1.1
Fundamental
p ar t ic l e s
▼ Table 1
The properties of the sub-atomic particles
Proper ty
Proton p
Neutron n
Electron e
Maths link
−30
0.911 × 10
−27
The masses and charges of
Mass / kg
−27
1.675 × 10
1.673 × 10
(very nearly 0)
sub-atomic par ticles are very
−19
Charge / C
+1.602 × 10
Position
in the nucleus
−19
0
−1.602 × 10
small so they are expressed in
around the
standard form. Refer to Section 8,
in the nucleus
nucleus
Mathematical skills, if you are not
sure about this notation.
These
mass
be
numbers
and
+1,
see
are
charge
so
the
Table
extremely
are
charge
used.
on
small.
The
an
In
relative
electron
is
practice,
charge
−1.
relative
on
a
values
proton
Neutrons
have
is
no
for
taken
to
charge,
2.
▼ Table 2
The relative masses and charges of the sub-atomic particles
Proton p
Neutron n
Electron e
Study tip
1
You must remember the relative
Relative mass
1
1
1840
masses and charges of a proton,
Relative charge
+1
0
−1
neutron, and an electron, as given
in T
able 2.
In
a
neutral
number
in
the
protons
number
because
of
their
electrons
charge
is
must
equal
be
in
the
size
same
and
as
the
opposite
sign.
The
electron
of
atom,
The
arrangement of the
sub-atomic
particles
sub-atomic
(protons,
neutrons,
particles
and
electrons)
are
+
+
proton
arranged
in
the
atom
as
shown
in
Figure
3.
+
neutron
The
by
a
the
▲ Figure 3
protons
force
and
called
neutrons
the
electrostatic
are
strong
forces
of
in
the
centre
nuclear
of
force.
attraction
that
the
This
hold
atom,
is
held
much
electrons
together
stronger
and
than
protons
The sub-atomic particles in
together
in
the
atom,
so
it
overcomes
the
repulsion
between
the
a helium atom (not to scale)
protons
within
+
The
in
the
the
nucleus.
It
acts
only
over
very
short
distances,
that
is,
nucleus.
nucleus
is
surrounded
by
electrons.
Electrons
are
found
in
a
series
Ex tension
of
shells,
sometimes
The diameter of the nucleus
and
further
of a hydrogen atom is about
will
develop
away
in
referred
from
Topic
the
to
as
orbits
nucleus.
or
This
is
levels,
a
which
simplied
get
further
picture
that
1.5.
–15
2 × 10
m, while the diameter
of the atom itself is about
–10
1 × 10
m, about 50 000 times
larger. This means that if the
Summary questions
nucleus were the size of a
y, the whole atom would be
1
a
Identify which of the following – protons, neutrons, or electrons:
roughly the size of a cathedral.
St Paul’s Cathedral is roughly
i
are nucleons
iv
have no charge
ii
have the same relative
v
are found outside the
200 m long. Estimate the
mass
nucleus
length of a y and, without
iii
have opposite charges
using a calculator, check that
b
Explain why we assume that there are the same number of protons
the analogy is realistic.
and electrons in an atom.
14
1.2
Mass
and
number,
atom ic
num be r,
isotopes
Learning
Mass
number
and
atomic
➔
Atomic
number
objective s:
number
Dene the terms mass
Z
number, atomic number,
As
you
have
seen
in
Topic
1.1,
atoms
consist
of
a
tiny
nucleus
made
up
and isotope.
of
protons
and
of
protons
in
number
neutrons
the
that
nucleus
is
is
surrounded
called
the
by
atomic
electrons.
number
The
or
the
number
proton
Z
➔
Explain why isotopes of the
same element have identical
chemical proper ties.
The
so
number
atoms
shell
of
(how
are
an
it
denes
of
electrically
atom
reacts)
the
of
atoms
of
different
Mass
The
total
number
that
weigh
mass
The
equal
chemical
an
the
of
proton
electrons
properties
it
is.
The
of
in
an
atomic
number,
the
outer
Specication reference: 3.1.1
element
number
element.
number)
element
to
number
element
of
(proton
have
of
have
different
protons
nucleons)
responsible
electrons
of
is
the
Z
=
number
same
atomic
atomic
of
protons
number.
Atoms
numbers.
A
number
of
are
same
atom
the
sort
identity
elements
number
the
neutral.
what
number
the
in
determines
and
chemical
atomic
All
electrons
is
for
A
=
neutrons
the
almost
virtually
number
plus
called
mass
all
of
in
the
number
the
mass
nucleus
A.
of
It
an
is
(the
the
atom
total
nucleons
because
nothing.
number
of
protons
+
number
of
neutrons
Isotopes
Every
single
atom
protons
in
But
number
•
the
Atoms
of
•
•
All
nucleus
with
of
are
same
isotopes
way
Atoms
of
number
their
nuclei.
is
as
of
of
the
what
respectively
example,
the
of
(Table
1).
in
of
the
same
protons
element
same
same
number
but
of
number
of
electrons.
different
the
for
All
same
numbers
three
oxygen
to
isotopes
vary
of
have
any
mass
will
exactly
in
neutrons
atomic
other
react
in
number
6.
element.
numbers
carbon
in
conguration.
element
than
with
form
chemically
number
example,
rather
isotopes
react
electron
different
carbon
three
burning
of
the
carbon,
them
has
the
the
vary.
same
have
isotopes
element
carbon
may
has
isotopes
because
makes
element
therefore
number
they
different
mass
atoms
and
called
the
same
particular
neutrons
the
neutrons
However,
for
any
Different
That
14
its
of
in
12,
the
13,
and
same
way,
dioxide.
Study tip
13
Isotopes
are
often
written
like
C .
this:
The
superscript
13
is
the
mass
6
number
of
the
isotope,
and
the
subscript
6
the
atomic
number.
The mass number of an isotope
must always be bigger than the
number
of
protons
and
1
neutrons
atomic number (except in
13
1
H).
C
Typically it is around twice as big.
6
number
of
protons
15
1.2
Mass
number,
at o m ic
num ber,
and
isot o pes
▼ Table 1
Isotopes of carbon
Name of isotope
carbon-12
12
carbon-13
13
C
Symbol
carbon-14
14
C
6
C
6
6
Number of protons
6
6
6
Number of neutrons
6
7
8
1.11%
trace
Abundance
98.89%
Summary questions
1
Isotopes
are
usually
Carbon dating
identified
the
by
element
number
of
carbon-13.
of
and
the
names.
often
called
hydrogen-3
However,
behave
the
name
the
have
protons,
as
isotopes
called
these
common
State
is
and
tritium .
isotopes
just
like
isotope,
how
neutrons,
Isotopes of an element have dierent numbers of neutrons in their nuclei and
most elements have some isotopes. Sometimes these isotopes are unstable
and the nucleus of the atom itself breaks down giving o bits of the nucleus
or energetic rays. This is the cause of radioactivity. Radioactive isotopes have
deuterium ,
is
in
their
chemically
hydrogen-1.
mass
Hydrogen-2
both
most
of
isotope,
However,
hydrogen
own
the
many
many uses. Each radioactive isotope decays at a rate measured by its half life.
This is the time taken for half of its radioactivity to decay.
One well-known radioactive isotope is carbon1
4. It has a half life of 5730 years
and is produced by cosmic-ray activity in the atmosphere. It is used to
date organic matter. Radiocarbon dating can nd the age of carbon-based
material up to 60 000 years old, though it is most accurate for materials up
to 2000 years old.
and
There is always a tiny xed propor tion of carbon14 in all living matter. All
electrons
following
the
atoms
of
the
have.
a
deuterium
b
tritium
living matter takes in and gives out carbon in the form of food and carbon
dioxide, respectively. As a result, the level of carbon14 stays the same. Once
the living material dies, this stops happening. The radioactive carbon breaks
down and the level of radioactivity slowly falls. So, knowing the half life of
carbon14, scientists work backwards. They work out how long it has taken
31
14
W ,
2
15
16
X ,
15
Y ,
7
8
Z
7
it is in the sample. So, a sample with half the level of radioactivity expected
(not their real symbols) is a
in a living organism would have been dead for 5730 years, while one with a
pair of isotopes.
3
for the level of radioactivity to fall from what it is in a living organism to what
Identify which of these atoms
quar ter of the expected level would have been dead for twice as long.
For each element in
1
The radioactivity in a wooden bowl was found to be
of that found in a
8
question 2, state:
sample of living wood.
a
the number of protons
b
the mass number
c
the number of neutrons
+
1
How old is the wood from the bowl?
2
Does this tell us the age of the bowl? Explain your answer.
Carbon-14
1
4
Radiocarbon dating was introduced in 1949 by the
If the half life of
American Willard Libby who won the Nobel Prize for the
is four half lives so the remaining radioactivity will be
1
technique. Carbon14 is produced in the atmosphere by a
1
x
2
1
x
2
1
1
x
2
C is taken to be 6000 years, 24 000 years
=
2
16
nuclear reaction in which a neutron (from a cosmic ray)
of the original activity.
hits a nitrogen atom and ejects a proton:
14
Suggest why 60 000 years is the practical limit for
14
1
N +
7
16
14
n ➝
0
1
C +
6
p
1
dating.
C
1.3
The
The
The
mass
mass
mass
spectrom e t e r
Learning
spectrometer
spectrometer
determination
of
is
the
relative
most
useful
atomic
instrument
masses
A
.
for
Relative
the
accurate
atomic
masses
➔
objective s:
Explain how a mass
spectrometer works and what
r
12
are
measured
dened
is
as
exactly
neutron
on
a
exactly
a
whole
has
a
scale
12.
on
No
which
other
number.
mass
of
This
exactly
is
atomic
mass
mass
has
of
a
because
an
atom
relative
neither
of
C
atomic
the
it measures.
is
mass
proton
that
nor
Specication reference: 3.1.1
the
1.
average
relative
the
isotope
mass
of
1
atom
of
an
element
=
A
r
1
12
mass
of
1
atom
of
C
12
average
relative
molecular
mass
M
mass
of
a
molecule
=
r
1
12
mass
of
1
atom
of
C
12
The
mass
spectrometer
molecules).
toolkit
to
help
There
of
are
The
to
layout
types
forming
the
called
an
of
For
of
ions
ratio
of
electro
this
type
are
such
mass
from
their
as
mass
an
are
illegal
but
and
their
ionisation
part
of
by
atoms
a
(or
chemist’s
forensic
scientists
drugs.
sample
to
separate
used
spectrometer,
the
mass
of
essential
they
charge
spray
of
the
example,
substances
several
of
according
is
spectrometers
equipment.
identify
principle
here
Mass
determines
mass.
time
spectrometer
all
then
of
is
work
on
the
separating
The
ight
type
in
ions
▲ Figure 1
A modern mass spectrometer
described
(TOF)
shown
the
instrument.
Figure
2.
+
+
+
solvent
+
evaporating
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Synoptic link
+
+
+
+
+
+
+
+
+
+
+
+
ne
Mass spectrometry can also be
+
hollow
needle
vacuum
pump
used to measure relative molecular
masses and much more, as
you will see in Topic 16.2, Mass
detector
droplets
beam
containing
sample
of
positive
ions
spectrometry.
ions
sample
signal
to
ight
high
tube,
length,
PC
d
voltage
electrons
+
▲ Figure 2
What
In
The layout of an electron spray ionisation time of ight mass spectrometer
happens
outline,
ions,
the
accelerated
charge
ratio),
in
a time of ight
substance(s)
to
and
high
arrive
in
the
speeds
at
a
mass
sample
(which
detector.
spectrometer?
are
converted
depend
The
on
steps
their
are
to
positive
mass
to
described
in
more
Synoptic link
detail
below.
You will nd out more about
•
Vacuum
The
whole
apparatus
is
kept
under
a
high
vacuum
to
relative atomic mass in Topic 2.1,
prevent
the
ions
that
are
produced
colliding
with
molecules
from
Relative atomic and molecular
the
air.
masses, the Avogadro constant,
•
Ionisation
The
sample
to
be
investigated
is
dissolved
in
a
volatile
and the mole.
solvent
and
forced
through
a
ne
hollow
needle
that
is
connected
17
1.3
The
mass
spect ro m e ter
to
the
positive
positively
+
terminal
charged
of
a
droplets
high
voltage
which
have
supply.
lost
This
produces
electrons
to
the
tiny
positive
The relationship
charge
between mass and
the
of
the
vacuum
contain
no
supply.
and
the
more
The
solvent
droplets
than
a
get
single
evaporates
smaller
from
and
positively
the
smaller
charged
ion
droplets
until
of
into
they
mass
may
m
time of ight of
•
Acceleration
The
positive
ions
are
attracted
towards
a
negatively
an ion
The kinetic energy of the ions in
charged
plate
charged
ions
and
accelerate
achieve
a
towards
higher
it.
Lighter
ions
and
more
highly
speed.
the ight tube is given by
•
1
KE =
mv
Ion
drift
through
d
2
and velocity, v =
The
a
ions,
hole
in
all
the
of
which
have
negatively
the
charged
same
kinetic
plate,
energy,
forming
a
KE,
beam
pass
and
,
2
t
travel
along
a
tube,
called
the
ight
tube,
to
a
detector
with
velocity,
v
where d is the length of the ight
•
Detection
When
ions
with
the
same
charge
arrive
at
the
detector,
tube.
the
2
1
m (
lighter
ones
are
rst
as
they
have
higher
velocities.
The
ight
2
d
So KE =
d
2
), so t
= m(
times
)
are
recorded.
The
positive
ions
pick
up
an
electron
from
the
2
2
t
2KE
detector,
which
causes
a
current
to
ow.
Since KE and d are constant,
•
2
m α
Data
analysis
The
signal
from
the
detector
is
passed
to
a
computer
t
which
Mass
The
a
spectra of
mass
make
generates
up
mass
element.
like
those
in
Figures
3
and
4.
elements
spectrometer
an
spectrum
can
It
be
used
detects
to
identify
individual
the
ions,
so
different
different
isotopes
isotopes
that
are
100
%
detected
/
data
for
ecnadnuba
were
separately
the
neon,
obtained.
isotope
and
because
germanium,
The
the
they
peak
and
height
horizontal
have
chlorine
gives
scale
different
gives
the
the
masses.
isotopes
relative
mass
to
This
in
is
how
Figures
abundance
charge
3,
of
ratio,
the
4,
and
5
each
m/z,
which,
50
for
a
singly
evitaler
Mass
charged
ion
spectrometers
places
of
an
atomic
is
can
numerically
measure
mass
unit
–
the
same
relative
this
is
as
atomic
called
the
mass
masses
high
number
to
ve
resolution
A
decimal
mass
0
0
20
40
spectrometry.
mass
/
charge
is
▲ Figure 3
However,
most
work
is
done
to
one
decimal
point
–
this
ratio
called
low
resolution
mass
spectrometry.
The mass spectrum of neon.
Low
resolution
mass
spectrometry
ven though the relative atomic mass is
The
low
resolution
mass
spectrum
of
neon
is
shown
in
Figure
3.
This
20.2, there is no peak at 20.2 because no
shows
that
neon
has
two
isotopes,
with
mass
numbers
20
and
22,
and
neon atoms actually have this mass
abundances
From
this
to
we
(90
×
the
can
20)
nearest
say
+
that
(10
×
whole
neon
has
an
of
90%
average
and
10%,
relative
respectively.
atomic
mass
of:
22)
=
100
number
20.2
100
When
%
80
take
calculating
account
of
the
the
relative
relative
atomic
mass
abundances
of
of
an
the
element,
isotopes.
you
The
must
relative
/
ecnadnuba
atomic
mass
of
neon
is
not
21
because
there
are
far
more
atoms
of
the
60
lighter
40
evitaler
36.5%
isotope.
Another
example
which
shown
is
the
mass
spectrum
of
the
element
germanium,
27.4%
is
in
Figure
4.
20.5%
20
7.8%
Isotopes of chlorine
7.8%
35
Chlorine
0
has
two
isotopes.
They
Cl ,
are
with
a
mass
number
of
35,
17
70
71
72
73
74
75
76
37
Cl ,
and
mass
/
charge
ratio
exactly
▲ Figure 4
with
mass
number
of
37.
They
occur
in
the
ratio
of
3
:
1.
The mass spectrum
35
of germanium (the percentage
abundance of each peak is given)
18
a
17
35
Cl
35
Cl
three
of
37
Cl
these
Cl
to
every
one
of
this
almost
Atomic
structure
1
35
So
there
are
75%
chlorine
gas
(see
37
Cl
and
Figure
25%
Cl
atoms
in
naturally
occurring
singly
charged
ions
5).
100
%
average
of
100
of
atoms
these
=
(35
×
is
35.5,
75)
+
as
(37
shown
×
25)
below.
=
ecnadnuba
Mass
mass
/
The
3550
3550
Average
mass
=
=
35.5
100
explains
approximately
why
the
relative
atomic
mass
of
chlorine
evitaler
This
50
is
35.5.
0
5
10
15
mass
+
▲ Figure 5
Identifying elements
/
20
25
charge
30
35
40
ratio
The mass spectrum of
chlorine
All elements have a characteristic pattern that shows the relative
abundances of their isotopes. This can be used to help identify any
par ticular element. Chlorine, for example, shows two peaks at mass 35 and
mass 37
. The peak of mass 35 is three times the height of the peak of mass
35
37 because there are three times as many
Study tip
Cl atoms in chlorine.
The spectrum will also show peaks caused by ionised Cl
molecules. These
Relative atomic masses are
2
are called molecular ions. There will be three of these:
weighted averages of the mass
numbers of the isotopes of the
35
•
at m/z 70, due to
•
at m/z 72, due to
•
at m/z 74, due to
35
Cl
Cl
element, taking account of both
35
37
Cl
Cl
the masses and their abundances,
37
37
Cl
12
Cl
relative to the
C isotope, which is
exactly 12. Chlorine has isotopes
High resolution mass spectrometers can measure the masses of atoms
of mass number 35 and 37 but the
to several decimal places. This allows us to identify elements by the exact
relative atomic mass of chlorine is
masses of their atoms that (apar t from carbon12 whose relative atomic
not 36, it is 35.5.
mass is exactly 12) are not exactly whole numbers.
+
What will be the relative abundances of the three Cl
ions of m/z 70, 72,
2
and 74 respectively? The relative abundances of the atoms are
35
3
37
Cl :
Cl = 3 : 1. i.e.,
1
:
4
Mass
Space
spectrometers
probes
such
spectrometers.
The
Huygens
Saturn,
and
in
landing,
it
They
the
also
in
the
are
spacecraft
January
measure
as
4
used
that
2005
Rover
to
of
Curiosity
identify
landed
carried
amounts
analysed
space
Mars
a
the
on
vaporised
one
of
spectrometer
in
Titan’s
samples
mass
elements
Titan,
mass
gases
the
carry
of
in
the
rock
used
to
atmosphere.
the
samples.
moons
of
identify
After
surface.
▲ Figure 6
The Mars Rover Curiosity
carries a mass spectrometer to look for
compounds of carbon that may suggest
that there was once life on Mars.
19
1.3
The
mass
spect ro m e ter
Mini-mass spectrometer
The
latest
carr y
10
as
kg ,
dr ugs,
development
a
back
light
pack.
enough
explosives,
chemical
to
or
in
The
be
mass
unit,
carried
chemical
spectrometr y
including
by
scene
weapons.
is
a
unit
rechargeable
of
crime
Other
small
of ficers
uses
enough
batteries,
looking
include
to
weighs
for
investigating
spills.
▲ Figure 7
The mini-mass spectrometer
The mass spectrometer includes software to match spectra of samples
investigated with a library of spectra and so identify them. The instrument
can be used by operators with little or no chemical knowledge.
Summary questions
1
Explain why the ions formed in a mass spectrometer
100
have a positive charge.
%
Explain what causes the ions to accelerate through
Describe what forms the ions into a beam.
4
State which ions will arrive at the detector rst.
5
Use the information about germanium in Figure 4 to
evitaler
3
ecnadnuba
the mass spectrometer.
80
/
2
60
40
20
calculate its relative atomic mass.
0
6
Figure 8 shows the mass spectrum of copper.
62
63
64
65
66
Calculate the relative atomic mass of copper.
mass
▲ Figure 8
20
/
charge
ratio
The mass spectrum of copper
1.4
The
atom
During
strides
the
The
the
in
and
early
arrangement
of
the
Learning
electrons
years
of
understanding
the
the
ele c trons
twentieth
structure
century,
of
the
physicists
atom.
These
made
are
great
some
➔
Describe how electrons are
arranged in an atom.
of
landmarks.
objective s:
➔
Recognise that the electron
can behave as a par ticle, a
1913
Niels
Bohr
positive
an
of
xed
like
size
next
was
forward
nucleus
atom
the
put
the
a
the
orbited
tiny
and
explained
beginning
by
solar
the
idea
that
of
atom
consisted
negatively-charged
system.
movement
how
the
atoms
what
is
The
of
absorbed
called
electrons
electrons
electrons
orbited
from
and
one
gave
quantum
of
a
to
in
wave, or a cloud of charge.
form
shells
shell
out
tiny
to
light.
This
➔
Describe how the structure
of an atom developed from
Dalton to Schr Ödinger.
theory.
Specication reference: 3.1.1
1926
Erwin
Schrödinger,
equation
that
properties
theory
James
the
used
mathematical
the
waves
called
predict
1932
of
a
as
idea
well
quantum
behaviour
Chadwick
that
as
physicist,
electrons
those
of
mechanics,
of
sub-atomic
discovered
the
worked
had
some
particles.
which
of
This
can
be
out
an
the
led
to
used
a
to
particles.
neutron.
Synoptic link
At
the
same
electrons
was
the
time,
chemists
allowed
atoms
American,
were
to
Gilbert
bond
developing
together.
Lewis.
He
put
their
One
ideas
about
important
forward
the
how
contributor
ideas
that:
This topic revises your knowledge
of electron arrangements from
GCSE. This will be useful when
•
the
inertness
of
the
noble
gases
was
related
to
their
having
full
you study Topic 1.5, More about
outer
shells
of
electrons
electron arrangements in atoms.
•
ions
were
outer
•
atoms
could
theories
explain
the
tend
Evolving
Early
of
a
to
gain
so
still
Dalton’s
•
Bohr’s
the
by
basis
of
the
you
losing
sharing
of
many
stable
the
can
can
can
or
gaining
electrons
modern
simple
electrons
ideas
of
to
form
to
full
chemical
compounds
electronic
use
model
electron
also
never
merely
volume
of
can
state
can
still
be
as
think
say
space
different
model
covalent
of
bond
model
you
You
•
The
atoms
attain
full
structure
using
of
the
outer
bonding,
the
idea
nearest
shells.
and
that
noble
gas.
ideas
particular
chemists
•
are
suggest
charge,
moment.
in
also
formulae
theories
theories
by
shells
atoms
Lewis’
formed
used
of
minute
the
for
has
of
a
where
probability
that
used
solid
electrons
exactly
models
be
a
a
the
to
smeared
electron
that
it
particular
atom
explain
simple
as
an
particle.
for
the
model
can
Later
out
is
be
at
clouds
any
found
shape.
However,
different
purposes.
geometries
of
ionic
of
crystals.
and
bonding
cloud
and
practice
Quantum theory makes
bonding.
charge
Quantum theory in
predictions that seem to
idea
the
is
used
shapes
of
for
a
more
sophisticated
explanation
contradict our everyday
experience, such as the fact that
molecules.
an electron can pass through two
•
The
simple
model
of
electrons
orbiting
in
shells
is
useful
for
many
dierent holes at once! However,
purposes,
particularly
for
working
out
bonding
between
atoms.
it is an extremely successful
You
will
be
familiar
with
the
electron
diagrams
in
this
section
from
theory and underlies electronic
GCSE.
They
lead
on
to
the
more
sophisticated
models
of
electron
gadgets such as computers,
structure
described
in
Topic
1.5.
However,
they
can
still
be
useful,
mobile phones, and DVD players.
for
example,
compounds
in
predicting
and
the
and
shapes
of
explaining
the
formulae
of
simple
molecules.
21
1.4
The
arrangeme nt
of
th e
el ec tron s
Electron
shells
Study tip
The
rst
shell,
which
is
closest
to
the
nucleus,
lls
rst,
then
the
You must remember the number of
2
second,
and
so
on.
The
number
of
electrons
in
each
shell
=
2n
,
where
electrons in each shell.
n
is
the
number
•
the
rst
•
the
second
•
the
third
shell
the
holds
shell
shell
shell,
so:
up
two
holds
holds
to
up
up
to
to
electrons
eight
18
electrons
electrons.
Electron diagrams
C
If
you
know
number
carbon
can
(2,4)
of
the
number
electrons
therefore
carbon
▲ Figure 1
of
has
draw
six
it
of
has.
an
protons
This
electron
electrons.
The
is
in
an
because
diagram
four
atom,
the
for
you
atom
any
electrons
in
also
is
neutral.
element.
the
know
outer
For
shell
the
You
example,
are
Electron diagram of carbon
usually
Sulfur
when
and
drawn
has
can
add
also
out
electrons.
drawing
then
You
16
spaced
bonding
the
next
draw
It
around
has
six
to
space
electrons
electron
atom
electrons
diagrams
two
the
to
diagrams
in
its
out
form
of
(Figure
outer
the
rst
pairs
ions,
1).
as
shell.
four
(Figure
long
as
It
helps
(as
in
carbon),
2).
you
know
S
the
number
of
electrons.
For
example,
a
sodium
atom,
Na,
has
11
+
electrons,
An
but
oxygen
its
atom
ion
has
has
10,
eight
so
it
has
electrons,
a
positive
but
its
ion
charge,
has
10,
Na
so
it
(Figure
has
3).
a
2–
negative
sulfur
O
(Figure
4).
(2,8,6)
You
▲ Figure 2
charge,
can
write
electron
diagrams
in
shorthand:
Electron diagram of sulfur
•
write
shell
+
•
the
and
separate
number
working
each
of
electrons
in
each
shell,
starting
with
the
inner
outwards
number
by
a
comma.
+
For
carbon
you
write
2,4;
for
sulfur
2,8,6;
for
Na
2,8.
Na
Summary questions
+
Na
11
sodium
protons,
ion
10
1
Draw the electron arrangement diagrams of atoms that have the
electrons
following numbers of electrons:
(2,8)
a
▲ Figure 3
3
b
9
c
14
Electron diagram of a
2
sodium ion
State, in shor thand, the electron arrangements of atoms with:
a
3
4 electrons
b
13 electrons
c
18 electrons
Identify which of the following are atoms, positive ions, or negative
2–
ions. Give the size of the charge on each ion, including its sign. Use the
Periodic T
able to identify the elements A–E.
O
×
Number of protons
Number of electrons
A
12
10
B
2
2
C
1
7
18
D
10
10
E
3
2
×
2–
O
8
oxygen
protons,
10
▲ Figure 4
oxygen ion
22
ion
electrons
(2,8)
Electron diagram of an
1.5
More
in
As
you
have
electrons
seen
are
about
electro n
ar r ang em e nt s
atoms
in
Topic
thought
of
1.4,
as
in
being
a
simple
model
arranged
in
of
shells
the
atom
around
Learning
the
➔
nucleus.
The
shells
can
hold
increasing
numbers
of
objective s:
the
electrons
as
Illustrate how the electron
they
congurations of atoms and
get
further
from
the
nucleus
–
the
pattern
is
2,
8,
18,
and
so
on.
ions are written in terms of s,
p, and d electrons.
Energy
levels
Specication reference: 3.1.1
Electrons
can
in
different
therefore
represent
Each
be
shell
have
represented
energy
main
shells
levels
can
and
hold
on
an
they
up
to
differing
energy
are
a
amounts
level
labelled
1,
maximum
2,
of
energy.
diagram.
3,
and
number
The
so
of
They
on
shells
(Figure
electrons
1).
z
given
y
2
by
the
formula
2n
,
where
n
is
the
number
of
the
main
shell.
So,
you
x
can
have
next,
two
and
Apart
so
from
electrons
in
the
rst
main
shell,
eight
in
the
next,
18
in
the
on.
the
rst
shell,
these
main
energy
levels
are
divided
s
into
z
sub-shells,
called
s,
p,
d,
and
f,
which
have
slightly
different
z
energies
y
(Figure
2).
Shell
s-sub-shell,
a
2
has
an
s-sub-shell
p-sub-shell,
and
a
and
a
p-sub-shell.
Shell
3
y
an
d-sub-shell.
x
energy
up
3
1
18
d
electrons
up
to
8
electrons
up
to
2
electrons
3
p
3
s
ygrene
2
to
x
p
p
z
p
2
y
2
s
x
1
electron
shells
main
s
1
shells
sub-shells
p
z
▲ Figure 1
Electron shells
▲ Figure 2
z
Main shells and sub-shells
y
Quantum
For
a
more
quantum
mechanics
complete
mechanics
x
description
is
used,
y
of
the
which
electrons
was
in
developed
atoms
a
during
theory
the
x
called
1920s.
This
d
describes
the
atom
mathematically
with
an
equation
(the
d
Schrödinger
z
z
equation).
The
solutions
to
this
equation
give
the
probability
of
nding
an
y
y
electron
in
a
given
volume
of
space
called
an
atomic
orbital.
x
x
Atomic orbitals
The
electron
is
no
longer
considered
to
be
a
particle
but
a
cloud
of
d
d
negative
orbital.
included
charge.
The
in
An
electron
concept
the
of
the
following
lls
main
a
volume
shells
in
and
space
the
called
sub-shells
its
is
atomic
z
then
y
way.
x
•
Different
a
atomic
number
to:
1,
2,
3,
that
and
orbitals
tells
so
us
have
the
different
main
energy
energies.
shell
Each
that
it
orbital
has
corresponds
on.
d
•
The
atomic
orbitals
of
each
main
shell
have
different
shapes,
which
▲ Figure 3
in
turn
have
slightly
different
energies.
These
are
the
The shapes of s-, p-, and
sub-shells.
d-orbitals
They
are
described
by
the
letters
s,
p,
d,
and
f.
23
1.5
More
about
ele c t ro n
arrang em ents
in
a to m s
The
shapes
of
the
s-,
p-,
and
d-orbitals
are
shown
in
Figure
3.
The
Study tip
shapes
You should know how many s-, p-,
•
These
f-orbitals
shapes
probability
and d-orbitals there are in each
to know the shapes of the s- and
are
even
represent
of
nding
a
an
more
volume
complicated.
of
electron
space
and
in
they
which
there
inuence
the
is
a
95%
shapes
of
molecules.
main shell. However, you only need
•
The
rst
shell
p-orbitals.
main
has
a
shell
single
energy,
the
slightly
higher
third
still,
and
so
•
Any
single
•
s-orbitals
•
p-orbitals
can
in
of
consists
s-orbital
main
energy,
on,
see
of
and
shell
and
Figure
a
single
three
has
ve
a
s-orbital.
p-orbitals
single
The
of
a
s-orbital,
d-orbitals
of
second
slightly
three
slightly
main
higher
p-orbitals
higher
of
energy
4.
3d3d3d3d3d
d
3p 3p 3p
p
3
of
3s
atomic
orbital
can
hold
a
maximum
of
two
electrons.
s
ygrene
can
hold
up
to
two
electrons.
2p 2p 2p
p
2s
2
hold
up
to
two
electrons
each,
but
always
come
s
groups
electrons
in
three
the
of
the
same
energy,
to
give
a
total
of
up
to
six
p-sub-shell.
1s
s
1
•
the
numbers
of
d-orbitals
ve
sub-levels
can
hold
up
to
two
electrons
each,
but
come
in
groups
of
different
of
the
same
energy
to
give
a
total
of
up
to
10
electrons
in
the
sub-levels
d-sub-shell.
▲ Figure 4
The subdivisions of orbitals
Table
1
summarises
the
number
of
electrons
in
the
different
shells
and
sub-shells.
▼ Table 1
The number of electrons in the dierent levels and sub-levels
Main energy level (shell)
1
sub-shell(s)
2
s
number of orbitals in
sub-shell
3
s
1
p
3
1
(2 electrons)
(2e
s
)
(6e
p
1
)
4
d
3
(2e
)
s
5
(6e
)
p
1
(10e
)
d
3
(2e
)
f
5
(6e
)
(10e
7
)
(14e
)
total number of electrons
2
8
18
32
in main shell
The
for
5p
4d
energy
the
main
level
rst
shell
few
has
diagram
in
elements
only
an
Figure
of
the
5
shows
Periodic
s-orbital.
The
the
Table.
second
energies
Notice
main
of
that
shell
the
the
has
an
orbitals
rst
s-
and
5s
p-sub-shell,
and
the
energy.
third
p-sub-shell
is
composed
of
three
p-orbitals
of
equal
4p
ygrenE
The
main
shell
has
an
s-,
p-,
and
d-sub-shell,
and
the
3d
d-sub-shell
4s
is
composed
of
ve
atomic
orbitals
of
equal
energy.
3p
•
3s
Each
‘box’
in
Figure
5
represents
an
orbital
of
the
appropriate
2p
shape
that
can
hold
up
to
two
electrons.
2s
•
1s
▲ Figure 5
Notice
that
neutral
The energies of the rst few
atomic orbitals
4s
is
atoms,
actually
though
of
this
slightly
can
lower
change
energy
when
than
ions
are
3d
for
formed.
Spin
Electrons
•
Two
•
The
also
have
electrons
electrons
down
to
in
the
the
are
show
property
same
usually
the
called
orbital
spin
must
represented
different
directions
have
by
of
opposite
arrows
spins.
pointing
up
or
spin.
Study tip
Putting
Although we use the term spin,
Remember
the electrons are not actually
(and
spinning.
means
is
24
electrons
that
shape)
the
of
an
main
spherical.
the
into
label
of
electron
shell
is
3
atomic orbitals
an
atomic
cloud.
and
the
For
orbital
tells
example,
sub-level
us
the
(and
about
atomic
therefore
the
energy
orbital
the
3s
shape)
Atomic
structure
1
There
are
three
rules
for
allocating
electrons
to
atomic
orbitals:
Study tip
1
Atomic
orbitals
of
lower
energy
are
lled
rst
–
so
the
lower
main
Practise working out the shor thand
shell
are
is
lled
lled
rst
and,
within
this
shell,
sub-shells
of
lower
energy
electronic structure of all the
rst.
elements at least up to krypton
2
Atomic
orbitals
of
the
same
energy
ll
singly
before
pairing
starts.
(atomic number 36).
This
3
No
The
is
because
atomic
electron
sodium
are
electrons
orbital
can
diagrams
shown
in
repel
hold
for
the
Figure
each
more
other.
than
elements
two
electrons.
hydrogen
to
6.
3p
3p
3s
3s
2p
2p
2s
2s
1s
1s
He
H
3p
3p
3p
3s
2p
3p
3p
3s
3s
2p
2p
3p
3s
3s
3p
3s
2p
2p
3p
3s
2p
3s
2p
2p
2s
2s
2s
2s
2s
2s
2s
2s
1s
1s
1s
1s
1s
1s
1s
1s
Li
Be
B
C
N
O
F
Ne
3p
3s
2p
2s
1s
Na
▲ Figure 6
The electron arrangements for the elements hydrogen to sodium – note
Synoptic link
how they obey the rule above
You will learn how electron
Writing
electronic
arrangements aect the proper ties
structures
of the transition metals in
A
shorthand
way
of
writing
electronic
structures
is
as
follows,
for
Topic 23.1, The general proper ties
example,
for
sodium
which
has
11
electrons:
of transition metals.
2
2
1s
2s
2
Note
6
1
2p
3s
8
how
Calcium,
this
1
matches
with
20
the
simpler
electrons
would
2,8,1
you
used
at
Summary questions
GCSE.
be:
1
2
2
1s
6
2s
2
2p
3s
6
a
Give the full electron
2
3p
4s
which
matches
2,8,8,2
arrangement for
Notice
lower
how
the
4s
orbital
is
lled
before
the
3d
orbital
because
it
is
phosphorus.
of
energy.
b
After
calcium,
electrons
2
23
electrons
is:
1s
begin
2
6
2s
2p
to
2
3s
ll
the
6
3p
3d
3
orbitals,
so
vanadium
arrangement for phosphorus
with
2
3d
Give the electron
using an inert gas symbol as
4s
a shorthand.
2
Krypton
with
36
electrons
is:
2
1s
2s
6
2
6
3p
10
2p
3s
the
previous
2
3d
4s
6
4p
2
Sometimes
it
simplies
things
to
use
noble
gas
a
Give the full electron
symbol.
arrangements of:
2
So
the
electron
arrangement
of
calcium,
Ca,
could
be
written
[Ar]
4s
2+
i
2
as
a
shorthand
for
[1s
2
2s
6
2p
2
3s
6
3p
2
]
4s
2
because
1s
2
2s
6
2p
2
3s
b
is
the
electron
arrangement
of
Ca
and
ii
F
6
3p
Give their electron
argon.
arrangements using an
+
You
can
use
the
same
notation
for
ions.
2
have
the
electron
arrangement
1s
So
2
2s
a
sodium
ion,
Na
,
would
inert gas symbol as a
6
2p
,
one
less
than
a
sodium
shorthand.
2
atom,
1s
2
2s
6
2p
1
3s
25
1.6
Electron
arrangem e nt s
ionisation
Learning
➔
objective s:
energy
The
patterns
evidence
State the denition of
an d
in
for
rst
ionisation
electron
energies
across
a
period
provide
sub-shells.
ionisation energy.
Ionisation
➔
energy
Describe the trend in
Electrons
ionisation energies a) down
them
can
can
be
removed
be
measured.
are
removed,
from
This
is
atoms
called
and
the
energy
ionisation
it
takes
energy
to
remove
because
as
the
a group and b) across a
electrons
the
atoms
become
positive
ions.
period in terms of electron
congurations.
•
Ionisation
energy
is
the
energy
required
to
remove
a
mole
of
electrons
–1
➔
from
Explain how trends in
•
ionisation energies
a
mole
Ionisation
of
atoms
energy
in
has
the
the
gaseous
state,
abbreviation
and
is
measured
in
kJ
mol
IE.
provide evidence for the
Removing the
electrons one
by one
existence of electron shells
You
can
measure
one
from
the
energies
required
to
remove
the
electrons
one
by
and sub-shells.
an
atom,
starting
from
the
outer
electrons
and
working
inwards.
Specication reference: 3.1.1
•
The
rst
being
•
electrons
The
second
shell
The
EI
shell
gol
main
The
1
2
total
3
4
number
third
from
electron
removed
fourth
a
+1
needs
from
are
needs
called
more
ion.
even
a
+2
yet
more,
successive
5
of
6
7
8
electrons
9
10
to
This
energy
remove
is
the
than
it
rst
the
because
it
is
IE.
This
is
the
second
rst
because
it
is
IE.
more
ion.
energy
This
is
the
to
remove
third
it
because
it
IE.
and
so
on.
ionisation
energies
example,
sodium:
11
removed
+
Na(g)
▲ Figure 1
needs
energy
atom.
3
For
0
removed
being
These
in
shell
least
2
•
electrons
the
neutral
in
is
main
electron
a
1
•
electrons
needs
from
in
being
main
electron
removed
➝
Na
➝
Na
➝
Na
+
The successive ionisation
Na
−1
(g)
+
e
rst
IE
(g)
+
e
second
(g)
+
e
third
=
+
496
kJ
mol
=
+
4563
kJ
mol
=
+
6913
kJ
mol
2+
(g)
−1
IE
energies of sodium against number of
2+
Na
3+
(g)
−1
IE
electrons removed. Note that the log of the
ionisation energy has been plotted in order
and
so
on,
see
Table
1.
to t the large range of values on the scale
▼ Table 1
Successive ionisation energies of sodium
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
11th
496
4563
6913
9544
13 352
16 6
11
20 115
25 491
28 934
141 367
159 079
Electron removed
Ionisation energy /
–1
kJ mol
Notice
that
the
second
IE
is
not
the
energy
change
for
Study tip
2+
Na(g)
➝
Na
(g)
+
2e
The shape of the graph in Figure 1
has to be thought about carefully.
The
energy
for
this
process
would
be
(rst
IE
+
second
IE).
The rst electron removed is in the
If
you
plot
a
graph
of
the
values
shown
in
Table
1
you
get
Figure
1.
outer main shell and the 10th and
11th electrons removed are in the
Notice
group
that
of
one
eight
electron
that
are
is
relatively
more
innermost main shell.
are
26
very
difcult
to
remove.
difcult
easy
to
to
remove,
remove,
and
then
comes
nally
two
a
that
Atomic
structure
1
This
suggests
that
sodium
has:
Study tip
•
one
electron
•
eight
•
two
furthest
away
from
the
positive
nucleus
(easy
to
remove)
The energy change for the
electrons
nearer
in
to
the
nucleus
(harder
to
remove)
formation of a negative ion is
electrons
very
close
to
the
nucleus
(very
difcult
to
remove
called the electron af nity. The
because
they
are
nearest
to
the
positive
charge
of
the
nucleus).
term ionisation energy is used only
This
tells
you
about
the
number
of
electrons
in
each
main
shell
or
for the formation of positive ions.
orbit:
2,8,1.
The
eight
electrons
in
shell
2
are
in
fact
2
two
further
second
groups
main
shell,
that
but
correspond
this
is
not
to
the
visible
2s
on
sub-divided
into
6
,
2p
the
electrons
scale
of
in
Figure
the
1.
1–
can
nd
by
number
looking
at
of
the
electrons
jumps
in
in
each
main
successive
shell
ionisation
of
any
1600
Ar
energies.
1400
Jk
element
the
lom
You
/
in
ionisation
energies
across
a
period
in the
The
trends
in
main
in
Table
shells
rst
ionisation
can
also
and
give
energies
moving
information
sub-shells.
across
about
Ionisation
the
energies
a
period
energies
generally
in
of
the
electrons
increase
tsr
Periodic
noit asinoi
Periodic Table
ygrene
Trends
1200
Cl
S
1000
P
Si
800
Mg
600
a
11
across
a
period
because
the
nuclear
charge
is
increasing
and
this
Al
400
12
1
14
atomic
it
more
difcult
to
remove
an
data
for
Period
3
are
16
1
numer
electron.
