Chapter
40
Equilibrium constant
40.1 Equilibrium Law and equilibrium constant
40.2 Equilibrium constants for homogeneous equilibrium
and heterogeneous equilibrium
40.3 What does an equilibrium constant tell us?
40.4 Calculations involving equilibrium constant
40.5 Determination of equilibrium constant from
experiments
Learning goal
After studying this chapter, you should be able to:
40.1-40.3
• express the mathematical relationship between concentrations of reactants and products at
equilibrium and equilibrium constant, Kc
�
• recognize that the value of Kc for an equilibrium system is a constant at constant temperature
irrespective of changes in concentration of reactants and products
40.4
• perform calculations related to equation stoichiometry and Kc by either finding Kc from
equilibrium concentrations or vice versa
40.5
• perform practical work on the determination of Kc
All answers
Chapter
40
Equilibrium constant
Haemoglobin, which is a protein in the red blood cells, transports oxygen from lungs to body tissues.
Red blood cells (× 5000)
The combination of oxygen with haemoglobin is a reversible reaction and can be represented by a
simplified equation:
Hb(aq) + O2(aq)
haemoglobin
oxygen
HbO2(aq)
oxyhaemoglobin
When this reaction reaches equilibrium, the ratio of the concentrations of products to the
[HbO2(aq)]
concentrations of reactants (i.e.
) is a constant.
[Hb(aq)] [O2(aq)]
Think about...
What determines the ratio of the concentrations of products to the concentrations of reactants
at equilibrium?
Does this ratio remain constant no matter what the initial concentrations of reactants or products
are?
After studying this chapter, you should be able to answer the above questions.
dynamic equilibrium 動態平衡
reversible reaction 可逆反應
haemoglobin 血紅蛋白
oxyhaemoglobin 氧合血紅蛋白
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40
Equilibrium constant
40.1 Equilibrium Law and equilibrium
constant
Defining
constant
Equilibrium
Law
and
equilibrium
In the previous chapter, we have learnt that the concentrations of the
reactants and products remain unchanged when the reaction reaches
equilibrium.
In the mid-nineteenth century, two Norwegian chemists, Cato
Maximilian Guldberg and Peter Waage (Figure 40.1), conducted detailed
studies of many equilibrium systems. They noticed that for a reversible
reaction, the ratio of the concentrations of reactants and products at
equilibrium was a constant at a particular temperature. It was true no
matter what the initial concentrations of the reactants were.
Figure 40.1 Cato Maximilian
Guldberg (1836–1902) (left)
and Peter Waage (1833–1900)
(right). They were brothers-inlaw.
Take the following reaction as an example,
aA + bB
cC + dD
the relationship between the concentrations of reactants and products at
equilibrium can be written as:
Kc =
c
d
a
b
[C] eqm[D] eqm
[A] eqm[B] eqm
where
•
[A]eqm, [B]eqm, [C]eqm and [D]eqm are the equilibrium concentrations
–3
(in mol dm ) of the substances within the brackets (i.e. their
concentrations at equilibrium)
•
a, b, c and d are the stoichiometric coefficients in the balanced
equation.
Learning tip
The value of equilibrium
constant will not change
as long as the
temperature remains
constant.
equilibrium concentration 平衡濃度
equilibrium constant 平衡常數
The above mathematical relationship between the equilibrium
concentrations of reactants and products is known as the Equilibrium
Law. Kc is called the equilibrium constant. It is a ratio with a constant
value at a given temperature. The subscript ‘c’ indicates that it is
expressed in concentrations.
Equilibrium Law 平衡定律
stoichiometric coefficient 計量系數
3
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Chemical equilibrium
Equilibrium constant — a constant value at a
given temperature
Consider the following reaction:
N2(g) + 3H2(g)
2NH3(g)
Three experiments were done by using N2, H2 and NH3 of different
initial concentrations. They were allowed to reach equilibrium at 500°C.
To calculate the value of Kc, we simply substitute the corresponding
equilibrium concentrations into the following expression:
2
Kc =
Experiment
Initial concentration
–3
(mol dm )
[NH3(g)]eqm
3
[N2(g)]eqm[H2(g)]eqm
2
[NH3(g)]eqm
Kc =
3
[N2(g)]eqm[H2(g)]eqm
Equilibrium concentration
–3
(mol dm )
–3 2
(0.16 mol dm )
[N2(g)] = 1.00
[H2(g)] = 1.00
[NH3(g)] = 0.00
[N2(g)]eqm = 0.92
[H2(g)]eqm = 0.76
[NH3(g)]eqm = 0.16
2
[N2(g)] = 0.00
[H2(g)] = 0.00
[NH3(g)] = 1.00
[N2(g)]eqm = 0.40
[H2(g)]eqm = 1.20
[NH3(g)]eqm = 0.20
Kc = 0.06 mol dm
3
[N2(g)] = 2.00
[H2(g)] = 1.00
[NH3(g)] = 3.00
[N2(g)]eqm = 2.59
[H2(g)]eqm = 2.77
[NH3(g)]eqm = 1.82
Kc = 0.06 mol dm
1
Kc =
–3
–3 3
(0.92 mol dm )(0.76 mol dm )
–2
6
= 0.06 mol dm
–2
6
–2
6
Table 40.1 Different initial concentrations and equilibrium concentrations of reactants and products of the reaction
N2(g) + 3H2(g)
2NH3(g) at 500°C. (Kc values are calculated.)
Table 40.1 shows that the initial concentrations of N2, H2 and NH3 in
the three experiments were different. Eventually, three different sets of
equilibrium concentrations of N2, H2 and NH3 were obtained. However, it
was found that the equilibrium constant is a constant value at a given
temperature, regardless of the initial concentrations.
Key point
The equilibrium constant, Kc, is a constant value for a reaction at
equilibrium at a given temperature, no matter what the initial
concentrations of the reactants or products are.
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All answers
40
Equilibrium constant
Skill corner 40.1
Writing the expression for equilibrium constant, Kc
Consider the following reaction:
2SO2(g) + O2(g)
2SO3(g)
The expression for the equilibrium constant, Kc, of the above reaction can be written as follows:
3
1 [Product] term is written
as the numerator.
a The indices are written according to the
stoichiometric coefficients of the
corresponding substances in the
balanced equation.
