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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
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.
Chemical Equilibrium Notes Index
Part 2a. How to
write equilibrium
expressions and the effect of temperature on the equilibrium
constant K
Part 2b. Writing K_{c}
equilibrium expressions, units and exemplar K_{c}
concentration calculations
including esterification
Part 2c.
Writing K_{p}
expressions, units, and exemplar K_{p}
partial pressure equilibrium calculations
Part 3.
Applying Le Chatelier's Principle
and equilibrium expressions to Industrial
Processes
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}) 


 

Examples
(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

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
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.


Part 2a. How to
write equilibrium
expressions and the effect of temperature on the equilibrium
constant K
Part 2b. Writing K_{c}
equilibrium expressions, units and exemplar K_{c}
concentration calculations
including esterification
Part 2c.
Writing K_{p}
expressions, units, and exemplar K_{p}
partial pressure equilibrium calculations
Part 3.
Applying Le Chatelier's Principle
and equilibrium expressions to Industrial
Processes
Chemical Equilibrium Notes Index
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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
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hydrogen iodine equilibrium problems using the equilibrium
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calculations, solving the equilibrium expression for the
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explain the effect of temperature on equilibrium constants,
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Chemical
Equilibrium Notes Index 