▲ Figure 2
The
1
makes
shown
in
Table
2.
Trends in rst ionisation
energies across Period 3
–1
▼ Table 2
The rst ionisation energies of the elements in Period 3 in kJ mol
Na
Mg
Al
Si
P
S
Cl
Ar
496
738
578
789
1012
1000
1251
152
1
nuclear charge increasing ➝
Plotting
a
graph
of
these
values
shows
that
the
2
(Figure
2
2).
In
2
2s
,
6
(1s
,
the
2p
increase
aluminium
3s
going
orbital.
,
in
is
It
2
3s
,
from
a
(1s
increase
2
,
2s
6
,
2p
is
not
regular
2
,
3s
)
to
aluminium
1
3p
),
nuclear
in
magnesium
3p
the
ionisation
charge.
orbital
therefore
needs
This
which
less
is
is
energy
because
of
energy
energy
actually
a
the
slightly
to
goes
outer
electron
higher
remove
it,
down,
energy
see
Figure
despite
in
than
the
3.
energy
complete
removal
complete
1st
1st
removal
IE
IE
3p
3p
3s
3s
2p
2p
2s
2s
1s
1s
2
magnesium
▲ Figure 3
1s
2
2s
6
2p
2
3s
2
aluminium
1s
2
2s
6
2p
2
3s
1
3p
The rst ionisation energy of aluminium is less than that of magnesium
27
18
1.6
Electron
arrang e m e nts
and
ionisation
en erg y
2
In
Figure
2
3s
notice
the
small
3
,
3p
three
3p
2,
2
)
and
3p
orbitals
orbitals
paired
sulfur
in
2
,
2s
contains
must
contain
electrons
increase
(1s
makes
nuclear
drop
6
,
2p
just
two
it
2
,
3s
one
see
phosphorus
3p
).
In
electron,
to
Figure
while
The
remove
2s
6
,
2p
in
sulfur,
repulsion
one
of
each
one
between
them,
of
of
the
these
despite
the
3p
3s
(sub-shells)
in
phosphorus
easier
2p
1s
▲ Figure 4
Both
that
3p
2s
orbitals
to
lose
3s
(sub-shells)
in
sulfur
Electron arrangements of phosphorus and sulfur
these
cases,
conrms
which
the
go
against
existence
predicted
by
quantum
Trends
in
ionisation
of
theory
s-
the
and
and
expected
trend,
p-sub-shells.
the
Schrödinger
energies down
a
are
These
evidence
were
equation.
group
in the
Periodic Table
Figure
5
energy
groups.
further
shows
going
This
that
down
is
from
because
the
900
there
is
Group
the
nucleus
a
2,
general
and
outer
in
the
decrease
same
electron
each
in
rst
pattern
is
in
a
is
ionisation
seen
main
in
shell
other
that
gets
case.
Be
Summary questions
1–
State why the second
ionisation energy of any
ygrene
atom is larger than the rst
ionisation energy.
noitasinoi
2
850
lom Jk /
1
Sketch a graph similar to
Figure 1 of the successive
tsr
ionisation energies of
800
Mg
750
700
650
Ca
600
aluminium (electron
Sr
550
arrangement 2,8,3).
Ba
3
500
An element X has the
0
10
0
0
0
50
60
–1
following values (in kJ mol
)
atomic
numer
for successive ionisation
▲ Figure 5
The first ionisation energies of the elements of Group 2
energies: 1093, 2359, 4627
,
6229, 37 838, 47 285.
a
b
28
Going
down
Identify which group in the
might
expect
a
Periodic T
able it is in.
electron.
group,
that
However,
the
outer
shell
the
effect
of
is
the
this
the
less
nuclear
would
actual
than
charge
make
the
it
increases.
more
positive
full
difcult
charge
nuclear
‘felt’
charge.
At
to
by
rst
sight
remove
an
This
is
inner
electrons
shielding
the
nuclear
you
an
electron
because
Explain your answer to a
the
,
the
4.
2s
orbitals
2
,
phosphorus,
2p
1s
(1s
4
,
electrons.
easier
charge,
between
charge.
in
of
Practice
The
diagram
energies
of
to
the
some
right
Period
shows
3
the
rst
/ygrene
1
questions
ionisation
elements.
1400
×
1200
lom
Jk
noitasinoi
1–
×
1000
800
×
×
×
tsr
600
×
400
Na
(a)
Draw
a
cross
on
the
diagram
to
show
the
rst
ionisation
Mg
energy
Al
of
Si
P
S
Cl
aluminium.
(1
(b)
Write
an
energy
equation
of
to
show
aluminium
is
the
process
that
occurs
when
the
rst
ionisation
measured.
(2
(c)
State
which
of
the
rst,
second,
or
third
ionisations
of
2
produce
an
ion
with
the
electron
conguration
aluminium
2
2s
1s
6
marks)
would
1
2p
3s
(1
(d)
Explain
the
(e)
Identify
and
(f)
why
value
the
give
State
its
the
Explain
of
the
the
value
rst
element
trend
your
in
of
the
rst
ionisation
in
electron
mark)
Period
ionisation
energy
2
that
of
has
energy
of
sulfur
is
less
mark)
than
phosphorus.
the
highest
rst
ionisation
(2
marks)
(2
marks)
energy
conguration.
rst
answer
ionisation
in
terms
of
energies
a
in
suitable
Group
model
2
of
from
beryllium
atomic
to
barium.
structure.
(3
marks)
AQA,
2
(a)
One
isotope
of
sodium
(i)
Dene,
in
terms
of
(ii)
the
term
Explain
has
of
a
the
relative
mass
fundamental
of
23.
particles
present,
the
meaning
same
chemical
isotopes
why
isotopes
of
the
same
element
have
the
properties.
(3
(b)
Give
the
2010
electronic
conguration,
showing
all
sub-shells,
for
a
sodium
marks)
atom.
(1
mark)
28
(c)
An
atom
has
half
as
many
protons
as
an
atom
Si
of
and
also
has
six
fewer
28
neutrons
and
the
than
an
atomic
atom
of
number,
Si.
of
this
Give
the
symbol,
including
the
mass
number
atom.
(2
marks)
AQA,
3
The
values
of
the
rst
ionisation
energies
of
neon,
sodium,
and
magnesium
2004
are
–1
2080,
(a)
(b)
494,
Explain
Write
when
(c)
that
of
second
of
the
,
mol
meaning
why
that
kJ
equation
the
Explain
736
the
an
Explain
than
(d)
and
of
respectively.
the
using
term
state
ionisation
value
of
the
rst
ionisation
symbols
energy
rst
of
to
energy
illustrate
the
magnesium
ionisation
of
is
energy
an
atom.
process
(2
marks)
(2
marks)
(2
marks)
(2
marks)
occurring
measured.
of
magnesium
is
higher
sodium.
why
the
value
of
the
rst
ionisation
energy
of
neon
is
higher
than
sodium.
AQA,
2004
29
Chapter
1
Atomic
structure
54
4
A
sample
(a)
The
of
iron
from
relative
a
meteorite
abundances
spectrometer.
(i)
State
(ii)
Explain
In
what
the
is
mass
meant
how,
abundance
of
is
in
a
was
these
found
isotopes
spectrometer,
by
the
mass
to
term
contain
can
the
be
the
isotopes
determined
sample
is
rst
56
Fe,
using
a
vaporised
57
Fe,
and
and
then
ionised.
isotopes
spectrometer,
ions
are
detected
and
how
their
measured.
(5
(b)
(i)
Dene
(ii)
The
the
term
relative
found
to
be
relative
atomic
abundances
as
of
the
data
your
the
of
an
isotopes
54
m/z
Give
mass
marks)
element.
in
this
sample
of
iron
were
follows.
Relative abundance %
Use
Fe.
mass
above
answer
to
to
56
5.80
91.60
calculate
the
57
the
2.60
relative
appropriate
atomic
number
of
mass
of
iron
signicant
in
this
sample.
gures.
(2
marks)
AQA,
5
The
(a)
(b)
diagram
Explain
Explain
shows
how
why
the
layout
positive
the
(c)
Explain
how
the
(d)
Explain
how
an
of
ions
are
instrument
ions
are
a
time
of
formed
is
kept
ight
from
under
accelerated
and
mass
the
spectrometer.
sample.
(1
mark)
(1
mark)
vacuum.
separated
by
mass
in
the
instrument.
(3
(e)
The
(i)
(ii)
low
electric
resolution
current
mass
is
produced
spectrum
of
when
magnesium
Mass / charge
Relative abundance / %
24
79.0%
25
10.0%
26
11.0%
Give
the
numbers
Calculate
Give
the
your
of
protons
relative
answer
to
atomic
the
and
neutrons
mass
of
appropriate
a
an
ion
shows
in
the
sample
number
of
of
arrives
three
nuclei
at
the
(1
mark)
(1
mark)
peaks.
of
each
isotope.
magnesium.
signicant
gures.
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
marks)
detector.
(2
30
2005
marks)
2
Amount
2.1
of
Relative
atomic
masses,
and
Relative
atomic
mass
substanc e
the
the
and
mo l ec ula r
Avogad ro
c on stan t ,
mole
A
Learning
objective s:
r
The
actual
nd
by
relative
mass
in
weighing.
masses
grams
of
Instead,
are
any
the
atom
masses
or
of
molecule
atoms
is
are
too
tiny
to
compared
➔
and
atomic mass.
used.
➔
This
was
done
hydrogen,
the
in
the
past
lightest
by
dening
element,
as
1.
the
The
relative
average
atomic
mass
mass
oxygen
(for
example)
is
16
times
heavier,
to
the
of
an
nearest
State the denition of relative
molecular mass.
of
atom
➔
of
State the denition of relative
State the meaning of the
whole
Avogadro constant.
number,
so
oxygen
has
a
relative
atomic
mass
of
16.
Scientists
now
➔
use
the
isotope
carbon-12
as
the
baseline
for
relative
atomic
State what the same
mass,
numbers of moles of dierent
because
the
mass
spectrometer
has
allowed
us
to
measure
the
masses
substances have in common.
of
individual
atomic
mass
standard
the
isotopes
of
extremely
carbon-12
(dened
below)
is
is
accurately.
given
now
a
value
One-twelfth
of
accepted
exactly
by
all
1.
of
The
chemists
the
relative
➔
carbon-12
throughout
Calculate the number of moles
present in a given mass of an
world.
element or compound.
The
relative
atomic
mass
A
is
the
weighted
Specication reference: 3.1.2
average
r
mass
of
an
atom
of
an
element,
taking
into
account
1
its
naturally
occurring
isotopes,
relative
to
the
12
relative
atomic
mass
of
an
average
relative
atomic
mass
A
atom
mass
of
of
carbon-12.
one
atom
of
an
element
=
r
1
12
mass
of
one
atom
of
C
12
average
mass
of
one
atom
of
an
element
×
12
=
12
mass
Relative
molecular
mass
of
one
atom
of
C
M
r
Molecules
can
be
handled
in
the
same
Study tip
way,
by
comparing
the
mass
of
The weighted average mass must
a
molecule
with
that
of
an
atom
of
carbon-12.
be used to allow for the presence
The
relative
molecular
mass,
M
,
of
a
molecule
is
the
of isotopes, using their percentage
r
1
mass
of
that
molecule
compared
to
the
relative
abundances in calculations.
12
atomic
mass
of
an
atom
of
average
relative
molecular
mass
M
carbon-12.
mass
of
one
molecule
=
r
1
12
mass
of
one
atom
of
C
12
average
mass
of
one
molecule
×
12
=
12
mass
You
nd
the
relative
molecular
mass
by
of
one
adding
atom
up
the
of
C
relative
Study tip
atomic
this
masses
from
the
of
all
the
formula.
atoms
present
in
the
molecule,
and
you
nd
It would be useful to learn the
exact denitions of A
r
and M
r
31
2.1
Relative
atomic
and
mol ec ular
m asse s ,
t he
Avog adro
▼ Table 1
Study tip
c on sta n t ,
an d
th e
m ole
Examples of relative molecular mass
Molecule
Formula
A
of atoms
M
r
In practice, the scale based on
water
H
carbon dioxide
CO
r
(2 × 1.0) + 16.0
O
18.0
2
1
2
C = 1
2 exactly is virtually the same
12.0 + (2 × 16.0)
44.0
2
as the scale based on hydrogen = 1.
methane
12.0 + (4 × 1.0)
CH
16.0
4
This is because, on this scale,
A
for
r
hydrogen = 1.000 7
.
Relative formula
The
term
they
relative
don’t
exist
mass
formula
as
mass
molecules.
is
used
However,
for
this
ionic
has
compounds
the
same
because
symbol
M
r
▼ Table 2
Some examples of the relative formula masses of ionic compounds
Ionic compound
Formula
A
of atoms
M
r
calcium uoride
r
40.1 + (2 × 19.0)
CaF
78.1
2
sodium sulfate
2
One
atom
of
microscope
chemists
count
any
and
must
money
Working
to
element
a
is
nearest
and the
too
to
large
bank
148.3
2
impossible
in
24.3 + (2 × (14.0 + (16.0 × 3))
constant
weigh
the
)
3
The Avogadro
142.1
4
Mg(NO
magnesium nitrate
(2 × 23.0) + 32.1 + (4 × 16.0)
SO
Na
small
weigh
whole
see
with
an
individually.
numbers
(Figure
to
mole
of
them.
optical
So,
This
is
to
count
how
atoms,
cashiers
1).
number,
a
helium
atom
(A
=
4)
is
four
r
times
heavier
than
an
atom
of
hydrogen.
A
lithium
atom
(A
=
7)
r
is
▲ Figure 1
seven
times
heavier
than
an
atom
of
hydrogen.
To
get
the
same
Large numbers of coins or
number
of
atoms
in
a
sample
of
helium
or
lithium,
as
the
number
of
bank notes are counted by weighing them
atoms
In
Study tip
in
fact,
1 g
if
amount
of
you
will
hydrogen,
weigh
also
out
you
the
contain
must
take
relative
this
same
4 g
of
atomic
helium
mass
number
of
of
or
any
7 g
of
lithium.
element,
this
atoms.
The Avogadro constant is the same
The
as the number of atoms in 1 g of
same
hydrogen H
logic
applies
to
molecules.
Water
H
O,
has
a
relative
molecular
2
mass
M
, not the number of
of
18.
So,
one
molecule
of
water
is
18
times
heavier
than
one
r
2
atom
of
hydrogen.
Therefore,
18 g
of
water
contain
the
same
number
hydrogen molecules
of
Study tip
as
is
dioxide
contain
you
44
there
dioxide
If
Entity is a general word for a
molecules
times
weigh
are
heavier
this
out
atoms
same
the
in
than
1 g
an
number
relative
or
of
hydrogen.
atom
of
of
A
molecule
hydrogen,
so
of
44 g
carbon
of
carbon
molecules.
formula
mass
M
of
a
compound
in
r
grams
par ticle. It can refer to an atom,
you
have
this
same
number
of
entities
molecule, ion, electron, or the
The Avogadro
constant
simplest formula unit of a giant
The
actual
number
of
atoms
in
1
g
of
hydrogen
atoms
is
ionic structure, such as sodium
unimaginably
huge:
chloride, NaCl.
23
602 200 000 000 000 000 000 000
The
difference
between
this
usually
scale,
written
based
on
H
=
6.022
1
and
×
10
the
scale
Study tip
12
today
based
on
C,
is
negligible,
for
most
purposes.
You can also use the term molar
The
Avogadro
constant
or
Avogadro
in
12 g
number
mass, which is the mass per mole of
is
the
number
of
atoms
of
carbon-12.
–1
a substance. It has units kg mol
–1
g mol
or
–1
The
mole
. The molar mass in g mol
23
The
is the same numerically as
amount
of
M
r
called
32
a
mole
substance
that
contains
6.022
×
10
particles
is
used
Amount
of
substan ce
2
The
of
relative
atoms.
substance
mole
It
is
of
For
in
the
to
or
or
any
contains
confuse
10
of
element
molecular
mass
one
in
(or
mole
of
grams
contains
relative
entities.
formula
You
can
one
mole
mass)
also
of
a
have
a
electrons.
formula
example,
atoms
mass
relative
grams
ions
easy
give
atomic
The
10
moles
when
moles
moles
of
of
atoms
working
of
and
out
hydrogen
hydrogen
moles
the
of
mass
could
molecules,
of
a
mean
molecules,
H
,
mole
10
so
of
moles
which
always
entities.
of
hydrogen
contains
twice
2
the
number
different
of
atoms.
particles
▼ Table 3
Using
that
take
the
part
mole,
in
you
can
chemical
compare
the
numbers
of
reactions.
Examples of moles
Mass of a mole / g
Entities
Formula
Relative mass
Summary questions
= molar mass
oxygen atoms
O
16.0
1
16.0
Calculate the M
for each of the
r
oxygen molecules
O
32.0
32.0
23.0
23.0
following compounds.
2
a
+
sodium ions
Na
sodium uoride
NaF
42.0
CH
b
Na
4
c
42.0
CO
2
Mg(OH)
d
3
(NH
2
)
4
SO
2
4
Use these values for the
Number of
moles
relative atomic masses
(A
):
r
If
you
mass
want
of
a
to
nd
out
substance
how
you
many
need
to
moles
know
are
the
present
in
substance’s
a
particular
formula.
C = 12.0, H =1.0, Na = 23.0,
From
O = 16.0, Mg = 24.3, N = 14.0,
the
formula
you
can
then
work
out
the
mass
of
one
mole
of
the
S = 32.1
substance.
2
You
use:
Imagine an atomic seesaw
with an oxygen atom on one
mass
number
of
moles
m
(g)
side. Find six combinations
n
mass
of
1
mole
M
(g)
of other atoms that would
make the seesaw balance.
For example, one nitrogen
atom and two hydrogen atoms
would balance the seesaw.
Worked
example:
Finding
the
number
of
moles
to the nearest
Use values of A
r
How
many
moles
are
there
in
0.53
g
of
sodium
carbonate,
Na
CO
2
?
3
whole number.
A
Na
=
23.0,
A
r
C
=
12.0,
A
r
O
=16.0,
r
3
so
M
of
Na
r
CO
2
=
(23.0
×
2)
+
12.0
+
(16.0
×
3)
=
Calculate the number of moles
106.0,
3
in the given masses of the
so
1
mole
of
calcium
carbonate
has
a
mass
of
106.0 g.
following entities.
0.53
Number
of
moles
=
=
0.0050
mol
106.0
a
32.0 g CH
b
5.30 g Na
4
CO
2
c
3
5.83 g Mg(OH)
2
4
Worked
You
have
contains
example:
3.94 g
the
of
Finding
gold,
greater
Au,
number
the
and
of
number
2.70 g
atoms?
of
of
atoms
aluminium,
(A
Au
=
Al.
197.0,
the fewest molecules: 0.5 g
of hydrogen H
Which
A
r
Identify which contains
Al
=
, 4.0 g of
2
oxygen O
27.0)
, or 11.0 g of carbon
2
r
dioxide CO
3.94
2
Number
of
moles
of
gold
atoms
=
=
0.020
mol
197.0
5
Identify the quantity in
2.70
Number
of
moles
of
aluminium
atoms
=
=
0.100
mol
Question 3 that contains the
27.0
greatest number of atoms.
There
are
more
atoms
of
aluminium.
33
2.2
Learning
➔
Moles
objective s:
in
solutions
Solutions
Calculate the number of
A
moles of a substance from
solution
(Figure
consists
of
a
solvent
with
a
solute
dissolved
in
it,
1).
the volume of a solution
and its concentration.
Specication reference: 3.1.2
solute
Study tip
To get a solution with a
−3
concentration of 1 mol dm
you
▲ Figure 1
A solution contains a solute and a solvent
have to add the solvent to the
3
solute until you have 1 dm
of
solution. You do not add 1 mol
The
3
of solute to 1 dm
The
This would give more than
3
1 dm
units of
concentration
of solvent.
a
of solution.
concentration
known
volume
of
of
a
solution
tells
us
how
much
solute
is
present
−3
Concentrations
means
there
is
of
1
in
solution.
solutions
mole
of
are
measured
solute
per
cubic
in
mol dm
decimetre
−3
.
of
1
mol
dm
solution;
2
−3
mol dm
means
solution,
and
so
there
are
2
moles
of
solute
per
cubic
decimetre
of
on.
Study tip
Worked
example:
Finding
the
concentration
−3
in
It is impor tant to state units in the
mol
dm
answers to numerical questions.
3
1.17
of
g
of
sodium
solution.
chloride
What
is
the
was
dissolved
concentration
in
of
water
the
to
make
solution
500 cm
in
−3
mol dm
?
A
Na
=
23.0,
r
A
Cl
=
35.5
r
The mass of 1 mole of sodium chloride, NaCl, is 23.0 + 35.5 = 58.5 g.
mass
number
of
moles
n
m
(g)
=
mass
of
1
mole
M
(g)
Study tip
1.17
So
1.17
g
of
NaCl
mol
contains
1 decimetre = 10 cm, so one
=
0.020
mol
to
3
s.f.
58.5
3
cubic decimetre, 1 dm
, is
3
This
is
dissolved
in
500
cm
3
,
so
1000 cm
3
(1 dm
)
would
3
10 cm × 10 cm × 10 cm = 1000 cm
.
contain
0.040
mol
of
NaCl.
This
This is the same as 1 litre (1 l or 1 L).
means
that
the
concentration
of
−3
the
solution
is
0.040
mol dm
If you are not condent about
conversion factors and writing
The
general
way
of
nding
a
concentration
is
to
remember
relationship:
units, see Section 8,
number
Mathematical skills.
of
moles
−3
concentration
c
(mol dm
) =
3
volume
The small negative sign in
Substituting
into
this
gives:
−3
mol dm
means per and is
0.020
sometimes written as a slash,
3
mol/dm
34
concentration
=
−3
=
0.500
0.040
mol dm
V
(dm
)
n
the
Amount
of
substan ce
2
The
number of
moles
in
a
given volume of
solution
Error in
You
often
volume
have
of
a
to
work
solution
of
out
how
known
many
moles
are
concentration.
present
The
in
general
a
particular
formula
measurements
for
−3
the
number
of
moles
in
a
solution
of
concentration
c
(mol dm
)
and
Every measurement has an
3
volume
V
(cm
)
is:
inherent uncer tainty (also
−3
number
of
concentration
moles
c
(mol dm
3
)
×
volume
V
(cm
)
known as error). In general,
=
in
solution,
n
1000
the uncer tainty in a single
measurement from an
Here
is
an
example
of
how
you
reach
this
formula
in
steps.
instrument is half the value
of the smallest division. The
uncer tainty of a measurement
Worked
example:
Moles
in
a
solution
may also be expressed by ± sign
3
How
many
moles
are
present
in
24.70 cm
of
a
solution
of
at the end. For example, the
−3
concentration
0.100
mol dm
?
mass of an electron is given
−31
as 9.109 382 91 × 10
From
the
kg
denition,
−31
± 0.000 000 40 × 10
3
1000
kg , that
−3
cm
of
a
solution
of
1.00
mol dm
contains
1
mol
is, it is between 9.109 383 31
−31
and 9.109 382 51 × 10
3
So
1000
So
1.0
kg
−3
cm
of
a
solution
of
0.100
mol dm
contains
0.100
mol
3
For example, a 100 cm
3
−3
cm
of
a
solution
of
0.100
mol dm
contains
3
measuring cylinder has 1 cm
0.10
mol
=
0.000 10
mol
as its smallest division so the
1000
3
So
24.7
cm
24.7
×
−3
of
a
solution
of
0.10
mol dm
contains
measuring error can be taken
3
0.000 10
=
0.002 47
as 0.5 cm
mol
. So if you measure
3
50 cm
Using
the
formula
gives
the
same
answer:
0.5
(
c
n
×
, the percentage error is
V
)
× 100% = 1%
50
=
1000
What is the percentage error if
0.10
×
i.e.,
24.7
0.0025
to
4
3
=
=
0.002 47 mol
you use a 100 cm
signicant
1000
measuring
gures
cylinder to measure
3
a 10 cm
3
b 100 cm
Summary questions
%5.0 b
%5 a
−3
1
Calculate the concentration in mol dm
of the following.
3
a
0.500 mol acid in 500 cm
of solution
3
b
0.250 mol acid in 2000 cm
of solution
3
c
2
0.200 mol solute in 20 cm
of solution
Calculate how many moles of solute there are in the following.
3
a
20.0 cm
−3
of a 0.100 mol dm
3
b
50.0 cm
of a 0.500 mol dm
3
c
3
25.0 cm
solution
−3
solution
−3
of a 2.00 mol dm
solution
0.234 g of sodium chloride was dissolved in water to make
3
250 cm
a
of solution.
State the M
for NaCl.
r
A
Na = 23.0, A
r
Cl = 35.5
r
b
Calculate how many moles of NaCl is in 0.234 g.
c
Calculate the concentration in mol dm
−3
35
2.3
Learning
➔
The
ideal
objective s:
gas
The
State the ideal gas equation.
to
Hindenburg
use
helium
source
➔
equa t io n
of
airship
as
large
its
(Figure
lifting
volumes
gas,
of
1)
was
rather
helium
originally
than
was
designed
hydrogen,
the
USA
and
but
in
the
they
the
1930s
only
refused
to
sell
Describe how it is used to
it
to
Germany
because
of
Hitler’s
aggressive
policies.
The
airship
was
calculate the number of
3
therefore
made
to
use
hydrogen.
It
held
about
210
000
m
of
hydrogen
moles of a gas at a given
gas,
but
this
volume
varied
with
temperature
and
pressure.
volume, temperature,
The
and pressure.
volume
pressure
of
and
a
given
mass
temperature.
of
any
gas
However,
is
not
there
xed.
are
a
It
changes
number
of
with
simple
Specication reference: 3.1.2
relationships
for
temperature,
a
and
given
mass
volume
of
of
a
gas
that
connect
the
pressure,
gas.
Boyle’s law
The
product
of
temperature
pressure
remains
and
volume
is
a
constant
as
long
as
the
constant.
pressure
P
×
volume
V
=
constant
Charles’ law
The
volume
remains
▲ Figure 1
is
proportional
to
the
temperature
as
long
as
the
pressure
constant.
The German airship
volume
3
Hindenburg held about 210 000 m
volume
of
V
∝
temperature
T
V
and
=
temperature
constant
T
hydrogen gas
Gay-Lussac’s law (also called the constant volume law)
The
pressure
remains
is
proportional
to
the
temperature
as
long
as
pressure
pressure
P
∝
temperature
T
volume
P
and
=
temperature
Combining
these
pressure
relationships
P
×
volume
gives
ideal
gas
us
the
constant
T
equation:
V
=
temperature
The
the
constant.
constant
for
a
xed
mass
of
gas
T
equation
Maths link
In
one
mole
of
gas,
the
constant
is
given
the
symbol
R
and
is
called
the
If you are not sure about
gas
constant.
For
n
moles
of
gas:
propor tionality and changing
pressure
×
volume
=
number
of
×
gas
constant
×
temperature
the subject of an equation, see
3
P
(Pa)
V
(m
−1
)
moles
n
R
(J
K
−1
mol
)
T
(K)
Section 8, Mathematical skills.
PV
−1
The
value
of
R
is
8.31
=
nRT
−1
mol
J K
Study tip
This
The units used here are par t of
is
the
ideal
temperature
useful
to
gas
and
equation.
pressure
imagine
a
gas
it
No
gases
holds
which
obey
quite
obeys
the
well
it
exactly,
for
many
equation
but
at
gases.
perfectly
–
room
It
is
an
often
ideal
the Système Internationale (SI) of
units. This is a system of units for
Notes on units
measurements used by scientists
When
using
the
ideal
gas
equation,
consistent
units
must
be
used.
throughout the world. The basic
If
you
want
to
calculate
n,
the
number
of
moles:
units used by chemists are:
−2
P
must
be
in
Pa
(N m
)
T
must
be
in
K
R
must
be
in
J K
metre m, second s, Kelvin K, and
3
kilogram kg.
36
V
must
be
in
m
−1
−1
mol
gas.
Amount
of
substan ce
2
Using the
Using
mole
in
the
of
the
for
the
ideal
gas
at
gas
any
equation
any
This
ideal
largest
equation
equation,
you
temperature
refers
to
a
can
and
calculate
pressure.
particular
gas,
the
volume
Since
this
none
volume
of
of
will
one
the
be
terms
the
same
gas.
may
gas
gas
seem
very
molecules
gas
between
particle
the
unlikely
that
is
at
rst
accounts
extremely
for
sight,
the
small
but
it
is
volume
the
of
compared
a
space
gas.
with
between
Even
the
the
space
in
particles.
Worked
Rearranging
the
ideal
gas
equation
to
nd
a
volume
example:
gives:
Volume from
the
ideal
nRT
V
=
gas
equation
P
The
worked
example
tells
you
that
the
volume
of
a
mole
of
any
gas
For
temperature
example,
one
and
pressure
mole
of
is
sulfur
approximately
dioxide
gas,
24
SO
000
cm
(mass
temperature
=
20.0 °C
3
(24
64.1
If
at
3
room
g)
dm
has
).
the
(293.0
K),
100 000
pressure
Pa,
and
n
=
=
1
for
2
same
volume
as
one
mole
of
hydrogen
gas,
H
(mass
2.0
one
g).
mole
of
gas
2
nRT
In
a
similar
way,
pressure
can
be
found
using
P
−1
8.31
=
V
−1
J K
mol
×
293
K
=
V
100 000
Finding the
If
you
number of
rearrange
the
moles n of
equation
PV
=
a
nRT
gas
so
Pa
3
that
n
is
on
the
=
0.024 3
m
=
0.024 3
×
=
24 300
left-hand
6
side,
you
10
3
cm
get:
3
cm
PV
n
=
RT
If
T,
P,
and
V
Worked
How
are
known,
example:
many
moles
of
then
you
Finding
can
the
hydrogen
nd
n
number
molecules
of
are
Study tip
moles
present
in
a
volume
3
of
100
cm
at
a
temperature
−1
R
=
8.31
First,
P
be
T
must
in
must
be
be
in
in
and
a
pressure
of
100
kPa?
and to cancel the units.
mol
to
the
Pa,
base
and
units:
100
kPa
3
V
20.0 °C
−1
J K
convert
must
of
Remember to conver t to SI units
m
K,
=
100 000
Pa
3
,
and
and
−6
100 cm
20 °C
=
=
293
100
K
×
(add
10
273
3
m
to
the
temperature
in
°C)
Study tip
Substituting
into
the
ideal
gas
equation:
To conver t °C to K add 273.
PV
n
=
RT
−6
100 000
×
100
×
10
=
8.31
=
0.004 11
Finding the
×
293
moles
relative
molecular
mass of
a
gas
Study tip
If
you
know
the
number
of
moles
present
in
a
given
mass
of
gas,
3
you
can
nd
molecular
the
mass.
mass
of
one
mole
of
gas
and
this
tells
us
the
relative
Using 24 000 cm
as the volume of
a mole of any gas is not precise, and
it is always necessary to apply the
ideal gas equation in calculations.
37
2.3
The
ideal
gas
eq uat ion
Study tip
Finding the relative molecular mass of lighter fuel
In reporting a measurement, you
The apparatus used to nd the relative molecular mass of lighter fuel is
should include the best value (e.g.,
shown in Figure 2.
the average) and an estimate of its
uncertainty. One common practice
is to round o the experimental
result so that it contains the digits
known with certainty plus the rst
uncertain one. The total number of
digits is the number of signicant
pressurised
gures used.
e.g.,
▲ Figure 2
Study tip
lighter
gas,
fuel
Measuring the relative molecular mass of lighter fuel
The lighter fuel canister was weighed.
3
In the equation PV = nRT, the units
1000 cm
of gas was dispensed into the measuring cylinder, until the levels
3
must be P in Pa (not kPa), V in m
,
of the water inside and outside the measuring cylinder were the same, so that
the pressure of the collected gas was the same as atmospheric pressure.
and T in K
The canister was reweighed.
Summary questions
Atmospheric pressure and temperature were noted.
The results were
1
a
Calculate approximately
how many moles of H
loss of mass of the can
=
2.29 g
temperature
=
14 °C = 287 K
atmospheric pressure
=
100 000 Pa
Volume of gas
=
1000 cm
n
=
2
molecules were contained
in the Hindenburg airship at
298 K.
b
The original design used
3
−6
= 1000 × 10
helium. State how many
PV
moles of helium atoms it
RT
−6
would have contained.
100 000 × 1000 × 10
=
2
a
8.31 × 287
Calculate the volume of 2
moles of a gas if the
=
0.042 mol
temperature is 30 °C, and
0.042 mol has a mass of 2.29 g
the pressure is 100 000
2.29
Pa.
= 54.5 g
So, 1 mol has a mass of
0.042 g
b
Calculate the pressure of
So, M
r
0.5 moles of a gas if the
3
volume is 11 000 cm
, and
the temperature is 25 °C.
3
Calculate how many moles
of hydrogen molecules are
present in a volume of 48 000
3
cm
4
, at 100 000 Pa and 25 °C.
State how many moles of
carbon dioxide molecules
would be present in Question 3.
Explain your answer.
38
= 54.5
3
m
2.4
The
The
Empirical
empirical
formula
number
compound.
,
mol e c ul ar
empirical formula
whole
CO
and
tells
ratio
For
us
of
is
example,
that
for
formula
atoms
the
every
Learning
the
the
of
empirical
carbon
that
each
represents
element
formula
atom
there
of
are
the
present
carbon
two
simplest
in
➔
a
oxygen
an
empirical
State the denitions of
molecular formula.
atoms.
➔
nd
objective s:
empirical formula and
dioxide,
2
To
for mu la e
Calculate the empirical
formula from the masses
formula:
or percentage masses of
1
Find
the
masses
of
each
of
the
elements
present
in
a
compound
the elements present in
(by
experiment).
a compound.
2
Work
out
the
number
of
moles
of
atoms
mass
number
3
of
Convert
the
number
ratio.
moles
number
of
of
each
element.
➔
element
Calculate the additional
=
of
mass
of
1
mol
moles
of
each
of
information needed to work
element
element
into
a
out a molecular formula from
whole
an empirical formula.
Specication reference: 3.1.2
Worked
of
example:
calcium
10.01 g
of
a
Finding
empirical formula
carbonate
white
solid
Study tip
contains
4.01 g
of
calcium,
1.20 g
of
The mass of 1 mole in grams is the
carbon,
and
4.80 g
of
oxygen.
What
is
its
empirical
formula?
same as the relative atomic mass
(A
Ca
=
40.1,
A
r
C
=
12.0,
A
r
Step
Step
1
2
Find
the
O
=
of
each
calcium
=
Mass
of
carbon
=
1.20
g
Mass
of
oxygen
=
4.80
g
A
Ca
=
number
4.01
element.
of
the
of the element.
r
masses
Mass
Find
16.0)
of
g
moles
of
atoms
of
each
element.
40.1
r
4.01
Number
of
moles
of
calcium
=
=
0.10
mol
40.1
A
C
=
12.0
r
1.2
Number
of
moles
of
carbon
=
=
0.10
mol
12.0
A
O
=
16
r
4.8
Number
of
moles
of
oxygen
=
=
0.30
mol
16.0
Step
3
Find
Ratio
the
in
simplest
moles
ratio.
of
calcium
0.10
So
The
the
formula
simplest
is
whole
therefore
number
ratio
is:
1
: carbon :
oxygen
:
0.10
:
0.30
:
1
:
3
CaCO
3
39
2.4
Empirical
Worked
copper
0.795 g
when
and
example:
for m ula e
Finding
the
empirical formula
of
black
of
copper
in
a
oxide
stream
copper
of
oxide?
is
reduced
hydrogen
A
Cu
=
to
1
Find
the
Mass
of
Started
copper
masses
copper
with
were
of
=
63.5,
A
left,
O
1).
of
copper
What
is
the
=16.0
r
each
0.635
0.795
0.635 g
(Figure
r
Step
of
oxide
heated
formula
m o l e c ular
g
of
element.
g
copper
oxide
and
0.635
g
of
so:
▲ Figure 1
Mass
Step
2
Find
of
oxygen
the
=
number
0.795
of
−
0.635
moles
of
=
0.160
atoms
of
g
each
element.
+
A
Cu
=
Finding the empirical
formula of copper oxide
Finding empirical
63.5
r
formula of copper
0.635
Number
of
moles
of
copper
=
=
0.01
63.5
A
O
oxide
=16.0
r
0.16
Number
of
moles
of
oxygen
=
In this experiment explain
=
0.01
16.0
Step
3
Find
the
simplest
why:
ratio.
1
The
ratio
of
copper
0.01
:
:
moles
of
copper
to
moles
of
oxygen
is:
of the tube
2
this ame goes green
3
droplets of water form
oxygen
0.01
near the end of the tube
4
So
the
there is a ame at the end
simplest
whole
number
ratio
is
1
:
the ame at the end of the
1
tube is kept alight until
The
Cu
simplest
to
one
O,
formula
CuO.
of
You
black
may
copper
nd
it
oxide
easier
is
to
therefore
make
Copper Cu
Oxygen O
0.160 g
63.5
of element
the apparatus is cool.
table.
0.635 g
mass of element
A
a
one
16.0
r
mass of element
0.635
0.160
= 0.01
number of moles =
A
63.5
= 0.01
16.0
r
1
ratio of elements
+
1
Erroneous results
One student carried out the experiment to nd the formula of black copper
oxide with the following results:
0.735 g of the oxide was reduced to 0.635 g after reduction.
1
Conrm that these results lead to a ratio of 0.01 mol, copper to
0.006 25 mol of oxygen, which is incorrect.
2
Suggest what the student might have done wrong to lead to this
apparently low value for the amount of oxygen.
40
Amount
of
substan ce
2
Finding the
Sometimes
that
are
simplest
you
not
will
easy
to
ratio of
end
up
elements
with
convert
to
ratios
whole
of
moles
numbers.
of
If
atoms
you
of
elements
divide
+
Another oxide of
each
copper
number
(or
by
ratios
the
you
smallest
can
number
recognise
you
more
will
end
easily).
up
Here
with
is
an
whole
numbers
There is another oxide of copper,
example.
which is red. In a reduction
experiment similar to that for
nding the formula of black
Worked
example:
Empirical formula
copper oxide, 1.43 g of red copper
Compound
X
contains
50.2 g
sulfur
and
50.0 g
oxygen.
What
is
its
oxide was reduced with a stream
empirical
formula?
A
S
=
32.1,
A
r
Step
1
Find
the
O
=
16.0
of hydrogen and 1.27 g of copper
r
number
of
moles
of
atoms
of
each
element.
were formed. Use the same steps
as for black copper oxide to nd
A
S
=
32.1
r
the formula of the red oxide.
50.2
Number
of
moles
of
sulfur
=
=
1.564
32.1
A
O
1
=16
Find the masses of each
r
element.
50.0
Number
of
moles
of
oxygen
=
=
3.125
16.0
Step
2
Find
the
simplest
2
ratio.
of atoms of each element.
3
Ratio
Now
of
sulfur
divide
Ratio
of
:
oxygen
each
sulfur
:
of
Find the number of moles
the
:
1.564
:
Find the simplest ratio.
3.125
numbers
by
the
smaller
number.
oxygen
Synoptic link
1.564
3.125
:
1.564
=
1:2
1.564
Having more than one oxide (and
other compounds) with dierent
The
empirical
formula
is
therefore
SO
.
Sometimes
you
may
end
2
formulae is a typical proper ty of
up
with
a
ratio
of
moles
of
atoms,
such
as
1:1.5.
In
these
cases
transition metals, see Topic 23.1,
you
must
nd
a
whole
number
ratio,
in
this
case
2:3.
The general proper ties of transition
metals.
Finding the
molecular formula
Study tip
The
molecular
each
element
formula
in
one
gives
molecule
the
of
actual
the
number
compound.
of
(It
atoms
applies
of
only
to
•
substances
that
exist
as
When calculating empirical
molecules.)
formulae from percentages,
check that all the percentages of
The
empirical
formula.
formula
There
the
molecular
For
example,
may
is
be
not
always
several
units
the
of
same
the
as
the
molecular
empirical
formula
the compositions by mass add
in
up to 100%. (Don’t forget any
formula.
oxygen that may be present.)
ethane
(molecular
formula
C
H
2
)
would
have
an
6
•
empirical
formula
of
Remember to use relative
CH
3
atomic masses from the Periodic
To
nd
the
number
of
units
of
the
empirical
formula
in
the
T
able, not the atomic number.
molecular
formula,
relative
divide
mass
the
of
relative
the
molecular
empirical
mass
by
the
formula.
Synoptic link
For
example,
28.0
but
its
ethene
is
empirical
found
to
formula,
have
,
CH
a
relative
has
a
molecular
relative
mass
of
mass
of
Once we know the formula of a
14.0.
2
Relative
molecular
mass
of
ethene
mass
of
empirical
formula
of
ethene
compound we can use techniques
28.0
=
=
Relative
2
such as infrared spectroscopy
14.0
and mass spectrometry to
So
there
must
be
two
units
of
the
empirical
formula
in
the
molecule
of
help work out its structure, see
ethene.
So
ethene
is
)
(CH
2
or
2
C
H
2
4
Chapter 16, Organic analysis.
41
2.4
Empirical
+
and
m o l e c ular
for m ula e
He,
O
2
Combustion analysis
Organic compounds are based on carbon and hydrogen. One method of nding
sample
in
empirical formulae of new compounds is called combustion analysis (Figure 2).
platinum
It is used routinely in the pharmaceutical industry. It involves burning the
capsule
He/H
unknown compound in excess oxygen and measuring the amounts of water,
O,
SO
2
,
CO
2
,
N
2
2
carbon dioxide, and other oxides that are produced. The gases are carried through
H
O
IR
2
the instrument by a stream of helium.