2
[SO3(g)]eqm
Kc =
2
[SO2(g)]eqm[O2(g)]eqm
b Do not put the stoichiometric
coefficients inside the square brackets.
2 [Reactant] terms are written
as the denominators.
H20
4 Include the state symbols
in the expression.
Example 40.1
Writing the expression for the equilibrium constant
Write the expression for the equilibrium constant, Kc, for the following reactions.
(a) N2(g) + O2(g)
2NO(g)
(b) P4O10(g) + 6PCl5(g)
3+
10POCl3(g)
–
2+
(c) Fe (aq) + SCN (aq)
FeSCN (aq)
Solution
2
(a) Kc =
[NO(g)]eqm
[N2(g)]eqm[O2(g)]eqm
10
(b) Kc =
[POCl3(g)] eqm
6
[P4O10(g)]eqm[PCl5(g)] eqm
2+
(c) Kc =
[FeSCN (aq)]eqm
3+
–
[Fe (aq)]eqm[SCN (aq)]eqm
Self-test 40.1
Write the expression for the equilibrium constant, Kc, for the following reactions.
(a) 2NO(g)
N2(g) + O2(g)
(b) 2SO3(g) + CO2(g)
CS2(g) + 4O2(g)
(c) CH3COOH() + CH3OH()
CH3COOCH3() + H2O()
Try Chapter Exercise Q8
denominator 分母
numerator 分子
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Chemical equilibrium
All answers
Class practice 40.1
1.
Find the mistakes in the expression for the equilibrium constant written
for the following reactions and correct them.
(a) 2H2S(g)
2H2(g) + S2(g)
(b) CO(g) + Cl2(g)
2.
[H2(g)]eqm[S2(g)]eqm
Kc =
COCl2(g)
Kc =
[H2S(g)]eqm
[CO(g)]eqm[Cl2(g)]eqm
[COCl2(g)]eqm
Write the balanced chemical equation that would give the following
expressions for equilibrium constants.
2
(a) Kc =
[NO2(g)]eqm
2
[NO(g)]eqm[O2(g)]eqm
3
(b) Kc =
[C2H2(g)]eqm[H2(g)]eqm
2
[CH4(g)]eqm
Relationship between equilibrium constant, Kc,
and chemical equation
The equation of an equilibrium reaction can be written in different ways.
The expression for the equilibrium constant, Kc, must be written according
to the particular equation concerned.
For example, three different equations representing the equilibrium
reaction that involves H2(g), I2(g) and HI(g) are shown below:
H2(g) + I2(g)
2HI(g)..................... Equation 1
2HI(g)
H2(g) + I2(g)..................... Equation 2
1
1
H2(g) + I2(g)
HI(g)................Equation 3
2
2
The equilibrium constants Kc, Kc’ and Kc” for Equations 1, 2 and 3
are:
2
Equation 1:
Kc =
Equation 2:
Kc’ =
Equation 3:
Kc’’ =
respectively.
40
6
[HI(g)]eqm
[H2(g)]eqm[I2(g)]eqm
[H2(g)]eqm[I2(g)]eqm
2
[HI(g)]eqm
[HI(g)]eqm
1
2
1
2
[H2(g)]eqm[I2(g)]eqm
All answers
Learning tip
The equilibrium
constant for the
backward reaction is
the reciprocal of the
equilibrium constant for
the forward reaction.
40
Equilibrium constant
It can be seen that when an equation is reversed (as shown by
Equations 1 and 2), Kc’ is the reciprocal of Kc.
i.e. Kc’ =
1
Kc
Furthermore, if an equation is halved (as shown by Equations 1 and 3),
Kc” is the square root of Kc.
i.e. Kc” =
Kc
Units of equilibrium constant
The units of equilibrium constant, Kc, depend on the powers of the
concentration terms in the expression. For the H2–I2–HI system, the
concentration units cancel out in the expression for Kc. Therefore, the Kc
has no units.
H2(g) + I2(g)
2HI(g)
2
Kc =
[HI(g)]eqm
[H2(g)]eqm[I2(g)]eqm
–3 2
The unit of Kc is
(mol dm )
–3
–3
(mol dm )(mol dm )
( no units)
Consider another example below:
2SO2(g) + O2(g)
2SO3(g)
2
Learning tip
Never try to remember
the unit of Kc for an
equilibrium reaction.
Work it out yourself.
Kc =
[SO3(g)]eqm
2
[SO2(g)]eqm[O2(g)]eqm
–3 2
The unit of Kc is
(mol dm )
–3 2
–1
–3
(mol dm ) (mol dm )
3
i.e. mol dm
Class practice 40.2
(a) Write the expressions for the equilibrium constants (Kc1 and Kc2) for the
following reactions:
(i)
N2(g) + 3H2(g)
(ii) 2NH3(g)
2NH3(g)
Kc1 = ?
N2(g) + 3H2(g)
Kc2 = ?
(b) What is the mathematical relationship between Kc1 and Kc2?
(c) What are the units of Kc1 and Kc2 respectively?
halve 減半
reciprocal 倒數
square root 平方根
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Chemical equilibrium
40.2 Equilibrium constants for
homogeneous equilibrium and
heterogeneous equilibrium
Homogeneous equilibrium
An equilibrium system in which all the reactants and products are in
the same phase is called a homogeneous equilibrium. For example,
the H2–I2–HI system is an example of homogeneous equilibrium because
all substances are in the gaseous phase.
H2(g) + I2(g)
2HI(g)
The expression for the equilibrium constant is shown below:
2
Kc =
[HI(g)]eqm
[H2(g)]eqm[I2(g)]eqm
Heterogeneous equilibrium
An equilibrium system in which two or more phases are present is a
Learning tip
Different phases are
separated by a distinct
phase boundary.
e.g. 1 solid and 1 gas — 2
phases
air
solid
phase
boundary
But different phases do
not necessarily mean
different physical states.
e.g. 2 immiscible liquids
— 2 phases
heterogeneous equilibrium. The expression for the equilibrium constant
in a heterogeneous equilibrium needs more considerations.
Let us consider the thermal decomposition of solid sodium
hydrogencarbonate. At 200°C, it rapidly decomposes to solid sodium
carbonate, carbon dioxide and steam. This reaction is reversible and can
reach an equilibrium in a closed container. The substances involved in
this equilibrium are not in the same phase (Figure 40.2).
