0.103/V) 
 Note that all the volume terms cancel out, so you can work purely
in moles to calculate K_{c}.

(e) Suggest
several reasons why it may take a week for the equilibrium point to
be reached and suggest ways of speeding up the reaction.

(i) The reaction
will be slow at room temperature, refluxing the mixture will speed
things up!

(ii) The
reaction is catalysed by hydrogen ions (H^{+}) and since
ethanoic acid is a weak acid, the hydrogen ion concentration will be
very low. Adding a strong mineral acid e.g. conc. sulphuric acid.

(iii) Not
surprisingly, esters are often prepared by refluxing the acid and
alcohol with a few drops of conc. sulphuric acid added to the
mixture.
2.2b
K_{p}, partial
pressure calculations and applying Le Chatelier's Principle

Example Q 2.2b.1
Nitrogen(IV) oxidedinitrogen tetroxide equilibrium
(nitrogen dioxide <=> dimer)

For the
equilibrium between dinitrogen tetroxide and nitrogen dioxide:

the value of the
equilibrium constant K_{p} = 0.664 atm at 45^{o}C.

(a) If the
partial pressure of dinitrogen tetroxide was 0.449 atm, what would be
the equilibrium partial pressure of nitrogen dioxide and the total
pressure of the gases? and what % of the dinitrogen tetroxide is
dissociated?

K_{p} =

p_{NO2}^{2}

 
p_{N204} 

Rearranging the

p_{NO2}^{2}
= K_{p} x p_{N204}, so, p_{NO2} =
√(K_{p}
x p_{N204} ) = √(0.664
X 0.449) = 0.546 atm

p_{tot}
= p_{NO2} + p_{N2O4} = 0.546 + 0.449 = 0.995 atm

The pressure of NO_{2}
is 0.546, which is equivalent to 0.273 atm of N_{2}O_{4}
before dissociation.

Therefore the
% N_{2}O_{4}
dissociated = 0.273 x 100 / (0.449 + 0.273) = 37.8%

(b) In
another experiment at 77^{o}C and a total pressure of 1.000
atm, the partial pressure of dinitrogen dioxide was found to be 0.175
atm. Calculate the value of the equilibrium constant at 77^{o}C
and the % dissociation of the dinitrogen tetroxide.

p_{tot} =
p_{NO2} + p_{N2O4}, so p_{NO2} = p_{tot}
 p_{N2O4} = 1.000  0.175 = 0.825 atm

From the
equilibrium expression above:

K_{p}
=
0.825^{2} / 0.175 = 3.89 atm

The pressure of NO_{2}
is 0.825, which is equivalent to 0.4125 atm of N_{2}O_{4}
before dissociation.

Therefore the %
N_{2}O_{4} dissociated = 0.4125 x 100 / (0.175 +
0.4125) = 70.2%

(c) At a
total pressure of 102000 Pa, dinitrogen tetroxide is 40%
dissociated at 50^{o}C.

(d) Explain
the effect, if any, of increasing the total pressure of the system on
the position of the N_{2}O_{4}/NO_{2}
equilibrium.

Increasing
pressure will favour the LHS, i.e. more N_{2}O_{4} and
less NO_{2}, because the system will move to the side with the
least number of gaseous molecules to try to minimise the increase in
pressure (1 mol gas <== 2 mol gas).

(e) From
information from parts (a)(c) is the dissociation exothermic or
endothermic? and give your reasoning.

The dissociation
is endothermic because on raising the temperature from 45^{o}C
to 77^{o}C the value of K_{p} has increased.

i.e. p_{NO2}
has increased in value and p_{N2O4} decreased in value from
more dissociation.

An equilibrium
system moves in the endothermic direction on raising the temperature
to absorb the 'added' heat to try to minimise the temperature rise.

Example Q
2.2b.2
Ammonia synthesis

Ammonia is
synthesised by combing nitrogen and hydrogen, but the reaction is
reversible.

N_{2(g)} + 3H_{2(g)
}
2NH_{3(g)} (ΔH = 92 kJ mol^{1})

In an experiment
starting with a 1:3 ratio H_{2}:N_{2} mixture,
at 400^{o}C and total pressure of 200 atm, the
equilibrium mixture was found to contain 36.3% by volume of ammonia.

(a) Write
out the equilibrium expression for K_{p}.

K_{p} =

p_{NH3}^{2}


(units atm^{2} or Pa^{2}) 
p_{N2}
p_{H2}^{3} 

(b)
Calculate the partial pressures of nitrogen, hydrogen and ammonia.

36% of ammonia
means its mole fraction is 0.36, so p_{NH3} = 0.36 x
200 = 72 atm

The other 64% of
gases must be split on a 1:3 ratio between nitrogen and hydrogen.

So mole fraction
of nitrogen x p_{tot} = p_{N2} = ^{64}/_{100}
x ^{1}/_{4} x 200 = 32 atm,

and mole fraction
of hydrogen x p_{tot} = p_{H2} = ^{64}/_{100}
x ^{3}/_{4} x 200 = 96 atm

check: p_{tot} = p_{N2}
+ p_{H2} + p_{NH3} = 32 + 96 + 72 = 200 atm!