The basic method measures carbon, hydrogen, sulfur, and nitrogen. It is assumed that
drying
oxygen makes up the dierence after the other four elements have been measured. Once
agent
the sample has been weighed and placed in the instrument, the process is automatic and
CO
,
N
2
2
He,
SO
2
controlled by computer.
SO
IR
2
The sample is burnt completely in a stream of oxygen. The nal combustion products are
CO
,
He,
N
2
SO
2
2
water, carbon dioxide, and sulfur dioxide. The instrument measures the amounts of these by
infrared absorption. They are removed from the gas stream leaving the unreacted nitrogen
CO
IR
2
which is measured by thermal conductivity. The measurements are used to calculate the
CO
,
He,
N
2
SO
2
2
masses of each gas present and hence the masses of hydrogen, sulfur, carbon, and nitrogen
soda
in the original sample. Oxygen is found by dierence.
lime
He,
N
2
Traditionally, the amounts of water and carbon dioxide were measured by absorbing them in
suitable chemicals and measuring the increase in mass of the absorbents. This is how the
composition data for the worked example below were measured. The molecular formula can
then be found, if the relative molecular mass has been found using a mass spectrometer.
out
▲ Figure 2
Soda lime is a mix ture containing mostly calcium hydroxide, Ca(OH)
Combustion analysis
. Construct a
2
balanced symbol equation for the reaction of calcium hydroxide with carbon dioxide.
2
O
Worked
An
example:
organic
carbon
and
OCaC
➝
2
2
OC +
)HO(aC
Molecular formula
compound
13.04%
3
H +
containing
hydrogen.
only
What
is
carbon,
its
hydrogen,
molecular
and
formula
oxygen
if
M
=
was
found
to
have
52.17%
46.0?
r
100.00 g
Step
1
of
this
Find
compound
the
empirical
would
contain
52.17 g
carbon,
hydrogen
and
(the
formula.
Carbon
mass of element/g
A
13.04 g
Hydrogen
52.1
7
of element
Oxygen
13.04
12.0
34.79
1.0
16.0
r
mass of element
52.1
7
number of moles =
13.04
= 4.348
A
12.0
34.79
= 13.04
= 2.1
74
1.0
16.0
r
divide through by the smallest
4.348
13.04
= 2
2.1
74
ratio of elements
So
the
formula
is
C
H
2
Step
2
Find
M
of
the
empirical
= 1
2.1
74
2
empirical
2.1
74
= 6
2.1
74
6
1
O.
6
formula.
r
(2
×
12.0)
C
So,
the
+
(6
×
1.0)
+
(1
H
molecular
×
16.0)
=
46.0
O
formula
is
the
same
as
the
empirical
formula,
C
H
2
42
O.
6
rest)
34.79 g
oxygen.
Amount
of
substan ce
2
Worked
0.53
and
g
example:
of
a
compound
0.54
g
of
molecular
To
Molecular formula
water
formula
calculate
the
X
on
if
containing
complete
its
relative
empirical
only
by
combustion
carbon,
combustion
molecular
in
hydrogen,
oxygen.
mass
is
analysis
and
What
oxygen,
is
its
gave
empirical
1.32
g
of
formula?
carbon
What
dioxide
is
its
58.0?
formula:
1.32
carbon
1.32
g
of
CO
(M
2
=
44.0)
is
=
0.03
mol
CO
r
2
44.0
As
each
mole
of
CO
has
1
mole
of
C,
the
sample
contained
0.03
mol
of
C
atoms.
2
0.54
hydrogen
0.54
g
of
H
O
(M
2
=
18.0)
is
=
0.03
mol
H
r
O
2
18.0
As
each
mole
of
H
O
has
2
moles
of
H,
the
sample
contained
0.06
mol
of
H
atoms.
2
oxygen
0.03
mol
of
carbon
atoms
(A
=
12.0)
has
a
mass
of
0.36
g
r
0.06
mol
of
hydrogen
atoms
(A
=
1.0)
has
a
mass
of
0.06
g
r
Total
The
rest
(0.58
−
mass
0.42)
of
carbon
must
be
and
hydrogen
oxygen,
so
the
is
0.42
sample
g
contained
0.16
g
of
oxygen.
0.16
0.16
g
of
oxygen
(A
=
16.0)
is
=
0.01
mol
oxygen
atoms
r
16.0
So
the
sample
Dividing
so
the
by
contains
the
0.03
smallest
empirical
number
formula
is
C
H
3
M
of
this
unit
is
58,
so
mol
the
C,
0.06
0.06
gives
mol
the
H,
and
0.01
ratio:
O.
mol
O
C
H
3
6
O
1
6
molecular
formula
is
also
C
r
H
3
O.
6
Summary questions
1
2
3
Calculate the empirical formula of each of the following compounds?
a
A liquid containing 2.0 g of hydrogen, 32.1 g sulfur, and 64.0 g oxygen.
b
A white solid containing 4.0 g calcium, 3.2 g oxygen, and 0.2 g hydrogen.
c
A white solid containing 0.243 g magnesium and 0.7
10 g chlorine.
3.888 g magnesium ribbon was burnt completely in air and 6.448 g of magnesium oxide was produced.
a
Calculate how many moles of magnesium and oxygen are present in 6.448 g of magnesium oxide.
b
State the empirical formula of magnesium oxide.
State the empirical formula of each of the following molecules.
a
cyclohexane, C
H
6
4
(You could try to name them too. )
M
b
dichloroethene, C
12
H
2
Cl
2
c
2
benzene, C
H
6
6
for ethane1,2-diol is 62.0. It is composed of carbon, hydrogen, and oxygen in the ratio by moles of 1 : 3 : 1.
r
Identify its molecular formula.
5
An organic compound containing only carbon, hydrogen, and oxygen was found to have 62.07% carbon and
10.33% hydrogen. Identify the molecular formula if
M
= 58.0.
r
6
A sample of benzene of mass 7
.8 g contains 7
.2 g of carbon and 0.6 g of hydrogen. If
M
is 78.0, identify:
r
a
the empirical formula
b
the molecular formula.
43
2.5
Balanced
equations
and
rel ate d
calculations
Learning
objective s:
Equations
➔
represent
what
happens
when
chemical
reactions
take
Demonstrate how an
place.
equation can be balanced
are
They
are
reactants.
based
After
on
experimental
these
have
evidence.
reacted
you
end
The
up
starting
with
materials
products.
if the reactants and products
reactants
products
➝
are known.
Word
➔
equations
only
give
the
names
of
the
reactants
and
products,
Calculate the amount of a
for
example:
product using experimental
hydrogen
+
oxygen
Once
the
idea
of
atoms
water
➝
data and a balanced equation.
had
been
established,
chemists
realised
that
Specication reference: 3.1.2
atoms
two
react
together
hydrogen
water
in
simple
molecules
react
hydrogen
molecules
+
2
in
ratio
by
to
in
simple
You
can
which
whole
build
working
a
Balanced
There
of
ratios.
oxygen
amounts
of
is
number
For
example,
molecule
to
give
two
the
2
products
are
stoichiometry
relationship
moles
that
molecules
:
and
the
water
from
react
of
produced,
the
reaction.
experimental
together.
This
data
leads
us
equations
equations
the
arrow.
use
same
the
(This
also
that
be
formulae
number
is
of
because
reactions.)
substances
can
react
2
➝
1
called
of
molecule
equation.
chemical
symbols
oxygen
stoichiometric
are
the
in
1
reactants
symbol
symbol
sides
a
symbol
products.
destroyed
the
the
Balanced
State
one
:
numbers,
up
out
balanced
both
with
number
molecules.
2
The
whole
react
atoms
Balanced
together
added.
These
of
atoms
reactants
of
each
are
never
equations
and
are
are
letters,
and
element
on
created
tell
us
or
about
the
produced.
in
brackets,
which
Study tip
can
Learn these four state symbols.
be
added
reactants
means
gas,
Writing
When
You
to
and
the
and
(aq)
balanced
aluminium
can
build
formulae
formulae
products
of
up
the
are
means
–
equations
(s)
means
aqueous
to
say
solid,
solution
what
(l)
(dissolved
in
oxygen
balanced
reactants
it
symbol
and
forms
solid
equation
Write
the
word
Write
in
the
in
product
+
correct
–
Al,
O
,
this
and
Al
and
O
2
3
oxygen
➝
aluminium
+
O
Al
➝
2
•
is
not
there
side)
•
there
side)
44
balanced
is
one
but
are
but
oxide.
the
two
on
the
on
atom
products
oxygen
three
O
2
3
because:
aluminium
two
oxide
formulae
Al
This
(g)
water).
equation
aluminium
2
the
liquid,
aluminium
from
2
1
state
means
equations
burns
a
in
in
the
atoms
on
the
side
on
products
reactants
(right-hand
the
side
reactants
side
(left-hand
side)
side
(right-hand
(left-hand
side).
Amount
of
substan ce
2
3
To
of
get
the
two
aluminium
atoms
on
the
left-hand
2Al
+
O
Al
➝
2
Now
4
If
the
you
side
aluminium
multiply
aluminium
the
oxide
is
correct
oxygen
by
2,
you
return
left-hand
not
the
left-hand
have
+
to
the
six
the
O
3O
on
side
each
6
of
is
+
balanced
each
numbers
called
by
need
and
the
O
2
You
3,
3
four
Al
on
the
side:
equation
atoms
The
front
side:
2Al
3O
2Al
➝
2
of
in
oxygen.
➝
aluminium.
4Al
The
2
3
2
Now
a
O
2
but
on
you
2Al
5
put
Al:
in
element
front
of
because
on
the
O
2
there
both
sides
formulae
are
of
(4,
the
the
3,
3
same
numbers
equation.
and
2)
are
coefcients
You
can
add
state
The
equation
symbols.
tells
you
the
numbers
of
moles
of
each
of
the
Hint
substances
masses
that
that
are
will
involved.
react
From
together:
this
(using
you
Al
=
can
work
27.0,
O
=
out
the
16.0)
Since we know that 1 mole of
4Al(s)
+
3O
(g)
2Al
➝
2
O
2
any gas at room conditions has a
(s)
3
3
volume of approximately 24 dm
4
moles
3
moles
2
moles
we can see that the volume of
108.0
g
96.0
g
204.0
g
oxygen required is approximately
The
total
mass
is
the
same
on
both
sides
of
the
equation.
This
is
3
3 x 24 dm
another
good
way
of
checking
whether
the
equation
is
3
, i.e., 72 dm
balanced.
Ionic equations
In
some
ions
reactions
present.
overall
reaction.
solution,
the
you
reaction
you
can
Sometimes
For
end
example,
up
with
between
HCl(aq)
simplify
there
a
are
when
salt
equation
that
any
(also
hydrochloric
+
the
ions
in
acid
NaOH(aq)
do
acid
by
not
reacts
solution)
and
considering
take
with
and
sodium
in
an
the
the
alkali
water.
in
Look
at
hydroxide:
NaCl(aq)
➝
part
+
H
O(l)
2
hydrochloric
The
ions
acid
present
+
sodium
hydroxide
➝
sodium
chloride
water
+
are:
+
HCl(aq)
H
(aq)
and
Cl
(aq)
+
NaOH(aq)
Na
NaCl(aq)
Na
(aq)
and
OH
(aq)
and
Cl
(aq)
+
If
you
that
write
appear
the
on
equation
each
side
+
using
we
these
(aq)
ions
and
then
strike
out
the
ions
have:
+
+
H
+
OH
(aq)
➝
+
H
O(l)
Study tip
2
Overall,
the
equation
The charges balance as well as the
is
elements. On the left +1 and −1
+
H
(aq)
+
OH
(aq)
➝
H
O(l)
2
(no overall charge) and no charge
+
Na
in
(aq)
the
and
Cl
(aq)
are
called
spectator
ions
–
they
do
not
take
part
on the right.
reaction.
45
2.5
Balanced
equat io ns
gas
and
rel at ed
ca lc u lation s
Whenever
syringe
same
as
an
the
acid
one
reacts
with
an
alkali,
the
overall
reaction
you
cannot
change
will
be
the
above.
Useful tips for balancing equations
•
dilute
hydrochloric
You
must
make
•
magnesium
You
the
the
correct
can
equation
formulae
–
them
to
only
balance.
change
the
numbers
of
atoms
by
putting
a
number,
ribbon
called
▲ Figure 1
use
acid
a
coefcient,
in
front
of
formulae.
Apparatus for collecting
•
The
coefcient
that
substance
in
front
of
the
symbol
tells
you
how
many
moles
of
hydrogen gas
•
It
often
many
•
takes
steps
When
are
reacting.
more
than
suggests
dealing
with
one
that
ionic
step
you
to
may
balance
have
equations
the
an
an
equation,
incorrect
total
of
the
but
too
formula.
charges
on
each
Synoptic link
side
must
also
be
the
same.
See Practical 1 on page 52
1.
Working out
You
can
product
use
is
a
Worked
How
example: Calculating
much
acid?
A
magnesium
Mg
=
24.3,
A
r
(The
1
H
=
1.0,
is
A
r
word
Step
chloride
‘excess’
Write
the
the
=
and
symbol
from
a
equation
reaction.
hydrochloric
acid
For
produces
mass
of
means
correct
there
is
atoms
the
on
equation.
the
by
0.120 g
of
magnesium
ribbon
and
than
enough
acid
to
react
with
all
the
+
HCl(aq)
MgCl
➝
(aq)
+
the
The
right-hand
hydrochloric acid
+
number
side
so
+
numbers
of
1
of
Mg
has
a
shown
excess
hydrochloric
mass
of
of
you
+
that
Mg
atoms
need
to
is
add
➝
correct.
a
2
in
There
front
MgCl
of
(aq)
are
the
+
two
g
2
because
mol
its
Cl
atoms
and
two
H
HCl.
(g)
H
2
react.
2HCl(aq)
mol
24.3
magnesium.)
hydrogen
+
➝
MgCl
(aq)
+
A
=
➝
1
(g)
H
2
mol
as
2
magnesium chloride
➝
2HCl(aq)
moles
Mg(s)
1
gas,
between
(g)
H
2
nd
hydrogen
much
reaction
equation.
Mg(s)
Now
how
the
35.5
more
formulae
magnesium
Balance
out
product
2
2
work
r
Mg(s)
Step
to
example,
1.
produced
Cl
balanced
produced
magnesium
Figure
amounts
2
mol
1
mol
24.3.
r
0.12
So,
0.12
g
of
Mg
is
=
0.0049
mol.
24.3
From
the
chloride.
M
MgCl
r
So
equation,
Therefore,
=
24.3
+
you
can
0.0049
(2
×
see
mol
35.5)
=
that
of
one
mole
magnesium
of
magnesium
produces
95.3
2
the
mass
of
MgCl
=
2
46
0.0049
×
95.3
=
0.47 g
to
2
s.f.
reacts
0.0049
to
mol
of
give
one
mole
magnesium
of
magnesium
chloride.
in
Amount
of
substan ce
2
Finding concentrations using titrations
burette
Titrations can be used to nd the concentration of a solution, for example, an
acid
in
burette
alkali by reacting an acid with the alkali using a suitable indicator.
You need to know the concentration of the acid and the equation for the
reaction between the acid and alkali.
The apparatus is shown in Figure 2.
The steps in a titration are:
alkali
1
Fill a burette with the acid of known concentration.
2
Accurately measure an amount of the alkali using a calibrated pipette
in
▲ Figure 2
and
indicator
ask
Apparatus for a titration
and pipette ller.
3
Add the alkali to a conical ask with a few drops of a suitable indicator.
4
Run in acid from the burette until the colour just changes, showing that
the solution in the conical ask is now neutral.
5
Summary questions
Repeat the procedure, adding the acid dropwise as you approach the end
point, until two values of the volume of acid used at neutralisation are
1
Balance the following equations.
the same, within experimental error.
a
Mg + O
➝ MgO
2
b
+ HCl
Ca(OH)
2
➝ CaCl
+ H
2
Worked
example:
Finding
concentration
c
Na
2
3
➝ NaNO
3
25.00
cm
of
a
solution
of
sodium
hydroxide,
NaOH,
of
unknown
3
concentration
was
neutralised
by
22.65
−3
cm
of
a
0.100
+ H
3
2
mol dm
O
2
O + HNO
O
2
State the concentration of
3
solution
of
hydrochloric
acid,
HCl.
What
is
the
concentration
hydrochloric acid if 20.0 cm
of
3
the
is neutralised by 25.0 cm
alkali?
of sodium hydroxide of
First
write
a
balanced
symbol
equation
and
then
the
numbers
of
3
concentration 0.200 mol dm
moles
that
react:
3
NaOH(aq)
+
HCl(aq)
NaCl(aq)
➝
+
H
sodium hydroxide
sodium chloride
hydrochloric acid
In the reaction
O(l)
Mg(s) + 2HCl(aq)
2
water
➝ MgCl
(aq) + H
2
1
mol
1
mol
1
mol
1
2
mol
2.60 g of magnesium
3
1
mol
of
sodium
hydroxide
reacts
with
1
mol
of
hydrochloric
acid.
was added to 100 cm
3
of 1.00 mol dm
c
number
of
moles
of
HCl
×
V
=
22.65
×
0.100
=
hydrochloric acid.
1000
From
the
sodium
equation,
hydroxide
there
and
must
be
an
hydrochloric
1000
equal
acid
number
for
of
moles
a
of
State if there is any
magnesium left when the
neutralisation:
reaction nished. Explain
number
of
moles
of
NaOH
=
number
of
moles
of
HCl
your answer.
22.65
×
0.100
3
So
you
must
mol
have
of
NaOH
in
the
25.00 cm
b
Calculate the volume of
1000
hydrogen produced at
of
sodium
hydroxide
solution.
25 °C and 100 kPa.
3
The concentration of a solution is the number of moles in 1000 cm
4
Therefore
the
concentration
of
the
a
Write the balanced
alkali
equation for the reaction
22.65
×
0.100
1000
−3
=
×
1000
mol
dm
−3
=
0.0906
mol
between sulfuric acid and
dm
25.00
sodium hydroxide
A
note
22.65
can
on
and
quote
signicant
25.00
the
both
answer
gures
have
to
3
4
s.f
s.f.
but
only.
0.100
So
has
only
rounding
up
3
s.f.
the
So
nal
we
digit
b
i
in full
ii
in terms of ions.
Identify the spectator ions
−3
gives
the
concentration
of
the
alkali
as
0.091
mol
dm
to
3
s.f.
in this reaction.
47
2.6
Balanced
equations,
economies,
Learning
objective s:
Once
can
➔
and
you
know
calculate
at o m
pe rc e nta g e
the
the
balanced
theoretical
equation
amount
for
that
a
yiel d s
chemical
you
should
reaction,
be
able
to
you
make
Describe the atom economy
of
any
of
the
products.
Most
chemical
reactions
produce
two
(or
more)
that
some
of a chemical reaction.
products
➔
but
often
only
one
of
them
is
required.
This
means
State how an equation is
of
the
products
will
be
wasted.
In
a
world
of
scarce
resources,
this
is
used to calculate an atom
obviously
not
a
good
idea.
One
technique
that
chemists
use
to
assess
a
economy.
given
➔
process
is
to
determine
the
percentage
atom
economy.
Describe the percentage yield
of a chemical reaction.
Atom
➔
economy
Calculate percentage yields.
The
Specication reference: 3.1.2
atom
equation.
economy
It
is
of
a
theoretical
reaction
rather
is
mass
%
atom
economy
found
than
of
can
see
real
reaction.
Chlorine,
what
Cl
,
atom
reacts
is
from
the
dened
mass
means
sodium
of
by
balanced
as:
product
×
economy
with
desired
It
=
total
You
directly
practical.
100
reactants
considering
hydroxide,
NaOH,
to
the
following
form
sodium
2
chloride,
NaCl,
water,
H
O,
and
sodium
chlorate,
NaOCl.
Sodium
2
chlorate
From
is
the
product
used
as
household
equation
you
can
bleach
work
–
out
this
the
is
the
mass
useful
of
each
reactant
and
involved.
2NaOH
+
Cl
NaCl
➝
+
H
2
2
product.
mol
1
80.0 g
Total
mol
➝
71.0 g
➝
1
Atom
mol
1
58.5 g
economy
NaOCl
+
of
desired
of
74.5 g
product
×
mass
mol
151.0 g
=
total
1
mol
18.0 g
Total
151.0 g
mass
%
O
2
100
reactants
74.5
=
×
100
151
=
So
only
49.3%
product,
It
may
the
of
rest
the
is
starting
be
easier
to
Those
coloured
those
two
of
in
red
NaOH
+
see
are
hydrogen,
materials
are
included
in
the
desired
wasted.
involved.
and
49.3%
what
in
wasted
and
one
NaOH
+
has
happened
green
–
of
one
are
if
you
included
atom
of
colour
in
the
sodium,
the
nal
one
of
atoms
product
chlorine,
oxygen.
ClCl
→
NaCl
+
H
O
+
NaOCl
2
Another
ethene
example
(the
C
H
2
is
product
OH
➝
the
reaction
wanted)
CH
5
46.0 g
ethanol
water
CH
2
➝
where
and
+
(which
H
2
O
2
28.0 g
18.0 g
28.0
%
Atom
economy
=
×
46.0
48
100
=
60.9%
breaks
is
down
wasted).
to
Amount
of
substan ce
2
Some
For
reactions,
example,
CH
in
theory
ethene
CH
2
at
reacts
+
least,
with
Br
2
have
bromine
CH
➝
2
28.0 g
wasted
to
form
BrCH
2
160.0 g
Total
no
atoms.
1,2-dibromoethane
Br
2
188.0 g
➝
188.0 g
Total
188.0 g
188.0
%
Atom
economy
=
×
(28
+
.0
+
100
=
100%
160.0)
Atom economies
Atom economy – a dangerous fuel
Hydrogen can be made by passing steam over heated coal, which is
There are clear advantages for
largely carbon.
industry and society to develop
chemical processes with
C(s) + 2H
O(g) ➝ 2H
2
(g) + CO
2
(g)
2
high atom economies. A good
example is the manufacture of
1
2.0 + (2 × 18.0) ➝ ( 2 × 2.0) + 44.0
the over-the-counter painkiller
As the only useful product is hydrogen, the atom economy of this reaction
and anti-inammatory
4.0
is
(
48.0
)
× 100% = 8.3% – not a very ecient reaction! The reason that it is so
drug ibuprofen. The original
manufacturing process had an
inecient is that all of the carbon is discarded as useless carbon dioxide.
atom economy of only 44%, but
However, under dierent conditions a mixture of hydrogen and carbon
a newly-developed process has
monoxide can be formed (this was called water gas or town gas).
improved this to 77%.
C(s) + H
O(g) ➝ H
2
(g) + CO(g)
2
Both hydrogen and carbon monoxide are useful fuels, so nothing is discarded
and the atom economy is 100%. You can check this with a calculation if you like.
Carbon monoxide is highly toxic. However, almost incredibly to modern eyes,
town gas was supplied as a fuel to homes in the days before the country
converted to natural gas (methane, CH
) from the North Sea.
4
Even methane is not without its problems. When it burns in a poor supply
of oxygen, carbon monoxide is formed and this can happen in gas res in
poorly-ventilated rooms. This has sometimes happened in student ats, for
example, where windows and doors have been sealed to reduce draughts and
cut energy bills resulting in a lack of oxygen for the gas re. Landlords are now
recommended to t a carbon monoxide alarm.
Write a balanced formula equation for the formation of carbon monoxide by
the combustion of methane in a limited supply of oxygen.
2
O
The
The
•
percentage yield of
yield
The
of
atom
wasted
•
The
a
in
yield
much
is
a
the
b
as
a
reaction
economy
a
is
a
different
tells
us
in
chemical
from
theory
the
4
O3 +
HC2
reaction
atom
how
2
H4 + OC2 ➝
economy.
many
atoms
must
be
reaction.
tells
lost
us
about
the
practical
efciency
of
the
process,
how
by:
practical
result
process
of
of
reactions
obtaining
that
do
a
not
product
go
to
and
completion.
49
2.6
Balanced
equat io ns,
atom
ec on o mi es ,
As
and
you
percen ta g e
have
chemical
you
the
seen,
once
reaction,
should
be
reaction
you
able
goes
yiel ds
to
to
2KI(aq)
you
can
get
know
from
+
Pb(NO
)
potassium iodide
2
1
Lime
For
(aq)
1
symbol
amount
amounts
of
of
equation
any
product
starting
for
a
that
materials
if
example:
PbI
➝
(s)
+
3
lead iodide
➝
1
mol
(aq)
2KNO
2
2
lead nitrate
mol
balanced
the
given
completion.
3
Summary questions
the
calculate
potassium nitrate
mol
2
mol
(calcium oxide, CaO )
332 g
461 g
331 g
202 g
is made by heating limestone
)
(calcium carbonate, CaCO
2
3
So
starting
from
3.32
(
g
to drive of f carbon dioxide
mol
)
of
potassium
iodide
in
solution
and
100
1
gas, CO
adding
3.31
(
g
2
mol
)
of
lead
nitrate
in
aqueous
solution
should
100
CaCO
1
→ CaO + CO
3
2
produce
4.61
(
g
mol
)
of
a
precipitate
of
lead
iodide
which
can
be
100
Calculate the atom economy
ltered
off
and
dried.
of the reaction.
However,
2
this
is
in
theory
only.
When
you
pour
one
solution
into
Sodium sulfate can be
another,
some
droplets
will
be
left
in
the
beaker.
When
you
remove
made from sulfuric acid and
the
precipitate
from
the
lter
paper,
some
will
be
left
on
the
paper.
This
sodium hydroxide.
sort
H
SO
2
of
problem
means
that
in
practice
you
never
get
as
much
product
+ 2NaOH
4
as
→ Na
SO
2
+ H
4
O
the
equation
laboratory
and
predicts.
in
Much
industry,
lies
of
in
the
skill
of
the
minimising
chemist,
these
sorts
both
of
in
the
losses.
2
If sodium sulfate is the
The
yield
of
the
a
number
of
moles
of
a
specied
product
=
required product, calculate the
chemical
×
theoretical
reaction
maximum
number
of
100%
moles
atom economy of the reaction.
of
3
Ethanol, C
H
2
the
product
O, can be made
It
6
H
by reacting ethene, C
2
can
equally
well
be
dened
as:
, with
4
the
number
of
grams
of
a
specied
O.
water, H
2
product
C
H
2
+ H
4
O → C
2
H
2
obtained
in
a
reaction
×
O
6
theoretical
Without doing a calculation,
maximum
grams
of
the
number
100%
of
product
state the atom economy
If
you
had
obtained
4.00 g
of
lead
iodide
in
the
above
reaction,
the
of the reaction. Explain
yield
would
have
been:
your answer.
4.00
×
4
100%
=
86.8%
Consider the reaction
4.61
CaCO
→ CO
3
a
+ CaO
A
2
Calculate the theoretical
maximum number of
moles of calcium oxide,
CaO, that can be obtained
further
go
to
process
is
it
problem
completion.
in
which
impossible
However,
to
arises
This
is
with
not
ammonia
get
chemists
a
can
is
yield
reactions
that
uncommon.
made
of
from
100%
improve
the
are
One
reversible
example
hydrogen
even
yield
with
by
and
the
is
do
not
Haber
nitrogen.
best
changing
and
the
Here
practical
the
skills.
conditions.
from 1 mole of calcium
carbonate, CaCO
3
b
Percentage yields
Starting from 10 g calcium
Yields of multi-step reactions can be surprisingly low because the overall
carbonate, calculate the
yield is the yield of each step multiplied together. So a four step reaction in
theoretical maximum
which each step had an 80% yield would be
number of grams of calcium
80% × 80% × 80% × 80% = 41%
oxide that can be obtained.
c
If 3.6 g of calcium oxide
What would be the overall yield of a three step process if the yield of
was obtained, calculate the
each separate step were 80%, 60%, and 75% respectively?
yield of the reaction.
50
%63
Practice
1
Potassium
questions
nitrate,
KNO
,
decomposes
on
strong
heating,
forming
oxygen
and
solid
Y
as
3
the
(a)
only
A
products.
1.00
g
sample
of
KNO
(M
3
into
=
101.1)
was
heated
strongly
until
fully
decomposed
r
Y
(i)
Calculate
(ii)
At
the
number
of
moles
of
KNO
in
the
1.00
g
sample.
3
298
K
and
100
kPa,
the
oxygen
gas
−4
occupied
a
State
ideal
the
oxygen
volume
gas
produced
of
1.22
×
equation
in
this
gas
constant
R
=
in
this
decomposition
3
10
m
and
use
it
to
calculate
the
number
of
moles
of
decomposition.
−1
(The
produced
8.31
−1
mol
J K
)
(5
(b)
Compound
remainder
(i)
State
(ii)
Use
Y
contains
being
what
the
45.9%
of
potassium
and
16.5%
of
nitrogen
by
mass,
the
oxygen.
is
data
meant
above
by
to
the
term
calculate
empirical
the
formula
empirical
formula
of
Y
(4
(c)
Deduce
an
marks)
equation
for
the
decomposition
of
into
KNO
Y
and
marks)
oxygen.
3
(1
mark)
AQA,
2
Ammonia
is
used
to
make
nitric
acid,
HNO
,
by
the
Ostwald
2006
process.
3
Three
reactions
occur
Reaction
1
4NH
Reaction
2
2NO(g)
(g)
in
+
this
5O
3
process.
(g)
➝
4NO(g)
O
3
3NO
(g)
+
➝
2NO
In
one
O(I)
➝
2HNO
2
production
O(g)
(g)
2
H
2
(a)
6H
2
(g)
2
Reaction
+
2
+
(aq)
+
NO(g)
3
run,
the
gases
formed
in
Reaction
1
occupied
a
total
volume
of
3
4.31
m
at
Calculate
25
the
Give
your
(The
gas
ºC
and
100
amount,
answer
to
in
the
kPa.
moles,
of
NO
appropriate
−1
constant
R
=
8.31
produced.
number
of
signicant
gures.
−1
J K
mol
)
(4
(b)
In
of
another
the
NO
production
gas
run,
produced
3.00
was
kg
used
of
to
ammonia
make
NO
gas
gas
were
in
used
in
Reaction
Reaction
1
marks)
and
all
2.
2
Calculate
the
assuming
an
mass
of
NO
formed
from
3.00
kg
of
ammonia
in
Reaction
2
2
Give
(c)
your
Consider
80.0%
answer
(g)
+
H
2
Calculate
in
Reaction
3NO
yield.
3
kilograms.
in
O(l)
this
➝
marks)
(2
marks)
process.
2HNO
2
the
(5
(aq)
+
NO(g)
3
concentration
of
nitric
acid
produced
when
0.543
mol
of
3
NO
is
reacted
with
water
and
the
solution
is
made
up
to
250 cm
2
(d)
Suggest
why
a
leak
of
NO
gas
from
the
Ostwald
process
will
cause
2
atmospheric
(e)
(f)
Give
one
reason
Ammonia
reacts
+
NH
pollution.
the
with
HNO
3
Deduce
why
➝
3
type
of
excess
nitric
NH
air
is
acid
used
as
in
the
shown
in
Ostwald
this
(1
mark)
(1
mark)
(1
mark)
process.
equation.
NO
4
reaction
3
occurring.
AQA,
2013
51
Chapter
2
Amount
of
3
substan ce
Zinc
(a)
forms
many
People
different
who
have
a
salts
zinc
including
deciency
zinc
can
sulfate,
take
zinc
hydrated
chloride,
zinc
and
sulfate,
zinc
uoride.
ZnSO
xH
4
as
A
a
dietary
student
heated
anhydrous
Use
these
zinc
data
4.38
g
of
hydrated
zinc
sulfate
and
obtained
2.46
your
g
of
sulfate.
to
calculate
the
value
of
the
integer
x
in
ZnSO
xH
4
Show
O,
2
supplement.
O.
2
working.
(3
(b)
Zinc
chloride
can
and
hydrochloric
The
equation
ZnO
+
for
be
prepared
the
laboratory
by
the
reaction
between
zinc
➝
reaction
+
ZnCl
is:
H
2
O
2
3
A
0.0830
mol
hydrochloric
Calculate
obtained
number
(c)
Zinc
zinc
Zn
the
pure
signicant
can
2HCl
impure
hydrogen
be
chloride
had
Calculate
the
signicant
was
a
of
+
added
to
100
cm
−3
of
1.20
mol dm
zinc
reaction.
chloride
Give
your
that
could
answer
to
be
the
appropriate
gas
in
the
laboratory
by
the
reaction
(4
marks)
(4
marks)
between
gas.
H
2
zinc
mass
anhydrous
this
prepared
chloride
ZnCl
sample
produced
3
oxide
gures.
also
➝
of
of
2
An
zinc
mass
products
hydrogen
+
of
maximum
from
chloride
and
sample
acid.
the
of
marks)
oxide
acid.
the
2HCl
in
powder
until
of
the
with
a
reaction
mass
was
of
5.68
g
complete.
was
The
reacted
zinc
with
chloride
10.7 g.
percentage
purity
of
the
zinc
metal.
Give
your
answer
to
gures.
AQA,
4
In
this
question
give
Magnesium
nitrate
and
as
oxygen
)
3
Thermal
your
answers
decomposes
shown
2Mg(NO
(a)
all
in
(s)
the
➝
the
heating
following
2MgO(s)
+
appropriate
to
form
4NO
(g)
+
O
2
of
a
number
magnesium
of
signicant
oxide,
nitrogen
dioxide,
sample
of
(g)
2
magnesium
nitrate
produced
0.741
g
of
oxide.
(i)
Calculate
the
amount,
(ii)
Calculate
the
total
in
moles,
of
MgO
in
0.741
g
of
magnesium
oxide.
(2
of
magnesium
amount,
in
moles,
of
gas
produced
sample
of
magnesium
from
this
marks)
sample
nitrate.
(1
(b)
2013
gures.
equation.
2
decomposition
magnesium
on
to
In
another
experiment,
to
produce
0.402
a
different
nitrate
mark)
decomposed
3
mol
of
gas.
Calculate
the
volume,
in
dm
,
that
this
gas
would
5
occupy
at
333
K
and
1.00
×
Pa.
10
−1
(The
gas
constant
R
=
8.31
−1
mol
J K
)
(3
(c)
A
of
0.0152
mol
magnesium
MgO
+
sample
of
nitrate,
2HCl
➝
magnesium
was
reacted
+
MgCl
H
2
oxide,
with
produced
from
hydrochloric
acid.
the
marks)
decomposition
O
2
3
This
for
0.0152
mol
complete
sample
reaction.
of
Use
magnesium
this
oxide
information
required
to
32.4
calculate
the
cm
of
hydrochloric
acid
concentration,
−3
in
mol
dm
,
of
the
hydrochloric
acid.
(2
marks)
AQA,
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
52
2010
3
Bonding
3.1
Why do
The
bonds
The
chemical
between
nature
of
io nic
bonding
Learning
bonds form?
atoms
always
involve
their
outer
➔
electrons.
objective s:
State how ions form and why
they attract each other.
•
Noble
are
gases
very
have
full
outer
main
shells
of
electrons
(Figure
1)
and
➔
unreactive.
State the proper ties of
ionically bonded compounds.
•
When
atoms
achieve
shell
•
of
There
and
a
bond
more
stable
electrons,
are
three
together
they
electron
like
types
the
of
share
or
transfer
arrangement,
noble
strong
electrons
often
a
full
to
outer
main
➔
gases.
Describe the structure of
ionically bonded compounds.
chemical
bonds
–
ionic,
covalent,
Specication reference: 3.1.3
metallic
Ionic
bonding
Metals
have
+
electrons
in
the
easiest
the
electron
gas
is
one,
their
way
two,
or
outer
for
shells,
them
to
compounds
so
attain
of
a
Ar
Ne
He
structure
Noble gas
three
The noble gases do form a few
noble
compounds although they are
to
lose
their
outer
electrons.
mostly unstable. The rst,
Non-metals
have
spaces
in
their
Xe PtF
helium
outer
shells,
so
that
the
neon
2
2
1s
way
for
them
to
attain
1s
is
to
structure
gain
Ionic
•
Electrons
•
Positive
•
a
noble
bonding
are
and
negative
Na,
has
11
2)
is
Chlorine,
has
Cl,
An
electron
atom
moves
is
6
17
2p
2
2s
6
2p
, was made in 196
1 by
6
2
3s
6
3p
Neil Bar tlett, pictured below.
Noble gases
2p
transferred.
into
the
non-metals.
atoms
to
non-metal
atoms.
bonding.
(and
11
protons).
The
electron
1
3s
6
2s
and
formed.
electrons
2
1s
metal
ionic
electrons
2s
2
is
are
has
2
1s
metals
from
ions
(Figure
arrangement
arrangement
•
2
1s
gas
between
transferred
chloride
Sodium,
occurs
2
•
6
2p
electrons.
•
Sodium
of
2
2s
the
▲ Figure 1
electron
argon
easiest
(and
2
3s
The
outer
17
protons).
The
electron
5
3p
single
shell
of
outer
the
electron
chlorine
of
the
sodium
atom.
There are as yet no compounds
of helium or neon. Xenon has
the largest number of known
•
Each
outer
main
shell
is
now
full.
compounds. In most of them
xenon forms a positive ion by
×
×
×
×
losing an electron.
×
Suggest why it is easier for
×
×
×
Cl
×
×
×
×
Na
×
sodium
11
11
2
2s
6
2p
×
×
×
chlorine
17
electrons
2
1s
atom
protons
17
1
3s
2
1s
than for helium or neon.
atom
protons
electrons
2
2s
xenon to form a positive ion
×
6
2p
2
3s
5
3p
1
▲ Figure 2
A dot-and-cross diagram to show the transfer of the 3s
electron from
the sodium atom to the 3p orbital on a chlorine atom. Remember that electrons are all
identical whether shown by a dot or a cross
53
3.1
The
nature
of
io nic
bonding
+
•
Both
sodium
and
arrangement.
chlorine
Sodium
has
now
the
have
neon
a
noble
noble
gas
gas
electron
arrangement
Na
whereas
the
The
ions
two
chlorine
in
has
Figure
charged
3
the
argon
with
particles
the
that
noble
noble
result
gas
gas
arrangement
atoms
from
the
in
(compare
Figure
transfer
of
1).
an
electron
+
Na
11
sodium
protons,
10
2
1s
ion
electrons
2
2s
are
called
ions
6
2p
•
×
The
sodium
negative
×
ion
is
positively
charged
because
it
has
lost
a
×
electron.
×
×
•
×
×
×
×
×
×
×
Cl
The
chloride
negative
ion
is
negatively
charged
because
it
has
gained
a
electron.
×
Cl
chlorine
×
×
×
×
ion
•
The
two
ions
oppositely
(called
are
charged
protons,
2
1s
18
2
2s
▲ Figure 3
in
each
the
other
sodium
and
to
chloride
other
compound
by
forces
electrons
6
2p
ions
to
chloride)
electrostatic
17
attracted
2
3s
6
3p
Therefore
ionic
oppositely
The ions that result from
Every
electron transfer
bonding
charged
positive
compounds
ion
always
Chloride
ion,
ion
Cl
formula
there
is
of
exist
result
a
for
electrostatic
extends
negative
structure
sodium
chloride
chloride
of
attraction
every
in
lattice
sodium
one
the
The
attracts
three-dimensional
The
is
ions.
is
ion
called
chloride
NaCl
and
a
attraction
throughout
vice
versa.
lattice.
with
because
its
compound.
Ionic
Figure
singly
for
between
the
4
shows
charged
every
one
the
ions.
sodium
ion.
Example: magnesium oxide
+
Magnesium,
2
+
Sodium
▲ Figure 4
ion,
Na
1s
2
2s
Mg,
6
has
12
electrons.
The
electron
arrangement
is
2
2p
3s
The sodium chloride
2
Oxygen,
O,
has
eight
electrons.
The
electron
arrangement
is
1s
2
2s
4
2p
structure. This is an example of a giant
×
ionic structure. The strong bonding
×
×
×
ex tends throughout the compound and
×
O
×
Mg
because of this it will be dicult to melt.
×
×
Study tip
magnesium
12
protons,
12
atom
oxygen
electrons
8
protons,
atom
8
electrons
Dot-and-cross diagrams can help
2
1s
2
2s
6
2p
2
2
3s
1s
2
2s
4
2p
you to understand the principles of
×
2+
bonding and to predict the shapes
×
2–
×
×
O
×
Mg
×
of molecules – see Topic 3.5, The
×
shapes of molecules and ions.
×
2–
2+
Mg
12
magnesium
protons,
10
2
1s
2
2s
▲ Figure 5
This
time,
two
magnesium
O
ion
oxygen
8
electrons
ion
protons,
2
6
1s
2p
(called
10
2
2s
oxide)
electrons
6
2p
Ionic bonding in magnesium oxide, MgO
electrons
atom.
Each
2p
orbital.
•
The
magnesium
two
negative
The
oxide
are
transferred
oxygen
atom
from
the
receives
3s
two
orbitals
electrons
on
each
into
its
2+
ion,
Mg
,
is
positively
charged
because
it
has
lost
electrons.
2–
•
negative
•
54
The
ion,
O
,
is
negatively
charged
because
electrons.
formula
of
magnesium
oxide
is
MgO.
it
has
gained
two
Bonding
3
The formulae of
ionic
compounds
+
Group
1
metals
non-metals
all
form
form
compounds
1
are
1
Study tip
+
ions,
ions
and
neutral,
and
Group
Group
we
can
6
2
metals
form
non-metals
predict
the
2
form
ions.
Group
ions.
Since
2
formulae
of
simple
7
Some ions consist of groups of
ionic
covalently bonded atoms which
2+
compounds.