2NaHCO3(s)
Na2CO3(s) + CO2(g) + H2O(g)
H2O
phase
boundary
water
Figure 40.2
A mixture of
NaHCO3(s), Na2CO3(s), CO2(g)
and
H2O(g)
forms
a
heterogeneous equilibrium.
40
8
CO2
oil
heterogeneous equilibrium 多相平衡
homogeneous equilibrium 均相平衡
a mixture of NaHCO3(s)
and Na2CO3(s)
phase boundary 相界
a mixture of gaseous products,
CO2(g) and H2O(g)
immiscible 不互溶的
All answers
40
Equilibrium constant
The equilibrium constant of the reaction is
[Na2CO3(s)]eqm[CO2(g)]eqm[H2O(g)]eqm
Kc’ =
2
[NaHCO3(s)]eqm
(Note: For reasons that will be apparent shortly, we have temporarily
indicated the equilibrium constant as Kc’.)
The ‘concentration’ of a pure solid is a constant. This does not depend
Think about
The ‘concentration’ of
a pure solid or liquid
amount (in mol)
=
3
volume (in dm )
Do you know why the
‘concentration’ of a
pure solid or liquid is
a constant?
on how much of the substance is present. For this reason, [NaHCO3(s)]eqm
and [Na2CO3(s)]eqm in the expression of Kc’ can be incorporated into the
equilibrium constant and the expression can thus be simplified.
2
Kc’
[NaHCO3(s)]eqm
[Na2CO3(s)]eqm
= [CO2(g)]eqm[H2O(g)]eqm
or Kc = [CO2(g)]eqm[H2O(g)]eqm
2
(
[NaHCO3(s)]eqm
is a constant.)
[Na2CO3(s)]eqm
Similarly, in heterogeneous equilibria in which a reactant or a product
is a pure liquid, the ‘concentration’ of that liquid is also a constant.
Therefore, the ‘concentrations’ of pure solids and liquids in
heterogeneous equilibria do not appear in the expression of Kc.
Key point
An equilibrium in which all the reactants and products are in the
same phase is a homogeneous equilibrium.
An equilibrium in which two or more phases are present is a
heterogeneous equilibrium.
Class practice 40.3
Which of the following reactions are heterogeneous equilibria? Write an
expression for the equilibrium constant, Kc, for each of these heterogeneous
equilibria:
(a) CaCO3(s)
CaO(s) + CO2(g)
+
–
(b) AgCl(s)
Ag (aq) + Cl (aq)
(c) H2O()
H (aq) + OH (aq)
2–
+
(d) Cr2O7 (aq) + H2O()
–
2–
+
2CrO4 (aq) + 2H (aq)
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Chemical equilibrium
40.3 What does an equilibrium constant
tell us?
Equilibrium constant and extent of reaction
An equilibrium constant tells us about the composition of the reaction
mixture. It provides a useful indication of the extent of a chemical
reaction i.e. how far a reaction proceeds to completion. Let us look at
two examples.
1. The reaction of hydrogen with chlorine at 25°C:
H2(g) + Cl2(g)
2HCl(g)
2
Kc =
[HCl(g)]eqm
[H2(g)]eqm[Cl2(g)]eqm
32
= 4.4 × 10
The very large value of Kc tells us that most of the reactants are
converted into products at equilibrium. The reaction proceeds almost
to completion.
2. The decomposition of water vapour at 25°C:
2H2O(g)
2
Kc =
2H2(g) + O2(g)
[H2(g)]eqm[O2(g)]eqm
2
eqm
[H2O(g)]
–81
= 1.1 × 10
–3
mol dm
The very small value of Kc tells us that only very small amounts of
products form at equilibrium. The reaction hardly occurs at all.
Table 40.2 below shows the relationship between the magnitude of an
equilibrium constant and the extent to which a reaction proceeds to
completion.
Value of Kc
Extent of reaction
Very large Kc
Most of the reactants are converted into products at
equilibrium. The reaction proceeds almost to completion.
Very small Kc
The reaction hardly occurs at all. Only very small
amounts of products form at equilibrium.
Table 40.2 Relationship between the magnitude of an equilibrium constant and the extent to
which a reaction proceeds to completion.
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All answers
40
Equilibrium constant
Class practice 40.4
Arrange the following reactions in order of their increasing tendency to proceed
towards completion.
(a) 4NH3(g) + 3O2(g)
(b) N2(g) + O2(g)
(c) 2HF(g)
(d) 2NOCl(g)
2N2(g) + 6H2O(g)
2NO(g)
H2(g) + F2(g)
2NO(g) + Cl2(g)
228
Kc = 1.0 × 10
–3
mol dm
–31
Kc = 5.0 × 10
–13
Kc = 1.0 × 10
–4
–3
Kc = 4.7 × 10 mol dm
Equilibrium constant and rate of reaction
Is it possible to know the rate of a chemical reaction from the magnitude
of the equilibrium constant, Kc? Let us take a look at the formation of
water vapour from hydrogen and oxygen at 25°C,
2H2(g) + O2(g)
2H2O(g)
80
–1
3
Kc = 9.09 × 10 mol dm
Almost all reactants react to form the product at equilibrium.
However, the reaction reaches equilibrium very slowly at 25°C. Although
the Kc for this reaction is very large, it tells us nothing about the rate of
the reaction.
Key point
The magnitude of Kc indicates the extent (how far) of a chemical
reaction at equilibrium, but not the rate (how fast) of the reaction.
40.4 Calculations involving equilibrium
e-Learning
Flipped classroom
Determining graphically
the equilibrium
constant of a reaction
constant
Determining equilibrium constant, Kc, from
equilibrium concentrations of reactants and
products
Equilibrium constant, Kc, of a reaction can be calculated when the
equilibrium concentrations of the substances involved in the reaction are
https://e-aristo.hk/r/
cm2fc40i01.e
known.
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Chemical equilibrium
H20
All answers
Example 40.2
Determining equilibrium constant from equilibrium concentrations of reactants and products
Consider the following equilibrium involving two nitrogen oxides.
2NO(g) + O2(g)
2NO2(g)
(a) Write the expression for the equilibrium constant for the above reaction.