(c)
Calculate the value of the equilibrium constant at 400^{o}C
and give its units.

K_{p} =

p_{NH3}^{2}

 
p_{N2}
p_{H2}^{3} 

K_{p} =

72^{2}


= 1.83 x 10^{4} atm^{2}

32 x 96^{3} 
 (d) What will be the
effect on ammonia yields and the value of K_{p} by (i)
raising the pressure and (ii) raising the temperature. Give
reasons for your answers.
 (i) The reaction to
form ammonia is exothermic, so higher temperature will favour its
endothermic decomposition back to nitrogen and hydrogen. So the
yield of ammonia and the value of K_{p} will be reduced as
the system will absorb heat energy to attempt to minimise temperature
rise.
 (ii) Higher pressure
will favour a higher yield of ammonia because 4 mol gaseous
reactants ==> 2 mol gaseous products, so if the system is subjected to
higher pressure it reduces the number of gas molecules to attempt to
minimise the enforced increase in pressure. Assuming the gases behave
ideally, the value of the equilibrium constant K_{p} is
unaffected by pressure changes, only temperature affects the value
of K_{p} in an ideal gas mixture.

Example Q
2.2b.3 Phosphorus(V) chloride
(phosphorus pentachloride)
dissociation

At high
temperatures vapourised phosphorus(V) chloride dissociates into
gaseous phosphorus(III) chloride (phosphorus trichloride) and
chlorine.

PCl_{5(g)}
PCl_{3(g)} + Cl_{2(g)}
(a) Write the
equilibrium expression for this reaction in terms of K_{p}
and partial pressures.
K_{p} =

P_{PCl3}
x P_{Cl2} 

(units atm, Pa, kPa etc.) 
P_{PCl5} 

(b) At a
particular temperature 40% of the PCl_{5} dissociates into
PCl_{3} and Cl_{2}. If the total equilibrium
pressure is 210 kPa calculate the value of K_{p} at this temperature
quoting appropriate units. The solution is set out below in a
series of logical steps, either thinking from the point of view of
starting with n/1 moles of PCl_{5} or
starting with 100/100% of PCl_{5}
molecules.

PCl_{5(g} 

PCl_{3(g)} 
Cl_{2(g)} 
comments, z
refers to any component in the mixture 
n(1x)
(1x)
can
think of as (100  x%) 

nx
x
can
think of as x% dissociated 
nx
x
can
think of as x% dissociated 
n = initial moles
of undissociated PCl_{5} for a general solution, but
consider 1 mole of PCl_{5} (n = 1) of which
fraction x dissociates giving a total of (1 + x)
moles of gas from 1 mol of PCl_{5}
This is
the logic thinking in terms of 100/100% undissociated P_{PCl5}
molecules and x% is the percentage of PCl_{5}
molecules dissociated. 
(1x)/(1 + x) 

x/(1+x) 
x/(1+x) 
calculation of mole
fractions mole
fraction z = moles z/total moles 
0.6/1.4 = 0.4286
can
think of as 60/140 = 0.4286 

0.4/1.4 = 0.2857
can
think of as 40/140 = 0.2857 
0.4/1.4 = 0.2857
can
think of as 40/140 = 0.2857 
since x = 0.4 (=
40% dissociated)
It works
out the same in terms of the original 100/100% undissociated
PCl_{5} 
P_{PCl5} =
0.4286 x 210 = 90.0 

P_{PCl3} =
0.2857 x 210 = 60.0 
P_{Cl2} = 0.2857
x 210 = 60 
partial pressure of
component z
P_{z} = mole fraction z x P_{tot}
note 90 + 60 + 60 =
210 check! 

K_{p} =

60.0 x 60.0 
 = 40.0 kPa 
90.0 



Example Q
WHAT NEXT?
Part 2 subindex: 2.1 K_{c} and K_{p} equilibrium expressions and Le
Chatelier's Principle *
2.2 Exemplar calculations using K_{c}
expressions or K_{p} expressions
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principlerules *
Part 3.
Equilibria and industrial processes * Part 4.
Partition,
solubility product and ionexchange * Part 5.
pH, weakstrong acidbase theory and calculations
* Part 6. Salt hydrolysis,
Acidbase titrationsindicators, pH curves and buffers * Part 7.
Redox equilibria, halfcell electrode potentials,
electrolysis and electrochemical series
*
Part 8. Phase equilibriavapour
pressure, boiling point and intermolecular forces
A level Revision notes for GCE Advanced
Subsidiary Level AS Advanced Level A2 IB
Revise AQA GCE Chemistry OCR GCE Chemistry Edexcel GCE Chemistry Salters
Chemistry CIE Chemistry, WJEC GCE AS A2 Chemistry, CCEA/CEA GCE AS A2 Chemistry revising courses for preuniversity students
(equal to US grade 11 and grade 12 and AP Honours/honors level courses)
Website
content copyright © Dr
Phil Brown 20002013 All rights reserved
on
revision notes, images, puzzles, quizzes, worksheets, xwords etc. * Copying of website
material is not permitted
*
chemhelp@tiscali.co.uk
Alphabetical Index for Science
Pages Content
A
B C D
E F
G H I J K L M
N O P
Q R
S T
U V W
X Y Z 