For
example,
calcium
uoride
is
Ca
and
2F
,
ie
CaF
,
have an overall charge. You will
2
+
lithium
oxide
is
2
2Li
and
O
,
ie
Li
O.
The
same
idea
applies
to
compound
2
+
ions,
see
the
Study
tip.
For
example,
sodium
sulfate
is
2Na
need to remember the formulae and
2
and
SO
,
4
charges of the following:
+
ie
Na
SO
2
and
ammonium
hydroxide
is
NH
4
and
OH
,
ie
NH
4
OH.
2
4
sulfate, SO
; hydroxide, OH
;
4
2
nitrate, NO
, carbonate CO
3
Properties of
ionically
bonded
compounds
,
3
+
ammonium NH
4
Ionic
compounds
are
always
solids
at
room
temperature.
They
have
When working out formulae of
giant
structures
and
therefore
high
melting
temperatures.
This
is
compounds these behave exactly
because
in
order
to
melt
an
ionic
compound,
energy
must
be
supplied
like ions with a single atom.
to
break
Ionic
the
lattice
compounds
water
the
up
(aqueous)
current
are
of
ions.
conduct
but
free
not
to
electricity
when
move
solid.
in
the
when
This
liquid
molten
is
or
because
state
but
dissolved
the
are
ions
not
in
that
free
in
carry
Hint
the
A current of electricity is a ow
solid
state
(Figure
6).
of charge. In metals, negative
electrons move. In ionic
liquid
compounds, charged ions move.
solid
+
+
+
1
+
+
–
–
+
–
+
cathode
+
+
+
+
+
cathode
–
+
+
+
+
ions
are
free
+
move
in
are
not
free
the
solid
to
and
state
in
the
the
liquid
Ionic
conducts
This
is
are
because
brittle
they
and
form
shatter
a
easily
lattice
of
small
displacement
between
when
+
electricity
Ionic liquids conduct electricity, ionic solids do not
compounds
blow.
+
state
compound
a
▲ Figure 6
+
to
+
move
ions
+
–
the
the
+
+
anode
+
+
anode
given
alternating
a
ions
with
the
causes
same
sharp
positive
and
2
+
negative
the
ions
ions,
and
see
Figure
produce
7.
A
contact
blow
in
the
between
direction
ions
with
shown
like
contact
charge...
may
+
+
move
+
+
+
+
+
+
charges.
+
+
+
Summary questions
1
Identify which of the following are ionic compounds and explain why.
... and
a
CO
b
KF
c
CaO
d
the
structure
shatters
HF
3
2
Explain why ionic compounds have high melting temperatures.
3
Describe the conditions where ionic compounds conduct electricity.
+
+
4
+
+
+
+
Draw dot-and-cross diagrams to show the formation of the following ions.
Include the electronic conguration of the atoms and ions involved.
shatters
a
the ions being formed when magnesium and uorine react
b
the ions being formed when sodium and oxygen react.
+
+
5
Give the formulae of the compounds formed in question
6
Look at the electron arrangements of the Mg
2+
noble gas they correspond to.
+
+
+
+
4
2–
and O
ions. State the
▲ Figure 7
The brittleness of ionic
compounds
55
3.2
Covalent
bonding
Non-metal
Learning
atoms
need
to
receive
electrons
to
ll
the
spaces
in
their
objective s:
outer
➔
Describe a covalent bond.
➔
Describe a co-ordinate bond.
➔
Describe the proper ties of
shells.
•
A
•
The
a
covalent
bond
atoms
stable
share
noble
forms
some
gas
between
of
their
a
pair
outer
of
non-metal
electrons
so
atoms.
that
each
atom
has
arrangement.
covalently bonded molecules.
•
A
covalent
bond
is
a
shared
pair
of
electrons.
Specication reference: 3.1.3
Forming
chlorine
2
1s
2
2s
atoms
6
2p
A
2
3s
small
molecules
group
of
by
covalently
covalent
bonded
bonding
atoms
is
called
a
molecule.
For
5
3p
example,
chlorine
exists
as
a
gas
that
is
made
of
molecules,
Cl
,
see
2
×
×
×
×
Figure
1.
×
×
×
×
×
Cl
×
×
×
Cl
Chlorine
2
has
2
1s
17
6
2s
2p
electrons
2
and
an
electron
arrangement
5
3s
3p
.
Two
chlorine
atoms
make
a
chlorine
molecule:
×
×
×
×
×
×
×
×
×
•
The
two
atoms
•
Each
•
The
•
Molecules
atom
share
now
formula
is
has
one
a
pair
stable
of
electrons.
noble
gas
arrangement.
Cl
2
×
×
×
×
×
Cl
×
×
×
Cl
from
one
are
neutral
atom
to
because
no
electrons
have
shared
electrons
in
been
transferred
another.
×
×
×
×
×
You
can
aline,
a
chlorine
represent
one
pair
of
a
covalent
bond
by
Cl—Cl.
molecule
Example: methane
▲ Figure 1
Formation of a chlorine
Methane
gas
is
a
covalently
bonded
compound
of
carbon
and
molecule – the two atoms share a 3p
hydrogen.
electron from each atom
2
1s
2
2s
Carbon,
C,
has
six
electrons
with
electron
2
2p
arrangement
1
and
hydrogen,
H,
has
just
one
electron
1s
×
H
C
carbon
Hint
2
1s
hydrogen
2
2s
2
1
2p
1s
Another way of picturing covalent
bonds is to think of electron
In
orbitals on each atom merging to
order
four
for
carbon
hydrogen
to
atoms
attain
to
a
stable
every
noble
carbon
gas
arrangement,
there
are
atom.
form a molecular orbital that holds
the shared electrons.
H
×
H
—
H
C
×
H — C — H
H
—
×
Hint
H
×
The hydrogen has a lled outer
H
2
shell with only two electrons (1s
).
methane,
It lls the rst shell to get the
CH
4
structure of the noble gas helium.
The
formula
of
methane
is
CH
.
The
four
2p
electrons
from
4
The carbon atoms have an electron
1
2
arrangement 1s
56
2
2s
6
2p
the
1s
electron
from
the
four
hydrogen
atoms
are
shared.
carbon
and
Bonding
3
How does
Atoms
with
attraction
place
sharing
covalent
between
within
the
molecule
The
forces
in
consists
when
the
and
held
of
together
the
two
repulsive
are
a
by
shared
held
shown
forces
particular
the
electrostatic
electrons.
example
protons
are
electron
atoms together?
simplest
forces
the
nuclei
hold
and
The
electrostatic
black
are
nuclei
molecule.
electrons.
balance
bonds
the
hydrogen
are
electrons
in
in
is
This
together
Figure
red.
distance
takes
hydrogen.
2.
These
by
a
The
The
pair
of
proton
+
+
proton
attractive
forces
just
apart.
electron
Double covalent bonds
In
a
double
oxygen
atoms
bond,
molecule
have
represent
adouble
a
four
share
double
the
two
line,
O
electrons
two
bond
pairs
of
are
pairs
of
between
shared
shared.
The
electrons
them
so
two
that
(Figure
electrons
in
a
atoms
the
3).
▲ Figure 2
an
The electrostatic forces
within a hydrogen molecule
oxygen
You
covalent
in
can
bond
by
O.
×
×
O
When
you
are
drawing
covalent
bonding
diagrams
you
may
O
O
=
O
leave
×
×
out
the
inner
examples
of
main
shells
molecules
because
with
these
covalent
are
not
bonds
involved
are
shown
at
in
all.
Table
Other
1.
oxygen,
O
2
All
the
examples
in
Table
1
are
neutral
molecules.
The
atoms
within
the
▲ Figure 3
molecules
are
strongly
bonded
together
with
covalent
bonds
within
An oxygen molecule has
the
a double bond which shares two 2p
molecule.
However,
the
molecules
are
not
strongly
attracted
to
each
other.
electrons from each atom
▼ Table 1
Examples of covalent molecules. Only the outer shells are shown.
Formula
Name
Formula
Name
NH
3
hydrogen
H
2
Each hydrogen atom
×
×
H
N
H
ammonia
has a full outer main
×
×
shell with just two
electrons
H
C
H
2
4
HCl
ethene
H
H
There is a carbon–
×
×
hydrogen chloride
×
×
H
Cl
C
carbon double bond in
C
×
×
×
this molecule
H
H
H
O
2
CO
carbon dioxide
2
O
There are two carbon–
×
×
×
×
×
×
×
×
water
oxygen double bonds in
×
×
H
this molecule
H
Properties of
Substances
low
substances
composed
melting
of
with
molecules
temperatures.
This
is
molecular
are
gases,
because
structures
liquids,
the
strong
or
solids
covalent
with
bonds
57
3.2
Covalent
bond ing
are
only
between
the
atoms
within
the
molecules.
There
is
only
weak
Hint
attraction
Some covalent compounds react
energy
to
with water to form ions. In such
They
cases the resulting solution will
neutral
conduct electricity. Hydrogen
the
chloride is an example of this:
If
between
are
move
poor
the
apart
molecules
from
conductors
overall.
This
each
of
means
so
the
do
not
need
much
other.
electricity
that
molecules
there
because
are
no
the
molecules
charged
are
particles
to
carry
current.
they
dissolve
in
water,
and
remain
as
molecules,
the
solutions
do
+
HCl(g) + aq ➝ H
(aq) + Cl
(aq)
not
conduct
electricity.
Again,
this
is
because
of
pair
there
are
no
charged
particles.
Co-ordinate
bonding
A
bond
Hint
Triple bonds are also possible
where three electrons from each
atom are shared, e.g., N
is N≡N
2
atoms.
But,
in
covalent
In
most
some
•
Synoptic link
a
a
•
atom
lled
or
that
outer
the
atom
not
being
atom
It
accepts
is
used
is
dative
shell
that
bonds,
one
bonding.
co-ordinate
the
consists
covalent
bonds,
co-ordinate
In
Co-ordinate (dative) bonding is
single
a
the
bond,
electrons
provides
both
the
dative
shared
one
of
the
electrons.
covalent
between
two
electrons.
This
is
called
bonding
bond:
electron
electrons
of
atom
called
covalent
donating
in
each
provides
also
the
of
a
–
pair
the
is
electrons
called
a
an
atom
has
lone
atom
is
a
that
does
not
have
electron-decient
pair
of
electrons
that
is
pair
very impor tant in the chemistry
of transition metal complexes.
Example: the ammonium ion
You will learn more about it in
For
example,
ammonia,
NH
,
has
a
lone
pair
of
electrons.
In
the
3
Chapter 23, The transition metals.
+
ammonium
ion,
NH
,
the
nitrogen
uses
its
lone
pair
of
electrons
4
+
to
form
a
electrons
co-ordinate
at
all
and
bond
with
therefore
an
H
ion
(a
bare
proton
with
no
electron-decient).
H
+
×
H
—
Summary questions
+
H
H —
written
H
N
—
×
1
H
State what a covalent bond is.
×
2
ammonium
Identify which of the following
ion
H
have covalent bonding and
explain your answer.
Co-ordinate
a
Na
covalent
bonds
are
represented
by
an
arrow.
The
arrow
O
2
points
b
towards
the
atom
that
is
accepting
the
electron
pair.
However,
CF
this
4
c
MgCl
d
C
is
is
only
to
completely
show
how
the
symmetrical
bond
and
all
was
the
made.
bonds
The
have
ammonium
exactly
the
ion
same
2
strength
2
3
and
length.
H
4
Draw a dot-and-cross diagram
for hydrogen sulde, a
•
Co-ordinate
ordinary
bonds
covalent
have
bonds
exactly
the
between
same
the
strength
same
pair
of
and
length
atoms.
compound of hydrogen and
The
ammonium
sulfur.
particle.
4
Draw a dot-and-cross diagram
to show a water molecule
forming a co-ordinate bond
+
with an H
58
ion.
ion
has
covalently
bonded
atoms
but
is
a
charged
as
3.3
Metals
to
are
three
Metallic
shiny
outer
elements
electrons,
Na,
2,8,1
2
Al,
(1s
2,8,3
2s
3s
2
)
loses
6
2s
atoms
that
metal
can
ions.
easily
For
lose
up
Learning
example,
objective s:
1
2p
(1s
of
positive
6
2
aluminium,
up
leaving
2
sodium,
made
bonding
2
2p
its
one
outer
electron,
➔
Describe the nature of
1
3s
3p
)
loses
its
three
outer
bonding in a metal.
electrons.
➔
Describe the proper ties of
metals.
Metallic
bonding
Specication reference: 3.1.3
The
in
atoms
ionic
them.
The
a
A
metal
a
metal
of
electrons
particular
tend
unless
attraction
delocalised
of
one
these
is
the
a
This
positive
in
is
and
ions
‘sea’
means
for
the
of
is
that
they
in
happens
to
atoms
one
metals
not
Figure
balanced
by
consist
1.
of
These
tied
to
The
the
receive
merge.
particular
electrons.
are
negatively
(as
present
the
any
outer
shown
is
atom
with
of
that
this
electrons
shells
bonding
a
metal
another
main
associated
metallic
existing
transfer
non-metal
outer
longer
Magnesium
repel
of
no
ions
cannot
there
delocalised.
atom.
to
are
picture
positive
are
element
element,
electrons
simple
lattice
ions
a
bonding)
In
outer
atom.
in
a
positive
electrostatic
charged
‘sea’
of
electrons.
e
e
e
e
Hint
2+
2+
Mg
2+
Mg
In Figure 1 the metal ions are
Mg
shown spaced apar t for clarity. In
e
e
e
‘sea’
of
fact, metal atoms are more closely
electrons
packed, and so metals tend to
e
e
have high densities.
2+
2+
Mg
Mg
e
The
number
The
e
The delocalised ‘sea’ of electrons in magnesium
electrons
•
Mg
e
▲ Figure 1
•
2+
of
have
metallic
delocalised
been
lost
bonding
electrons
by
each
spreads
depends
metal
on
how
many
atom.
throughout
so
metals
have
giant
structures.
Properties of
Metals
The
good
delocalised
explain
why
electron
at
are
one
leaves
the
conductors of
electrons
metals
from
end
metals
of
the
a
wire
are
such
negative
metal
at
that
wire
the
can
electricity
move
good
conductors
terminal
while
positive
at
of
the
the
as
of
time
heat
the
structure
electricity.
supply
same
terminal,
and
throughout
a
shown
joins
the
different
in
An
electron
sea
electron
Figure
Hint
2.
The word ‘delocalised’ is often
Metals
are
also
conductivities.
property,
of
the
with
closely
good
The
conductors
sea
energy
packed
of
of
electrons
also
spread
heat
is
by
–
they
partly
have
high
responsible
increasingly
thermal
for
vigorous
this
vibrations
used to describe electron clouds
that are spread over more than
two atoms.
ions.
59
3.3
Metallic
bonding
metal
+
+
+
M
e
e
M
e
M
electron
electron
e
e
+
e
in
+
M
+
▲ Figure 2
The
–
–
+
In
+
+
The conduction of electricity by a metal
strength of
general,
the
e
M
e
out
M
metals
strength
of
any
metallic
bond
depends
on
the
+
following:
–
–
push
+
+
–
+
–
+
+
•
+
the
–
+
charge
greater
on
the
+
+
+
•
+
–
electron
‘sea’
the
size
the
positive
Metals
These
to
+
+
+
+
+
–
the
greater
delocalised
the
charge
electrons
on
and
the
the
ion,
the
stronger
the
tend
attraction
ion
to
–
the
nucleus
be
between
smaller
and
strong.
The
throughout
the
the
the
positive
ion,
the
stronger
solid
so
closer
the
delocalised
the
ions
and
the
electrons.
electrons
are
to
bond.
electrons
there
the
are
no
also
explain
individual
this.
bonds
break.
–
–
+
+
+
+
+
+
+
+
are
malleable
and ductile
+
–
Metals
can
is
be
still
are
malleable
pulled
in
into
exactly
retained,
▲ Figure 3
–
of
+
+
–
of
extend
Metals
–
ion
+
electrostatic
–
the
number
see
(they
thin
the
Figure
can
wires).
same
be
beaten
After
a
into
small
environment
as
shape)
and
distortion,
before
so
ductile
each
the
(they
metal
new
ion
shape
is
3.
–
The malleability and
Contrast
this
with
the
brittleness
of
ionic
compounds
in
Topic
3.1.
ductility of metals
Metals
Metals
giant
have
high
generally
structures.
delocalised
sea
melting
have
There
of
high
is
points
melting
strong
electrons.
and
boiling
attraction
This
makes
points
between
the
atoms
because
metal
ions
difcult
to
they
and
have
the
separate.
Summary questions
1
Give three dierences in physical proper ties between metals and
non-metals.
2
Write the electron arrangement of a calcium atom, Ca.
3
Which electrons will a calcium atom lose to gain a stable noble gas
conguration?
4
State how many electrons each calcium atom will contribute to the
delocalised sea of electrons that holds the metal atoms together.
5
Sodium forms +1 ions with a metallic radius of 0.191 nm. Magnesium
forms +2 ions with a metallic radius of 0.160 nm. How would you expect
the following proper ties of the two metals to compare? Explain your
answers.
60
a
the melting point
b
the strength of the metals.
3.4
One
of
mass,
the
is
Bonding
key
made
ideas
of
of
tiny
science
particles
is
–
and
that
it
is
matter,
physic al
which
particulate.
is
These
anything
particles
p rope r t ie s
with
Learning
are
➔
in
motion,
which
means
they
have
kinetic
energy.
To
objective s:
understand
State the energy changes
the
that occur when solids melt
differences
between
the
three
states
of
matter
–
gas,
liquid,
and
solid
–
and liquids vaporise.
you
need
to
be
able
to
explain
the
energy
changes
associated
with
➔
changes
between
these
physical
Explain the values of
states.
enthalpies of melting (fusion)
and vaporisation.
The three
states of
matter
➔
Table
1
sets
out
the
simple
model
used
for
the
three
states
of
Explain the physical
matter.
proper ties of ionic solids,
metals, macromolecular
▼ Table 1
The three states of matter
solids, and molecular solids
Solid
Liquid
Gas
in terms of their detailed
arrangement
structures and bonding.
regular
random
random
of par ticles
➔
List the three types of
None direct but a
Crystal shapes have
strong bonds.
None direct but
liquid changes shape
straight edges. Solids
evidence
a gas will ll its
➔
List the three types of
to ll the bottom of
have denite shapes.
container.
intermolecular forces.
its container.
➔
spacing
close
close
far apar t
Solids are not easily
Liquids are not
Gases are easily
compressed.
easily compressed.
compressed.
vibrating about a point
rapid ‘jostling’
rapid
Describe how melting
temperatures and structure
evidence
are related.
➔
movement
Describe how electrical
conductivity is related
Diusion is very slow.
Diusion is
Diusion is slow.
to bonding.
Solids expand on
evidence
rapid. Gases
Liquids evaporate.
heating.
exer t pressure
melting
boiling
vibration
evaporation
point
point
T
T
m
b
heat
heat
models
cool
particles
about
vibrate
a
cool
particles
point
close
Energy
changes on
Heating
a
you
makes
the
In
particles
still
rst
average
order
–
close
needed
them
heat
a
to
distance
solid to
turn
into
While
a
the
far
are
too
particles
except
and
surface
are
have
random
free
rapid
motion
heating
solid
is
a
solid
more
and
between
liquid
liquid
–
where
you
the
or
have
forces
solid
more
melting,
its
state.
act
The
and
also
energy
the
solid
packed,
the
it
increases
expands.
This
is
energy
of
change
but
is
holding
called
change
not
vibrating
randomly
particles,
needed
does
slightly
moving
enthalpy
temperature
the
energy.
between
particles,
called fusion)
are
more
the
This
so
closely
particles
supply
correctly
the
–
ordered,
that
to
position.
particles
the
to
energy
xed
(melting
with
the
a
the
–
–
supply
about
solid
in
melting,
a
a
weaken
together
of
heat
vibrate
together
to
but
travel
solid
them
Turning
move
to
at
When
Specication reference: 3.1.3
the
latent
melting.
because
the
61
3.4
Bonding
and
p hysic al
prop erties
heat
energy
provided
is
absorbed
as
the
forces
between
particles
are
Hint
weakened.
The enthalpy change of melting
Enthalpy
is
the
heat
energy
change
measured
under
constant
pressure
is sometimes called the enthalpy
whilst
temperature
depends
on
the
average
kinetic
energy
of
the
particles
change of fusion.
and
is
they
Synoptic link
therefore
related
to
their
speed
–
the
greater
the
energy,
the
faster
go.
Heating
When
a
you
liquid
heat
a
liquid,
you
supply
energy
to
the
particles
which
You will learn more about enthalpy
makes
them
move
more
quickly
–
they
have
more
kinetic
energy.
On
in Topic 4.2, Enthalpy.
average,
on
the
particles
move
a
little
further
apart
so
liquids
also
expand
heating.
Turning
a
In
order
to
to
break
all
liquid to
turn
a
gas
liquid
(boiling
into
a
gas,
–
also
you
called vaporisation)
need
to
supply
enough
energy
Hint
It is through an understanding of
the relationship between bonding
and physical proper ties that
material scientists can engineer
consists
The
of
the
particles
energy
correctly
there
is
intermolecular
the
no
that
needed
is
are
far
called
enthalpy
forces
apart
the
and
latent
change
temperature
between
of
change
the
moving
heat
of
particles.
the
gas
independently.
vaporisation
vaporisation.
during
A
As
process
of
or
with
more
melting,
boiling.
new materials with exciting
Heating
a
gas
proper ties. Examples include:
As
•
carbon bres
•
materials based on carbon
nanotubes
•
materials that can self-repair.
you
get
heat
much
a
gas,
further
the
particles
apart
and
so
gain
kinetic
gases
energy
expand
a
and
great
move
deal
on
faster.
They
heating.
Crystals
Crystals
held
are
solids.
together
by
such
as
covalent,
such
as
van
der
The
forces
ionic,
Waals,
particles
of
have
attraction.
or
metallic,
a
regular
These
or
dipole–dipole,
weaker
or
arrangement
could
be
strong
and
intermolecular
hydrogen
bonds.
are
bonds,
forces,
The
strength
Synoptic link
of
The forces between molecules,
the
the
forces
physical
force,
the
of
attraction
properties
higher
the
of
between
the
melting
the
crystals.
particles
For
temperature
in
the
example,
and
the
crystal
the
affects
stronger
greater
the
the
enthalpy
including Van der Waals forces
of
fusion
(the
more
difcult
they
are
to
melt).
There
are
four
basic
are discussed in Topic 3.7
, Forces
crystal
types
–
ionic,
metallic,
molecular,
and
macromolecular.
acting between molecules.
Ionic
Ionic
crystals
compounds
oppositely
crystal,
This
is
see
a
Topic
result
throughout
order
for
melting
the
Metals
of
electrons,
result
of
ions
of
as
a
see
these
Molecular
to
chloride,
compounds
require
apart
from
chloride
is
attractions
NaCl,
have
electrostatic
These
move
sodium
lattice
Topic
of
positive
3.3.
the
strong
is
high
a
melting
attractions
a
lot
each
801 °C
of
between
typical
which
energy
other.
For
to
ionic
points.
extend
break
example,
in
the
(1074 K).
Again
crystal.
metallic
ions
the
The
embedded
attraction
high
bonds.
of
melting
Magnesium
in
a
delocalised
positive
to
negative
temperature
is
a
typical
sea
is
a
example.
crystals
crystals
intermolecular
62
Ionic
strong
structure.
throughout
Molecular
atoms
the
electrostatic
Sodium
crystals
exist
extends
strong
ions.
3.1.
of
the
point
Metallic
have
charged
together
consist
forces.
but
of
molecules
Covalent
they
do
not
bonds
act
held
in
within
between
a
the
the
regular
array
molecules
molecules.
by
hold
the
Bonding
3
Intermolecular
forces
are
much
weaker
than
covalent,
ionic
or
distance
between
covalently
metallic
bonds,
so
molecular
crystals
have
low
melting
atoms
and
low
Iodine
enthalpies
(Figure
covalent
bond
1)
of
is
pair
of
iodine
=
0.267 nm
melting.
an
holds
a
bonded
temperatures
example
pairs
of
of
a
iodine
molecular
atoms
crystal.
together
to
A
strong
form
I
2
molecules.
the
van
Since
der
Waals
together
as
covalent
bonds,
•
crystals
•
low
•
does
to
a
are
carry
But
giving
soft
are
van
iodine
and
conduct
a
large
enough
Waals
the
break
iodine
have
strong
der
temperature
gaseous
not
molecules
forces
solid.
melting
form
iodine
to
forces
following
number
hold
are
the
much
of
electrons,
molecules
weaker
than
properties:
easily
(114 °C,
387 K)
and
sublimes
readily
to
molecules.
electricity
because
there
are
no
charged
particles
charge.
distance
between
molecules
Macromolecular
compounds
are
not
always
made
up
of
small
molecules.
=
substances
pair
van
of
iodine
der
Waals
the
covalent
bonds
typical
property
of
extend
throughout
the
0.354 nm
In
▲ Figure 1
some
a
by
crystals
forces)
Covalent
(held
The arrangement of an
compound
iodine crystal
and
have
strong
of
the
bonds
–
a
high
macromolecular
melting
crystals,
a
giant
structure
temperature.
including
held
There
diamond
are
and
together
many
with
examples
graphite.
Diamond and graphite
109.5°
Diamond
They
are
different
and
graphite
called
both
polymorphs
materials
arranged.
are
They
because
are
or
made
the
allotropes
their
examples
of
of
atoms
of
are
element
carbon.
carbon
They
differently
macromolecular
only.
are
very
bonded
and
structures.
Diamond
Diamond
every
consists
carbon
is
why
it
A
carbon
is
a
of
atom.
giant
atom
pure
The
carbon
bonds
with
spread
covalent
bonding
throughout
the
between
structure,
which
structure.
has
four
electrons
forms
four
single
in
its
outer
shell.
In
diamond,
each
C
carbon
atom
atoms,
other,
Topic
of
a
shown
3.5.
In
carbon
four
in
three
rules
2.
of
These
the
dimensions
(with
atom
other
Figure
the
tetrahedron
Each
by
as
following
covalent
is
in
carbon
bond
an
four
the
bonds
angles
Figure
of
with
electron
electron
identical
atoms.
bonds
pair
other
pairs
repel
repulsion
actually
point
C
C
carbon
each
theory.
to
the
▲ Figure 2
See
A dot-and-cross diagram
showing the bonding in diamond
corners
109.5°).
position
in
3
this
shows
the
structure,
surrounded
three-dimensional
arrangement.
The
atoms
bonds,
form
which
is
a
giant
why
diamond
•
very
hard
material
•
very
high
melting
•
does
not
particles
conduct
to
carry
three-dimensional
(one
of
has
the
the
electricity
following
hardest
temperature,
lattice
over
because
of
strong
covalent
properties:
known)
3700
there
K
are
no
free
charged
charge.
Graphite
Graphite
arranged
strong
also
consists
differently
covalent
and
of
pure
from
the
carbon,
diamond.
weaker
but
the
Graphite
van
der
atoms
has
Waals
are
two
bonded
sorts
of
▲ Figure 3
and
bonding
–
A three-dimensional
diagram of diamond
forces.
63
3.4
Bonding
and
p hysic al
prop erties
In
graphite,
other
120°
120°
each
carbon
theory,
As
form
at
these
trigonal
carbon
atoms.
planar,
a
with
atom
forms
predicted
trigonal
a
bond
three
by
single
bonding
arrangement,
angle
of
120°
covalent
electron
bonds
pair
sometimes
(Figure
4).
to
repulsion
called
This
leaves
120°
each
of
carbon
the
This
three
atom
with
single
arrangement
hexagons
of
a
‘spare’
covalent
atoms,
in
a
p-orbital
that
is
not
part
bonds.
produces
carbon
electron
a
two-dimensional
rather
like
a
layer
of
chicken-wire
linked
fence
(Figure5).
C
The
p-orbitals
plane
of
the
with
the
carbon
‘spare’
atoms
in
electron
each
merge
layer.
above
These
and
electrons
below
can
the
move
C
anywhere
strength
▲ Figure 4
within
of
the
the
layer.
bonding
They
and
is
are
delocalised.
rather
like
the
This
adds
to
delocalised
the
sea
of
A dot-and-cross diagram
electrons
in
a
metal,
but
in
two
dimensions
only.
showing the three covalent bonds
in graphite
These
(very
strong
within
covalent
the
delocalised
rare
for
a
electrons
are
non-metal).
what
They
make
can
graphite
travel
freely
conduct
through
electricity
the
bonds
material,
layers
planes,
There
They
is
layers
It
is
5.
the
graphite
right
angles
covalent
held
weak
slide
‘lead’
from
•
Graphite
•
It
in
is
a
to
by
only
conduct
between
the
one
pencils.
pencil
much
another
The
to
along
the
hexagonal
them.
intermolecular
across
the
will
bonding
together
This
can
transfer
at
no
are
Figure
though
not
the
force
soft
material.
high
melting
of
making
akiness
the
layers
weaker
of
van
carbon
der
attraction
graphite
allows
the
atoms.
Waals
forces,
means
soft
and
graphite
that
see
the
aky.
layers
to
paper.
weak
van
der
bonds
Waals
has
a
before
the
very
temperature
and
in
fact
it
breaks
down
between
it
melts.
This
is
because
of
the
strong
network
of
covalent
layers
bonds,
▲ Figure 5
which
make
it
a
giant
structure.
Van der Waals forces between
•
It
conducts
electricity
along
the
planes
of
the
hexagons.
the layers of carbon atoms in graphite
Giant footballs
More
Hint
recently
discovered.
a
number
Chemists
of
other
found
the
forms
rst
of
one
pure
whilst
carbon
they
have
were
been
looking
It is now believed that molecules
for
molecules
in
outer
space.
The
structures
of
these
new
forms
of
such as oxygen can slide in
carbon
between the layers of carbon.
include
nanotubes.
closed
The
most
cages
of
famous
carbon
is
atoms
and
also
tubes
buckminsterfullerene,
C
,
called
in
which
60
atoms
are
arranged
colleagues
are
received
investigating
Bonding
There
ionic,
the
are
–
a
the
many
football-like
Nobel
uses
Prize
for
shape
for
these
the
new
(Figure
6).
discovery.
Harry
Now,
Kroto
and
scientists
materials.
summary
three
covalent,
atoms
in
types
and
of
strong
metallic.
bonding
All
three
that
hold
involve
atoms
the
together
outer
–
electrons
of
concerned.
•
In
covalent
•
In
ionic
bonding,
bonding,
the
electrons
electrons
non-metal
atoms.
In
bonding,
are
are
shared
transferred
between
from
metal
atoms.
atoms
to
C
60
•
▲ Figure 6
Buckminsterfullerene –
also called buckyballs
64
metallic
form
a
lattice
of
ions
electrons
held
are
spread
together
by
between
delocalised
metal
atoms
electrons.
to
Bonding
3
If
you
know
bonding
what
from
the
•
metal
atoms
•
metal
and
•
non-metal
The
three
The
compound
–
of
metallic
non-metal
atoms
of
atoms
–
only
that
best
give
tells
conductivity.
conduct
electricity
metallic
bonding.
that
hold
us
rise
can
usually
tell
the
type
of
contains:
bonding.
to
what
Metals
to
electrons
it
bonding
covalent
bonding
metals)
their
you
bonding
ionic
–
is,
that
different
properties.
conductivity
property
electrical
only
types
Electrical
the
types
the
and
well,
The
metal
sort
in
both
current
ions
of
alloys
bonding
(an
the
is
alloy
solid
carried
together,
see
you
is
a
and
by
have
is
mixture
liquid
the
Figure
of
states
due
delocalised
7.
metal
e
+
+
e
M
electron
e
+
M
M
electron
e
e
in
out
e
+
▲ Figure 7
+
M
e
+
e
M
M
The conduction of electricity by a metal
solid
Ionic
compounds
only
conduct
electricity
in
the
liquid
state
(or
when
cathode
dissolved
current
in
is
water).
carried
They
by
the
do
not
conduct
movement
of
when
ions
they
towards
are
the
solid.
anode
The
electrode
of
+
opposite
is
charge.
The
liquid
or
dissolved
position
in
the
ionic
ions
in
are
free
water.
lattice,
In
to
move
the
Figure
solid
when
state
the
they
ionic
are
+
compound
xed
rigidly
+
+
in
+
8.
+
Generally,
convalently
bonded
substances
do
not
conduct
electricity
the
in
either
the
solid
or
liquid
state.
This
is
because
there
are
no
ions
to
in
but
water
carry
the
some
current.
react
to
Covalent
form
ions,
compounds
for
example,
are
often
ethanoic
are
not
the
solid
free
to
charged
move
particles
+
in
state
insoluble
acid
liquid
(present
in
vinegar).
The
solutions
can
then
conduct
electricity.
cathode
You
can
looking
therefore
at
how
it
decide
what
conducts
type
of
electricity.
bonding
This
is
a
substance
summarised
has
in
anode
by
Table
2.
+
+
▼ Table 2
The pattern of electrical conductivity tells us about the type of bonding
+
+
Type of bonding
Electrical conductivity
+
solid
liquid
aqueous solution
the
does not dissolve
metallic
✓
ions
are
compound
free
to
move
conducts
and
the
electricity
✓
but may react
▲ Figure 8
ionic
✗
✓
covalent
✗
✗
✓
Ionic liquids conduct
electricity, ionic solids do not
✗ (but may react)
Hint
Structure
–
summary
Note there are some covalently
Structure
are
held
describes
together
molecular,
the
in
arrangement
space.
There
macromolecular
are
(giant
in
which
four
atoms,
main
covalent),
types
giant
ions,
–
or
molecules
simple
ionic,
and
bonded substances that do conduct
electricity, for example, graphite.
metallic.
65
3.4
Bonding
and
p hysic al
prop erties
•
A
simple
groups
forces
50
of
atoms
structure
strongly
attraction
times
forces.
molecular
of
weaker
between
than
Examples
held
of
a
is
composed
together
molecules
covalent
molecules
by
are
bond)
include
of
much
and
Cl
,
A
macromolecular
atoms
are
covalent
(silica),
•
A
•
linked
bonds.
the
giant
in
structure
in
A
structure
metallic
a
metal
giant
The
and
property
of
boiling
that
molecular
is
•
Simple
molecular
•
Giant
a
the
best
compound
has
of
a
a
together
and
lattice
by
(often
over
,
and
NH
4
large
3
numbers
arrangement
and
silicon
by
a
positive
lattice
cloud
metallic
to
of
negative
regular
they
of
by
dioxide
of
of
regular
simple
and
each
vice
versa.
positively
delocalised
structures
have
ions
ions
are
electrons.
often
three-dimensional
molecular
structures.
points
us
(or
if
a
generally
low
structure
boiling)
compounds
a
which
small
intermolecular
SO
2
–
The
sand.
of
contrast
tells
melting
structures
of
because
in
in
H
2
diamond
arrangement
ionic,
atoms
one
called
O,
three-dimensional
consists
held
giant
is
include
consists
structures
arrangements
Melting
regular
ions
Macromolecular,
called
regular
constituent
surrounded
charged
If
a
Examples
main
ionic
structure
molecules
bonding.
weaker
are
H
2
•
small
covalent
have
have
melting
is
giant
or
simple
point.
low
high
(and
melting
melting
boiling)
(and
(and
point,
boiling)
boiling)
it
has
a
points.
points.
simple
Hint
molecular
Generally any substance with a
So
high melting point also has a high
covalent
all
structure.
compounds
All
with
molecular
low
compounds
melting
(and
are
boiling)
covalently
points
must
bonded.
have
bonding.
boiling point. However, there are
However,
a
compound
with
covalent
bonding
may
have
either
a
giant
some substances, such as iodine,
structure
or
a
simple
molecular
structure
and
therefore
may
have
that sublime – they turn directly
either
a
high
or
low
melting
(and
boiling)
point.
from solid to vapour.
Intermolecular forces
When
you
breaking
covalent
melt
the
and
within
determines
There
are
three
these
der
types
van
Waals,
•
dipole–dipole
•
hydrogen
highly
66
of
So
between
the
(and
compounds,
the
strength
boiling)
intermolecular
which
you
molecules,
of
the
are
not
the
intermolecular
points.
force.
In
order
of
increasing
act
between
which
act
all
atoms
between
molecules
with
permanent
δ
−Y
bonds,
which
electronegative
hydrogen
forces
melting
forces,
δ+
X
the
them.
molecular
are:
•
dipoles:
simple
intermolecular
bonds
forces
strength,
boil
atoms
are
act
between
atoms
the
(oxygen,
covalently
molecules
nitrogen,
bonded.
formed
and
when
uorine)
and
Bonding
3
Table
3
is
covalent,
▼ Table 3
a
summary
ionic,
and
of
the
metallic
different
properties
of
substances
with
bonding.
Summary of properties of substances with covalent, ionic, and metallic bonding
Electrical conductivity
Melting
Structure
Bond
point, T
Solid
Liquid
Aqueous solution
no
yes
yes
no
no
no (but may react)
m
+
–
+
+
–
+
giant
+
ionic
high
+
+
+
no
giant
(except
covalent
high
(macromolecular)
graphite and
graphene)
+
+
+
+
+
+
+
+
+
simple molecular
covalent
low
no
no
giant
metallic
high
yes
yes
does not dissolve but
may react
Summary questions
1
Describe what is the dierence between a macromolecular crystal and a
molecular crystal in terms of the following.
a
bonding
b
proper ties
2
Explain why graphite can be used as a lubricant.
3
Explain how graphite conducts electricity. How does it conduct dierently from metals?
4
Explain why both diamond and graphite have high melting points.
5
The table below gives some information about four substances.
a
Identify which substances have giant
Electrical
Melting point /
structures.
Boiling point /
conductivity
Substance
K (°C)
b
K (°C)
Identify which substance is a gas at room
solid
temperature.
A
c
Identify which substance is a metal.
d
Identify which substances are covalently
bonded.
e
Identify which substance has ionic bonding.
f
Identify which substance is a macromolecule.
B
C
D
1356 (1083)
91 (−182)
2840 (2567)
109 (−164)
1996 (1
723)
2503 (2230)
1266 (993)
1968 (1695)
liquid
good
good
poor
poor
poor
good
poor
poor
67
3.5
Learning
The
shapes
objective s:
Molecules
shapes
➔
of
mole c ul e s
are
three-dimensional
(Figure
and
and
they
come
ions
in
many
different
1).
State the rules that
govern the shapes of
➔
simple molecules.
Electron
Describe how the number
You
of electron pairs around an
space
atom aects the shape of
molecule,
for
a
of
the molecule.
➔
Describe what happens to
have
seen
called
number
•
pair
each
repulsion theory
that
electrons
orbitals.
example,
other
pair
You
of
in
can
one
atoms,
electrons
molecules
predict
the
consisting
by
using
around
of
the
an
exist
shape
a
pairs
a
central
ideas
atom
in
of
in
volumes
simple
atom
of
covalent
surrounded
by
that:
will
repel
all
other
electron
pairs
the shape of a molecule
when a bonding pair of
•
the
electrons is replaced by
pairs
possible
of
to
electrons
minimise
will
therefore
take
up
positions
as
far
apart
as
repulsion.
a non-bonding pair.
This
is
called
the
electron
pair
repulsion
theory.
Specication reference: 3.1.3
Electron
The
pairs
shape
electrons
any
of
may
a
that
molecule
be
simple
a
pair
molecule
surround
you
shared
rst
the
need
a
lone
depends
central
to
or
on
atom.
draw
a
pair.
the
To
number
work
out
dot-and-cross
of
the
pairs
of
shape
diagram
to
of
nd
water
the
number
Two
If
pairs of
there
be
of
are
linear.
pairs
electrons.
electrons
two
The
of
pairs
of
furthest
electrons
away
from
around
each
the
other
atom,
the
the
two
molecule
pairs
can
will
get
is
methane
180°
in
apart.
the
gas
example
Beryllium
phase,
of
chloride,
despite
being
which
a
is
a
covalently
metal–non-metal
bonded
molecule
compound,
is
an
this.
ammonia
two
▲ Figure 1
groups
of
electrons
The shapes of water,
methane, and ammonia molecules
180°
×
×
— Be — Cl
Hint
It is acceptable to draw electron
Three
pairs of
electrons
diagrams that show electrons in
If
there
are
three
pairs
of
electrons
around
the
central
atom,
they
the outer shells only.
will
be
Boron
120°
apart.
triuoride
The
is
an
molecule
example
is
of
planar
and
is
called
trigonal
planar.
this.
Hint
F
F
You can also think of electron pairs
×
120°
as clouds of negative charge.
B
B
F
F
F
F
Hint
Four
Notice that in neither BeCl
nor BF
2
does the central atom have a full
outer main electron shell.
If
are
electrons
four
pairs
of
electrons,
they
are
furthest
apart
when
they
3
are
arranged
This
see
68
pairs of
there
shape,
Figure
so
that
with
2.
they
one
point
atom
to
the
positioned
four
at
corners
the
of
centre,
is
a
tetrahedron.
called
tetrahedral,
Bonding
3
Methane,
CH
,
is
an
example.
The
carbon
atom
is
situated
at
the
4
centre
of
the
tetrahedron
with
the
hydrogen
atoms
at
the
vertices.
each
The
angles
here
arrangement
so
are
109.5°.
the
sum
This
of
the
is
a
three-dimensional,
angles
can
H
be
more
—
×
H
C
—
×
ammonium
surrounding
ion
the
is
109.5°
C
H
H
also
nitrogen
360°.
H
H
The
than
angle = 109.5°
planar,
H
×
×
H
not
tetrahedral.
atom.
The
It
fact
has
four
that
the
groups
ion
has
of
an
electrons
▲ Figure 2
overall
A tetrahedron has four
points and four faces
charge
does
not
affect
the
shape.