(b) Calculate the equilibrium constant, Kc, from the given equilibrium concentrations at 25°C.
–11
[NO(g)]eqm = 1.5 × 10
–3
mol dm
–3
–3
[O2(g)]eqm = 8.9 × 10 mol dm
–6
–3
[NO2(g)]eqm = 2.2 × 10 mol dm
Solution
2
(a) Kc =
[NO2(g)]eqm
2
[NO(g)]eqm[O2(g)]eqm
(b) By substituting the equilibrium concentration values into the expression for Kc,
–6
Kc =
–3 2
(2.2 × 10 mol dm )
–11
(1.5 × 10
12
–3 2
–3
Think about
–3
mol dm ) (8.9 × 10 mol dm )
–1
–1
3
Why is the unit of Kc in mol dm ?
3
= 2.4 × 10 mol dm
Self-test 40.2
3
A mixture of nitrogen and chlorine gases was kept in a 5 dm reaction flask at a certain temperature.
N2(g) + 3Cl2(g)
2NCl3(g)
The equilibrium mixture was found to contain 0.0070 mol of N2(g), 0.0022 mol of Cl2(g) and 0.9500
mol of NCl3(g). Calculate the equilibrium constant for this reaction.
e-Learning
Flipped classroom
Determining equilibrium
constant from changes
in concentrations
Determining equilibrium constant from changes
in concentrations
If the initial concentration and equilibrium concentration of a reactant are
known, we can find the equilibrium concentrations of other reactant(s)
and the product(s). As a result, we can calculate the equilibrium constant.
https://e-aristo.hk/r/
cm2fc40i02.e
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40
Equilibrium constant
Problem-solving strategy 40.1
Determining equilibrium constant from changes in concentrations
The colourless gas N2O4 and the brown gas NO2 exist
in equilibrium as shown by the following equation:
N2O4(g)
2NO2(g)
In an experiment, 0.80 mol of N2O4 was introduced
3
into a 5 dm container and allowed to reach equilibrium
with NO2 at a certain temperature. At equilibrium, the
–3
concentration of N2O4 was 0.09 mol dm . Calculate
the Kc for this reaction.
1
Try it now
3
A 1 dm container contains 1.00 mol of
phosgene, which decomposes at a certain
temperature according to the following
equation:
COCl2(g)
CO(g) + Cl2(g)
(a) At equilibrium, the concentration of Cl2 is
–3
0.028 mol dm . What are the concentrations
of CO and COCl2 respectively?
(b) Calculate the equilibrium constant.
Calculate the initial concentrations, if needed.
0.80 mol
3
5 dm
–3
= 0.16 mol dm
Initial concentration of N2O4(g) =
2
Set up an ICE (initial, change, equilibrium)
table of concentrations. Record the given
concentrations
and
determine
other
concentrations.
Consider the equilibrium,
Concentration
–3
(mol dm )
N2O4(g)
0.16
0
0.09 – 0.16
= –0.07
2 × (+0.07) = +0.14
(equilibrium conc.
1 mol of N2O4
– initial conc.)
decomposes.)
0.09
0.14
Initial
Change
Equilibrium
3
2NO2(g)
(2 mol of NO2 forms as
Write the expression for equilibrium constant.
2
Kc =
4
[NO2(g)]eqm
[N2O4(g)]eqm
Substitute the equilibrium concentrations into
the expression for equilibrium constant.
–3 2
Kc =
(0.14 mol dm )
–3
0.09 mol dm
–3
= 0.218 mol dm
phosgene 光氣
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Chemical equilibrium
All answers
Class practice 40.5
1.
COCl2 decomposes to CO and Cl2 according to the following equation:
COCl2(g)
CO(g) + Cl2(g)
–6
–3
At 900°C, the equilibrium concentration of COCl2 is 3.0 × 10 mol dm
–4
–3
while those of CO and Cl2 are both 5.0 × 10 mol dm . Calculate the
value of Kc.
2.
3
A mixture of 2.5 mol H2(g) and 5.0 mol I2(g) is placed in a 5 dm container
at 450°C and allowed to reach equilibrium with HI(g). At equilibrium, the
–3
concentration of HI(g) is 0.934 mol dm . Calculate the equilibrium
constant for the reaction.
Determining equilibrium concentrations from
initial concentrations and the equilibrium
constant
We can also use the equilibrium constant to find the equilibrium
concentrations of reactants and products.
Problem-solving strategy 40.2
Determining equilibrium concentrations from initial concentrations and the equilibrium constant
Try it now
At a given temperature, the equilibrium constant, Kc,
–1
3
for the following reaction is 137.5 mol dm .
3+
–
At 450°C, the equilibrium constant for the
following reaction is 50.0.
2+
Fe (aq) + SCN (aq)
FeSCN (aq)
3
H2(g) + I2(g)
–3
In an experiment, 500 cm of 0.200 mol dm
3
–3
Fe(NO3)3(aq) was mixed with 500 cm of 0.200 mol dm
KSCN(aq) at the above given temperature. Calculate
the equilibrium concentrations of all species in the
mixture.
1
2HI(g)
In an experiment, 0.15 mol of H2 and 0.15 mol
3
of I2 were placed in a 5 dm container. The
mixture was allowed to react at 450°C. Calculate
the concentration of each substance at
equilibrium.
Calculate the initial concentrations, if needed.
After mixing the two solutions,
3+
initial concentration of Fe (aq)
–3
3
3
500
(500 + 500)
= 0.200 mol dm ×
dm ÷
dm
1000
1000
–3
= 0.100 mol dm
–
initial concentration of SCN (aq)
–3
3
3
500
(500 + 500)
= 0.200 mol dm ×
dm ÷
dm
1000
1000
–3
= 0.100 mol dm
cont’d
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40
Equilibrium constant
Set up an ICE table of concentrations.
• Record the initial concentrations.
• Let the change in concentration of one of the
reactants be x.
• Based on the stoichiometric coefficients of
the equation, work out the equilibrium
concentrations.
–3
Let x mol dm be the change in concentration of
3+
Fe (aq).
Concentration Fe3+(aq) + SCN–(aq)
–3
(mol dm )
3
2+
FeSCN (aq)
Initial
0.100
0.100
0
Change
–x
–x
+x
Equilibrium
0.100 – x
0.100 – x
0+x=x
Write the expression for equilibrium constant.
