Doc Brown's Chemistry Advanced A Level Notes 
TheoreticalPhysical Advanced Level Chemistry Equilibria Chemical
Equilibrium Revision Notes PART 2
2b. K_{c} chemical equilibrium expression, K_{c}
equilibrium constant and K_{c} equilibrium calculations
email doc
brown  comments  query?
Index of ALL my chemical equilibrium
context revision notes Index
ALL my advanced A
level theoretical
chemistry revision study notes
Use your
mobile phone or ipad etc. in 'landscape' style
This is a BIG
website, you need to take time to explore it [SEARCH
BOX]
How do we write out K_{c} chemical equilibrium
expressions. What is the K_{c} equilibrium constant? How do you do K_{c} equilibrium
calculations using concentrations? All is explained with examples including the units for
concentrations (mol dm^{3}) and
how to solve equilibrium problems using concentration units.
The equilibrium constant K_{c} is deduced from the equation for
a reversible reaction, NOT experimental data as for rate expressions in
kinetics. The concentration, in mol dm^{3}, of a species X
involved in the expression for K_{c} is represented by in square
brackets i.e. [X] The value of the equilibrium constant is not affected
either by changes in concentration or addition of a catalyst. You need
to be able to construct an expression for K_{c} for a homogeneous system in
equilibrium, calculate a value for K_{c} from the equilibrium
concentrations for a homogeneous system at constant temperature, perform
calculations involving K_{c} and predict the qualitative effects
of changes of temperature on the value of K_{c}.
How to
write equilibrium expressions and work out the units of K, the
equilibrium constant
Some 'VERY
rough rules of thumb' for an equilibrium K value
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} 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} is approx. 1
the equilibrium is more evenly distributed between the RHS and LHS.

If K_{c} 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. likely to be less
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'.

Finally a catalyst may
speed up getting to the equilibrium but a catalyst cannot
affect the position of the equilibrium constant or the value of
the equilibrium constant K (K_{c} or K_{p}).

Equilibrium expression
example 2b.1 The formation of hydrogen iodide

Equilibrium expression
example 2b.2 The formation of the ester ethyl ethanoate

Equilibrium
expression
example 2b.3 The formation of phosphorus(V) chloride
(gaseous phase)

Equilibrium expression
example 2b.4 The synthesis of ammonia

Equilibrium expression
example 2b.5 The synthesis of phosphorus(V) fluoride

Equilibrium expression
example 2b.6 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 expression
example 2b.7 The
acid behaviour of high oxidation state hexaaqua ion

The
hexaaquairon(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}.
TOP OF PAGE
2b
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
2b.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
2b.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}) 


 

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

PLEASE NOTE  for UK A level
chemistry courses you do NOT need to solve quadratic equations!

HI/H_{2}/I_{2} 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, K_{c} = K_{p} even if you were working in partial
pressures.

However,
if you were given K instead, and had 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
2b.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 tetrachloroferrate(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
2b.4
Iodineiodide equilibrium in aqueous solution

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 aqueous 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
2b.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. sulfuric acid.

(iii) Not
surprisingly, esters are often prepared by refluxing the acid and
alcohol with a few drops of conc. sulfuric acid added to the
mixture.
advanced level
chemistry how to write equilibrium expressions and do
equilibrium calculations
for AQA AS chemistry, how to write equilibrium expressions and do
equilibrium calculations
for Edexcel A level AS chemistry, how to write equilibrium expressions and
do equilibrium calculations for A level OCR AS chemistry A,
how to write equilibrium expressions and do equilibrium
calculations for OCR Salters AS chemistry B,
how to write equilibrium expressions and do equilibrium
calculations for AQA A level chemistry, how to write
equilibrium expressions and do equilibrium calculations for A level Edexcel A level chemistry,
how to write equilibrium expressions and do equilibrium
calculations for OCR A level chemistry
A, how to write equilibrium expressions and do equilibrium calculations for A level OCR Salters A
level chemistry B how to write equilibrium expressions and do equilibrium
calculations for US Honours grade 11 grade 12 how to write
equilibrium expressions and do equilibrium calculations for
preuniversity chemistry courses preuniversity A level revision
notes for how to write equilibrium expressions and do equilibrium
calculations A level guide
notes on how to write equilibrium expressions and do equilibrium
calculations for schools colleges academies science course tutors images
pictures diagrams for how to write equilibrium expressions and do
equilibrium calculations A level chemistry revision notes on
how to write equilibrium expressions and do equilibrium
calculations for revising module topics notes to help on understanding of
how to write equilibrium expressions and do equilibrium
calculations university courses in science
careers in science jobs in the industry laboratory assistant
apprenticeships technical internships USA US grade 11 grade 11 AQA A
level chemistry
notes on how to write equilibrium expressions and do equilibrium
calculations Edexcel
A level chemistry notes on how to write equilibrium
expressions and do equilibrium calculations for OCR A level chemistry
notes WJEC A level chemistry notes on how to write
equilibrium expressions and do equilibrium calculations CCEA/CEA A level
chemistry notes on how to write equilibrium expressions and
do equilibrium calculations for university entrance examinations
with advanced level chemistry practice questions for solving
equilibrium problems, how do you write Kc equilibrium
expressions? how do you write Kp equilibrium expressions?
what are the units for Kc equilibrium expressions? what are
the units for Kp equilibrium expressions? how to do Kc
equilibrium calculations using molarity concentrations, how
to do Kp equilibrium calculations using partial pressures,
how to use partial pressure and molarities in equilibrium
expressions, explaining the carboxylic acid alcohol ester
water equilibrium experimental procedure, explaining methods
of solving equilibrium problems, solving hydrogen iodide
hydrogen iodine equilibrium problems using the equilibrium
expression, how to solve PCl5 phosphorus(V) chloride PCl3
phosphorus(III) chloride Cl2 chlorine problems calculations
from the equilibrium expression, solving the ammonia
hydrogen nitrogen equilibrium expression, how to do
calculations from the formation of ammonia equilibrium
expression, solving the phosphorus fluorine phosporus(5)
fluoride equilibrium expression units calculations, how to
write equilibrium expressions for complex ion formation, how
to work out units for equilibrium expressions, solving the
nitrogen oxygen nitrogen(II) oxide equilibrium expression
calculations, solving the equilibrium expression for the
contact process, solving the equilibrium expression for the
synthesis of methanol CH3OH from hydrogen H2 and carbon
monoxide CO
explain the effect of temperature on equilibrium constants,
how do you write Kc equilibrium expression? how do you solve
numerical equilibrium problems? how do you write out Kp
equilibrium expression in partial pressures? how do you
solve equilibrium problems using partial pressures? how do
you calculate the Kc equilibrium constant from
concentrations in mol/dm3? how do you calculate Kp
equilibrium constants using partial pressures atm Pa?
WHAT NEXT?
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principlerules
* Part 2. K_{c} and K_{p} equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product K_{sp}, common ion effect,
ionexchange systems *
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 watch out for subindexes
to multiple sections or pages
TOP OF PAGE
Chemical
Equilibrium Notes Index
















Website content © Dr
Phil Brown 2000+. All copyrights reserved on revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries & references to science course specifications
are unofficial. 
Chemical
Equilibrium Notes Index 