+
+
Hint
H
H
109.5°
×
×
H
H
N
—
In three-dimensional
N
×
H
H
representations of molecules, a
H
H
wedge is used to represent a bond
coming out of the paper and a
Five
pairs of
electrons
dashed line represents one going
If
there
are
ve
pairs
of
electrons,
the
shape
usually
adopted
is
that
of
into the paper, away from the reader,
a
trigonal
bipyramid.
Phosphorus
pentachloride,
PCl
,
is
an
example.
5
as shown in these gures.
Cl
×
Cl
Cl
Cl
×
Cl
P
×
Cl
×
120°
P
×
Cl
Cl
Cl
Cl
Six
If
pairs of
there
with
are
bond
electrons
six
pairs
angles
of
of
electrons,
90°.
The
the
sulfur
shape
adopted
hexauoride,
is
SF
,
octahedral,
molecule
is
an
6
example
of
this.
F
×
F
F
—
×
F
×
F
×
F
F
F
×
—
×
F
90°
S
S
F
Hint
F
F
T
ake care. Octahedral sounds as
if there should be eight electron
Molecules
with
lone
pairs of
electrons
groups, not six. Remember that an
Some
molecules
electrons
shape
of
that
the
lone
are
part
diagram
Ammonia
pairs
not
unshared
molecule.
dot-and-cross
effect.
have
affect
and
the
of
a
pairs
covalent
Always
because
water
(lone)
are
watch
bond.
out
otherwise
good
of
for
electrons.
The
the
you
lone
lone
might
examples
of
These
pairs
pairs
affect
in
overlook
molecules
are
octahedron has eight faces but six
the
points.
your
their
where
shape.
Study tip
Remember that electron pairs will
get as far apar t as possible.
69
3.5
The
shapes
of
m o l e c ul es
and
ion s
Ammonia,
NH
3
Study tip
Ammonia
has
four
pairs
of
electrons
and
one
of
the
groups
is
a
lone
pair.
×
Draw structures showing bonds
×
and lone electron pairs.
H
H
N
×
H
With
its
four
molecule
three
pairs
has
‘arms’
a
so
of
electrons
shape
the
based
shape
around
on
is
a
that
the
nitrogen
tetrahedron.
of
a
atom,
However,
triangular
the
ammonia
there
are
only
pyramid
N
—
H
H
H
Another
but
the
vertex
way
of
bonds
but,
looking
form
unlike
a
at
this
is
triangular
the
that
the
electron
pyramid.
tetrahedral
(There
arrangement,
pairs
is
an
no
form
atom
atom
in
a
tetrahedron
at
each
the
centre.)
Bonding pair –lone pair repulsion
The
angles
of
a
regular
tetrahedron,
see
Figure
2,
are
all
109.5°
but
Study tip
lone
It is dicult to fully appreciate
the 3-D shapes of molecules
without using models. Professional
pairs
pairs
the
of
affect
these
electrons
hydrogen
are
nucleus.
nitrogen
nucleus
chemists routinely use powerful
pairs.
repulsion
computer software (based on the
of
principles described in this topic)
squeezes
to accurately model the shapes
The
of complex molecules before they
in
So
electrons
is
the
and
is
rule
ammonia,
towards
the
therefore
than
hydrogen
are
In
However,
between
greater
approximate
ammonia
angles.
attracted
a
atoms
of
lone
pulled
lone
that
pair
together,
is
2°
for
example,
nitrogen
pair
is
closer
of
between
thumb
approximately
the
attracted
to
than
and
bonding
reducing
per
it
electrons
two
lone
the
all
pair,
bonding
nucleus
the
a
the
shared
This
H–N–H
the
also
by
bonding
pairs.
the
so
and
only
bond
pair
effect
angles.
angles
107°:
have even been synthesised. This
can help to predict how they might
—
N
act as drugs, for example.
H
H
H
107°
Water,
H
O
2
at
the
dot-and-cross
diagram
for
water.
×
Look
O
×
There
are
four
pairs
of
electrons
around
the
oxygen
atom
so,
as
with
You might consider buying a kit
ammonia,
the
shape
is
based
on
a
tetrahedron.
However,
two
of
the
of molecular models (par ticularly
‘arms’
of
the
tetrahedron
are
lone
pairs
that
are
not
part
of
a
bond.
if you intend to continue to study
This
results
in
a
V-shaped
or
angular
molecule.
As
in
ammonia
the
chemistry) or alternatively you
electron
pairs
form
a
tetrahedron
but
the
bonds
form
can do a lot with matchsticks
two
lone
pairs,
the
H–O–H
angle
is
reduced
and a modelling material such as
©
Plasticene
. Fur thermore, there
O
are websites and programmes
—
—
where you can draw 3-D models
104.5°
of molecules and rotate them to
view from dierent points in space.
70
to
104.5°.
a
V-shape.
With
Bonding
3
–
Chlorine tetrauoride
ion, ClF
4
diagram
for
this
ion
×
dot-and-cross
Cl
×
F
There
are
four
bonding
pairs
of
as
shown:
F
×
F
is
×
The
F
electrons
and
two
lone
pairs.
One
of
Hint
the
lone
charge
in
the
pairs
on
the
contains
ion
is
around
electron
negative
dot-and-cross
electrons
an
(–1).
diagram.
the
chlorine
that
This
This
has
electron
means
atom
–
been
four
that
donated
is
shown
there
bonds
to
are
and
as
six
two
it,
a
so
the
square
pairs
lone
of
Some ions contain more than one
atom. These are called compound
pairs.
ions. The atoms within the ion are
The
shape
is
therefore
based
on
an
octahedron
in
which
two
arms
are
covalently bonded but the ion has
not
part
of
a
bond.
an overall charge. Ones you will
As
lone
leaves
pairs
a
at
repel
the
most,
square-shaped
they
ion
adopt
a
described
position
as
square
furthest
planar.
apart.
The
This
come across include:
lone
+
ammonium
NH
4
pairs
are
above
and
below
the
plane,
as
shown
here.
2
carbonate
CO
3
2
sulfate
SO
4
F
F
Cl
nitrate
NO
3
F
F
hydroxide
A
summary of the
bonding
repulsion
pair–bonding
lone
pair–bonding
lone
pair–lone
between
electron
OH
pairs
pair
pair
repulsion
increases
pair
Summary questions
1
Draw a dot-and-cross diagram for NF
and predict its shape.
3
2
Explain why NF
has a dierent shape from BF
3
3
3
Draw a dot-and-cross diagram for the molecule silane, SiH
, and
4
describe its shape.
4
State the H—Si—H angle in the silane molecule.
5
Predict the shape of the H
S molecule without drawing a dot-and-cross
2
diagram.
71
3.6
Electronegativity
in
Learning
covalent
objective s:
The
of
➔
–
bo nd
p o lar it y
bonds
forces
that
positive
hold
charges
atoms
to
together
negative
are
all
charges.
In
about
ionic
the
attraction
bonding
there
is
State what is meant by the
complete
transfer
of
electrons
from
one
atom
to
another.
But,
even
in
term electronegativity.
➔
covalent
bonds,
spread
one
the
electrons
shared
by
the
atoms
will
not
be
evenly
State what makes one atom
if
of
the
atoms
is
better
at
attracting
electrons
than
the
more electronegative than
other.
This
atom
is
more
electronegative
than
the
other.
another.
➔
State what the symbols
δ+
Electronegativity
and δ– mean when placed
Flourine
is
better
at
attracting
electrons
than
hydrogen.
Fluorine
is
above atoms in a covalent
said
to
be
more
electronegative
than
hydrogen.
bond.
Electronegativity
is
the
power
of
an
atom
to
attract
the
Specication reference: 3.1.3
electron
When
chemists
electron
Study tip
charge
is
density
consider
density
is
distributed
in
the
often
in
a
a
covalent
electrons
used
to
bond
as
towards
charge
describe
the
clouds,
way
the
itself.
the
term
negative
molecule.
Learn the denition of
The
Pauling
scale
is
used
as
a
measure
of
electronegativity.
It
runs
electronegativity.
from
0
atom,
▼ Table 1
not,
Some values for Pauling
to
4.
see
in
The
Table
general,
greater
1.
The
form
the
number,
noble
gases
covalent
the
have
more
no
electronegative
number
because
the
they
do
bonds.
electronegativity
Electronegativity
H
depends
on:
He
1
the
nuclear
charge
2
the
distance
3
the
shielding
2.1
Li
Be
B
C
N
O
F
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Na
Mg
Al
Si
P
S
Cl
0.9
1.2
1.5
1.8
2.1
2.5
3.0
Note
the
of
the
the
nucleus
nuclear
and
charge
by
the
outer
electrons
in
shell
electrons
inner
shells.
following:
Ar
•
The
smaller
outer
Br
between
Ne
main
the
atom,
shell
the
closer
electrons
and
the
the
nucleus
greater
is
its
to
the
shared
electronegativity.
Kr
•
The
larger
the
nuclear
charge
(for
a
given
shielding
effect),
the
2.8
greater
Trends
Going
in
up
a
the
electronegativity.
electronegativity
group
in
the
Periodic
Table,
electronegativity
increases
Hint
(the
The Pauling scale is named after the
atoms
get
Going
the 1954 Nobel Prize in Chemistry for
increases.
his work on chemical bonding.
shells
So,
across
the
most
of
the
compounds).
and
having more ‘electron-pulling power’.
72
a
The
remain
corner
Think of electronegative atoms as
and
there
is
less
shielding
by
electrons
in
inner
shells.
US chemist Linus Pauling, who won
Hint
smaller)
nitrogen
period
in
nuclear
the
the
Periodic
charge
same
and
the
electronegative
Periodic
The
Table
most
followed
Table,
increases,
atoms
atoms
chlorine.
electronegativity
number
found
the
electronegative
by
the
become
are
(ignoring
the
at
noble
atoms
of
inner
main
smaller.
the
top
gases
are
right-hand
which
uorine,
form
few
oxygen,
Bonding
3
×
×
×
×
×
9+
1+
F
9+
×
×
9+
H
F
F
×
×
▲ Figure 1
×
Electron diagram
▲ Figure 2
of uorine molecule
▼ Table 2
Electron cloud
▲ Figure 3
around uorine molecule
Electron diagram of
hydrogen uoride molecule
▲ Figure 4
Electron cloud around
hydrogen uoride molecule
Trends in electronegativity
Increasing electronegativity
Li
Be
B
C
N
O
F
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Cl
Increasing
3.0
electronegativity
Br
2.8
Polarity of
covalent
bonds
Hint
Polarity
is
about
that
bonded
the
unequal
sharing
of
the
electrons
between
atoms
δ+ and δ− are pronounced ‘delta
are
together
covalently.
It
is
a
property
of
the
bond
plus’ and ‘delta minus’.
Covalent
bonds
between two
atoms that
are the
same
The + and – signs represent one
When
both
atoms
are
the
same,
for
example,
in
uorine,
F
,
the
2
‘electron’s wor th’ of charge.
electrons
(Figure
the
in
1)
bond
the
–
is
bond
both
must
atoms
completely
be
shared
have
equally
exactly
the
between
same
the
atoms
electronegativity
and
δ+ and δ− represent a small
non-polar.
charge of less than one ‘electron’s
wor th’.
If
you
cloud
think
is
of
the
electrons
uniformly
Covalent
spread
bonds
as
being
between
in
the
between two
a
cloud
two
of
charge,
atoms,
atoms that
as
then
shown
the
in
Figure
2.
are different
Summary questions
In
a
the
covalent
bond
electrons
in
between
the
bond
two
will
atoms
not
be
of
different
shared
electronegativity,
equally
between
the
1
atoms.
For
example,
the
molecule
hydrogen
uoride,
HF
,
shown
Explain why uorine is more
in
electronegative than chlorine.
Figure
3.
2
Hydrogen
has
an
electronegativity
of
2.1
and
uorine
of
4.0.
Write δ+ and δ− signs to show
This
the polarity of the bonds in a
means
that
the
electrons
in
the
covalent
bond
will
be
attracted
more
hydrogen chloride molecule.
by
the
uorine
than
the
hydrogen.
The
electron
cloud
is
distorted
3
towards
the
uorine,
as
shown
in
Figure
Identify if these covalent
4.
bonds is/are non-polar, and
The
the
uorine
end
hydrogen
show
this
by
of
end
the
molecule
relatively
adding
partial
is
therefore
positive,
charges
that
to
δ+
the
is,
relatively
electron
negative
decient.
and
explain your answer.
You
a
H—H
b
F —F
c
H—F
a
Arrange the following
formula:
δ
H—F
Covalent
bonds
like
this
are
said
to
be
polar.
The
greater
the
4
difference
in
electronegativity,
You
say
the
more
polar
is
the
covalent
bond.
covalent bonds in order
could
ionic
that
character.
It
although
is
going
the
some
H—F
way
bond
is
towards
covalent,
the
it
has
separation
some
of
of increasing polarity:
the
H—O,
atoms
into
charged
ions.
It
is
also
possible
to
have
ionic
bonds
b
some
covalent
H—F
,
H—N
with
Explain your answer.
character.
73
3.7
Learning
➔
Forces
objective s:
State the three types of
intermolecular force.
acting
Atoms
in
forces.
➔
molecules
covalent,
attracted
betw e e n
ionic,
to
or
one
‘Inter’
and
in
metallic
another
means
giant
structures
bonds.
by
If
are
Molecules
other,
between.
m ol ec ule s
weaker
the
held
and
forces
together
separate
called
intermolecular
by
strong
atoms
are
intermolecular
forces
are
strong
Describe how dipole–dipole
enough,
then
molecules
are
held
closely
enough
together
to
be
liquids
and van der Waals forces
or
even
solids.
arise.
➔
Describe how van der
Intermolecular forces
Waals forces aect boiling
There
are
three
types
of
intermolecular
forces:
temperatures.
➔
State what is needed for
•
van
act
der
Waals
between
all
forces
atoms
and
molecules.
weakest
hydrogen bonding to occur.
•
➔
Explain why NH
, H
3
Dipole–dipole
forces
O, and
2
act
only
between
certain
types
of
molecules.
HF have higher boiling
temperatures than might be
•
Hydrogen
acts
only
bonding
between
certain
types
of
molecules.
strongest
expected.
Specication reference: 3.1.3
Dipole–dipole forces
Hint
Dipole
moments
Polarity
is
with
of
the
polar
the
property
bonds
polarity
of
may
all
of
a
particular
have
the
a
dipole
bonds
in
bond,
see
moment.
the
Topic
This
3.6,
sums
but
up
molecules
the
effect
molecule.
van der Waals is spelt with a
small v, even at the beginning of
In
a sentence. van der Waals forces
may
cancel,
are sometimes called dispersion
may
also
forces or London forces
of
molecules
the
For
with
more
leaving
add
up
a
and
than
one
molecule
so
polar
with
reinforce
bond,
no
each
the
dipole
other.
effects
moment.
It
depends
of
each
The
on
bond
effects
the
shape
molecule.
example,
carbon
dioxide
is
a
linear
δ+
δ
O
Tetrachloromethane
is
molecule
the
dipoles
cancel.
δ
C
tetrahedral
and
O
and
here
too
the
dipoles
cancel.
δ
Cl
δ +
δ
δ
tetrachloromethane
C
δ
Cl
But
of
in
the
dichloromethane
the
dipoles
do
not
cancel
because
of
the
shape
molecule.
H
δ +
δ
dichloromethane
Cl
δ
Cl
Dipole–dipole
forces
act
between
molecules
that
have
permanent
Study tip
dipoles.
Do not confuse intermolecular
is
more
towards
For
example,
in
electronegative
the
chlorine
the
than
atom
hydrogen
chloride
hydrogen.
rather
than
So
the
the
molecule,
electrons
hydrogen
atom.
forces with covalent bonds, which
δ+
molecule
are at least 10 times stronger.
74
therefore
has
a
dipole
and
is
written
H
δ
–Cl
chlorine
are
pulled
The
Bonding
3
Two
molecules
Figure
which
both
have
dipoles
will
attract
one
another,
see
attraction
δ +
1.
δ
δ +
H — Cl
Whatever
their
to
arrangement
give
an
starting
positions,
where
the
the
molecules
two
with
molecules
dipoles
will
δ
H — Cl
‘ip’
attract.
attraction
δ +
H
δ
—
δ +
van der Waals forces
H — Cl
rotates
δ
All
atoms
and
molecules
are
made
up
of
positive
and
negative
Cl
charges
repulsion
even
though
weak
are
they
are
electrostatic
called
van
neutral
attractions
der
Waals
overall.
These
between
all
charges
atoms
produce
and
very
molecules.
These
forces.
δ +
δ +
δ
H — Cl
How do van der Waals forces
δ
H — Cl
work?
attraction
Imagine
two
a
helium
negatively
atom.
charged
It
has
two
electrons.
positive
The
charges
atom
as
a
on
its
whole
nucleus
is
neutral
and
repulsion
but
δ +
at
any
moment
in
time
the
electrons
could
be
anywhere,
see
Figure
δ
H — Cl
This
means
Any
of
the
distribution
of
charge
is
changing
at
every
δ +
δ
2.
Cl
— H
instant
rotates
the
moment.
almost
arrangements
An
instant
certainly
though
any
the
in
later,
atom
particular
Figure
the
will
dipole
2
mean
dipole
have
will
may
a
be
the
be
dipole
just
for
atom
in
at
a
a
different
any
an
has
point
instant
–
dipole
that
direction.
in
a
at
time,
But,
even
δ +
temporary
δ +
δ
H — Cl
dipole.
This
so
they
dipole
then
affects
the
electron
distribution
in
nearby
δ
H — Cl
atoms,
attraction
that
are
attracted
to
the
original
helium
atom
for
that
instant.
▲ Figure 1
The
original
atom
has
induced
dipoles
in
the
nearby
atoms,
as
shown
Two polar molecules, such
in
as hydrogen chloride, will always attract
Figure
3
in
which
the
electron
distribution
is
shown
as
a
cloud.
one another
δ +
δ
2+
2+
▲ Figure 2
the
nucleus
electrons
in
will
attract
nearby
the
these
atoms
electrons
electrons
in
will
nearby
repel
2+
2+
These are just a few of
the
atoms
the possible arrangements of the
two electrons in helium. Remember,
δ +
δ
δ +
2+
δ
2+
▲ Figure 3
the
to
the
new
van
original
name
der
•
van
•
They
•
The
so
one.
is
in
These
dipole
van
the
der
of
the
atoms
forces
forces,
Waals
original
around
are
but
is
rather
after
changes,
which
sometimes
this
forces
atom
it,
the
will
called
a
be
it
instantaneous
mouthful.
Dutch
will
attracted
The
scientist,
more
Johannes
Waals.
der
Waals
are
in
dipole
the
will
attraction
distribution
dipoles
dipole–induced
usual
electrons are never in a xed position
Instantaneous dipoles induce dipoles in nearby atoms
electron
induce
δ
2+
attraction
As
δ +
addition
is
more
forces
caused
act
between
to
any
by
the
electrons
there
other
all
atoms
molecules
intermolecular
changing
are,
or
the
position
larger
the
of
at
all
times.
forces.
electron
cloud,
instantaneous
the
dipole
be.
75
3.7
Forces
acting
be t w e en
m ol ec ule s
Therefore
number
large
•
than
or
the
van
molecular
or
der
Waals
This
forces
means
masses
molecules
points
of
boiling
chain
the
present.
produce
with
increases
that
atoms
stronger
small
atomic
or
with
van
or
the
molecules
der
with
Waals
molecular
masses.
why:
boiling
numbers
•
of
atoms
explains
the
size
electrons
atomic
forces
This
the
of
the
of
the
noble
points
of
noble
gases
gases
increase
as
the
atomic
increase
hydrocarbons
increase
with
increased
length.
δ
O
Hydrogen
δ +
bonding
δ +
H
—
—
H
Hydrogen
dipole–dipole
bonding
is
a
special
type
of
intermolecular
force
with
some
interaction
characteristics
bond.
It
of
dipole–dipole
consists
of
a
attraction
hydrogen
atom
and
some
‘sandwiched’
of
a
covalent
between
two
very
δ
δ +
O
—
—
δ +
electronegative
H
for
hydrogen
atoms.
bonding
There
to
are
occur.
conditions
You
need
a
that
very
have
to
be
present
electronegative
H
atom
▲ Figure 4
with
a
lone
pair
of
electrons
covalently
bonded
to
a
hydrogen
Dipole attraction between
atom.
Water
molecules
full
these
conditions.
Oxygen
is
much
more
water molecules
electronegative
You
would
between
expect
The
2
In
because
the
positively
very
the
strong
is
very
have
1)
polar,
but
very
The
small.
because
case
their
4.
(as
shown
the
reasons:
of
electrons.
and
hydrogen
These
of
Figure
electron
electronegative
it.
this
two
pairs
highly
see
attractions
in
for
lone
are
towards
eld
is
stronger
atoms
and
electric
Figure
water
bond
charged
in
water
dipole–dipole
much
hydrogen
oxygen
in
is
in
so
weak
chloride
atoms
the
electrons
nd
bonding
oxygen
water
hydrogen
to
hydrogen
intermolecular
1
than
decient.
attracts
atoms
exposed
small
the
in
This
shared
water
protons
is
have
are
a
size.
δ +
The
lone
pair
of
electrons
on
the
oxygen
atom
of
another
water
H
δ
δ +
molecule
is
strongly
attracted
to
the
electron
decient
hydrogen
atom.
H
O
—
—
—
O
—
δ
This
strong
intermolecular
force
is
called
a
hydrogen
bond.
Hydrogen
δ +
δ +
bonds
are
considerably
stronger
than
dipole–dipole
attractions,
though
H
H
much
▲ Figure 5
weaker
than
a
covalent
bond.
They
are
usually
represented
by
Hydrogen bond between
dashes
–
–
–,
as
in
Figure
5.
water molecules
When do
hydrogen
Water
the
a
•
is
a
•
a
a
example
there
atom
produce
very
only
bond
hydrogen
will
δ +
not
hydrogen
bonds form?
must
that
strong
is
of
be
bonded
partial
electronegative
hydrogen
the
atom
bonding.
In
order
to
form
following:
to
a
positive
with
a
very
electronegative
charge
lone
on
pair
the
of
atom.
hydrogen
electrons.
This
atom.
These
H
δ
will
be
attracted
to
the
partially
charged
hydrogen
atom
in
another
δ
δ +
δ +
molecule
—
—
δ +
H
N
form
the
bond.
H
—
O
and
δ +
The
only
atoms
that
are
electronegative
enough
to
form
hydrogen
H
H
bonds
▲ Figure 6
Hydrogen bond between
are
ammonia
oxygen,
O,
molecules,
nitrogen,
NH
,
3
a water molecule and an ammonia
molecule
76
molecules,
see
Figure
6.
form
N,
and
uorine,
hydrogen
F
.
bonds
For
with
example,
water
Bonding
3
The
the
nitrogen–hydrogen–nitrogen
pair
of
hydrogen
always
The
The
bond
the
effect
at
of
the
plotted
in
the
between
case
boiling
look
7
electrons
with
O—H
nitrogen
hydrogen
points of the
hydrogen
boiling
against
of
period
is
and
linear.
bond
This
repels
hydrogen.
This
is
because
those
in
the
linearity
is
bonds.
hydrides
bonding
points
the
system
covalent
between
hydrides
number,
of
molecules
elements
see
Figure
of
can
be
Group
seen
4,
if
5,
you
6,
and
7.
400
H
O
2
300
K /
H
HF
Te
2
tniop
H
Se
SbH
2
3
NH
3
H
HI
S
2
AsH
gniliob
3
200
SnH
HBr
4
HCl
PH
3
GeH
4
Xe
SiH
4
CH
4
Kr
100
Ar
Ne
He
0
1
2
3
4
5
period
▲ Figure 7
Boiling points of the hydrides of Group 4, 5, 6, and 7 elements with the
noble gases for comparison
The
noble
only
these
The
gases
forces
show
acting
increase
boiling
with
points
a
gradual
between
the
of
the
increase
atoms
number
water,
H
of
O,
in
are
boiling
van
electrons
hydrogen
der
point
Waals
because
forces,
the
and
present.
uoride,
HF
,
and
ammonia,
2
NH
,
are
all
higher
than
those
of
the
hydrides
of
the
next
elements
in
3
their
group,
Waals
whereas
forces
between
the
were
molecules
intermolecular
separate.
covalent
a
lot
bonds,
are
of
hydrogen
of
and
them.
can
to
unaffected,
is
are
very
these
make
fact
make
the
the
that
lower
if
only
bonding
and
three
bonding
these
more
van
is
der
present
stronger
difcult
elements
that
to
are
possible.
bonding
about
10%
signicant
they
under
be
molecules
are
hydrogen
be
to
hydrogen
compounds
uorine
only
can
them
because
hydrogen
very
or
and
make
effect
The
break
of
is
attraction
bonds
their
expect
This
each
nitrogen,
enough
bonds,
would
in
importance of
Although
are
forces
Oxygen,
electronegative
The
you
operating.
are
–
of
the
strength
especially
weaker
conditions
than
where
of
when
there
covalent
covalent
bonds
signicant.
77
3.7
Forces
acting
be t w e en
m ol ec ule s
The
structure
In
water
as
the
in
the
In
Topic
order
packed
water
▲ Figure 8
longer
shown
to
t
than
and
in
in
into
in
the
xed
hydrogen
about.
free
Figure
on
ponds
have
under
this
liquid
forms
the
must
water
no
state,
moving
ice
to
move
positions.
8,
bonds
When
about
The
resembles
break
water
and
the
resulting
the
and
freezes,
reform
the
easily
water
hydrogen
bonds
three-dimensional
structure
of
diamond,
3.4.
insulates
This
are
are
molecules
structure,
see
liquid
molecules
molecules
hold
its
and density of
structure,
water.
top
and
helped
the
ice,
of
the
This
ponds
enables
life
to
during
molecules
means
rather
sh
to
Ice
ice
at
survive
continue,
the
that
than
in
are
the
is
slightly
less
the
less
dense
bottom.
through
relative
the
closely
than
This
winter.
warmth
of
the
Ages.
The three-dimensional
network of covalent bonds (grey) and
hydrogen bonds (red) in ice. The blue
lines are only construction lines
Living with hydrogen bonds
Proteins are a class of important biological molecules that full a wide variety
of functions in living things, including enzyme catalysts. The exact shape of
a protein molecule is vital to its function. Proteins are long chain molecules
with lots of C
O and N—H groups which can form hydrogen bonds. These
hydrogen bonds hold the protein chains into xed shapes. One common
shape is the protein chain that forms a spiral (helix), as shown here.
C
=
H
O
N
C
C
=
H
O
=
O
H
N
N
C
=
O
H
N
▲ Figure 9
Another example is the beta-pleated sheet. Here protein chains line up side
by side, held in position by hydrogen bonds to form a two-dimensional
sheet. The protein that forms silk has this structure.
H
H
—
O
R
=
H
—
=
O
N
N
—
=
N
H
—
=
N
H
O
O
R
R
H
H
H
R
O
R
Synoptic link
H
H
—
=
H
=
—
R
O
The bonding in proteins and in
H
=
78
N
—
O
H
proteins.
H
=
—
in Chapter 30, Amino acids and
N
H
N
N
DNA is discussed in more detail
O
R
H
R
Bonding
3
The relative weakness of hydrogen bonds means that the shapes of proteins
can easily be altered. Heating proteins much above body temperature star ts
to break hydrogen bonds and causes the protein to lose its shape and thus
its function. This is why enzymes lose their eect as catalysts when heated
– the protein is denatured. You can see this when frying an egg. The clear
liquid protein albumen is transformed into an opaque white solid.
Ironing
When you iron clothes, the iron provides heat to break hydrogen bonds in the
crumpled material and pressure to force the molecules into new positions
so that the material is at. When you remove the iron, the hydrogen bonds
reform and hold the molecules in these new positions, keeping the fabric at.
DNA
Another vital biological molecule is DNA (deoxyribonucleic acid) (Figure 10).
It is the molecule that stores and copies genetic information that makes
ospring resemble their parents. This molecule exists as a double-stranded
helix. The two strands of the spiral are held together by hydrogen bonds. When
cells divide or replicate, the hydrogen bonds break (but the covalently bonded
main chains stay unchanged). The two separate helixes then act as templates
▲ Figure 10
The DNA double helix is
for a new helix to form on each, so you end up with a copy of the original helix.
held together by hydrogen bonds
Summary questions
1
Place the following elements in order of the strength of the van der
Waals forces between the atoms
(weakest rst): Ar, He, Kr, Ne. Explain
your answer.
2
Identify which one of the following molecules
2
3
Explain
is
4
a
hexane
is
a
liquid
2
at
room
temperature
whereas
butane
gas.
Explain
melting
5
why
cannot have dipole–dipole
O, HCl, H
forces acting between them: H
why
covalent
molecules
are
gases,
liquids,
or
solids
with
low-
temperature.
Draw two hydrogen bromide molecules to show how they would be
attracted together by dipole–dipole forces.
6
Identify in which of the following does hydrogen bonding
between molecules: H
O, NH
2
7
not occur
, HBr, HF
3
Explain why hydrogen bonds do not form between:
a
methane molecules, CH
b
tetrachloromethane molecules, CCl
4
4
8
Draw a dot-and-cross diagram for a molecule of water.
a
State how many lone pairs it has.
b
State how many hydrogen atoms it has.
c
Explain why water molecules form on average two hydrogen bonds
per molecule, whereas the ammonia molecule, NH
, forms only one.
3
79
Practice
questions
1
Phosphorus
exists
phosphorus.
in
White
several
different
phosphorus
forms,
consists
of
two
P
of
which
molecules,
are
and
white
melts
phosphorus
at
44
and
red
°C.
4
Red
phosphorus
Explain
what
bonding
so
is
is
macromolecular,
meant
present
in
by
these
the
two
term
and
has
a
melting
macromolecular .
forms
of
point
By
phosphorus,
above
considering
explain
why
550
the
°C.
structure
their
melting
and
points
are
different.
(5
marks)
AQA,
2
(a)
Predict
the
species
to
shapes
of
the
molecule
SF
and
the
AlCl
6
for
the
show
bond
their
angles.
Draw
diagrams
of
these
4
three-dimensional
Explain
ion.
your
shapes.
Name
the
shapes
and
suggest
values
reasoning.
(8
(b)
Perfume
When
time.
is
a
mixture
applied
After
a
to
of
the
while,
fragrant
skin
the
2006
the
compounds
solvent
fragrance
dissolved
evaporates,
may
be
in
a
causing
detected
some
volatile
the
skin
distance
marks)
solvent.
to
cool
away.
for
a
short
Explain
these
observations.
(4
marks)
AQA,
3
Fritz
Haber,
soluble
(a)
in
German
chemist,
rst
manufactured
ammonia
in
1909.
Ammonia
is
very
water.
State
and
a
the
one
strongest
molecule
type
of
of
intermolecular
force
between
one
molecule
of
ammonia
water.
(1
(b)
Draw
of
a
diagram
water.
to
Include
show
all
how
partial
one
molecule
charges
and
all
of
ammonia
lone
pairs
of
is
attracted
electrons
in
to
one
your
Phosphine,
PH
,
has
a
structure
similar
to
mark)
molecule
diagram.
(3
(c)
2003
marks)
ammonia.
3
In
is
terms
of
almost
intermolecular
insoluble
in
forces,
suggest
the
main
reason
why
phosphine
water.
(1
mark)
AQA,
2013
+
4
The
following
equation
shows
the
reaction
of
a
phosphine
molecule,
PH
,
with
an
H
ion.
3
+
PH
+
H
+
➝
PH
3
(a)
Draw
the
shape
of
the
PH
4
molecule.
Include
any
lone
pairs
of
electrons
that
3
inuence
the
shape.
(1
mark)
+
(b)
State
the
type
of
bond
that
is
formed
between
the
molecule
PH
and
the
H
ion.
3
Explain
how
this
bond
is
formed.
(2
marks)
+
(c)
Predict
the
bond
angle
in
the
ion.
PH
4
(1
(d)
Although
bonding
Suggest
phosphine
between
an
molecules
phosphine
explanation
for
contain
hydrogen
atoms,
there
is
no
mark)
hydrogen
molecules.
this.
(1
mark)
AQA,
5
There
are
several
types
of
crystal
structure
and
bonding
shown
by
elements
2012
and
compounds.
(a)
(i)
Name
(ii)
Use
the
type
of
bonding
in
the
element
sodium.
(1
your
how
You
the
knowledge
particles
should
are
identify
two-dimensional
of
structure
arranged
the
and
in
particles
a
bonding
crystal
and
of
show
a
to
draw
a
diagram
that
sodium.
minimum
of
six
particles
in
a
diagram.
(2
marks)
AQA,
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
80
mark)
shows
2011
4
Energetics
4.1
Endothermic
an d
e xot her mic
reactions
Most
The
chemical
amount
reactions
of
energy
give
out
involved
or
take
when
a
in
energy
chemical
as
they
reaction
Learning
proceed.
takes
place
➔
is
important
for
many
reasons.
For
objective s:
Dene the terms endothermic
example:
and exothermic.
•
you
can
measure
•
you
can
calculate
•
you
can
bonds
work
and
the
out
the
energy
values
the
energy
the
theoretical
amount
of
of
fuels
requirements
energy
Specication reference: 3.1.4
for
industrial
amount
of
released
when
energy
processes
required
bonds
are
to
break
made
Hint
•
it
helps
to
predict
whether
or
not
a
reaction
will
take
place.
The unit of energy is the joule, J.
The
energy
involved
may
be
in
different
forms
–
light,
electrical,
or
One joule represents quite a
most
usually
heat.
small amount of heat energy. For
example, in order to boil water for
Thermochemistry
a cup of tea you would need about
Thermochemistry
chemical
•
When
new
•
a
Energy
energy
The
are
must
At
the
At
is
changes
during
80 000 J which is 80 kJ.
reaction
takes
place,
chemical
bonds
break
and
be
put
are
in
to
formed,
break
so
bonds
most
and
energy
chemical
is
given
reactions
out
involve
an
may
result
in
energy
being
given
out
or
end
the
is
end
of
the
reaction,
if
energy
has
been
given
if
energy
has
been
taken
out,
the
exothermic
of
the
reaction,
in,
the
reaction
endothermic
Some
reactions
exothermic
an
and
give
reactions.
exothermic
Some
reactions
reaction
of
heat
formed.
change
Exothermic
of
of
in.
reaction
•
study
change.
overall
taken
•
the
chemical
ones
whenbonds
•
is
reactions.
going.
limestone
dioxide
is
an
endothermic
out
heat
as
they
Neutralising
an
reactions
proceed.
acid
with
These
an
are
alkali
called
is
an
example
reaction.
take
in
These
heat
are
(calcium
example
from
called
an
surroundings
endothermic
carbonate)
of
their
to
lime
reactions.
(calcium
endothermic
to
reaction
keep
The
oxide)
–
it
the
breakdown
and
needs
carbon
heat
to
proceed.
Another
Blue
example
copper
of
sulfate
an
endothermic
crystals
have
the
reaction
formula
is
heating
CuSO
4
molecules
bonds
and
supplied
are
bonded
make
(Figure
CuSO
.5H
4
O
2
blue copper sulfate
to
white,
1).
copper
anhydrous
This
➝
the
reaction
sulfate.
copper
takes
in
In
order
sulfate,
heat
so
CuSO
is
+
4
white anhydrous copper sulfate
O.
sulfate.
The
water
2
to
heat
it
copper
•5H
break
these
energy
must
be
endothermic.
5H
O
2
water
▲ Figure 1
Heating copper sulfate
81
4.1
Endothermic
an d
e xot her mic
reacti on s
When
out
you
add
water
to
anhydrous
copper
CuSO
+
5H
4
the
reaction
this
direction
It
is
is
exothermic
always
the
the
case
in
O
CuSO
➝
2
white anhydrous copper sulfate
In
sulfate,
gives
heat.
reaction
that
the
a
is
O
2
blue copper sulfate
exothermic.
reaction
reverse
.5H
4
water
that
is
endothermic
in
one
direction
direction.
Quantities
The
amount
of
heat
given
out
or
taken
in
during
a
chemical
Hint
reaction
depends
on
the
quantity
of
reactants.
energy
is
usually
–1
–1
The expression mol
This
measured
is a
shor thand for ‘per mole’ and could
also be written /mol. So kJ/mol has
–1
the same meaning as kJ mol
.
about
in
kilojoules
quantities
combustion
of
you
per
need
methane,
mole,
to
kJ mol
give
CH
,
an
one
.
To
avoid
equation.
mole
of
For
any
confusion
example,
methane
reacts
in
the
with
two
4
moles
of
oxygen:
Also note that the state symbols
CH
(g)
+
2O
4
(g)
➝
CO
2
(g)
+
2H
2
O(l)
2
such as (g), meaning the gaseous
890 kJ
are
given
out
when
one
mole
of
methane
burns
in
two
moles
state, are used. These are also
of
oxygen.
impor tant here.
Useful
When
heat
fuels
exothermic
For
mole,
product
Carbon
is
burnt
changes
there
is
a
large
heat
output.
coal
12 g,
is
is
formed.
dioxide
is
mostly
burnt
This
the
carbon.
completely
is
carbon
only
Carbon
so
that
dioxide
you
saw
is
above,
burnt
Physiotherapists
produce
natural
completely
often
‘coldness’
by
+
O
gas,
to
treat
an
NH
(g)
➝
very
and
out
most
not
393.5 kJ
highly
carbon
when
oxidised
monoxide.
CO
methane,
carbon
sports
(s)
+
(g)
2
gives
dioxide
injuries
endothermic
NO
4
gives
the
2
mole
are
product.
C(s)
As
These
reactions.
example,
one
energy
are
(aq)
out
and
with
reaction
➝
3
NH
cold
such
NO
4
890 kJ
when
one
water.
packs.
These
as:
(aq)
3
−1
This
absorbs
26 kJ mol
of
heat
energy.
The energy values of fuels
One important practical application of the study of thermochemistry is that
it enables us to compare the efciency of dierent fuels. Most of the fuels
used today for transport (petrol for cars, diesel for cars and lorries, kerosene
for aviation fuel, etc.) are derived from crude oil. This is a resource that will
eventually run out so chemists are actively studying alternatives. Possible
replacements include ethanol and methanol, both of which can be made
from plant material, and hydrogen, which can be made by the electrolysis
of water.
82
Energetics
4
Theoretical chemists refer to the energy given out when a fuel burns completely
as its heat (or enthalpy) of combustion. They measure this energy in kilojoules
–1
per mole (kJ mol
) because this compares the same number of molecules of
each fuel. For use as fuels, the energy given out per gram of fuel burned, or the
energy density of a fuel, is more important.
Some approximate values are given in the T
able 1.
▼ Table 1
Enthalpy of combustion for various fuels
Enthalpy of
Energy
Mass of
Fuel
combustion
density /
1 mole / g
–1
–1
kJ g
/ kJ mol
petrol (pure octane)
–5500
114
48.2
–1370
46
29.8
methanol
–730
32
22.8
hydrogen
–242
2
12
1.0
ethanol
Notice that petrol stores signicantly more energy per gram than either
ethanol or methanol. This is a factor that will be signicant for vehicles
fuelled by either of these alcohols.
At rst sight, hydrogen’s energy density seems amazing. However, there is a
catch. The other three fuels are liquids, whereas hydrogen is a gas. Although
hydrogen stores lots of energy per gram, a gram of gaseous hydrogen takes
up a lot of space because of the low density of gases. How to store hydrogen
efciently is a challenge for designers.
1
Write a balanced symbol equation for the combustion of methanol,
2
How do the product(s) of combustion vary between hydrogen and
CH
OH.
3
the other fuels?
3
What environmental signicance does this have?
Summary questions
1
Natural gas, methane, CH
, gives out 890 kJ when one mole is
4
burnt completely.
CH
(g) + 2O
4
(g) ➝ CO
2
(g) + 2H
2
O(l)
2
Calculate how much heat would be given out when 8 g of methane is
burnt completely.
2
The following reaction does not take place under normal conditions.
CO
(g) + 2H
2
O(l) ➝ CH
2
(g) + 2O
4
(g)
2
If it did, would you expect it to be exothermic or endothermic?
3
Explain your answer to question
2
4
Approximately how much methane would have to be burnt to provide
enough heat to boil a cup of tea? Choose from
a
16 g
b
1.6 g
a, b, or c
c
160 g
83
4.2
Enthalpy
Learning
objective s:
The
the
➔
amount
of
conditions
heat
–
given
out
or
temperature,
taken
in
pressure,
by
a
reaction
concentration
varies
of
with
solutions,
Dene what an enthalpy
and
so
on.
This
means
that
you
must
state
the
conditions
under
which
change is.
measurements
➔
are
made.
For
example,
you
normally
measure
heat
Describe what an enthalpy
changes
at
constant
atmospheric
pressure.
level diagram is.
Specication reference: 3.1.4
Enthalpy
When
you
enthalpy
change,
measure
a
ΔH
heat
change
at
constant
pressure,
it
is
called
an
change
Hint
Enthalpy
has
the
symbol
H
so
enthalpy
changes
are
given
the
symbol
Chemists often use asks open
ΔH.
The
Greek
letter
Δ
(delta)
is
used
to
indicate
a
change
in
any
quantity.
to the atmosphere to measure
heat changes. The reaction is then
There
are
standard
conditions
for
measuring
enthalpy
changes:
carried out at atmospheric pressure.
•
pressure
of
100 kPa
•
temperature
(approximately
normal
atmospheric
pressure)
This varies slightly from day to
of
298 K
(around
normal
room
temperature,
25 °C).
day. Because these slight daily
variations are small, this is only a
(The
small source of systematic error.
standard
298 K
and
When
298 K
an
state
of
an
element
is
the
state
in
which
it
exists
at
100 kPa.)
enthalpy
change
is
measured
under
standard
conditions,
298 K
mix
it
is
written
ΔH
as
although
usually
the
298
is
left
out.