2+
Kc =
4
[FeSCN (aq)]eqm
3+
–
[Fe (aq)]eqm[SCN (aq)]eqm
Substitute the equilibrium concentrations into
the expression for equilibrium constant.
x
137.5 =
(0.100 – x)(0.100 – x)
x
137.5 =
2
(0.100 – x)
Solving for x,
2
137.5 × (0.01 – 0.2x + x ) = x
x = 0.0764 or x = 0.1308 (rejected)
 equilibrium concentrations:
3+
–3
[Fe (aq)]eqm = (0.100 – 0.0764) mol dm
–3
= 0.0236 mol dm
–
–3
[SCN (aq)]eqm = (0.100 – 0.0764) mol dm
–3
= 0.0236 mol dm
2+
–3
[FeSCN (aq)]eqm = 0.0764 mol dm
Class practice 40.6
1.
Consider the following reaction:
2CH4(g)
C2H2(g) +3H2(g)
A mixture of CH4(g), C2H2(g) and H2(g) was allowed to reach equilibrium at
1700°C. The equilibrium concentrations of CH4(g), C2H2(g) and H2(g) were
–3
–3
–3
0.0203 mol dm , 0.0451 mol dm and 0.1120 mol dm respectively.
Calculate the equilibrium constant at that temperature.
(Cont’d)
15
40
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Chemical equilibrium
2.
All answers
A mixture of CO(g) and H2(g) was allowed to reach equilibrium at a
certain temperature:
CO(g) + 2H2(g)
CH3OH(g)
–3
–3
The initial concentrations of CO and H2 were 0.50 mol dm and 1.00 mol dm
–3
respectively. At equilibrium, the concentration of CO was 0.15 mol dm .
Calculate the equilibrium constant at this temperature.
3.
3
4.00 mol of HI(g) were placed in an evacuated 5.00 dm flask and then
heated to 800 K. The system was allowed to reach equilibrium.
2HI(g)
H2(g) + I2(g)
Kc = 0.016 (at 800 K)
Calculate the equilibrium concentration of each species.
40.5 Determination of equilibrium
constant from experiments
We can perform simple experiments in the school laboratory to determine
the equilibrium constants of some chemical reactions. An example is the
esterification between ethanol and ethanoic acid in the presence of
concentration sulphuric acid as a catalyst.
conc. H2SO4
CH3CH2OH() + CH3COOH()
ethanol
ethanoic acid
CH3COOCH2CH3() + H2O()
ethyl ethanoate
water
The procedure for determining the equilibrium constant, Kc, of this
esterification reaction is shown below:
Step 1: Mix known amounts of ethanol, ethanoic acid and a small but
known amount of concentrated sulphuric acid.
Step 2: Heat the reaction mixture under reflux for at least 30 minutes so
that the mixture reaches an equilibrium. See Figure 40.3.
Step 3: Determine the equilibrium concentration of ethanoic acid by
titrating a known amount of the equilibrium mixture against
standard sodium hydroxide solution. See Figure 40.4.
Learning tip
The equilibrium system is
a homogeneous
equilibrium, so [H2O()]eqm
should be counted in the
calculation of Kc.
Step 4: Based on the balanced equation, calculate the equilibrium
concentrations of ethanol, ethyl ethanoate and water.
Step 5: Calculate the equilibrium constant, Kc, by substituting the
equilibrium concentrations of all species into the expression:
Kc =
40
16
esterification 酯化作用
reflux 回流
[CH3COOCH2CH3()]eqm[H2O()]eqm
[CH3CH2OH()]eqm[CH3COOH()]eqm
TE
40
standard
sodium
hydroxide
solution
water out
Learning tip
Before determining the
equilibrium concentrations
of the substances
involved, we have to stop
the chemical reaction. This
can be done by cooling
the equilibrium mixture in
an ice bath.
reflux
condenser
water in
equilibrium
mixture of
esterification
a mixture of ethanol
and ethanoic acid
and a few drops of
concentrated
sulphuric acid
water bath
Figure 40.3 Heating a mixture of
ethanol and ethanoic acid under
reflux until an equilibrium is
reached.
H20
Equilibrium constant
Figure 40.4 Titrating the equilibrium
mixture of esterification against
standard sodium hydroxide solution
(in burette) in order to determine the
equilibrium
concentration
of
ethanoic acid.
Example 40.3
Determining the equilibrium constant of esterification between ethanol and ethanoic acid
An experiment was carried out to determine the equilibrium constant, Kc, of an esterification reaction.
The following steps were used.
Step
Procedure
1
Mix 0.25 mol of ethanoic acid and 0.25 mol of ethanol in a dry, clean pear-shaped flask in
ice-water.
2
Pipette 1.0 cm of the mixture to 25 cm distilled water in a conical flask. Immediately titrate it
3
against 0.50 M sodium hydroxide solution. Record the volume (V1 cm ) of alkali used.
3
Add a few drops of concentrated sulphuric acid to the reaction mixture and pipette 1.0 cm of
3
the reaction mixture to 25 cm distilled water in a conical flask. Immediately titrate it against
3
0.50 M sodium hydroxide solution. Record the volume (V2 cm ) of alkali used.
4
Add a few anti-bumping granules to the reaction mixture in the pear-shaped flask and heat the
content under reflux using a hot water bath for about 30 minutes.
5
Rapidly cool the pear-shaped flask in ice-water. Pipette 1.0 cm of the reaction mixture to
3
25 cm distilled water in a conical flask. Immediately titrate it against 0.50 M sodium hydroxide
3
solution. Record the volume (V3 cm ) of alkali used.
3
3
3
3
The titration results are tabulated below:
Titration
3
Volume of 0.50 M NaOH(aq) used (cm )
1
2
3
V1 = 15.10
V2 = 15.80
V3 = 7.15
cont’d
17
40
X
Chemical equilibrium
(a) Write a chemical equation for the esterification reaction.
(b) State ONE reason of immersing the reaction flask in ice-water when ethanoic acid and ethanol are
mixed in Step 1.
(c) (i) What is the purpose of adding a few drops of concentrated sulphuric acid to the reaction
mixture in Step 3?
(ii) Explain why the titration in Step 3 should be carried out immediately.
(d) Calculate
(i) the concentrations of ethanoic acid and ethanol in the original mixture.