ΔH
is
298
pronounced
‘delta
It
strange
may
seem
temperature
reactant
H
standard’.
to
because
talk
heat
about
measuring
changes
normally
heat
cause
changes
at
a
constant
temperature
changes.
reactant
1
2
The
way
Figure
reactants
at
room
temperature,
to
1.
Mix
A
rises
to
about
the
this
reactants
is
to
and
imagine
heat
is
the
reactants
produced
(this
at
is
298 K,
an
see
exothermic
298 K
reaction).
T
emperature
think
This
reaction
is
heat
not
is
given
thought
out
of
as
to
the
being
surroundings.
over
until
the
products
have
328 K
cooled
back
to
298 K.
The
heat
given
out
to
the
surroundings
while
the
328 K
reaction
•
In
an
mixture
cools
exothermic
energy
than
the
energy
when
is
the
enthalpy
reaction
starting
they
the
products
materials
heated
up
change
end
because
their
for
the
up
reaction,
with
they
less
have
surroundings.
heat
lost
This
ΔH
heat
means
that
product
ΔH
heat
transferred
from
product
Some
is
negative.
It
endothermic
is
therefore
reactions
given
that
a
take
negative
place
in
sign.
aqueous
solution
to
absorb
heat
from
the
water
and
cool
it
down,
for
example,
dissolving
surroundings
ammonium
being
they
over
nitrate
until
in
the
water.
products
Again
have
you
don’t
warmed
up
think
to
the
of
the
reaction
temperature
at
as
which
started.
298 K
In
do
this
case
this.
that
is
the
Unless
solution
you
absorbing
has
to
remember
heat,
take
this,
initially
in
it
gets
heat
can
from
seem
the
surroundings
strange
that
a
to
reaction
cold.
product
•
In
an
than
product
back
at
endothermic
the
▲ Figure 1
at 298 K
84
reaction
materials,
the
so
products
ΔH
is
end
positive.
up
It
with
is
more
therefore
energy
given
a
room
positive
temperature,
starting
sign.
298 K
A reaction giving out heat
Pressure
involve
affects
gases.
If
the
a
amount
gas
is
of
given
heat
out,
energy
some
given
energy
out
is
by
reactions
required
to
that
push
Energetics
4
away
the
atmosphere.
The
greater
the
atmospheric
pressure,
Hint
the
to
more
be
energy
given
have
a
out
is
as
standard
used
heat
of
for
by
this.
the
pressure
This
means
reaction.
for
This
measuring
that
is
less
why
energy
energy
it
is
remains
important
to
changes.
Don’t be confused by the dierent
terms. Heat is a form of energy,
so a heat change can also be
The
physical
states of the
reactants
and
products
described as an energy change.
The
physical
also
affect
states
(gas,
liquid,
or
solid)
of
the
reactants
and
products
An enthalpy change is still an
the
enthalpy
change
of
a
reaction.
For
example,
heat
energy change, but it is measured
must
be
put
in
to
change
liquid
to
gas
and
is
given
out
when
a
gas
is
under stated conditions of
changed
symbols
For
in
a
liquid.
your
example,
two
1
to
This
means
that
you
must
always
include
state
temperature and pressure.
equations.
hydrogen
burns
in
oxygen
to
form
water
but
there
are
possibilities:
forming
liquid
water
Hint
1
(g)
H
+
2
−1
O
(g)
➝
H
2
2
O(l)
ΔH
−285.8
kJ mol
2
One way of making sure that
2
forming
steam
both reactants are at the same
temperature is simply to leave
1
(g)
H
2
+
−1
O
(g)
➝
H
2
2
O(g)
ΔH
−241.8
kJ mol
2
them in the same room for
The
difference
one
mole
of
Enthalpy
Enthalpy
in
ΔH
water
represents
into
the
amount
of
heat
needed
to
turn
some time.
steam.
level diagrams
level
diagrams,
sometimes
called
energy
level
diagrams,
are
enthalpy
used
to
levels
represent
of
vertical
the
axis
enthalpy
reactants
(starting
represents
of
the
reaction.
You
of
the
reaction,
100%
changes.
show
materials)
enthalpy,
are
They
usually
and
only
the
and
the
the
products.
horizontal
interested
relative
in
axis,
the
enthalpy
The
the
reactants
extent
beginning
∇
negative
reactants
without
Figure
(the
2
(and
100%
reactants,
products),
and
so
the
the
end
of
the
horizontal
reaction,
axis
is
usually
left
units.
shows
products
products
a
general
have
less
enthalpy
enthalpy
diagram
than
the
for
an
exothermic
reactants)
and
reaction
Figure
extent
an
endothermic
reaction
(the
products
have
more
of
reaction
3
▲ Figure 2
shows
H
0%
Enthalpy diagram for an
enthalpy
exothermic reaction
than
the
reactants).
Summary questions
enthalpy
1
Consider this reaction:
products
–1
CH
(g) + 2O
4
a
(g) ➝ CO
2
(g) + 2H
2
O(l)
ΔH
2
= –890 kJ mol
298
State what the symbol Δ means.
∇
positive
b
State what the symbol H means.
c
State what the 298 indicates.
d
State what the minus sign indicates.
e
Explain whether the reaction is exothermic or endothermic.
f
Draw an enthalpy diagram to show the reaction.
H
reactants
extent
▲ Figure 3
of
reaction
Enthalpy diagram for an
endothermic reaction
85
4.3
Learning
Measuring enthalpy changes – calorimetry
objective s:
The
general
standard
➔
name
molar
for
the
enthalpy
enthalpy
change
change
of
for
reaction
any
ΔH
.
reaction
It
is
is
the
measured
in
Describe how enthalpy
–1
kilojoules
per
mole,
kJ mol
(molar
means
‘per
mole’).
You
write
change is measured in
a
balanced
symbol
equation
for
the
reaction
and
then
find
the
heat
a reaction.
change
➔
for
enthalpy changes
more accurately.
For
example,
quantities
in
moles
given
ΔH
for
2NaOH
+
H
SO
2
enthalpy
of
H
by
this
equation.
change
when
two
➝
Na
4
moles
of
SO
2
NaOH
+
2H
4
O
is
the
2
react
with
one
mole
SO
2
➔
the
Describe how you measure
4
Describe how you measure
enthalpy changes in solution.
Standard
enthalpies
Specication reference: 3.1.4
Some
the
commonly
enthalpy
used
change
enthalpy
of
changes
Δ
formation
H
are
given
and
the
names,
for
enthalpy
example,
change
of
f
Δ
combustion
H
.
Both
of
these
quantities
are
useful
when
calculating
c
Synoptic link
enthalpy
changes
for
reactions.
In
addition,
Δ
H
s
are
relatively
easy
c
to
measure
for
compounds
that
burn
readily
in
oxygen.
Their
formal
See Practical 2 on page 522.
definitions
The
are
as
standard
follows:
molar
enthalpy
of
Δ
formation,
H
,
is
the
enthalpy
f
Hint
change
when
elements
one
under
mole
of
standard
substance
conditions,
is
formed
all
from
reactants
its
and
constituent
products
being
The apparatus used to measure
in
their
standard
states.
enthalpy changes is called a
The
calorimeter and the process of
measuring enthalpy changes is
called calorimetry
standard
molar
enthalpy
oxygen
change
under
when
standard
one
mole
Heat
Δ
combustion,
a
system.
increases
many
a
is
given
and
a
low
the
the
are
of
of
so
the
as
enthalpy
the
is
reactants
H
,
is
completely
the
and
burnt
products
in
being
in
states.
so
total
It
much
will
has
much
a
heat
it.
doesn’t
the
than
is
no
the
a
is
matter
of
the
energy
how
number
heat
a
hot
the
So
a
bath
because
from
the
in
the
bath
high
to
water,
nail.
reaction
given
instrument
of
nail
flows
into
than
present
much
included.
red
nail
heat
the
how
always
change of
is
particles
kinetic
particles
on
particle
Heat
from
the
thermometer.
all
more
reaction
There
a
depend
in
flow
it
of
average
independent
every
more
enthalpy
of
of
of
energy
their
But
is
with
does
particles
proceeds.
up.
energy
energy
heat
water
goes
temperature
The
has
kinetic
faster,
measured
the
change
reaction
is
more
Measuring the
The
average
move
substance.
water
many
though
the
are,
present.
temperature,
even
all
standard
temperature
measure
is
to
particles
there
amount
lukewarm
there
related
Temperature
substance
of
is
As
particles
present.
Heat
substance
and temperature
Temperature
in
of
conditions,
their
Hint
of
c
enthalpy
out
that
or
taken
measures
in
heat
The words heat and temperature
directly.
To
measure
the
enthalpy
change
you
arrange
for
the
heat
to
are often used to mean the same
be
transferred
into
a
particular
mass
of
a
substance,
often
water.
Then
thing in daily conversation, but in
you
need
to
know
three
things:
science they are quite distinct
and you must be clear about the
dierence.
86
1
mass
2
temperature
of
the
substance
change
3
specific
capacity
heat
that
of
is
the
being
heated
substance.
up
or
cooled
down
Energetics
4
The
specific
heat
capacity
c
is
the
amount
of
heat
needed
to
raise
Hint
the
temperature
of
1 g
of
substance
–1
gram
per
kelvin,
or
–1
water
is
of
g
1 K.
Its
units
are
joules
per
–1
K
.
For
example,
the
specific
heat
capacity
of
The size of a kelvin is the same
–1
4.18 J g
temperature
J
by
K
1
.
This
gram
of
means
water
that
by
1
it
takes
kelvin.
4.18
This
joules
is
often
to
raise
the
rounded
as the size of a degree Celsius. Only
up
the starting point of the scale is
–1
to
4.2 J g
–1
K
dierent. A temperature change is
numerically the same whether it is
Then:
measured in Celsius or kelvin. To
mass
enthalpy
change
q
of
specific
=
You
calorimeter or q
simple
can
use
enthalpy
the
apparatus
change
temperature
×
substance
The
heat
when
a
in
capacity
c
change
ΔT
mcΔT
=
Figure
fuel
convert °C to K add 273.
×
m
1
to
find
the
Hint
approximate
burns.
Chemists normally repor t enthalpy
–1
You
burn
the
fuel
the
temperature
the
fuel
The
goes
meaning
heat
rise
into
apparatus
to
of
the
used
a
known
the
water.
mass
You
of
water
assume
and
that
then
all
the
measure
heat
changes in kJ mol
from
water.
is
called
a
calorimeter
(from
the
Latin
calor
heat).
200 g
Worked
The
example: Working
calorimeter
change
of
in
Figure
combustion
of
1
out
was
the
used
to
enthalpy
measure
water
change
the
enthalpy
methanol.
spirit
burner
1
ethanol
CH
OH(l)
+
O
1
3
(g)
➝
CO
2
2
(g)
+
2H
2
O(l)
2
▲ Figure 1
0.32 g
the
(0.01 mol)
200.0 g
of
of
methanol
water
rose
by
was
burnt
and
the
temperature
A simple calorimeter
of
4.0 K.
Maths link
Heat
change
=
q
=
m
×
c
×
=
200.0
ΔT
×
4.2
×
4.0
=
200.0 is 4 signicant gures (s.f.),
3360 J
0.32 is 2 signicant gures, 4.0 is
0.01 mol
gives
3360 J
2 signicant gures. So you can
So
1
mol
would
give
336 000 J
or
336 kJ
only give the answer to 2 s.f. You
–1
Δ
H
=
–340 kJ
mol
(negative
because
heat
is
given
out)
round it up rather than down as
c
336 is nearer to 340 than 330. See
Section 8, Mathematical skills, if
The
simple
calorimeter
can
be
used
to
compare
the
Δ
H
values
of
c
you are not sure about signicant
a
series
of
similar
compounds
because
the
errors
will
be
similar
for
gures.
every
down
experiment.
the
heat
combustion
by
However,
loss,
as
you
shown
burning
the
in
fuel
can
improve
Figure
in
2,
and
oxygen
the
results
reducing
rather
than
by
cutting
incomplete
air.
lid
The ame calorimeter
The
flame
version
of
changes
calorimeter,
the
of
designed
simple
shown
reduce
Figure
calorimeters
combustion.
to
in
heat
It
used
incorporates
loss
even
3
overleaf,
for
the
is
an
measuring
following
improved
mineral
features
that
are
further:
draught
•
the
spiral
•
the
ame
•
the
fuel
chimney
is
is
made
of
wool
enthalpy
screen
copper
enclosed
burns
in
pure
oxygen,
rather
than
air.
▲ Figure 2
An improved calorimeter
87
4.3
Measuring
to
ent h al py
lter
c h ange s – cal or im e tr y
Measuring
pump
It
is
enthalpy
relatively
place
in
easy
solution.
to
changes of
measure
The
heat
is
heat
reactions
changes
generated
in
for
the
in
solution
reactions
solutions
that
take
themselves
stirrer
and
only
beakers
has
are
to
be
often
kept
used
in
for
the
the
calorimeter.
Expanded
calorimeters.
These
polystyrene
are
good
insulators
copper
(this
reduces
heat
loss
through
their
sides)
and
they
have
a
low
heat
spiral
capacity
so
they
absorb
very
little
heat.
The
specific
heat
capacity
of
chimney
dilute
water
solutions
–1
4.2 J g
is
usually
taken
to
be
the
–1
–1
K
(or
more
precisely
same
as
that
of
water,
–1
4.18 J g
K
).
Neutralisation reactions
Neutralisation
heat.
When
reactions
an
acid
is
in
solution
neutralised
are
by
exothermic
an
alkali
the
–
they
give
equation
out
is:
oxygen
acid
+
alkali
➝
salt
+
water
ethanol
To
▲ Figure 3
find
an
enthalpy
change
for
a
reaction,
you
use
the
quantities
in
A ame calorimeter
moles
given
enthalpy
acid
by
by
the
change
sodium
equation
needs
balanced
of
reaction
hydroxide,
to
be
HCl(aq)
equation.
for
the
the
For
example,
neutralisation
heat
given
out
by
to
of
find
the
molar
hydrochloric
the
quantities
in
the
found:
+
NaOH(aq)
NaCl(aq)
➝
+
H
O(l)
2
hydrochloric acid
sodium hydroxide
sodium chloride
water
1 mol
1 mol
1 mol
1 mol
Hint
Worked
Remember to use the total volume
example:
3
50 cm
Enthalpy
change for
a
reaction
–3
of
3
1.0 mol dm
hydrochloric
acid
and
50 cm
of
3
of the mixture, 100 cm
.
A common
–3
1.0 mol dm
sodium
hydroxide
solution
were
mixed
in
an
3
mistake is to use 50 cm
expanded
polystyrene
beaker.
The
temperature
rose
by
6.6 K.
3
The
total
volume
approximately
of
the
100 g
mixture
because
is
the
100 cm
density
.
of
This
has
water
a
and
mass
of
of
dilute
–3
aqueous
solutions
enthalpy
is
approximately
mass
of
specic
=
change
number
of
heat
capacity
×
q
water
q
1 g cm
=
m
×
=
100
c
×
×
temperature
×
m
of
solution
c
change
ΔT
ΔT
4.2
×
6.6
=
2772 J
moles
−3
concentration
of
acid
(and
also
c
(mol
dm
3
)
×
volume
=
1000
of
alkali)
n
50
=
1.0
×
=
0.05 mol
1000
2772
so
1
mol
would
give
J
=
55 440 J
=
55.44 kJ
0.05
−1
ΔH
=
–55.44
kJ
mol
−1
ΔH
The
88
sign
of
ΔH
is
=
–55 kJ mol
negative
(to
because
2 s.f.)
heat
is
given
out.
V
(cm
)
Energetics
4
Displacement reactions
A
metal
one
that
from
reaction
For
a
is
reactive
compound.
can
be
example,
sulfate.
more
The
If
the
investigated
zinc
will
reaction
Zn(s)
another
compound
using
displace
is
+
than
a
will
will
copper
from
CuSO
(aq)
in
less
water,
reactive
this
beaker
as
before.
solution
of
copper
one
mole
(aq)
+
Cu(s)
4
1 mol
equation
a
ZnSO
➝
4
the
dissolve
polystyrene
the
exothermic.
1 mol
From
displace
1 mol
of
zinc
reacts
1 mol
with
one
mole
of
copper
sulfate.
Study tip
Worked
example:
Enthalpy
change
in
a
displacement
Rearrange q = m × c × ΔT to nd
reaction
any of the quantities in terms of
3
0.50 g
of
zinc
was
added
to
–3
25.0 cm
of
0.20 mol dm
copper
the others:
sulfate
solution.
The
temperature
rose
by
10 K.
q
ΔT =
q
=
m
×
=
25
c
×
ΔT
mc
q
×
4.2
×
10
=
1050 J
m =
zinc
=
65.4,
so
0.50 g
of
zinc
is
ΔT
c
0.50
A
moles
=
0.0076
moles
r
q
65.4
c =
c
number
of
moles
of
copper
sulfate
in
solution
×
V
ΔT
m
=
1000
–3
where
c
is
concentration
in
mol
3
dm
and
V
is
volume
=
0.005
in
cm
25.0
=
0.20
×
mol
1000
This
means
reactant
zinc
that
has
the
taken
zinc
part
was
in
in
the
excess;
0.005
reaction,
mol
leaving
of
each
some
unreacted
behind.
1050
Therefore,
1
mole
of
zinc
would
produce
J
=
210
000 J.
0.005
–1
So,
ΔH
The
for
sign
this
of
ΔH
Allowing for
Although
heat
will
by
of
stand
the
the
in
a
from
curve.
of
temperature,
as
Place
50 cm
cups
sides
by
As
is
2 s.f.).
given
out.
this
an
are
and
good
top
method.
example,
acid
insulators,
leading
This
the
and
to
can
low
be
some
values
allowed
measurement
sodium
of
for
for
the
hydroxide
is
curve.
all
for
the
apparatus
some
that
of
time.
the
and
This
both
solutions
ensures
laboratory
that
are
they
all
left
to
reach
itself.
follows:
3
1
(to
heat
hydrochloric
cooling
laboratory
proceed
the
measured
experiment,
the
because
polystyrene
lost
cooling
using
same
Then
a
–210 kJ mol
loss
neutralisation
repeated
Before
be
is
negative
heat
changes
plotting
heat
is
expanded
still
enthalpy
reaction
–3
of
1.0 mol dm
hydrochloric
acid
in
one
polystyrene
cup
▲ Figure 4
3
and
50 cm
Polystyrene beakers make
–3
of
1.0 mol dm
sodium
hydroxide
solution
in
another.
good calorimeters because they are
2
Using
a
thermometer
that
reads
to
0.1 °C,
take
the
temperature
good insulators and have low heat
of
each
solution
every
30
seconds
for
four
minutes
to
confirm
capacities
89
4.3
Measuring
ent h al py
time
of
c h ange s – cal or im e tr y
that
mixing
both
solutions
remain
at
the
same
temperature,
that
of
the
28
laboratory.
27
likely
C°
26
temp.
of
A
there
indicating
line
will
of
be
‘best
very
random
fit’
is
small
drawn
through
variations
these
around
the
points.
line
of
It
is
best
fit,
errors.
/
erutarepmet
25
temp.
products
rise
–
3
on
slow
mixing
Now
pour
one
solution
into
the
other
and
stir,
continuing
to
cooling
24
record
the
temperature
every
30
seconds
for
a
further
six
minutes.
6.9°C
23
The
results
also
be
are
shown
on
the
graph
in
Figure
5.
The
experiment
can
22
temp.
of
done
using
an
electronic
temperature
sensor
and
data
logging
reactants
21
software
remains
to
plot
the
graph
directly.
steady
20
0
1
2
3
4
time
▲ Figure 5
5
/
6
7
8
9
10
On
mixing,
the
temperature
rises
rapidly
as
the
reaction
gives
min
Graph to show temperature
as a neutralisation reaction proceeds
out
heat,
the
polystyrene
and
then
drops
cup.
To
slowly
find
immediately
after
mixing,
graph
after
mixing
This
The
points
gives
temperature
calculation
q
The
a
=
m
number
×
is
c
of
×
as
the
you
and
rise
and
regularly
best
draw
estimate
the
best
extrapolate
of
as
of
heat
the
straight
back
to
is
lost
from
temperature
line
the
through
time
of
the
mixing.
6.9 °C.
before.
ΔT
=
moles
100
of
×
acid
4.2
×
(and
6.9
=
alkali)
2898 J
was
0.05 mol
(as
before).
2898
So
1
mol
would
give
J
=
57
960 J
=
57.96 kJ
0.05
–1
Δ
H
=
−58 kJ mol
(to
2 s.f.)
neut
The
sign
of
ΔH
is
negative
because
heat
is
given
out.
Summary questions
1
0.74 g (0.010 mol) of propanoic acid was burnt in the simple
calorimeter like that described above for the combustion of methanol.
The temperature rose by 8.0 K. Calculate the value this gives for the
enthalpy change of combustion of propanoic acid.
3
2
50.0 cm
–3
of 2.00 mol dm
3
sodium hydroxide and 50.0 cm
of
–3
2.00 mol dm
hydrochloric acid were mixed in an expanded
polystyrene beaker. The temperature rose by 11.0 K.
a
Calculate ΔH for the reaction.
b
Describe how this value will compare with the accepted value for
this reaction.
c
3
90
Explain your answer to b
Consider the expression
q = mcΔT
a
State what the term q represents.
b
State what the term m represents.
c
State what the term c represents.
d
State what the term ΔT represents.
4.4
The
To
Hess’s
enthalpy
nd
changes
these
you
use
for
an
some
law
reactions
indirect
cannot
approach.
be
measured
Chemists
use
directly.
Learning
enthalpy
➔
changes
that
they
can
measure
to
work
out
enthalpy
objective s:
changes
Describe how to nd enthalpy
that
changes that cannot be
they
cannot.
In
particular,
it
is
often
easy
to
measure
enthalpies
measured directly.
of
combustion.
To
do
this,
chemists
use
Hess’s
law,
rst
stated
by
Specication reference: 3.1.4
Germain
Hess,
Hess’s
law
Hess’s
is
the
This
law
a
consequence
the
would
see
what
C
2
How
can
Route
moles
chemist,
enthalpy
is
more
which
the
states
is
law
to
look
at
ethane,
enthalpy
reaction
to
H
2
The
Ethyne,
takes
give
C
can
same.
to
the
Law
never
points
If
not,
reaction
products.
of
be
of
created
a
process
energy
(g)
of
H
,
following
H
,
by
two
example
different
where
routes.
6
reaction?
place
directly
–
ethyne
reacts
with
two
ethane.
+
2H
2
(g)
➝
C
2
H
2
(g)
ΔH
6
=
?
1
ethane
reaction
2
C
law,
finishing
the
the
C
ethyne
a
be
2
the
hydrogen
2:
chemical
destroyed.
means,
converted
find
The
for
a
1802.
reactants
energy
and
must
in
scientific
that
starting
born
law
C
Route
from
general
change
or
change
taken
2
we
1:
of
,
a
created
Hess’s
H
of
energy
been
Russian
route
provided
the
Hess’s
ethyne,
the
Energy,
So,
same,
have
Using
To
of
destroyed.
are
that
whatever
Conservation
or
Swiss-born
states
same,
is
a
takes
reacts
place
with
one
in
two
mole
stages.
of
hydrogen
to
give
ethene,
2
H
2
4
–1
H
C
2
(g)
+
H
2
(g)
➝
C
2
ethyne
b
H
2
(g)
ΔH
4
=
–176 kJ mol
2
ethene
Ethene,
C
H
2
ethane,
C
,
then
reacts
with
a
second
mole
of
hydrogen
to
give
4
H
2
6
–1
H
C
2
(g)
+
H
4
(g)
➝
C
2
ethene
Hess’s
route
law
you
Youcan
The
H
2
(g)
ΔH
6
=
–137 kJ mol
3
ethane
tells
take
show
us
–
that
the
direct
this
on
thermochemical
a
or
total
via
energy
ethene
diagram
cycle
for
change
(or,
called
a
in
is
fact,
the
by
same
any
whichever
other
thermochemical
converting
ethyne
to
ethane
route).
cycle.
is
shown
overleaf.
91
4.4
Hess’s
law
H
(?)
—
H
H
—
∇
1
H — C = C — H (g)
+
2H
(g)
—
H
2
C
—
C
—
H (g)
—
—
1.
H
H
2.
∇
∇
3.
H
2
H
3
–1
( –176 kJ mol
)
–1
( –137 kJ mol
)
—
—
H
H
—
C
(g)
—
=
C
H
H
+
H
(g)
2
Hess’s
law
means
that:
ΔH
=
ΔH
1
+
ΔH
2
3
–1
The
actual
gures
are:
ΔH
=
–176 kJ mol
=
–137 kJ mol
2
–1
ΔH
3
–1
So
ΔH
=
(–176)
+
(–137)
=
–313
kJ
mol
1
This
method
of
calculating
ΔH
is
ne
if
you
know
the
enthalpy
1
changes
for
that
be
the
can
the
other
looked
enthalpy
two
up
change
for
of
reactions.
a
large
There
range
Δ
formation,
of
H
,
are
certain
enthalpy
compounds.
and
These
enthalpy
changes
include
change
of
f
Δ
combustion,
H
.
In
practice,
many
values
Δ
of
c
from
Δ
H
via
H
are
calculated
f
Hess’s
law
cycles.
c
Using the
enthalpy
changes of formation
Δ
H
f
The
enthalpy
of
Δ
formation,
H
,
is
the
enthalpy
change
when
f
one
mole
under
of
compound
standard
is
formed
conditions,
their
Another
theoretical
elements
•
carbon
Ethyne
is
of
the
is
•
has
Hess’s
Then
the
enthalpy
Hess’s
law
to
of
standard
the
its
constituent
and
elements
products
being
in
states.
ethyne
to
ethane
could
be
via
the
hydrogen.
to
formation
of
from
reactants
convert
converted
enthalpy
reaction
of
and
rst
reverse
the
way
all
its
and
formation.
negative
elements,
of
the
This
its
ΔH
carbon
enthalpy
is
a
change
general
value.
It
and
is
rule.
in
hydrogen.
is
the
The
fact
a
This
negative
reverse
of
a
consequence
law.
carbon
of
tells
and
hydrogen
formation
us
that
for
ΔH
=
react
to
form
ethane.
This
is
the
ethane.
ΔH
1
+
ΔH
4
5
H
(?)
H
H
—
—
∇
1
H — C = C — H (g)
+
2H
(g)
H — C — C — H (g)
2
H
Graphite is the most stable form of
4.
∇
5.
H
∇
4
H
5
carbon (another form is diamond).
It has a special state symbol
(s, graphite).
92
2C (s,
graphite)
+
(g)
3H
2
—
—
1.
Hint
H
Energetics
4
ΔH
is
the
enthalpy
of
Δ
formation,
H
,
of
ethane
whilst
reaction
4
is
f
5
the
reverse
of
the
formation
of
ethyne.
–1
The
values
you
need
Δ
are:
H
(C
f
H
2
)
=
+228 kJ mol
2
–1
and
Δ
H
(C
f
H
2
)
=
–85 kJ mol
=
–228 kJ mol
6
–1
So
ΔH
4
(remember
to
change
the
sign)
–1
ΔH
=
–85
kJ
=
–228
mol
5
–1
Therefore
ΔH
+
–85
=
–313 kJ mol
1
This
was
the
expect
from
Notice
that
only
of
is
one
result
Hess’s
in
of
reaction
the
hydrogen
you
got
from
the
previous
method,
as
you
should
law.
three
remain
4
there
moles
in
are
of
their
two
moles
hydrogen
standard
is
of
hydrogen
involved.
states
and
so
‘spare’
These
no
two
enthalpy
as
moles
change
invoved.
C
H
2
but
(g)
➝
2C(s,
graphite)
+
H
2
(g)
is
the
reaction
you
are
considering,
2
you
have:
C
H
2
However,
involved
this
in
(g)
2H
(g)
➝
2C(s,
graphite)
+
2
makes
the
+
2
no
reaction
3H
(g)
2
difference.
and
it
The
does
not
‘extra’
affect
hydrogen
is
not
ΔH
Summary questions
1
Use the values of Δ
H
in the table to calculate
ΔH
for each of the
f
reactions below using a thermochemical cycle.
a
CH
COCH
3
b
C
H
2
c
C
(l) + H
3
(g) + Cl
4
H
2
(g) ➝ CH
2
(g) ➝ C
2
H
2
(g) + HCl(g) ➝ C
4
CH(OH)CH
3
Cl
4
H
2
(l)
3
(l)
2
Cl(l)
5
d
Zn(s) + CuO(s) ➝ ZnO(s) + Cu(s)
e
Pb(NO
1
)
3
(s) ➝ PbO(s) + 2NO
2
(g) +
O
2
2
(g)
2
–1
Compound
H
Δ
/ kJ mol
f
CH
COCH
3
CH
(l)
CH(OH)CH
3
C
–318
(g)
+52
4
H
2
C
(l)
3
H
2
C
–248
3
Cl
4
H
2
(l)
–165
2
Cl(l)
–137
5
HCl(g)
–92
CuO(s)
–157
ZnO(s)
–348
Pb(NO
)
3
PbO(s)
NO
(g)
(s)
–452
2
–2
1
7
+33
2
93
4.5
Enthalpy
Learning
changes of
objective s:
The
➔
combustion
Describe how the enthalpy
enthalpy
change
when
change of combustion can
change
one
of
mole
oxygen
Δ
combustion,
of
substance
under
standard
is
c
H
,
is
the
enthalpy
completely
burnt
in
conditions.
be used to nd the enthalpy
change of a reaction.
Thermochemical
Specication reference: 3.1.4
of
cycles
using
enthalpy
changes
combustion
Look
again
reaction
at
the
between
thermochemical
ethyne
and
cycle
used
hydrogen
to
to
find
form
ΔH
for
the
ethane.
Study tip
C
H
2
(g)
+
2H
2
(g)
➝
C
2
H
2
(g)
6
Remember to multiply by the
This
time
use
enthalpy
changes
of
combustion.
In
this
case
you
can
go
number of moles of reagents
via
the
combustion
products
of
the
three
substances
–
carbon
dioxide
involved in each step.
and
All
water.
three
This
substances
means
measured.
their
The
–
ethyne,
enthalpy
hydrogen,
changes
thermochemical
of
cycle
and
ethane
combustion
–
can
burn
be
readily.
easily
is:
H
H
H
1
—
—
∇
1
3
1
O
(g) +
2
H — C = C — H (g) + 2H
(g)
H — C — C — H (g)
2
1.
+ 3
O
H
(g)
2
2
—
—
2
H
∇
2
(H
H
×
∇
H
(C
H
2
c
)
2
c
)
∇
2
7.
6.
8.
H
(C
c
2CO
(g)
+
3H
2
Putting
in
the
H
2
)
6
O (l)
2
values:
H
H
H
1
—
—
∇
1
3
1
O
(g) +
2
H — C = C — H (g) + 2H
(g)
H — C — C — H (g)
2
1.
+ 3
O
2
—
—
2
H
(g)
2
H
–1
Hint
×
–286 kJ mol
=
–572 kJ mol
–1
–1
–1301 kJ mol
•
2
7.
6.
8.
Both reactions 6 and 7 have
–1
–1560 kJ mol
to occur to get from the
2CO
(g)
+
3H
2
–1
O (l)
2
–1
–1873 kJ mol
+1560 kJ mol
star ting materials to the
combustion products.
Do not forget the hydrogen.
To
•
In this case there are two
in
moles of hydrogen so you need
twice the value of Δ
H
get
the
the
you
enthalpy
direction
must
of
change
change
the
its
red
for
reaction
arrows.
This
1
you
means
must
go
round
reversing
the
reaction
cycle
8
so
sign.
which
C
–1
So
ΔH
=
–1873
+
1560 kJ mol
=
–313 kJ mol
1
refers to one mole of hydrogen.
–1
–1
Δ
H
C
(C
H
2
ΔH
) = –1301 kJ mol
once
again,
the
–1
Δ
H
(H
C
same
answer
as
before
1
2
1
) = –286 kJ mol
Notice
2
that
in
reaction
1
there
are
3
moles
of
oxygen
on
either
side
2
–1
Δ
H
C
(C
H
2
) = –1560 kJ mol
6
of
the
value
94
equation.
of
ΔH
They
take
no
part
in
reaction
1
and
do
not
affect
the
Energetics
4
Finding
Δ
H
from
Δ
f
Enthalpy
changes
impossible
to
not
react
For
example,
ethanol
of
formation
measure
directly
to
the
from
H
c
its
of
directly.
form
the
following
compounds
This
is
compound
equation
are
because
that
often
the
you
represents
difficult
reactants
are
the
or
often
interested
formation
do
in.
of
elements.
1
2C(s,
graphite)
+
3H
(g)
+
O
2
This
does
readily
be
not
burn
take
in
measured.
place.
oxygen
The
However,
so
their
all
the
enthalpy
thermochemical
(g)
➝
C
2
2
cycle
H
2
species
concerned
changes
you
OH(l)
5
of
need
will
combustion
can
is:
∇
H
(g) + 2C (s, graphite) + 3H
2
(ethanol)
f
1
3O
(g) +
O
2
(g)
C
2
H
2
OH (l) + 3O
5
(g)
2
2
9.
∇
3
11.
(g))
(H
H
×
2
c
∇
2
×
(C
H
(s,
graphite))
10.
c
∇
2CO
(g)
+
3H
2
Putting
in
the
(C
H
12.
c
H
2
OH(l))
5
O (l)
2
values:
Hint
∇
H
(g) + 2C (s, graphite) + 3H
2
(g) +
(ethanol)
f
1
3O
O
2
(g)
C
2
H
2
2
OH (l) + 3O
5
(g)
2
9.
The values we need are:
11.
12.
–1
3
10.
×
–1
–285.8 kJ mol
Δ
–1367.3 kJ mol
H
(C(s, graphite)) =
C
–1
=
–1
–857.4 kJ mol
–393.5 kJ mol
–1
2
×
–393.5 kJ mol
–1
=
–787 kJ mol
–1
Δ
H
(H
C
2CO
(g)
+
3H
2
–1
O (l)
2
–1
–1
–1644.4 kJ mol
+1367.3 kJ mol
Δ
H
(C
C
Note
of
that
the
do
not
Note
in
reaction
equation
affect
also
that
the
that
Δ
9
value
H
there
take
of
(C(s,
no
are
part
three
in
moles
the
of
oxygen
reaction.
This
H
(H
c
To
in
(g))
is
graphite))
is
the
same
as
Δ
get
the
same
as
you
H
(H
f
the
enthalpy
direction
must
H
of
change
change
the
its
ΔH
=
–1644.4
+
OH(l)) = –1
36
7
.3 kJ mol
5
side
that
they
red
(g))
and
2
O(l)).
2
for
reaction
arrows.
This
9,
you
means
must
go
round
reversing
the
reaction
cycle
12
so
sign.
–1
So,
either
(CO
f
Δ
2
the
on
means
H
2
ΔH
c
Δ
(g)) = –285.8 kJ mol
2
1367.3 kJ mol
–1
=
–277.1 kJ mol
9
–1
So,
Δ
H
(C
f
H
2
OH(l))
=
–277.1 kJ mol
5
Summary questions
Calculate ΔH
for the reaction by thermochemical
a
via Δ
H
values
b
via Δ
f
H
values
c
cycles:
–1
Compound
ΔH
/ kJ mol
f
H
H
=
H — C — C
(l)
+
H
(g)
H — C — C — O — H (l)
—
—
—
2
–1
ΔH
/ kJ mol
c
H
—
—
—
O
—
1
CH
CHO
–192
–1166
–
–286
–277
–1367
3
H
2
H
H
H
H
CH
CH
3
OH
2
95
4.6
Representing
Learning
objective s:
You
to
➔
can
ther m o c h emi ca l
use
enthalpy
represent
the
diagrams
enthalpy
rather
changes
than
cycl es
thermochemical
in
chemical
of
the
reactions.
cycles
These
Describe what an enthalpy
show
the
energy
(enthalpy)
levels
reactants
and
products
of
diagram is.
a
➔
chemical
reaction
on
a
vertical
scale,
so
you
can
compare
their
State what is used as the zero
energies.
If
a
substance
is
of
lower
energy
than
another,
you
say
it
is
for enthalpy changes.
energetically
more
stable.
Specication reference: 3.1.4
The
So
enthalpy of
far
you
drawing
give
have
The
diagrams
numbers
enthalpies
states
in
(298 K
This
considered
enthalpy
absolute
which
and
is
of
H
to
all
they
100 kPa
convention
example,
elements
you
the
are
not
changes,
need
a
zero
enthalpies
elements
exist
means
and
enthalpy
at
in
of
and
approximately
that
H,
the
standard
because
absolute
work
different
their
298 K
not
to
standard
normal
hydrogen
of
states
are
Pure
carbon
as
H
can
exist
These
called
taken
symbol
as
the
(s,
in
a
number
diamond,
allotropes.
standard
graphite),
C(s,
and
at
of
forms
of
is
carbon.
graphite)
cycles
at
room
the
zero.
for
room
temperature
buckminsterfullerene
Graphite
state
so
Thermochemical
H
as
conditions.)
hydrogen,
exists
(i.e.,
pressure.
graphite,
is
then
2
and
including
are
When
can
taken
room
2
temperature
You
substances.
100 kPa)
state
values.
from.
and
the
It
is
most
stable
given
represents
the
(buckyballs).
of
these
special
and
state
graphite.
enthalpy diagrams
H
—
—
Here
H — C — O — C — H
are
two
presented
examples
both
as
of
reactions,
thermochemical
with
cycles
their
and
enthalpy
as
changes
enthalpy
diagrams.
—
—
H
Example
H
What
is
1
ΔH
for
the
change
from
methoxymethane
to
ethanol?
(The
methoxymethane
compounds
different
H
— C —
The
C — O — H
—
—
H
a
pair
structures,
standard
compounds
see
isomers
Figure
–
they
have
the
same
formula
but
1.)
molar
enthalpy
changes
of
formation
of
the
two
are:
–1
H
CH
Δ
OCH
3
H
=
–184 kJ mol
H
=
–277 kJ mol
f
3
–1
ethanol
C
H
2
▲ Figure 1
of
H
—
—
H
are
Isomers of C
H
2
Δ
OH
f
5
O
6
Using a thermochemical cycle
The
following
steps
1
Write
an
2
Write
down
3
are
equation
the
for
of
each.
Put
Δ
H
the
the
elements
quantities
in
shown
values
in
red
on
the
thermochemical
cycle.
reaction.
in
with
the
two
arrows
compounds
showing
the
with
the
direction
correct
–
from
f
elements
4
Put
in
the
elements
96
to
compounds.
arrows
(the
red
to
go
from
arrows).
starting
materials
to
products
via
the
Energetics
4
5
Reverse
the
sign
Δ
of
H
if
the
red
arrow
is
in
the
opposite
f
direction
6
Go
round
the
Hess’s
to
ΔH
law
the
the
cycle
values
states
black
as
that
arrow.
in
the
you
this
direction
of
the
red
arrows
and
add
up
go.
is
the
same
as
ΔH
for
the
direct
reaction.
∇
H
CH
OCH
3
(g)
C
3
H
2
OH (l)
5
1.
methoxymethane
ethanol
3.
3.
∇
∇
H
H
f
f
–1
–1
–184 kJ mol
–277 kJ mol
5.
1
–1
2C (s,
graphite)
+
3H
(g)
+
O
2
+184 kJ mol
–1
(g)
– 277 kJ mol
2
2
elements
2.
∇
H
6.
=
+ 184
=
– 93 kJ mol
–
277
–1
▲ Figure 2
Thermochemical cycle for the formation of ethanol from methoxymethane
Using an enthalpy diagram
The
following
1
Draw
2
Look
a
steps
line
up
at
the
are
shown
level
values
0
to
of
Δ
in
red
on
represent
H
for
the
the
each
enthalpy
diagram.
elements.
compound
and
enter
these
f
on
the
values
3
4
Find
enthalpy
are
the
diagrams,
below
0,
difference
in
represents
the
ΔH
difference
Up
to
is
is
the
positive
ethanol
is
down
methoxymethane
in
is
the
the
their
negative.
of
is
of
the
signs
–
negative
above.
the
two
compounds.
This
enthalpies.
taking
sign
sign
are
between
levels
down
so
in
account
values
levels
difference
and
taking
positive
account
From
negative.
ΔH
would
of
the
direction
of
change.
methoxymethane
From
be
ethanol
to
positive.
1
1.
2C (s,
graphite)
+
3H
(g)
+
O
2
(g)
2
2
0
–1
–184 kJ mol
–1
enthalpy
–277 kJ mol
2.
CH
OCH
3
(g)
3
–184
–1
3.
2.
C
H
2
–93 kJ mol
OH (l)
5
–277
Study tip
▲ Figure 3
The enthalpy diagram for the formation of ethanol from methoxymethane
Remember that you do not need to
Notice
than
how
the
the
enthalpy
thermochemical
methoxymethane.
compound.
method
The
you
This
values
level
cycle
means
of
diagram
ΔH
that
that
for
makes
ethanol
it
the
is
the
it
has
much
less
more
reaction
clearer
energy
energetically
are
the
same
draw these diagrams accurately
than
stable
whichever
to scale, but a rough scale is
impor tant to ensure that the
relative levels are correct.
use.
97
4.6
Representing
t h e r m oc he mic al
cycl e s
Example
2
Hint
To
find
ΔH
for
the
reaction
NH
(g)
+
HCl(g)
➝
NH
3
Cl(s)
4
You can use a short cut to save
The
standard
molar
enthalpy
changes
of
formation
of
the
compounds
are:
drawing an enthalpy diagram or a
–1
Δ
NH
thermochemical cycle. The enthalpy
H
=
–46 kJ mol
H
=
–92 kJ mol
H
=
–314 kJ mol
f
3
change of a reaction is the sum of
–1
Δ
HCl
f
the enthalpies of formation of all
–1
NH
the products minus the sum of the
Δ
Cl
f
4
enthalpies of formation of all the
Using a thermochemical cycle
reactants. In this example
The
ΔH
thermochemical
cycle
for
the
formation
of
ammonium
chloride
is
= −31
4 −(−46 + (−92))
shown
in
Figure
4.
= −314 – (−138)
−1
= −1
76 kJ mol
1
Write
an
2
Write
down
equation
the
for
the
elements
reaction.
that
make
up
the
two
compounds
with
If you use this short cut, you must
be very careful of the signs.
3
the
correct
Put
in
the
quantities
Δ
H
of
values
each.
with
arrows
showing
the
direction,
that
is,
f
from
4
5
elements
Put
in
the
the
elements
Reverse
to
compounds.
arrows
the
going
(the
sign
red
of
from
the
starting
materials
to
products
via
arrows).
Δ
H
if
the
red
arrow
is
in
the
opposite
f
direction
6
Go
to
round
the
values
the
the
of
black
cycle
ΔH
arrow(s).
in
as
the
you
direction
of
the
red
arrows
and
add
up
go.
1.
∇
H
NH
(g)
+
HCl (g)
NH
3
Cl (s)
4
3.
3.
3.
∇
∇
∇
H
H
f
f
H
f
–1
–1
–92 kJ mol
–1
–314 kJ mol
–46 kJ mol
–1
–1
+138 kJ mol
– 314 kJ mol
1
1
(g)
N
+
2H
2
(g)
+
Cl
2
(g)
2
2
2
2.
∇
–1
H
=
+ 46
+
=
– 176 kJ mol
92
+
– 314 kJ mol
–1
▲ Figure 4
+
Thermochemical cycle for the formation of ammonium chloride
Using an enthalpy diagram
The following steps are shown in red on the enthalpy diagram.
1
Draw a line at level 0 to represent the elements.
2
Draw in NH
–1
Cl at the enthalpy 314 kJ mol
below this.
4
–1
3
Draw a line representing ammonia 46 kJ mol
1
H
elements. (There is still
2
below the level of the
1
2
and
Cl
2
left unused.)
2
–1
4
Draw a line 92 kJ mol
below ammonia. This represents hydrogen
chloride.
5
Find the dierence in levels between the (NH
+ HCl) line and the NH
3
one. This represents ΔH
for the reaction. As the change from (NH
3
to NH
Cl is down, ΔH
4
98
must be negative.
Cl
4
+ HCl)
Energetics
4
1
1
N
1.
(g)
+
2H
2
(g)
+
Cl
2
(g)
2
2
2
0
1
–1
–46
kJ
3.
NH
(g)
+
1
H
3
mol
(g)
+
Cl
2
2
(g)
2
2
–46
–1
–92 kJ mol
4.
NH
(g)
+
HCl (g)
3
–138
–1
–314 kJ mol
–1
5.
–176 kJ mol
enthalpy
2.
NH
Cl (s)
4
–314
Notice how the enthalpy level diagram makes it much clearer than the
thermochemical cycle that ammonium chloride is more energetically
stable than the gaseous mixture of ammonia and hydrogen chloride. This
is part of the reason why ammonia and hydrogen chloride react readily to
form ammonium chloride. The values of ΔH
for the reaction are the same
whichever method you use.
What would be the enthalpy change when solid ammonium chloride
decomposes into the gases ammonia and hydrogen chloride?
lom Jk 67
1+
1−
Summary questions
1
Use the values of Δ
H
in the table to calculate ΔH
for each of the
f
reactions below using enthalpy diagrams.
a
CH
COCH
3
b
C
H
2
c
C
(l) + H
3
(g) + Cl
4
H
2
(g) ➝ CH
2
(g) ➝ C
2
H
2
(g) + HCl(g) ➝ C
4
CH(OH)CH
3
Cl
4
H
2
(l)
3
(l)
2
Cl(l)
5
d
Zn(s) + CuO(s) ➝ ZnO(s) + Cu(s)
e
Pb(NO
1
)
3
(s) ➝ PbO(s) + 2NO
2
(g) +
O
2
2
(g)
2
–1
Compound
H
Δ
/ kJ mol
f
CH
COCH
3
CH
(l)
CH(OH)CH
3
C
–318
(g)
+52
4
H
2
C
(l)
3
H
2
C
–248
3
Cl
4
H
2
(l)
–165
2
Cl(l)
–137
5
HCl(g)
–92
CuO(s)
–157
ZnO(s)
–348
Pb(NO
)
3
PbO(s)
NO
(g)
(s)
–452
2
–2
1
7
+33
2
99
4.7
Bond
Learning
enthalpies
objective s:
Δ
H
is
the
enthalpy
change
of
combustion.
If
you
Δ
plot
H
c
➔
State what the denition of a
the
bond enthalpy is.
alkanes,
number
one
➔
against
c
of
you
carbon
get
Δ
carbon),
a
atoms
straight
H
is
the
in
line
the
molecule,
graph,
enthalpy
see
for
Figure
change
straight
1.
For
chain
methane
(with
for:
c
Describe how mean bond
enthalpies are worked out
CH
(g)
+
2O
4
(g)
➝
CO
2
(g)
+
2H
2
O(l)
2
from given data.
The
straight
line
means
Δ
that
H
changes
by
the
same
amount
for
c
➔
Demonstrate how bond
each
extra
carbon
atom
in
the
chain.
enthalpies are used
Each
in calculations.
alkane
differs
from
the
previous
one
by
one
CH
group,
that
is,
2
there
is
one
extra
C
C
bond
in
the
molecule
and
two
extra
C
H
Specication reference: 3.1.4
bonds.
to
a
This
suggests
particular
Bond
that
bond.
you
This
is
can
assign
called
the
a
denite
bond
amount
of
energy
enthalpy.
enthalpies
6000
You
have
to
put
endothermic
in
energy
change.
to
Bond
break
a
covalent
dissociation
bond
–
enthalpy
this
is
is
an
dened
as
the
5000
1–
lom Jk
enthalpy
required
to
break
a
covalent
bond
with
all
species
in
4000
the
/
gaseous
bond
3000
H
is
bond,
c
∇
change
state.
formed
for
–
The
this
example
same
is
C
an
H,
amount
of
exothermic
may
have
energy
is
change.
slightly
given
out
However,
different
when
the
bond
the
same
enthalpies
2000
in
1000
different
value
is
molecules,
called
energy).
The
the
fact
but
mean
that
you
usually
bond
you
get
use
enthalpy
out
the
the
average
(often
same
value.
called
amount
of
the
This
bond
energy
when
0
1
2
3
4
5
6
7
8
9
10
you
number
in
of
the
carbon
make
molecule
As
mean
specic
▲ Figure 1
H
Δ
a
bond,
as
you
put
in
to
break
it,
is
an
example
of
Hess’s
law.
atoms
bond
enthalpies
compounds
will
are
only
averages,
give
calculations
approximate
using
answers.
them
for
However,
plotted against the
c
they
are
useful,
and
quick
and
easy
to
use.
Mean
bond
enthalpies
have
number of carbon atoms in the alkane
been
calculated
books
and
H
from
Hess’s
law
cycles.
They
can
be
looked
up
in
data
databases.
The
H
in
hydrogen
bond
energy
is
the
energy
required
to
separate
the
two
atoms
Hint
•
a
molecule
in
the
gas
phase
into
separate
gaseous
atoms.
If the bonds in the methane
–1
H
(g)
➝
2H(g)
ΔH
=
+436 kJ mol
2
are broken one at a time, the
The
energy required is not the
for
same for each bond.
C
the
H
mean
bond
following
energy
process,
in
in
methane
which
four
is
one
bonds
quarter
are
of
energy
broken.
−1
•
the
–1
The value of +416 kJ mol
CH
(g)
➝
C(g)
+
ΔH
4H(g)
=
+1664 kJ mol
4
is the C
H bond energy in
1664
So
the
mean
(or
average)
C
H
bond
energy
in
methane
=
methane. The value in other
4
–1
=
compounds will vary slightly.
+416 kJ mol
The average over many
−1
•
compounds is +413 kJ mol
Using
All mean bond energies are
changes of
positive because we have
to put energy in to break
You
can
mean
use
reactions,
bond
calculate
enthalpy
reaction
mean
for
enthalpies to
bond
enthalpies
to
work
H
Cl(g)
out
the
enthalpy
example:
bonds – they are endothermic
C
H
2
(g)
6
+
Cl
(g)
2
➝
C
2
+
HCl(g)
5
processes.
ethane
10 0
chlorine
chloroethane
hydrogen chloride
change
of
Energetics
4
The
mean
Table
bond
enthalpies
you
will
need
for
this
example
are
given
in
▼ Table 1
Mean bond enthalpies
1.
Bond
Bond enthalpy /
–1
The
1
steps
First
are
as
draw
drawn
kJ mol
follows:
out
the
showing
molecules
all
the
and
show
are
called
bonds
all
the
bonds.
displayed
C
H
413
C
C
347
(Formulae
formulae.)
Cl
H
H
H
H
H
imagine
that
all
the
bonds
in
the
Cl
H
reactants
them
to
atoms.
Look
all
up.
This
break
the
bonds
You
need
to
up
will
break
the
give
and
bond
you
form
these
enthalpy
break
the
total
separate
for
×
C
H
6
×
×
C
C
1
×
Cl
432
Br
193
Br
H
366
leaving
bond
must
and
be
Br
285
add
put
in
–1
413 kJ mol
=
2478 kJ mol
1
×
347 kJ mol
=
347 kJ mol
1
×
243 kJ mol
=
243 kJ mol
=
3068 kJ mol
–1
–1
Cl
that
H
bonds:
–1
1
346
atoms.
–1
6
each
energy
Cl
Br
C
separate
243
H — Cl (g)
+
—
—
Now
C
H — C — C — Cl (g)
—
—
H
2
Cl — Cl (g)
+
—
—
—
—
H — C — C — H (g)
Cl
H
–1
–1
–1
So
3068 kJ mol
separate
3
Next
imagine
Add
up
give
you
You
must
hydrogen,
the
to
put
separate
bond
the
need
the
be
make
enthalpy
these
to
and
of
convert
carbon
atoms
enthalpies
total
in
chlorine,
join
the
given
together
bonds
out
ethane
by
that
the
to
×
C
H
5
×
must
×
C
C
1
=
2065
kJ
mol
×
C
1
×
Cl
the
form.
products.
This
will
forming.
=
347
kJ
mol
=
346
kJ
mol
=
432
kJ
mol
–1
347 kJ mol
–1
1
to
–1
413 kJ mol
×
chlorine
bonds:
–1
1
give
bonds
–1
5
and
atoms.
Cl
1
×
346 kJ mol
H
1
×
432 kJ mol
–1
–1
–1
–1
=
3190 kJ mol
–1
So
3190 kJ mol
chlorine,
The
given
is
given
carbon
difference
energy
the
and
between
out
to
out
atoms
the
form
when
to
you
convert
chloroethane
energy
bonds
is
put
the
in
the
and
to
separate
hydrogen
break
approximate
the
hydrogen,
chloride.
bonds
enthalpy
and
the
change
of
reaction.
–1
The
4
difference
Finally
was
work
put
(the
than
is
in
3190
out
than
reaction
was
put
is
–
the
was
3068
=
sign
of
given
122 kJ mol
the
out,
endothermic).
in
the
enthalpy
enthalpy
the
If
change.
enthalpy
more
change
energy
is
If
change
was
negative
more
is
given
(the
energy
positive
out
reaction
is
exothermic).
In
this
case,
more
enthalpy
is
given
out
than
put
in,
so
the
reaction
–1
is
Note
like
exothermic
that
this.
answer
in
and
practice
However,
whatever
ΔH
it
=
–122 kJ mol
would
Hess’s
route
law
you
be
impossible
tells
take,
us
that
real
or
for
you
the
will
reaction
get
the
to
happen
same
theoretical.
101
4.7
Bond
enthalpies
A
short
cut
Synoptic link
You
can
often
shorten
mean
bond
enthalpy
calculations:
Bond enthalpies give a measure
H
H
H
H — C — C — Cl (g)
H
H — Cl (g)
+
—
H
—
—
—
molecule is most likely to break.
Cl — Cl (g)
+
—
H — C — C — H (g)
help to predict which bond in a
H
—
—
—
of the strength of bonds, and can
H
H
However, this is not the only
Only
the
bonds
drawn
in
red
make
or
break
during
the
reaction
so
factor, the polarity of the bond is
–1
you
only
need
to
break:
1
×
C
H
=
413 kJ mol
Cl
=
243 kJ mol
in
=
656 kJ mol
also impor tant – see Topic 3.6,
–1
1
Electronegativity – bond polarity
×
Cl
–1
in covalent bonds, and Topic 13.2,
Total
energy
put
Nucleophilic substitution in
–1
You
only
need
to
make:
1
×
C
Cl
=
346 kJ mol
1
×
H
Cl
=
432 kJ mol
out
=
778 kJ mol
656
=
122 kJ mol
halogenoalkanes.
–1
–1
Total
energy
given
–1
The
Summary questions
difference
More
energy
is
is
778
given
out
–
than
taken
in
so
–1
ΔH
=
–122 kJ mol
(as
before)
These questions are about the
reaction:
Comparing the
CH
CH
3
1
+ Br
3
➝ CH
2
CH
3
result
with that from
a
Br + HBr
2
thermochemical
cycle
Draw out the displayed
This
is
only
an
approximate
value.
This
is
because
the
bond
enthalpies
structural formulae of all the
are
averages
whereas
in
a
compound
any
bond
has
a
specic
value
products and reactants so that
for
its
enthalpy.
You
can
nd
an
accurate
value
for
ΔH
by
using
a
all the bonds are shown.
thermochemical
2
a
cycle
as
shown
here:
Identify the bonds that
have to be broken to
∇
conver t the reactants into
H
C
H
2
6
(g)
+
Cl
2
(g)
C
H
2
Cl (g)
5
+
HCl (g)
separate atoms.
b
How much energy does
this take?
∇
∇
H
3
a
∇
H
f
Identify the bonds that
H
f
–1
f
–1
–85 kJ mol
–1
–137 kJ mol
–92 kJ mol
have to be made to conver t
separate atoms into the
–1
–1
+ 85 kJ mol
2 C (s,
graphite)
+
Cl
2
(g)
+
3H
2
(g)
products.
b
How much energy does
this take?
Remember
Cl
(g)
is
an
element
so
its
Δ
H
is
zero.
f
2
4
Describe what the dierence
–1
ΔH
=
85
ΔH
=
–144 kJ mol
–
229 kJ mol
is between the energy put in
to break bonds and the energy
–1
given out when the new bonds
from
bond
–1
(compared
with
–122 kJ mol
calculated
enthalpies)
are formed.
This
5
a
State what is ΔH
difference
is
typical
of
what
might
be
expected
using
mean
bond
for the
enthalpies.
The
answer
obtained
from
the
thermochemical
cycle
is
the
from
the
reaction (this requires
‘correct’
one
because
all
the
Δ
H
values
have
been
obtained
f
a sign)
actual
b
compounds
involved.
Identify if the reaction in
Mean
bond
enthalpy
calculations
also
allow
us
to
calculate
an
par t a is endothermic or
approximate
value
for
Δ
H
f
exothermic.
102
for
a
compound
that
has
never
been
made.
Practice
1
A
student
used
anhydrous
This
questions
Hess’s
law
copper(II)
enthalpy
to
determine
sulfate
change
was
is
a
value
for
the
enthalpy
change
that
occurs
when
hydrated.
labelled
ΔH
by
the
student
in
a
scheme
of
reactions.
exp
∇
H
exp
anhydrous
copper(II)
sulfate
hydrated
copper(II)
water
water
sulfate
water
∇
∇
H
H
1
2
copper(II)
sulfate
(a)
State
(b)
Write
Hess’s
a
solution
law.
(1
mathematical
expression
to
show
how
ΔH
,
ΔH
exp
to
(c)
each
Use
other
the
data
by
Hess’s
mathematical
book
values
for
,
a
value
for
ΔH
1
are
2
(1
expression
the
mark)
related
law.
two
that
you
enthalpy
have
changes
written
and
Δ H
in
part
ΔH
1
calculate
and
(b),
and
shown,
mark)
the
to
2
ΔH
exp
−1
ΔH
=
−156
kJ
mol
1
−1
ΔH
=
+12
kJ
mol
2
3
(d)
The
student
deionised
the
added
water
0.0210
in
temperature
an
of
mol
open
the
of
pure
anhydrous
polystyrene
water
increased
cup.
by
An
14.0
copper(II)
exothermic
sulfate
to
reaction
25.0
cm
occurred
of
and
ºC.
−1
(i)
Use
of
these
data
copper(II)
to
calculate
sulfate.
This
the
is
enthalpy
the
student
change,
value
in
for
kJ
,
mol
for
this
reaction
Δ H
1
In
this
raise
experiment,
the
you
temperature
−1
water
(ii)
is
Suggest
4.18
one
J
should
of
the
assume
25.0
g
of
that
all
water.
of
The
the
heat
specic
released
heat
is
used
capacity
−1
g
K
to
of
reason
(3
why
the
student
value
for
calculated
ΔH
in
part
(d)
marks)
(i)
1
is
(e)
less
Suggest
accurate
one
reason
than
the
data
book
why
the
value
for
value
given
in
cannot
ΔH
part
be
(c)
(1
measured
mark)
directly.
exp
AQA,
Hydrazine,
N
H
2
Write
an
nitrogen
(b)
State
(c)
Some
decomposes
with
in
hydrogen
an
exothermic
equation
for
the
peroxide
when
reaction.
decomposition
used
of
as
a
Hydrazine
rocket
also
reacts
hydrazine
fuel.
into
ammonia
and
only.
the
(1
meaning
mean
bond
of
the
term
enthalpies
mean
are
bond
given
N
in
H
enthalpy
the
N
(2
mark)
marks)
table.
N
N
N
O
H
O
O
–1
Mean bond enthalpy/ kJ mol
388
data
and
to
calculate
hydrogen
the
163
enthalpy
change
944
for
the
463
gas-phase
146
reaction
between
peroxide.
H
—
—
these
hydrazine
+
—
—
Use
H
(a)
,
4
exothermically
H
2
2013
2
H
— O
— O
—
— O
— H
H
AQA,
2013
103
Chapter
4
Energetics
3
Hess’s
to
law
is
used
determine
(a)
(b)
(c)
State
State
a
the
calculate
the
enthalpy
change
in
reactions
for
which
it
is
difcult
experimentally.
meaning
Hess’s
Consider
to
value
of
the
term
enthalpy
change
(1
mark)
(1
mark)
law.
the
following
table
of
data
and
the
scheme
of
reactions.
−1
Reaction
Enthalpy change/kJ mol
+
HCI(g) ➝ H
(aq) + CI
(aq)
–75
H(g) + CI(g) ➝ HCI(g)
–432
+
H(g) + CI(g) ➝ H
(g) + CI
(g)
+963
Δ
Use
the
data
in
the
(g)
+
Cl
H(g)
+
Cl(g)
table,
H
r
+
H
the
(g)
scheme
+
H
(aq)
+
Cl
(aq)
HCl(g)
of
reactions,
and
Hess’s
law
to
calculate
a
value
for
Δ
H
r
(3
marks)
AQA,
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
104
2010
5
Kinetics
5.
1 Collision theory
Kinetics
how
is
the
quickly
‘Popping’
a
study
they
test
of
the
take
tube
factors
place.
full
of
that
There
is
hydrogen
affect
rates
a
large
is
over
of
chemical
variation
in
a
in
reactions
reaction
fraction
of
a
–
rates.
Learning
➔
objective s:
Describe what must
second,
happen before a reaction
whilst
the
complete
rusting
away
of
an
iron
nail
could
take
several
years.
will take place.
Reactions
can
be
speeded
up
or
slowed
down
by
changing
the
conditions.
➔
The
rate
of
a
chemical
concentration
of
one
of
reaction
the
is
dened
reactants
or
as
the
products
change
with
in
unit
Explain why all collisions
do not result in a reaction.
time.
It
is
Specication reference: 3.1.5
−3
usually
measured
in
mol
−1
dm
s
Collision theory
Hint
For
a
with
reaction
enough
between
the
orientation
to
take
energy
parts
also
to
of
has
place
break
the
a
between
to
particles,
The
collision
bonds.
molecule
part
two
play.
that
To
are
get
a
going
lot
of
they
must
must
to
also
react
collide
take
place
together,
collisions
you
A rough rule for many chemical
so
need
a
reactions is that if the temperature
lot
goes up by 10 K (10 °C), the rate
of
particles
in
a
small
volume.
For
the
particles
to
have
enough
energy
to
of reaction approximately doubles.
break
need
bonds
plenty
Most
they
of
need
rapidly
collisions
to
be
moving
moving
between
fast.
particles
molecules
or
So,
in
a
for
a
small
other
fast
reaction
rate
you
volume.
particles
do
not
lead
to
transition
reaction.
wrong
They
either
have
enough
energy,
or
they
are
in
state
the
following
factors
Increasing
which
in
affect the
the
turn
will
rate of
increase
temperature
increases
both
chemical
the
This
their
rate
of
increases
energy
reactions
a
reaction.
the
and
speed
the
E
yplahtne
•
not
orientation.
Factors that
The
do
of
the
number
molecules,
of
a
reactants
collisions.
products
•
Increasing
particles
and
the
the
concentration
present
in
reaction
a
rate
given
of
volume
would
be
a
solution
then
faster.
If
there
collisions
However,
are
as
a
are
more
more
likely
reaction
extent
proceeds,
in
most
the
reactants
reactions
the
are
rate
used
of
up
and
reaction
their
drops
concentration
as
the
reaction
falls.
goes
of
reaction
So,
on.
▲ Figure 1
An exothermic reaction with
a large activation energy, E
a
•
Increasing
effect
as
or
Increasing
total
pressure
increasing
molecules
•
the
the
surface
the
atoms
in
of
a
a
gas
reaction
concentration
a
given
surface
area
of
area
solid,
of
volume
of
the
solid
more
a
so
This
solution
–
collisions
reactants
of
has
its
the
there
are
are
more
The
particles
same
more
likely.
greater
are
the
available
to
transition
with
a
lump
solid
because
molecules
into
there
are
in
smaller
more
a
gas
or
pieces
sites
a
liquid.
increases
for
This
the
means
rate
of
that
its
state
breaking
reaction
reaction.
yplahtne
collide
E
a
reactants
•
Using
of
a
a
catalyst
chemical
A
catalyst
reaction
is
a
without
substance
being
that
can
chemically
change
changed
the
rate
itself.
products
Activation
Only
For
a
a
a
very
small
collision
certain
energy
to
proportion
result
minimum
in
a
energy,
of
collisions
reaction,
enough
the
to
actually
result
molecules
start
in
must
breaking
a
reaction.
have
bonds.
extent
▲ Figure 2
of
reaction
An exothermic reaction with
a small activation energy, E
a
105
5.1
Collision
theor y
The
minimum
activation
energy
energy
needed
and
has
to
the
start
a
reaction
abbreviation
is
called
the
E
a
You
can
include
that
shows
the
Exothermic
Figure
a
with
low
activation
1
at
room
Figure
small
about
The
the
a
activation
energy
on
an
enthalpy
diagram
reaction.
reaction
a
activation
energy.
temperature
to
2
bring
prole
shows
the
a
a
for
reaction
an
very
will
exothermic
few
take
place
reaction
with
collisions
extremely
a
will
have
slowly
sufcient
reaction.
reaction
energy.
because
prole
This
many
for
reaction
collisions
an
exothermic
will
take
will
place
have
reaction
rapidly
enough
at
with
a
room
energy
to
bring
reaction.
situation
large
This
because
about
activation
amount
of
is
a
little
energy
amount
of
is
like
a
needed
energy
is
ball
in
on
a
Figure
needed
in
hill,
3a,
see
to
Figure
set
Figure
the
3.
A
ball
small
rolling,
whilst
3b.
energy
The
▲ Figure 3
of
of
reactions
shows
temperature
high
course
activation
energy
with
idea
energy
large
b
the
species
that
exists
at
the
top
of
the
curve
of
an
enthalpy
diagram
is
Ball on a mountainside models
called
the
a
transition
process
of
being
state
or
made
activated
and
some
complex.
bonds
are
in
hill,
it
Some
the
bonds
process
are
of
in
being
Hint
broken.
‘Species’ is a term used by chemists
is
Like
the
ball
at
the
very
top
of
the
has
extra
energy
and
unstable.
to refer to an atom, molecule, or ion.
Endothermic
Endothermic
energy
than
energy
E
,
is
reactions
reactions
the
are
reactants.
shown
in
those
An
Figure
in
which
the
endothermic
4.
The
products
reaction,
transition
state
have
with
has
more
activation
been
labelled.
a
transition
state
Notice
top
of
that
the
the
activation
energy
is
measured
from
the
reactants
to
the
curve.
yplahtne
E
a
products
Summary questions
reactants
1
List ve factors that aect the speed of a chemical reaction.
Use the reaction prole in Figure 5 to answer questions
extent
of
2
▲ Figure 4
2 and 3:
reaction
a
What is A?
b
What is B?
c
What is C?
d
What is D?
a
Identify whether the enthalpy prole represents an endothermic or
An endothermic reaction
with activation energy E
a
C
3
D
yplahtne
A
an exothermic reaction.
b
B
extent
▲ Figure 5
10 6
of
reaction
A reaction prole
Explain your answer to par t
a
5.2
The
a
particles
few
in
The
are
the
in
moving
middle.
particles
any
also
Maxwell–Boltzmann distribution
gas
(or
slowly,
The
have
a
energy
a
solution)
few
of
range
a
of
are
very
all
fast
particle
moving
but
most
depends
energies.
If
you
at
different
are
on
plot
its
a
speeds
–
Learning
objective s:
somewhere
speed
graph
so
of
➔
Dene activation energy.
➔
Explain how temperature
the
energy
aects the number of
against
the
fraction
of
particles
that
have
that
energy,
you
end
up
molecules with energy equal
with
the
curve
shown
in
Figure
1.
This
particular
shape
is
called
the
to or more than the activation
Maxwell–Boltzmann
distribution
–
it
tells
us
about
the
distribution
energy.
of
energy
amongst
the
particles.
➔
•
No
particles
•
Most
have
zero
Explain why a small increase
energy.
in temperature has a large
particles
have
intermediate
energies
–
around
the
peak
of
eect on the rate of a reaction.
the
curve.
Specication reference: 3.1.5
•
A
In
•
few
have
fact
Note
very
there
also
probable
is
that
high
no
energies
upper
the
(the
right-hand
side
of
the
curve).
limit.
average
energy
is
not
the
same
as
the
most
energy.
most
probable
energy
E
ygrene
average
energy
htiw
selcitrap
fo
noitcarf
energy
▲ Figure 1
E
The distribution of the energies of particles. The area under the graph
represents the total number of particles
Activation
E
energy
a
For
a
reaction
enough
of
to
energy
energy
is
take
to
place,
start
called
the
a
collision
breaking
between
bonds,
activation
see
energy
E
particles
Topic
.
If
5.1.
you
must
This
mark
E
a
Maxwell–Boltzmann
the
graph
number
The
of
need
takes
to
the
particles
for
place
the
For
example,
station.
But
combustion
enough
the
to
a
energy
is
at
fuels
the
a
provided
all
mostly
The
to
to
be
2,
then
line
the
the
area
represents
under
the
react.
present
reactions
safe
may
heat
head
at
that
room
provide
given
activation
match
by
energy
not
Figure
energy
energy
on
a
before
are
a
reaction
exothermic
occur
temperature.
spark
reaction.
graph,
activation
enough
why
are
small
in
the
with
room
supply
chemicals
of
activation
explains
spontaneously
distribution
right
have
amount
out
energy
are
temperature,
enough
quite
by
for
energy
the
initial
further
stable
as
to
in
start
the
petrol
the
reaction
reactions.
until
a
is
Similarly
activation
friction.
107
5.2
The
Maxwell–Bo l t z m ann
d istr ibut ion
E
Hint
ygrene
The rate of a chemical reaction
htiw
is dened as the change in
concentration of one of the
−3
It is usually measured in mol dm
−1
s
of
particles
sufcient
energy
fo
per second or mol dm
with
selcitrap
reactants or products with unit time.
−3
number
to
react
E
a
noitcarf
energy
▲ Figure 2
Only particles with energy greater than E
E
can react
a
Even
This
the
is
why
electrical
energy
The
The
high
if
to
begin
smell
an
of
gas,
a
single
you
provided
spark
must
by
the
not
shape
of
the
as
The
total
area
set
even
off
turn
could
a
on
reaction.
a
light.
produce
The
enough
explosion.
Maxwell–Boltzmann
shown
in
temperatures
right.
can
switch
effect of temperature on
higher
the
you
connection
temperature,
At
temperature
number
Figure
the
of
reaction
graph
rate
changes
with
3.
peak
of
particles
the
curve
with
very
is
lower
high
and
energy
moves
to
increases.
Summary questions
The
because
1
it
under
the
represents
curve
the
total
is
the
same
number
for
of
each
temperature
particles.
Use Figure 4 to answer the
The
following questions:
shaded
areas
to
the
right
of
the
E
line
represent
the
number
of
a
molecules
a
What is the axis labelled A?
b
What is the axis labelled B?
that
have
greater
energy
than
E
at
each
temperature.
a
The
graphs
have
show
energy
that
greater
at
higher
than
E
so
temperatures
a
higher
more
percentage
of
of
the
molecules
collisions
a
c
What does area C
will
result
in
reaction.
This
is
why
reaction
rates
increase
with
represent?
temperature.
d
If the temperature is
increase
in
In
the
fact,
a
small
number
of
increase
particles
in
temperature
with
energy
produces
greater
than
a
large
E
a
increased, what happens
Also,
the
total
number
of
collisions
in
a
given
time
increases
a
little
as
to the peak of the cur ve?
the
e
If the temperature is
increased, what happens
of
particles
reaction
with
move
as
energy
the
faster.
However,
increase
greater
than
in
E
the
this
is
number
not
of
as
important
effective
to
collisions
the
rate
(those
).
a
to
E
?
a
T
>
T
2
E
T
ygrene
htiw
A
a
with
E
at
a
are
a
more
greater
T
than
2
particles
energy
at
than
T
1
T
2
selcitrap
C
E
1
there
1
E
a
fo
▲ Figure 4
The Maxwell–Boltzmann
noitcarf
B
distribution of energies of particles
energy
E
at a particular temperature, with the
activation energy, E
marked
a
▲ Figure 3
The Maxwell–Boltzmann distribution of the energies of the same number
of particles at two temperatures
108
5.3
Catalysts
Catalysts
without
are
substances
being
chemically
that
affect
changed
the
rate
of
themselves
chemical
at
the
Learning
reactions
end
of
the
➔
reaction.
Catalysts
are
usually
used
to
speed
up
reactions
so
objective s:
they
State the denition of a
are
catalyst.
important
in
industry.
It
is
cheaper
to
speed
up
a
reaction
by
using
a
➔
catalyst
than
by
using
high
temperatures
and
pressures.
This
is
Describe how a catalyst
true,
aects activation energy.
even
if
the
catalyst
is
expensive,
because
it
is
not
used
up.
➔
Describe how a catalyst
aects enthalpy change.
How
catalysts
work
Specication reference: 3.1.5
Catalysts
work
reaction,
one
the
with
activation
that
is
in
a
to
of
start
Figure
they
lower
energy
needed
diagrams
because
a
activation
the
the
provide
reaction
reaction).
different
energy.
(the
You
pathway
Therefore
minimum
can
see
for
the
they
amount
this
on
the
reduce
of
energy
enthalpy
1.
transition
state
(no
catalyst)
–1
183
kJ
mol
ygrene
–1
105 kJ
mol
–1
5 8 kJ
2HI
mol
(g)
–1
53
kJ
mol
H
(g)
+
I
2
extent
▲ Figure 1
For
of
(g)
2
reaction
The decomposition of hydrogen iodide with dierent catalysts
example,
for
the
decomposition
2HI(g)
➝
of
H
hydrogen
(g)
+
I
2
iodide:
(g)
2
–1
E
=
183
kJ
mol
=
105
kJ
mol
=
58
(without
a
catalyst)
a
–1
E
(with
a
gold
catalyst)
a
–1
E
kJ
mol
(with
a
platinum
catalyst)
a
You
can
you
look
see
The
area
collisions
plus
the
at
what
the
that
area
collisions
is
that
shaded
can
that
that
happens
when
you
Maxwell–Boltzmann
can
pink
happen
is
takes
represents
without
shaded
lower
pink,
place
a
the
a
activation
curve
number
catalyst.
represents
with
the
distribution
The
the
of
area
in
energy
Figure
if
2.
effective
shaded
number
of
blue,
effective
catalyst.
109
5.3
Catalysts
fraction
of
particles
with
energy
E
activation
with
energy
catalyst
activation
without
E
energy
catalyst
E
cat
a
energy
▲ Figure 2
Catalysts
affect
▼ Table 1
do
the
not
affect
position
of
With a catalyst the ex tra particles in the blue area react
the
enthalpy
equilibrium
change
in
a
of
the
reversible
reactions,
reaction,
nor
see
do
Topic
they
6.1.
Examples of catalysts
Reaction
N
E
(g) + 3H
2
(g) ➝ 2NH
2
Catalyst
Use
(g)
3
iron
making fer tilisers
Haber process
4NH
+ 5O
3
➝ 4NO + 6H
2
making fer tilisers and
O
2
platinum and rhodium
explosives
Ostwald process for making nitric acid
H
C
CH
2
+ H
2
➝ CH
2
CH
3
3
nickel
making margarine
hardening of fats with hydrogen
aluminium oxide and silicon dioxide
cracking hydrocarbon chains from crude oil
making petrol
zeolite
catalytic conver ter reactions in car exhausts
platinum and rhodium
removing polluting gases
making ethanol – a fuel
+
H
C
CH
2
+ H
2
O ➝ CH
2
CH
3
H
OH
absorbed on solid silica
2
additive, solvent, and
phosphoric acid, H
hydration of ethene to produce ethanol
PO
3
4
chemical feedstock
CH
CO
3
H(l) + CH
2
OH(l) ➝ CH
3
CO
3
CH
2
(aq) + H
3
O(l)
+
2
H
making solvents
esterication
Different
trial
and
catalysts
in
new
their
exhaust
ways
–
most
were
discovered
by
with
honeycomb
reactions
cars
systems.
are
now
These
equipped
reduce
the
with
levels
catalytic
of
a
converters
number
of
gases.
catalytic
coated
different
converters
petrol-engine
polluting
The
in
error.
Catalytic
All
work
converter
platinum
shape
take
is
a
and
honeycomb,
rhodium
provides
place,
so
a
an
little
made
metals
enormous
of
these
–
of
the
a
ceramic
catalysts.
surface
expensive
area,
on
metals
material
The
which
goes
a
the
long
way.
Synoptic link
As
You will learn more about
they
other
to
pass
over
form
less
the
catalyst,
harmful
the
polluting
products
by
the
gases
react
following
with
each
reactions:
catalytic conver ters in Topic 12.4,
carbon
monoxide
+
nitrogen
oxides
➝
nitrogen
+
carbon
dioxide
Combustion of alkanes.
hydrocarbons
110
+
nitrogen
oxides
➝
nitrogen
+
carbon
dioxide
+
water
Kinetics
5
The
1
reactions
The
take
gases
catalyst
–
rst
this
molecules
2
in
The
gases
The
products
is
called
more
The
long
is
form
on
just
the
react
then
is
of
the
critical.
enough
to
take
react,
bonds
but
be
weak
the
for
from
frees
place
bonds
must
with
the
catalyst
metal
to
two
atoms
This
them
in
holds
react
of
steps:
the
the
gas
together.
surface.
away
This
their
of
adsorption.
position
the
break
weak
They
surface
called
right
on
desorption.
to
the
weak
process
then
gases
strength
surface
place
up
and
metal
room
on
atoms
the
–
this
catalyst
process
surface
for
react.
holding
strong
the
the
gases
enough
enough
to
to
onto
hold
release
the
the
the
metal
gases
products
for
easily.
Zeolites
Zeolites are minerals that have a very open pore structure that ions or
molecules can t into. Zeolites conne molecules in small spaces, which
causes changes in their structure and reactivity. More than 150 zeolite
types have been synthesised and 48 naturally occurring zeolites are known.
Synthetic zeolites are widely used as catalysts in the petrochemical industry.
▲ Figure 3
Part of the structure of a
synthetic zeolite
Hardening fats
Unsaturated fats, used in margarines for example, are made more solid or
hardened when hydrogen is added across some of the double bonds. This
is done by bubbling hydrogen into the liquid fat which has a nickel catalyst
mixed with it. The nickel is ltered o after the reaction. This allows the
manufacturer to tailor the spreadability of the margarine.
▲ Figure 4
Margarine
111
5.3
Catalysts
Catalysts and the ozone layer
Until recently, a group of apparently unreactive compounds called
chlorouorocarbons (CFCs) were used for a number of applications such
as solvents, aerosol propellants, and in expanded polystyrene foams.
They escaped high into the atmosphere where they remain because
they are relatively unreactive. This is par tly due to the strength of the
carbon–halogen bonds.
CFCs do eventually decompose to produce separate chlorine atoms. These
act as catalysts in reactions that bring about the destruction of ozone, O
.
3
Ozone is impor tant because it forms a layer in the atmosphere of the Ear th
that acts as a shield. The layer prevents too much ultraviolet radiation from
reaching the Ear th’s surface.
The overall reaction is shown below:
chlorine atom catalyst
O
(g) + O(g)
3
2O
(g)
2
Nitrogen monoxide acts as a catalyst in a similar way to chlorine atoms.
International agreements, such as the 1987 Montreal Protocol, have resulted
in CFCs being phased out. Unfor tunately there is still a reser voir of them
remaining from before these agreements. Chemists have developed, and
continue to work on, suitable substitutes for CFCs that do not result in
damage to the upper atmosphere. These include hydrochlorouorocarbons
and hydrouorocarbons. Former United Nations Secretary General, Ko
Annan, has referred to the Montreal Protocol as ‘perhaps the single most
successful international agreement to date’.
Summary questions
1
D
The following questions refer to
Figure 5.
C
a
What are labels A, B, C, D, R,
R
and P?
A
b
What do the distances
P
from D to R and from
C to R represent?
B
c
Is the reaction exothermic
▲ Figure 5
A prole for a reaction
or endothermic?
with and without a catalyst
112
Practice
1
The
questions
gas-phase
reaction
between
hydrogen
H
(g)
+
Cl
2
(a)
Dene
the
term
activation
and
chlorine
➝
(g)
is
very
slow
at
room
temperature.
2HCl(g)
2
energy .
(2
(b)
Give
one
room
reason
why
the
reaction
between
hydrogen
and
chlorine
is
very
marks)
slow
at
temperature.
(1
(c)
Explain
why
reaction
an
increase
between
in
pressure,
hydrogen
and
at
constant
temperature,
increases
the
rate
Explain
why
reaction
a
small
between
increase
hydrogen
in
temperature
and
can
lead
to
a
large
increase
in
the
marks)
rate
Give
the
(f)
Suggest
meaning
of
the
term
marks)
catalyst.
(1
of
a
one
reason
why
a
solid
catalyst
for
a
gas-phase
reaction
is
often
in
the
mark)
form
powder.
(1
mark)
AQA,
2
The
diagram
in
sample
a
of
chlorine.
(2
(e)
of
chlorine.
(2
(d)
mark)
below
of
a
represents
gas
at
a
given
a
Maxwell–Boltzmann
temperature.
The
distribution
questions
below
curve
refer
for
to
the
this
2006
particles
sample
of
particles.
(a)
Label
(b)
On
the
axes
on
a
copy
of
the
diagram.
(2
the
diagram
draw
a
curve
to
show
the
distribution
for
this
sample
at
a
marks)
lower
temperature.
(2
(c)
In
order
not
for
result
two
in
a
particles
to
react
they
must
collide.
Explain
why
most
marks)
collisions
(1
(d)
State
one
increased
way
in
which
without
the
changing
collision
the
frequency
between
particles
in
a
gas
can
Suggest
reaction
(f)
Explain
why
rate
in
a
small
increase
between
general
in
colliding
terms
how
a
mark)
be
temperature.
(1
(e)
do
reaction.
temperature
can
lead
to
a
large
increase
in
mark)
the
particles.
catalyst
(2
marks)
(2
marks)
works.
AQA,
2004
113
Chapter
5
Kinetics
3
The
diagram
two
different
shows
the
Maxwell–Boltzmann
distribution
of
molecular
energies
in
a
gas
at
temperatures.
V
W
X
Y
energy
(a)
One
of
the
axes
(b)
State
the
effect,
(c)
State
the
letter,
is
labelled.
Complete
the
diagram
by
labelling
the
other
axis.
(1
if
any,
of
a
solid
catalyst
on
the
shape
of
either
of
these
mark)
distributions.
(1
molecules
at
V,
the
W,
X,
lower
or
Y,
that
represents
the
most
probable
energy
of
mark)
the
temperature.
(1
(d)
Explain
what
must
happen
for
a
reaction
to
occur
between
molecules
of
two
mark)
different
gases.
(2
(e)
Explain
why
a
small
increase
in
temperature
has
a
large
effect
on
the
initial
marks)
rate
of
a
reaction.
(1
mark)
AQA,
4
The
diagram
shows
the
Maxwell–Boltzmann
distribution
for
a
sample
of
gas
at
a
2012
xed
temperature.
E
is
the
activation
energy
for
the
decomposition
of
this
gas.
a
selucelom
fo
rebmun
E
energy
E
is
the
most
probable
value
for
the
energy
of
a
the
molecules.
mp
(a)
On
the
On
this
appropriate
axis
of
this
diagram,
mark
the
value
of
E
for
this
distribution.
mp
diagram,
sketch
a
new
distribution
for
the
same
sample
of
gas
at
a
lower
temperature.