(ii) the equilibrium concentration of ethanoic acid.
(iii) the equilibrium constant, Kc, under the experimental conditions.
(e) Suggest what further actions should be taken after Step 5 to make sure that equilibrium had been
attained.
Solution
(a) CH3CH2OH() + CH3COOH()
CH3COOCH2CH3() + H2O()
(b) To stop or slow down the esterification reaction at the start by lowering the temperature, so that
the initial concentration of ethanoic acid in the mixture can be found by titration.
(c) (i) Concentrated sulphuric acid acts as a catalyst, which can speed up the esterification reaction.
(ii) In the presence of concentrated sulphuric acid, the esterification reaction will proceed rapidly.
If the titration is not carried out immediately, some of the ethanoic acid will react with ethanol
to give the ester.
(d) (i) From Step 2,
NaOH(aq) + CH3COOH(aq)
3
CH3COONa(aq) + H2O()
3
15.10 cm
1.0 cm
0.50 M
?M
3
Number of moles of CH3COOH in 1.0 cm of the original mixture
= number of moles of NaOH used
–3
= 0.50 mol dm ×
–3
= 7.55 × 10 mol
[CH3COOH()] =
15.10
3
dm
1000
–3
7.55 × 10 mol
3
1.0
dm
1000
= 7.55 mol dm
–3
As the number of moles of CH3CH2OH used in the experiment was the same as that of
CH3COOH,
–3
[CH3CH2OH()] = 7.55 mol dm
3
3
(ii) The titration results V1 and V2 indicated that (15.80 – 15.10) cm = 0.70 cm NaOH(aq) was
used to neutralize the concentrated sulphuric acid.
�  volume of NaOH(aq) used to neutralize the CH3COOH after reflux
3
3
= (7.15 – 0.70) cm = 6.45 cm
cont’d
40
18
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All answers
40
Equilibrium constant
3
Number of moles of CH3COOH in 1.0 cm of the equilibrium mixture
= number of moles NaOH used
6.45
–3
3
= 0.50 mol dm ×
dm
1000
–3
= 3.225 × 10 mol
–3
3.225 × 10 mol
[CH3COOH()]eqm =
3
1.0
dm
1000
–3
= 3.225 mol dm
(iii) Consider the following equilibrium:
–3
Concentration (mol dm )
Kc =
CH3CH2OH() + CH3COOH()
CH3COOCH2CH3() + H2O()
Initial
7.55
7.55
0
0
Change
–4.325
–4.325
+4.325
+4.325
Equilibrium
3.225
3.225
4.325
4.325
[CH3COOC2H5()]eqm[H2O()]eqm
[C2H5OH()]eqm[CH3COOH()]eqm
=
–3 2
(4.325 mol dm )
–3 2
(3.225 mol dm )
= 1.80
(e) Repeat Step 5 of the experiment until the volumes of NaOH(aq) used in two consecutive titrations
are about the same.
Experiment 40.1
Experiment Workbook 4A
Determining the equilibrium constant of esterification between ethanol and
ethanoic acid
In this experiment, you are going to design and perform an experiment for
determining the equilibrium constant of the esterification reaction between
ethanol and ethanoic acid.
Class practice 40.7
The equilibrium constant, Kc, for the following reaction is 4.0 at a certain
temperature.
CH3COOH() + CH3CH2OH()
CH3COOCH2CH3() + H2O()
(a) Write an expression for the equilibrium constant, Kc’, of the backward
reaction, i.e. the hydrolysis of ethyl ethanoate. Hence, calculate the
equilibrium constant for the hydrolysis of ethyl ethanoate at that
temperature.
(b) In an experiment, one mole of ethyl ethanoate and one mole of water are
mixed at the above temperature. Calculate the number of moles of each
species in the equilibrium mixture.
19
40
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Chemical equilibrium
All answers
Key terms
English term
Chinese translation
Page
1.
equilibrium constant
平衡常數
3
2.
Equilibrium Law
平衡定律
3
3.
heterogeneous equilibrium
多相平衡
8
4.
homogeneous equilibrium
均相平衡
8
Progress check
Can you answer the following questions? Put a ‘✓’ in the box if you can. Otherwise, review the relevant
part on the page as shown.
Page
40
20
1.
What are Equilibrium Law and the equilibrium constant, Kc?
3
2.
What is a homogeneous equilibrium?
8
3.
What is a heterogeneous equilibrium?
8
4.
In a heterogeneous equilibrium, why are the ‘concentrations’ of pure solids and liquids
not included in the equilibrium constant expression?
9
5.
How does the magnitude of Kc relate to the extent of chemical reaction?
10
6.
Is there any relationship between the Kc value and the rate of reaction?
11
7.
What are the essential steps to determine Kc experimentally?
16
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40
Equilibrium constant
Summary
40.1 Equilibrium Law and equilibrium constant
1.
For a reversible reaction, aA + bB
cC + dD, there is an expression written as
Kc =
c
d
a
b
[C]eqm[D]eqm
[A]eqm[B]eqm
at a given temperature where Kc is called the equilibrium constant. This is known as the
Equilibrium Law.
2.
The equilibrium constant, Kc, is a constant value for a reaction at equilibrium at a given
temperature, no matter what initial concentrations of the reactants or products are. Kc is
temperature dependent.
3.
Numerical value and unit of equilibrium constant, Kc, depend on how the equation of an
equilibrium reaction is written.
40.2 Equilibrium constants for homogeneous equilibrium and heterogeneous equilibrium
4.
An equilibrium in which all substances are in the same phase is referred to as a homogeneous
equilibrium.
5.
An equilibrium in which two or more phases are present is referred to as a heterogeneous
equilibrium.
6.
For a heterogeneous equilibrium, ‘concentrations’ of pure solids and liquids remain constant
throughout the reaction, so they do not appear in the expression for Kc.
40.3 What does an equilibrium constant tell us?
7.
A very large value of Kc tells us that the reaction proceeds almost to completion. A very small
value of Kc tells us that the reaction hardly occurs.
8.
The magnitude of the equilibrium constant of a reaction provides information about the extent
of a reaction at equilibrium, but gives no information about the rate of the reaction.
40.4 Calculations involving equilibrium constant
9.
To find the numerical value of Kc, it is essential to find the equilibrium concentrations of all
substances and substitute them into the equilibrium constant expression.
10.
Equilibrium constants can be used to calculate the equilibrium concentrations of the substances,
provided that their initial concentrations are known.
40.5 Determination of equilibrium constant from experiments
11.
To determine the equilibrium constant, it is essential to find the equilibrium concentrations of the
substances by performing experiments.
21
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Chemical equilibrium
All answers
Concept map
Complete the following concept map.
Equilibrium Law
EQUILIBRIUM CONSTANT (Kc )
Unit of Kc: depends
on how the
is written
Kc:
at constant
temperature
equilibrium
equilibrium
Kc includes
equilibrium
concentrations
of all substances
Kc does not
include equilibrium
‘concentrations’ of
pure solids and liquids
Kc indicates the
of a chemical reaction
(but not the
of reaction)
reaction
proceeds
almost to
completion
Kc value
(Hints: constant, equation, extent, heterogeneous, homogeneous, large, rate, small)
40
22
reaction
hardly
occurs
Kc value
All answers
40
Equilibrium constant
Chapter exercise
Fill in the blanks
4.
Section 40.1
1.
For the general equilibrium reaction:
aA + bB
Kc =
[C]eqm
2.
and
cC + dD
where Kc is called the
, while [A]eqm, [B]eqm,
[D]eqm are
the
equilibrium
of A, B, C and D respectively.
Kc is a constant value at a given
no matter what
concentrations
of the reactants or products are.
Section 40.2
3.
H
equilibrium is the one in which
two or more phases are present. The
of pure solids and liquids remain constant
throughout the reaction and they do not appear
in the expression of Kc.
Section 40.3
5.
The magnitude of the equilibrium constant, Kc,
indicates the
of the chemical
reaction when the state of equilibrium is reached,
but not the
of the reaction.
6.
A very
Kc value indicates that the
reaction proceeds almost to completion. A very
Kc value indicates that the
reaction hardly occurs.
H
equilibrium is the one in which
all the reactants and products are in the same
phase.
Practice questions
Section 40.1
7.
For each of the following equilibrium constant expressions, write the balanced chemical equation for the
reaction involved.
2
(a) Kc =
(b) Kc =
[HF(g)]eqm[Cl2(g)]eqm
2
[HCl(g)]eqm[F2(g)]eqm
[CH4(g)]eqm[H2O(g)]eqm
3
[CO(g)]eqm[H2(g)]eqm
2
(c) Kc =
[SO3(g)]eqm[CO2(g)]eqm
4
[CS2(g)]eqm[O2(g)]eqm
Section 40.2
8.
Write the expression for the equilibrium constant, Kc, for each of the following reversible reactions.
(a) 4NH3(g) + 5O2(g)
(b) CO(g) + H2O(g)
4NO(g) + 6H2O(g)
CO2(g) + H2(g)
(c) 2N2(g) + O2(g)
2N2O(g)
(d) Ni(s) + 4CO(g)
Ni(CO)4(g)
(e) NH3(aq) + H2O()
(f)
+
–
Ag (aq) + Cl (aq)
+
–
NH4 (aq) + OH (aq)
AgCl(s)
23
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29
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All answers
Chemical equilibrium
Multiple-choice questions
12. Consider the following reactions:
2SO2(g) + O2(g)
Section 40.1
9.
Consider the following reaction:
2+
–
2–
Co (aq) + 4Cl (aq)
CoCl4 (aq)
Which of the following equilibrium constant
expressions for the above reaction is correct?
2+
A.
–
[Co (aq)]eqm[Cl (aq)]eqm
2–
[CoCl4 (aq)]eqm
2+
B.
–
4
[Co (aq)]eqm[Cl (aq)]eqm
2–
[CoCl4 (aq)]eqm
2–
C.
[CoCl4 (aq)]eqm
2+
–
–
[Co (aq)]eqm[Cl (aq)]eqm
A. C2H4(g) + H2(g)
10. Consider the following equilibrium constant
expression:
2–
Kc =
[Cr2O7 (aq)]eqm
2–
2
+
2
Which of the following equations gives the above
equilibrium constant expression?
B.
2–
+
Cr2O7 (aq)
2–
+
Cr2O7 (aq) +
H2O()
2CrO4 (aq) + 2H (aq)
2–
7
C. Cr2O (aq) + H2O()
2–
+
D. CrO4 (aq) + H (aq)
2–
2–
2–
2CrO4 (aq) +
+
2H (aq)
2–
Cr2O7 (aq)
11. Consider the following reaction:
CO(g) + 2H2(g)
CH3OH(g)
What is the unit of Kc for the above equilibrium
system?
A.
B.
C.
D.
40
24
No units
2
–6
mol dm
–3
mol dm
–2
6
mol dm
B.
C2H6(g)
CH3OH() + HCOOH()
C. 5CO(g) + I2O5(s)
+
[CrO4 (aq)]eqm[H (aq)]eqm
A. 2CrO4 (aq) + 2H (aq)
K1 =
13. Which of the following equilibrium systems is a
heterogeneous equilibrium?
[CoCl4 (aq)]eqm
2+
1
K2
1
B. K1 =
K2
1
C. K1 =
K2
D. K1 = K2
A.
Section 40.2
4
[Co (aq)]eqm[Cl (aq)]eqm
2–
D.
2SO3(g)
K1
1
SO3(g)
SO2(g) + O2(g)
K2
2
Which
of
the
following
mathematical
relationships between K1 and K2 is correct?
–
D. H (aq) + OH (aq)
HCOOCH3()
+ H2O()
I2(g) + 5CO2(g)
H2O()
14. Consider the following reaction:
Ca(HCO3)2(aq)
CaCO3(s) + H2O() + CO2(g)
Which of the following equilibrium constant
expressions for the above reaction is correct?
A.
B.
C.
D.
[CaCO3(s)]eqm[H2O()]eqm[CO2(g)]eqm
[Ca(HCO3)2(aq)]eqm
[Ca(HCO3)2(aq)]eqm
[CaCO3(s)]eqm[H2O()]eqm[CO2(g)]eqm
[H2O()]eqm[CO2(g)]eqm
[Ca(HCO3)2(aq)]eqm
[CO2(g)]eqm
[Ca(HCO3)2(aq)]eqm
All answers
40
Section 40.3
15. The equilibrium constants of four different
equilibrium systems are shown below.
Reaction
Kc
I
II
–34
2.3 × 10
III
16
1.6 × 10
IV
–8
5.0 × 10
112
1.0 × 10
Which of the following is the correct order for the
tendencies of the reactions to proceed towards
completion?
A.
B.
C.
D.
I < II < III < IV
I < III < II < IV
IV < II < III < I
III < II < I < IV
16. Which of the following statements concerning
the magnitude of equilibrium constant, Kc, is
correct?
A. It tells us the rate of the forward reaction.
B. It tells us whether the reaction is exothermic
or endothermic.
C. It tells us the extent of the reaction.
D. It tells us the percentage yield of the reaction.
Section 40.4
17. Consider the following reaction:
CaCO3(s)
Equilibrium constant
Questions 19 and 20 refer to the following equilibrium
system:
N2O4(g)
2NO2(g)
colourless
brown
19. Which of the following methods can be used to
show that the equilibrium has been established?
(1) Measuring the colour intensity of the reaction
mixture
(2) Measuring the pressure of the reaction
mixture
(3) Measuring the mass of the reaction mixture
A.
B.
C.
D.
(1) and (2) only
(1) and (3) only
(2) and (3) only
(1), (2) and (3)
20. In an experiment, 0.125 M of N2O4 was
introduced into a closed vessel at a certain
temperature. At equilibrium, the concentration
of N2O4 was 0.075 M. What is the value of Kc for
this reaction at that temperature?
A.
B.
C.
D.
0.045 M
0.133 M
1.333 M
7.500 M
21. Consider the following reaction:
CaO(s) + CO2(g)
H2(g) + I2(g)
2HI(g)
Kc = 50
At equilibrium, the concentration of CO2(g) at
a certain temperature was found to be 0.548 M.
What is the value of the equilibrium constant at
this temperature?
At 500°C, 0.08 M of HI was allowed to decompose
in a sealed container until an equilibrium was
established. What is the equilibrium concentration
of HI?
A.
B.
C.
D.
A.
B.
C.
D.
0.274 M
0.300 M
0.548 M
1.096 M
18. Consider the following reversible reaction:
2SO2(g) + O2(g)
2SO3(g)
1.0 mol of SO2 and 0.5 mol of O2 were mixed
with a catalyst and compressed to a volume of
3
250 cm at room temperature. When equilibrium
was reached, it was found that 80% of SO2 was
oxidized. What is the Kc of the reaction?
A.
B.
C.
D.
–1
3
0.039 mol dm
–1
3
0.125 mol dm
–1
3
32 mol dm
–1
3
40 mol dm
0.07 M
0.06 M
0.03 M
0.008 M
22. Consider the following reaction:
A(g) + B(g)
C(g) + D(g)
Kc = 4
How many moles of B are present in the
equilibrium mixture formed when one mole of A
is mixed with one mole of B?
A. 1 mol
B. 2 mol
1
C.
mol
3
2
D.
mol
3
25
40
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Chemical equilibrium
All answers
Structured questions
Section 40.4
23. Consider the oxidation of sulphur dioxide to sulphur trioxide:
2SO2(g) + O2(g)
2SO3(g)
Calculate the equilibrium constant for the reaction if the equilibrium concentrations of SO2(g), O2(g) and
–3
–3
–3
SO3(g) are 1.75 mol dm , 1.50 mol dm and 2.25 mol dm respectively.
24. The equilibrium constant, Kc, for the following reaction at 450°C is 0.0198.
2HI(g)
3
(a) HI(g) is introduced into a 2.0 dm
1.87 mol of HI(g) at 450°C.
(i)
H2(g) + I2(g)
sealed container. At equilibrium, the container contains
Calculate the concentration of HI(g) at equilibrium.
(ii) Calculate the equilibrium concentrations of H2(g) and I2(g) respectively.
(iii) Calculate the concentration of HI(g) initially introduced into the container.
(b) Calculate the value of equilibrium constant, K1, at 450°C for the following reaction.
H2(g) + I2(g)
2HI(g)
25. Carbon dioxide decomposes to carbon monoxide and oxygen as shown in the equation below:
2CO2(g)
2CO(g) + O2(g)
3
Initially, there was 0.001 mol of CO2(g) in a 1.0 dm reaction vessel at a certain temperature. When an
–4
–3
equilibrium was established, it was found to contain 1 × 10 mol dm of CO(g). Calculate the equilibrium
constant, Kc, of the reaction at this temperature.
26. At 373 K, N2O4(g) decomposes to NO2(g) and NO2(g) combines to form N2O4(g) until an equilibrium is
–3
–3
reached. The equilibrium concentration of NO2(g) is found to be 3.0 × 10 mol dm and the equilibrium
–3
constant, Kc = 0.2 mol dm .
(a) Write a chemical equation for the reaction.
(b) Calculate the initial concentration of N2O4(g).
27. Hydrogen and bromine react to give hydrogen bromide as shown in the following equation:
H2(g) + Br2(g)
2HBr(g)
At a certain temperature, the equilibrium constant, Kc, for the above reaction is 12.0. Calculate the
3
concentrations of all species at equilibrium if the following amounts of reactants are mixed in a 2.0 dm
reaction vessel.
(a) 4.0 mol of H2(g) and 4.0 mol of Br2(g)
(b) 6.0 mol of H2(g) and 4.0 mol of Br2(g)
40
26
All answers
40
9
–1
Equilibrium constant
3
28. At 110°C, the equilibrium constant, Kc, for the following reaction is 4.7 × 10 mol dm .
CO(g) + Cl2(g)
COCl2(g)
3
If 0.20 mol COCl2 was placed in an 8.0 dm reaction vessel at 110°C, calculate the concentrations of all
substances at equilibrium.
–4
–3
29. At 1000 K, the Kc value for the dissociation of F2(g) is 1.2 × 10 mol dm .
F2(g)
2F(g)
(a) Write the expression for the equilibrium constant, Kc, of the reaction.
–3
(b) If the initial concentration of F2(g) is 0.2 mol dm , calculate the equilibrium concentrations of F2(g) and
F(g) at 1000 K.
(c) Calculate the percentage of F2(g) dissociated at 1000 K.
27
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29
All answers
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