(3
(b)
With
reference
temperature
to
the
Maxwell–Boltzmann
decreases
the
rate
of
distribution,
decomposition
of
this
explain
why
a
decrease
marks)
in
gas.
(2
marks)
AQA,
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
114
2013
6
Equilibria
6.1
Chemists
ending
usually
with
the
The
think
of
a
idea
reaction
as
of
starting
equil ibr ium
with
the
reactants
Learning
and
➔
reactants
some
reactions
are
reversible.
For
State the denition of a
reversible reaction.
products
➝
➔
However,
objective s:
products.
example,
when
you
State what is meant by
heat
chemical equilibrium.
blue
as
hydrated
the
water
returns
to
copper
of
sulfate
it
crystallisation
blue
if
you
add
becomes
is
driven
white
off.
anhydrous
The
white
copper
copper
sulfate
➔
sulfate
➔
CuSO
.5H
4
Explain why all reactions do
not go to completion.
water.
O
CuSO
2
+
5H
4
Explain what happens when
O
2
equilibrium has been reached.
blue
hydrated
copper
However,
reaction
react
or
something
in
a
together
products
which
the
mixture
is
can
and
you
different
form
get
called
up
sulfate
a
an
an
the
As
of
happen
soon
reactants
mixture
all
anhydrous
copper
would
container.
proportions
Setting
You
closed
white
of
the
again,
both.
three
as
if
you
that
Eventually
components
equilibrium
were
products
so
Specication reference: 3.1.6
sulfate
to
are
instead
you
get
remain
do
this
formed
of
a
they
reactants
mixture
constant.
in
This
mixture
equilibrium
understand
how
an
equilibrium
mixture
is
set
up
by
thinking
a
about
what
water.
This
happens
is
easier
in
to
a
physical
picture
process,
than
a
like
chemical
the
evaporation
of
change.
vacuum
First
imagine
a
puddle
molecules
at
liquid
evaporate.
and
the
of
surface
water
will
out
move
Evaporation
in
the
fast
will
open.
enough
Some
to
continue
of
escape
until
all
the
water
from
the
evaporation
the
water
water
isgone.
level
But
think
about
putting
some
water
into
a
closed
container.
At
rst
water
the
water
will
get
will
go
will
begin
smaller
up.
molecules
But
will
and
as
to
evaporate
the
more
start
to
number
as
of
molecules
re-enter
before.
vapour
enter
the
The
molecules
the
liquid,
volume
vapour,
see
in
of
the
some
Figure
the
liquid
gas
phase
gas-phase
1.
b
After
a
time,
become
same
same
of
In
equal.
and
liquid.
so
The
rate.
the
rate
The
will
of
the
you
of
could
of
the
number
situation
ideas
evaporation
level
evaporation
This
key
fact,
the
this
have
and
is
and
liquid
of
the
water
molecules
condensation
called
a
rate
will
in
condensation
then
the
are
dynamic
of
stay
vapour
still
going
exactly
and
on
equilibrium
in
but
and
will
the
water
vapour
condensation
the
at
is
the
one
topic.
started
by
lling
the
empty
container
with
the
water
same
mass
vapour
of
water
would
begin
vapour
to
as
you
condense
originally
and,
in
had
time,
liquid
would
water.
reach
The
exactly
the
▲ Figure 1 a
same
equilibrium
Water will evaporate into
position.
an empty container. Eventually the rates
of evaporation and condensation will be
the same
b
Equilibrium is set up
115
6.1
The
idea
of
equil ibr ium
The
conditions for
equilibrium
Study tip
Although
the
system
used
here
is
very
simple,
you
can
pick
out
four
Remember that at equilibrium,
conditions
that
apply
to
all
equilibria:
both for ward and backward
•
Equilibrium
can
only
be
reached
in
a
closed
system
(one
where
reactions occur at the same rate
the
reactants
and
products
can’t
escape).
The
system
does
not
have
so the concentrations of all the
to
be
sealed.
For
example,
a
beaker
may
be
a
closed
system
for
reactants and products remain
a
reaction
that
takes
place
in
a
solvent,
as
long
as
the
reactants,
constant.
products,
•
and
Equilibrium
from
will
liquid
be
the
pressure,
•
•
can
or
Equilibrium
is
a
opposing
1,
These
are
pressure
of
vapour)
long
as
and
either
nal
conditions,
direction
(in
equilibrium
such
of
which
properties
has
the
such
as
that
It
is
as
are
reached
going
condensation),
equilibrium
properties
reversible
,
reaction
for
that
been
Figure
1,
position
temperature
not
do
when
on
are
not
all
the
reached
system
density,
do
can
and
the
on
when
time
(in
the
change
the
rates
same.
concentration,
depend
the
with
time.
colour,
total
and
quantity
reach
equilibrium
water
H
Summary questions
water
O(l)
H
2
Chemical
is
denoted
by
the
example:
liquid
For each of the following
the
matter.
symbol
1
from
and
process.
processes,
properties
–
evaporate.
same).
evaporation
that
not
dynamic
Figure
know
do
approached
(as
the
oftwo
You
be
from
same
stay
macroscopic
A
solvent
vapour
O(g)
2
equilibria
statements about all
The
same
principles
that
you
have
found
for
a
physical
change
also
equilibria, say whether it is
apply
to
chemical
equilibria
such
as:
true or false.
A
a
+
B
C
+
D
Once equilibrium
reactants
products
is reached the
concentrations of the
b
•
Imagine
reactants and the products
the
do not change.
reverse
At equilibrium the for ward
•
starting
forward
Then
reaction
as
the
reaction
and the backward
so
is
A
fast,
and
because
the
up.
At
B
only.
because
there
concentrations
speeds
decrease
with
rate
the
forward
of
A
is
C
At
and
no
C
and
same
start
are
and
D
time
reaction
the
B
slows
the
reaction
There
is
no
D.
build
the
of
plentiful.
up,
the
reverse
concentrations
of
A
and
B
down.
reactions come to a halt.
•
c
A
point
is
reached
where
exactly
the
same
number
of
particles
are
Equilibrium is only reached
changing
from
A
+
B
to
C
+
D
as
are
changing
from
C
+
D
to
A
+
B.
in a closed system.
Equilibrium
d
has
been
reached.
An equilibrium mix ture
One
important
point
to
remember
is
that
an
equilibrium
mixture
can
always contains half
have
any
proportions
of
reactants
and
products.
It
is
not
necessarily
reactants and half
half
reactants
and
half
products,
though
it
could
be.
The
proportions
products.
may
2
What can be said about the
rates of the for ward and the
backward reactions when
equilibrium is reached?
116
be
changed
temperature,
conditions
depending
pressure,
the
and
proportions
on
the
conditions
concentration.
of
reactants
of
But
and
the
at
reaction,
any
products
given
do
not
such
as
constant
change.
6.2
Changing
an
Some
industrial
sulfuric
these
acid,
products
and
condit io ns
equilibrium
processes,
have
reactions
the
such
reversible
would
In
the
as
a
key
equilibrium
principle,
rea c t io n
production
reactions
produce
reactants.
as
you
of
ammonia
step.
In
mixtures
would
of
like
closed
systems
containing
to
Learning
or
increase
objective s:
➔
State Le Châtelier ’s principle.
➔
Explain how an equilibrium
both
the
position is aected by
proportion
of
products.
For
this
reason
it
is
important
to
understand
concentration, temperature,
how
to
control
equilibrium
reactions.
pressure, or a catalyst.
Specication reference: 3.1.6
The
It
is
equilibrium
possible
to
equilibrium
of
the
•
If
mixture.
the
If
we
forward
the
You
we
varying
often
or
move
the
Châtelier’s
whether
the
conditions
It
a
of
So
in
the
other
oppose
Le
so
the
the
such
in
the
able
the
to
to
obtain
position
of
equilibrium
is
products
a
an
greater
yield
equilibrium.
mixture
moved
in
to
the
is
right,
or
in
the
in
the
equilibrium
equilibrium
is
mixture
moved
to
the
is
left,
or
in
equilibrium
as
position
temperature,
(in
the
case
the
of
to
the
left
or
right
concentration
reactions
of
involving
by
species
gases).
principle
is
useful
because
moves
equilibrium
at
to
the
mixture
you
it
gives
right
are
or
us
to
a
rule.
the
left
It
tells
when
us
the
changed.
that
any
factor
the
A.
by
the
principle
predict
does
the
more
A
of
the
reduce
changed
not
equilibrium
the
which
equilibrium
tell
quantities
the
concentration
that
the
add
only
of
uses
equilibrium
up
to
some
way
A
that
more
+
of
moves
disturbance.
affects
will
the
shift
so
as
to
us
how
far
the
equilibrium
moves
involved.
B,
right.
this
one
will
this
A.
B
This
(so
produces
You
end
the
C(aq)
system
with
of
shift
reactant.
B(aq)
extra
reacting
the
of
equilibrium
concentration
you
The
is
to
concentrations
says
some
disturbed,
position
A(aq)
Suppose
is
tends
change.
increase
reduce
if
mixture,
cannot
principle
equilibrium
direction
Châtelier’s
you
Changing
is
are
equilibrium
reactants
pressure
words,
equilibrium
of
the
that
principle
an
system
in
to
you
changing
products
that
equilibrium
of
way
reactants
states:
If
If
this
called
of
direction.
Le Châtelier ’s
Le
of
say
conditions
involved,
In
proportion
direction.
backward
can
the
is
say
proportion
increased,
the
This
proportion
increased,
•
change
products.
the
mixture
would
can
more
the
Look
+
C
with
at
the
increase
the
more
and
a
Le
Châtelier’s
direction
that
tends
reaction:
D(aq)
reduce
forming
up
reactants,
in
D,
the
concentration
C
and
and
greater
concentration
D).
So,
moves
the
proportion
of
A,
adding
of
▲ Figure 1
Henri-Louis Le Châtelier
was a French chemist who rst put
forward his ‘Loi de stabilité d’équilibre
chimique’ in 1884
117
6.2
Changing
the
c o nd itions
of
an
e quil i br ium
rea c ti on
products
equilibrium
thing
You
can
pulling
decrease
out
the
the
syringe
pressure
the
would
could
move
You
in
reaction
mixture
by
same
to
happen
also
the
thing
if
you
remove
right
to
would
C
overall
side
moment
the
because
you
mixture
have
of
you
added
A.
The
same
as
it
happen
if
was
formed.
more
you
B.
C
The
(and
D)
removed
D
equilibrium
using
as
up
soon
A
as
would
and
it
B.
was
The
formed.
brown
only
will
reaction
the
if
pressure
affect
only
there
reactions
change
are
a
the
involving
position
different
gases.
of
number
Changing
equilibrium
of
of
molecules
the
a
on
either
equation.
becomes
reduced
the
An
concentration
of
changes
pressure
gaseous
a
before
more
barrel.
Pressure
paler
added
produce
Changing the overall
For
than
mixture
NO
example
of
a
such
a
reaction
is:
.
2
N
O
2
(g)
dinitrogen
1
2NO
4
tetraoxide
nitrogen
mole
2
colourless
After
a
few
becomes
equilibrium
more
moments
darker
moves
brown
the
brown
NO
to
is
dioxide
moles
brown
mixture
as
the
(g)
2
Increasing
the
right
the
pressure
of
a
gas
means
that
there
are
more
molecules
and
of
it
of
a
in
a
given
volume
–
it
is
equivalent
to
increasing
the
concentration
formed.
2
If
▲ Figure 2
N
O
2
(g)
2NO
4
solution.
you
increase
the
pressure
on
this
system,
Le
Châtelier’s
principle
tells
(g)
2
us
that
the
position
of
equilibrium
will
move
to
decrease
the
pressure.
The equilibrium moves to the right as
This
means
that
it
will
move
to
the
left
because
fewer
molecules
you decrease the pressure
exert
less
pressure.
equilibrium
will
In
the
move
to
same
the
way
right
if
–
you
decrease
molecules
of
the
N
O
2
Hint
to
form
molecules
of
NO
,
thereby
increasing
the
pressure,
will
the
decompose
4
pressure.
2
Increasing the pressure or
Dinitrogen
tetraoxide
is
a
colourless
gas
and
nitrogen
dioxide
is
brown.
decreasing the volume of a
You
can
investigate
this
in
the
laboratory,
by
setting
up
the
equilibrium
mix ture of gases increases the
mixture
in
a
syringe.
If
you
decrease
the
pressure,
by
pulling
out
the
concentration of all the reactants
syringe
barrel,
you
can
watch
as
the
equilibrium
moves
to
the
right
and products by the same amount,
because
the
colour
of
the
mixture
gets
browner,
see
Figure
2.
not just one of them.
Note
that
sides
of
if
the
position.
there
is
the
equation,
For
same
then
number
pressure
of
has
moles
no
of
effect
gases
on
on
the
both
equilibrium
example:
Hint
H
The rate at which equilibrium is
(g)
+
2
reached will be speeded up by
I
(g)
2HI(g)
2
2
moles
2
moles
increasing the pressure, as there
The
equilibrium
position
will
not
change
in
this
reaction
when
the
will be more collisions in a given
pressure
is
changed
so
the
proportions
of
the
three
gases
will
stay
time.
thesame.
Changing temperature
Study tips
Reversible
direction
•
reactions
are
that
are
endothermic
exothermic
(take
in
heat)
(give
in
out
the
heat)
other
in
one
direction,
see
The terms ‘move for wards’ and
Topic
4.4.
The
size
of
the
enthalpy
is
the
same
in
both
directions
but
‘move to the right’ mean the
the
sign
changes.
same thing in this contex t.
Example 1
•
The terms ‘move backwards’
Suppose
you
increase
the
temperature
of
an
equilibrium
mixture
and ‘move to the left’ mean the
is
exothermic
in
the
forward
direction.
An
example
is:
same thing in this contex t.
–1
2SO
(g)
2
118
+
O
(g)
2
2SO
(g)
3
ΔH
=
–197
kJ
mol
that
Equilibria
6
The
negative
dioxide
and
direction.
reverse
Le
This
this
it
In
the
the
a
right
move
that
greater
same
to
is,
heat
to
in
to
if
we
us
the
the
the
increase
cool
the
is
the
tells
in
that
form
heat
sulfur
is
given
trioxide
absorbed
as
out
in
the
when
the
sulfur
forward
reaction
goes
in
the
left.
that
if
you
direction
direction
left.
proportion
way,
and
is,
moves
will
means
that
that
principle
endothermic),
contain
ΔH
react
means
equilibrium
do
of
direction,
Châtelier’s
the
To
sign
oxygen
of
The
cools
which
proportion
of
the
temperature,
system
absorbs
dioxide
mixture
the
the
equilibrium
sulfur
the
increase
that
heat
oxygen
equilibrium
sulfur
(is
mixture
and
down.
will
then
than
will
before.
move
to
trioxide.
Summary questions
Example 2
The
effect
of
temperature
on
the
dinitrogen
tetraoxide/nitrogen
1
dioxide
equilibrium
can
also
be
investigated
using
the
same
In which of the following
apparatus
reactions will the position of
you
used
to
investigate
the
effect
of
pressure
on
this
reaction.
The
equilibrium be aected by
reaction
is
endothermic
as
it
proceeds
from
dinitrogen
tetraoxide
to
changing the pressure?
nitrogen
dioxide
(the
forward
direction).
Explain your answers.
–1
N
O
2
(g)
2NO
4
ΔH
(g)
=
+58
kJ
mol
2
a
2SO
(g) + O
2
The
gas
mixture
is
contained
in
a
syringe
as
before.
The
syringe
(g)
2
is
2SO
(g)
3
then
immersed
in
warm
water
along
with
another
syringe
containing
b
CH
CO
3
the
same
volume
of
air
for
comparison.
The
plunger
of
the
H(aq)
2
syringe
+
CH
CO
3
containing
air
will
containing
the
rise
as
the
air
expands.
The
plunger
of
the
O
2
/
NO
4
mixture
will
also
rise
but
by
a
(aq) + H
H
(g) + CO
2
greater
(g)
2
2
H
amount.
in
this
each
This
indicates
syringe.
molecule
This
of
is
N
is
consistent
that
more
because
O
2
This
that
molecules
the
of
equilibrium
disappears
gas
has
produces
have
been
moved
two
to
the
molecules
2
right;
of
NO
4
with
.
Consider the following
equilibrium reaction.
2
Le
Châtelier’s
principle.
When
the
mixture
is
(g) + 3H
2
up,
the
equilibrium
moves
in
the
endothermic
direction,
it
absorbs
heat
which
tends
to
cool
the
mixture
should
during
were
this
be
able
to
predict
experiment
repeated
in
ice
and
the
also
colour
what
change
would
2NH
(g)
3
–1
= –92 kJ mol
down.
a
You
(g)
2
that
ΔH
is,
O(g) + CO(g)
2
formed
N
warmed
(aq)
syringe
c
N
2
that
happen
you
if
would
the
see
What would be the eect
on the equilibrium position
experiment
of heating the reaction?
water.
Choose from ‘move to the
right’, ‘move to the left’, and
Catalysts
‘no change’.
Catalysts
alter
an
have
no
composition
alternative
energy
of
reactions
the
effect
of
route
the
for
reaction,
on
the
position
equilibrium
the
see
reaction,
Topic
5.3.
of
equilibrium
mixture.
which
This
They
has
a
affects
so
work
lower
the
they
by
do
not
producing
b
on the equilibrium position
activation
forward
and
What would be the eect
back
of adding an iron catalyst?
equally.
Choose from ‘move to the
right’, ‘move to the left’, and
Although
catalysts
have
no
effect
on
the
position
of
equilibrium,
that
‘no change’.
is,
the
more
yield
of
quickly
the
and
reaction,
are
they
therefore
do
allow
important
equilibrium
in
industry.
to
be
reached
c
What eect would an
iron catalyst have on the
reaction?
d
To get the maximum
yield of ammonia in this
reaction would a high or
low pressure be best?
Explain your answer.
119
6.3
The
equilibrium
K
const an t
c
Learning
objective s:
As
you
have
completion,
➔
seen,
but
many
instead
reactions
end
up
as
are
an
reversible
and
equilibrium
do
not
mixture
go
of
to
reactants
Dene the expression
and
products.
A
reversible
reaction
that
can
reach
equilibrium
is
reversible reaction.
indicated
➔
by
the
symbol
.
In
this
topic
you
see
how
you
can
Dene the term chemical
tackle
equilibrium
reactions
mathematically.
You
will
deal
only
with
equilibrium.
homogeneous
➔
State the denition of an
are
in
the
systems
same
–
phase,
those
for
where
example,
all
all
the
reactants
and
products
liquids.
equilibrium constant and
describe how it is determined.
The
equilibrium
constant
K
c
Specication reference: 3.1.6
Many
The
reactions
reaction
are
reversible
between
and
ethanol,
C
will
H
2
to
produce
ethyl
ethanoate,
CH
reach
OH,
equilibrium
and
ethanoic
with
acid,
5
CO
3
time.
CH
CO
3
C
2
H
2
,
(an
ester)
and
water
H,
2
is
5
typical.
Synoptic link
If
ethanol
and
evaporation)
ethanoic
and
left
acid
for
are
mixed
several
days
in
a
with
ask
a
(stoppered
strong
acid
to
prevent
catalyst,
an
There is more about titrations
equilibrium
mixture
is
obtained
in
which
all
four
substances
are
in Topic 2.5, Balanced equations
present.
You
can
write:
and related calculations.
C
H
2
OH
(l)
+
CH
5
The
mixture
standard
is
may
alkali
possible
to
do
this
titration
reaction.
titration
in
the
number
of
If
several
allows
moles
if
of
it
is
for
CH
by
to
work
other
total
always
reaction
out
mixture.
the
the
are
titrating
amount
the
From
the
H
2
(l)
+
H
5
number
this
you
done
of
the
with
that
the
the
slower
of
(and
with
added).
of
is
this
the
ethanoic
calculate
from
It
equilibrium
than
moles
can
mixture
different
acid
catalyst
much
O(l)
2
water
ethanoic
acid
disturbing
is
components
volume
found
C
2
of
signicantly
reversible
us
CO
ethyl ethanoate
the
without
the
experiments
materials,
(l)
3
analysed
equilibrium
concentrations
▲ Figure 1
be
(allowing
because
acid
H
2
ethanoic acid
mixture
The
CO
3
ethanol
the
their
known).
quantities
of
starting
ratio:
Titrating the ethanoic acid
[CH
to investigate the equilibrium position
CO
3
[CH
C
2
CO
3
H
2
(l)]
5
H(l)]
2
[H
eqm
[C
eqm
O(l)]
2
H
2
eqm
OH(l)]
5
eqm
Hint
has
The concentration of a solution
a
constant
temperature.
been
value,
The
measured
provided
subscript
when
the
‘eqm’
experiments
means
equilibrium
has
that
been
are
the
done
at
the
same
concentrations
have
reached.
is the number of moles of solute
3
dissolved in 1 dm
of solution. A
For
square bracket around a formula
any
reaction
equation
in
the
that
reaches
an
equilibrium
we
can
write
the
form:
is shor thand for ‘concentration of
aA
–3
that substance in mol dm
+
bB
+
cC
xX
+
yY
+
zZ
’.
x
[X]
y
[Y]
eqm
Then
the
z
[Z]
eqm
eqm
is
expression
a
[A]
b
[B]
eqm
constant,
provided
the
c
[C]
eqm
eqm
Study tip
temperature
is
constant.
We
call
this
constant,
K
.This
expression
c
Practise calculating K
from given
can
be
applied
to
any
reversible
reaction.
data.
120
K
is
called
the
equilibrium
c
c
constant
and
is
different
for
different
reactions.
It
changes
with
Equilibria
6
temperature.
The
units
of
K
vary,
and
you
must
work
them
out
for
c
Maths link
each
reaction
by
cancelling
out
the
units
of
each
term,
for
example:
See Section 8, Mathematical
2
[C]
2A
+
B
2C
K
=
skills, if you are not sure about
c
2
[A]
[B]
cancelling units.
2
)
1
–1
Units
are:
=
2
)
The
value
of
K
is
found
=
–3
(mol
by
dm
mol
3
dm
–3
)
mol
experiment
for
dm
any
particular
reaction
at
a
c
given
Study tip
temperature.
It is acceptable to omit the ‘eqm’
subscripts unless they are
K
To nd the value of
for the
reaction
between
c
specically asked for.
ethanol
0.10
mol
allowed
and
of
to
ethanoic
ethanol
reach
is
acid
mixed
with
equilibrium.
The
3
madeup
that
to
0.033
From
this
20.0
mol
you
components
(0.020
ethanoic
at
dm
acid
work
present
mol
total
of
ethanoic
volume
of
the
acid
and
system
is
3
cm
can
0.10
out
is
)
with
water.
present
the
once
number
of
By
titration,
equilibrium
moles
of
the
it
is
is
found
reached.
other
equilibrium:
At start
Summary questions
C
H
2
OH
(l)
+
CH
5
CO
3
0.10
mol
H
(l)
CH
2
0.10
CO
3
C
2
mol
0
H
2
(l)
+
H
5
O(l)
2
mol
0
mol
1
You
know
that
there
are
0.033
mol
of
CH
CO
3
means
•
at
equilibrium.
for the equilibrium constant
This
2
for the following:
that:
there
must
also
be
0.033
mol
of
C
H
2
equation
with
•
H
the
(0.10
–
Write down the expressions
tells
you
same
0.033)
that
they
number
=
0.067
of
react
moles
mol
of
OH
1:1
of
CH
at
equilibrium.
and
you
know
we
started
CO
you
that
when
1
b
2A + B
c
2A + 2B
C
C
H
has
been
used
up.
2C
The
2
2
tells
A + B
each.)
3
equation
a
(The
5
mol
of
CH
CO
3
H
is
used
up,
1
Work out the units for K
for
c
mol
2
question 1a to c
each
of
CH
CO
3
0.067
mol
C
2
of
H
2
each
and
H
5
of
O
are
produced.
So,
there
must
be
2
3
these.
For the reaction between
ethanol and ethanoic acid, at
At equilibrium
a dierent temperature to the
C
H
2
OH
(l)
+
CH
5
CO
3
0.033
mol
H
(l)
CH
2
0.033
CO
3
mol
C
2
0.067
H
2
(l)
+
H
5
O(l)
example above, the equilibrium
2
mol
0.067
mol
mixture was found to contain
You
need
the
concentrations
of
the
components
at
equilibrium.
As
the
3
volume
of
the
system
is
0.020
dm
these
0.11
7 mol of ethanoic acid,
0.01
7 mol of ethanol, 0.083 mol
are:
ethyl ethanoate and 0.083 mol
C
H
2
OH
(l)
+
CH
5
CO
3
0.033
H
(l)
CH
2
0.033
C
2
H
2
(l)
H
mol
O(l)
2
of water.
0.067
–3
dm
+
5
0.067
–3
mol
CO
3
–3
dm
mol
–3
dm
mol
dm
a
0.020
0.020
0.020
0.020
c
b
Enter
the
concentrations
into
the
equilibrium
Calculate K
Why do you not need to
equation:
know the volume of the
[CH
CO
3
K
C
2
H
2
(l)][H
5
O(l)]
2
c
[CH
CO
3
H(l)][C
2
H
2
OH(l)]
this example?
5
–3
[0.067/0.020
mol
dm
–3
][0.067/0.020
mol
dm
]
=
K
in
system to calculate K
=
c
c
=
Is the equilibrium fur ther
4.1
c
–3
[0.033/0.020
mol
dm
–3
][0.033/0.020
mol
dm
to the right or to the left
]
compared with the worked
3
The
units
all
cancel
out,
and
the
volumes
(0.020 dm
)
cancel
out,
example above?
so
in
so
K
this
=
c
case
4.1.
In
you
this
didn’t
case,
need
K
has
to
know
no
the
volume
of
the
system,
units.
c
121
6.4
Calculations
constant
Learning
➔
objective s:
Describe how K
is used to
A
using
e q uil ibr ium
expressio ns
reaction
be
a
that
mixture
has
of
reached
reactants
equilibrium
and
products.
at
a
given
You
can
temperature
use
the
will
equilibrium
c
work out the composition of
expression
to
calculate
the
composition
of
this
mixture.
an equilibrium mixture.
Specication reference: 3.1.6
Worked
a
example: Calculating
reaction
The
H
2
OH
of
(l)
ethanol
+
CH
5
and
CO
3
ethanol
You
know
composition
of
mixture
reaction
C
the
ethanoic
H
(l)
acid
CH
2
at
CO
3
ethanoic acid
that
is:
C
2
H
2
(l)
+
H
5
O(l)
2
ethyl ethanoate
water
equilibrium:
[CH
CO
3
K
C
2
H
2
(l)][H
5
O(l)]
2
=
c
CO
[CH
3
Suppose
that
K
=
4.0
at
H(l)][C
2
the
H
2
OH(l)]
5
temperature
of
our
experiment
and
c
you
by
want
to
mixing
out
the
know
one
how
mol
of
information
Equation:
C
H
2
as
OH
much
ethanol
shown
(l) +
CH
5
start:
At
equilibrium:
You
do
1
not
produced,
of
water
ethanol
of
these
(1
know
x)
CO
how
so
you
call
also
be
(1
many
this
x.
–
H
(l)
acid
at
you
could
ethanoic
CH
produce
acid.
Set
x)
of
ethyl
equation
be
in
(l)
+
H
5
mol
x
mol
is
tells
(1
up.
–
So
x)
us
so,
x
that
mol
the
O(l)
2
water
1
x
ethanoate
doing
used
equilibrium
H
2
1
mol
and,
C
2
ethyl ethanoate
mol
The
will
CO
3
moles
produced
ethanoic
remaining
of
2
1
mol
will
and
mol
ethanoic acid
mol
–
ethanoate
one
below:
3
ethanol
At
ethyl
and
will
x
mol
mol
be
mol
of
both
amount
of
each
mol.
–3
These
to
gures
substitute
are
in
in
the
moles,
but
you
equilibrium
need
law
concentrations
expression.
in
mol
Suppose
dm
the
–3
volume
of
the
system
at
equilibrium
H
OH(l)]
was
(1
–
V
dm
.
Then:
x)
–3
[C
2
5
mol dm
=
eqm
V
(1
–
x)
–3
[CH
CO
3
H(l)]
2
=
mol dm
eqm
V
x
–3
[CH
CO
3
C
2
H
2
(l)]
5
mol dm
=
eqm
V
x
–3
[H
O(l)]
2
mol dm
=
eqm
V
These
gures
may
now
be
put
into
the
expression
for
K
:
c
x/V
×
x/V
x)/V
×
(1
=
K
c
(1
–
–
x)/V
➔
122
Equilibria
6
The
Vs
cancel,
volume
of
the
so
in
this
case
you
do
not
need
to
know
the
Study tip
actual
system.
The volume of the reaction
4.0
x
×
x
x)
×
(1
mix ture will cancel for all systems
=
(1
–
–
with equal numbers of moles of
x)
2
products and reactants, so
x
4.0
=
V is sometimes omitted. But, it
2
(1
–
x)
is always better to include V and
Taking
the
square
root
of
both
sides,
you
get:
cancel it out later, so you will not
x
2
forget it for systems where the Vs
=
(1
2(1
2
–
–
x)
=
x
2x
=
x
2
=
3x
x
=
–
x)
do not cancel out.
2
3
2
So
2
mol
of
ethyl
ethanoate
and
mol
3
the
of
water
are
produced
if
3
reaction
reaches
equilibrium,
and
the
composition
of
the
1
equilibrium
mixture
would
be:
ethanol
mol,
ethanoic
acid
3
1
2
mol,
ethyl
ethanoate
3
You
2
mol,
water
mol.
3
can
also
use
K
to
nd
3
the
amount
of
a
reactant
needed
to
give
a
c
required
amount
Worked
of
product.
example: Calculating
the
amount
of
Hint
a
Notice that this equilibrium
reactant
needed
constant does have units, which
–1
For
the
following
reaction
in
ethanol
solution,
K
=
30.0
mol
3
dm
:
c
CH
COCH
3
+
HCN
CH
3
propanone
can be deduced by cancelling the
C(CN)(OH)CH
3
hydrogen cyanide
units of the concentrations in the
3
2-hydroxy-2-methylpropanenitrile
expression for K
c
[CH
C(CN)(OH)CH
3
]
3
–1
K
=
=
30.0
mol
3
dm
c
[CH
COCH
3
][HCN]
3
3
Suppose
How
of
you
much
product
Set
out
as
are
carrying
hydrogen
if
out
cyanide
you
start
with
before
with
the
this
is
reaction
required
in
to
2.00 dm
produce
4.00 mol
of
propanone?
quantities
at
the
start
and
of
ethanol.
1.00 mol
at
equilibrium.
At
of
equilibrium,
moles
of
you
HCN
Equation:
want
CH
start:
At
equilibrium:
mol
COCH
3
At
1
of
product.
Let
x
be
the
number
required.
+
HCN
3
(4.00
–
1.00) mol
3.00 mol
CH
C(CN)(OH)CH
3
4.00 mol
x
mol
3
0 mol
(x
–
1.00) mol
1.00 mol
(x
–
1.00) mol
1.00 mol
➔
123
6.4
Calculations
using
e qui librium
con s t a n t
e xp ressi on s
These
to
are
put
in
the
the
numbers
of
equilibrium
moles,
law
but
we
need
expression.
the
The
concentrations
volume
of
the
3
solution
is
2.00 dm
–3
and
the
units
for
concentration
are
mol dm
3
so
you
next
divide
each
quantity
by
2.00 dm
3.00
–3
So,
at
equilibrium
[CH
COCH
3
]
3
mol dm
=
eqm
2.00
(x
–
1.00)
–3
[HCN]
mol dm
=
eqm
2.00
1.00
–3
[CH
C(CN)(OH)CH
3
]
3
mol dm
=
eqm
2.00
Putting
the
gures
into
the
equilibrium
expression:
–3
3
30.0
–1
mol
3
dm
=
–3
–3
×
Cancelling
through
3/2(x
30
(
–
1)
2
–
rearranging
we
–
1.00)/2.00
mol dm
have:
1
)
45(x
and
(x
=
2
1)
=
1
45x
=
46
x
=
46
=
1.02
45
So,
to
obtain
1 mol
of
product
you
must
start
with
1.02 mol
3
hydrogen
In
this
cyanide,
example
because
this
products
if
the
volume
volume
reaction
and
the
does
of
not
of
the
the
system
system
have
the
does
same
is
2.00 dm
make
a
number
difference,
of
moles
reactants.
Summary questions
1
Try reworking the problem above with the same conditions but:
3
124
a
with a volume of 1.00 dm
b
star ting with 2.0 mol of propanone
c
to produce 2.0 mol of product.
of ethanol
of
6.5
The
on
As
at
you
have
seen
equilibrium
direction
that
principle
to
is
Le
effect
Châtelier’s
reduce
predict
the
chang ing
c on dit ion s
equilibria
disturbed,
will
of
the
the
principle
states
equilibrium
disturbance.
qualitative
effect
when
position
You
of
that
can
a
moves
use
changing
Le
Learning
system
in
the
➔
Châtelier’s
temperature
objective s:
Explain how Le Châtelier ’s
principle can predict how
and
changes in conditions aect
concentration
on
the
position
of
equilibrium.
the position of equilibrium.
In
this
topic
changing
you
look
conditions
at
on
what
the
underlies
equilibrium
this
by
examining
constant
the
effect
of
K
➔
Describe how the equilibrium
constant is aected by
c
changing the conditions
The
effect of
equilibrium
Changing
constant,
the
K
.
temperature
Whether
K
changes
increases
the
or
value
of
the
decreases
equilibrium
depends
on
whether
c
reaction
Table
Specication reference: 3.1.6
constant
c
the
of a reaction.
changing temperature on the
is
exothermic
or
endothermic,
see
the
summary
in
1.
▼ Table 1
The eect of changing temperature on equilibria
Direction of
Type of
Temperature
reaction
change
Eect on
Eect on
Eect on
products
reactants
change of
K
c
equilibrium
If
endothermic
decrease
decrease
decrease
increase
moves left
endothermic
increase
increase
increase
decrease
moves right
exothermic
increase
decrease
decrease
increase
moves left
exothermic
decrease
increase
increase
decrease
moves right
the
equilibrium
constant
increases
K
in
value,
the
equilibrium
c
moves
it
to
the
decreases
backward
right,
in
that
value,
direction
is,
the
(less
the
forward
equilibrium
direction
moves
to
(more
the
product).
left,
that
is,
If
the
product).
[products]
This
is
because
the
expression
for
K
is
always
of
the
form
c
[reactants]
The
•
general
For
an
rule
For
an
for
an
reaction
decreases
endothermic
temperature
So
that:
exothermic
temperature
•
is
the
equilibrium
the
temperature
to
the
left
–
move
is
negative)
(ΔH
is
increasing
for
the
an
increasing
the
constant.
positive)
equilibrium
reaction,
the
will
(ΔH
equilibrium
reaction
increases
exothermic
the
increasing
the
constant.
the
temperature
endothermic
equilibrium
to
reaction,
the
will
move
increasing
right.
Study tip
When the value for ΔH is given for
The
effect of
changing
concentration on the
a reversible reaction, it is taken to
position of
equilibrium
refer to the for ward reaction, that
First
the
remember
that
temperature
the
equilibrium
constant
does
not
change
unless
is, left to right.
changes.
125
6.5
The
effect
of
c h ang ing
c ondition s
on
equi li bria
Look
at
the
following
example:
Study tip
C
H
2
Practise applying Le Châtelier ’s
OH
(l)
+
CH
5
CO
3
ethanol
H
(l)
CH
2
CO
3
ethanoic acid
C
2
H
2
(l)
+
H
5
O(l)
2
ethyl ethanoate
water
principle for all changes in
Le
Châtelier’s
principle
tells
you
that
the
equilibrium
will
react
to
any
conditions.
disturbance
Imagine
The
you
only
ethanol
and
add
way
of
the
forward
Let
us
see
this
such
a
ethanol,
way
ethanoic
a
new
The
as
to
thereby
concentration
with
products.
can
acid
be
reduce
increasing
reduced
producing
equilibrium
equilibrium
the
will
has
is
its
set
moved
concentration.
by
more
be
disturbance.
some
ethyl
up
to
of
with
the
the
ethanoate
relatively
right
(or
in
direction).
how
know
in
more
Eventually
the
You
moving
reacting
water.
more
by
this
works
mathematically.
that:
[CH
CO
3
K
C
2
H
2
(l)][H
5
O(l)]
2
=
c
[CH
CO
3
Remember
that
K
remains
H(l)][C
2
H
2
constant,
OH(l)]
5
provided
that
temperature
c
remains
constant.
equilibrium
the
in
ethanol
the
and
law
reacts
bottom
water,
combined
line
thus
Adding
ethanol
expression
with
of
is
to
acid
fraction.
increasing
effect
larger.
ethanoic
the
the
restore
makes
bottom
restore
reducing
This
value
the
To
the
both
produces
in
the
fraction
to
top
the
the
line
the
more
line
of
the
situation,
of
of
concentrations
ethyl
the
original
some
ethanoate
fraction.
value
of
The
K
c
K
and the
position of
equilibrium
c
The
size
of
the
equilibrium
constant
K
can
tell
us
about
the
c
composition
of
the
equilibrium
mixture.
The
equilibrium
expression
is
[products]
always
of
the
general
form
.
So:
[reactants]
•
If
K
is
much
greater
than
1,
products
predominate
over
reactants
c
and
•
If
K
the
is
equilibrium
much
less
position
than
1,
is
over
reactants
to
the
right.
predominate
and
the
c
equilibrium
position
is
over
to
the
left.
Synoptic link
10
Reactions
regarded
where
as
the
going
to
equilibrium
completion,
constant
whilst
is
greater
those
with
than
an
10
are
usually
equilibrium
Look back at Topic 5.1, Collision
–10
constant
of
less
than
10
are
regarded
as
not
taking
place
at
all.
theory, to revise activation energy
for reactions.
Catalysts
K
and the value of
c
Catalysts
have
no
effect
whatsoever
on
the
value
of
K
and
therefore
c
the
position
of
equilibrium.
This
is
because
they
affect
the
rates
of
Synoptic link
both
forward
The equilibrium constant can also
the
be calculated for gases using
rate
par tial pressures. This equilibrium
processes.
constant has the symbol K
and
activation
at
which
back
energy
reactions
for
the
equilibrium
is
equally.
reactions.
attained
–
They
They
this
is
do
do
this
by
however
important
in
reducing
affect
the
industrial
. The
p
Gaseous
equilibrium constant K
equilibria
is covered
p
in Chapter 19, Equilibrium constant
K
Reversible
solution.
reactions
These
may
include
take
many
place
in
the
reactions
of
gas
phase
industrial
as
well
as
in
importance
such
p
the
126
manufacture
of
ammonia
by
the
Haber
process
and
a
key
stage
as
Equilibria
6
of
the
obey
in
a
Contact
the
process
equilibrium
different
way
Reversible
Many
key
its
One
making
using
but
industrial
example
is
sulfuric
usually
partial
reactions
important
step.
for
law,
in
pressures
Gaseous
equilibria
concentrations
rather
than
are
also
expressed
concentrations.
industry
processes
the
acid.
their
Haber
involve
process
reversible
for
reactions
making
as
ammonia
a
from
elements:
–1
N
(g)
+
3H
2
To
(g)
gain
tells
low
us
the
high
temperature
is
hydrogen
Similar
acid
the
medium
catalyst
will
are
to
gases.
=
92
kJ
mol
to
the
ammonia,
low
down
a
and
speed
set
the
of
it
reaction.
and
the
also
(by
reaction.
Châtelier’s
are
High
principle
required.
pressure
incurs
compromise
temperature
up
Le
temperature
withstand
So
pressure
used
and
slow
energy
conditions
industrial
is
However,
requires
costs
standards)
Unconverted
in
reached
nitrogen
and
and
recycled.
considerations
and
ΔH
conversion
pressure
equipment
compressing
with
(g)
3
maximum
that
expensive
a
2NH
2
apply
hydration
of
to
the
ethene
Contact
to
give
process
for
making
sulfuric
ethanol.
Summary questions
1
A + B
C + D represents an exothermic reaction and
[C][D]
K
=
c
2
In the above expression, what would happen to
:
K
c
[A][B]
a
if the temperature were decreased
b
if more A were added to the mix ture
c
if a catalyst were added?
The reaction of ethanol with ethanoic acid produces ethyl ethanoate
and water.
C
H
2
OH(l) + CH
5
COOH(l)
CH
3
COOC
3
H
2
(l) + H
5
O(l)
2
A student suggested that the yield of ethyl ethanoate, CH
COOC
3
H
2
, could
5
be increased by removing the water as it was formed.
Explain, using the idea of
K
, why this suggestion is sensible.
c
3
These questions are about reversible reactions. Give the correct word
from increases/decreases/does not change to ll in the blank for
each statement.
a
In an endothermic reaction
K
when the temperature is
c
increased.
b
In an endothermic reaction
when the concentration of the
K
c
reactants
c
is decreased.
In an exothermic reaction
when the temperature is
K
c
decreased.
d
In an exothermic reaction
when the concentration of the
K
c
reactants
e
is increased.
If a suitable catalyst is added to the reaction
K
c
127
Practice
questions
1
Methanol
can
be
synthesised
from
carbon
monoxide
by
the
reversible
reaction
shownbelow.
–1
CO(g)
+
2H
(g)
CH
2
The
process
of
copper-containing
a
(a)
By
operates
reference
to
at
a
pressure
catalyst.
rates
ΔH
OH(g)
=
–91
kJ
mol
3
and
of
This
5
MPa
and
reaction
a
can
concentrations,
temperature
reach
explain
of
dynamic
the
700
K
in
the
presence
equilibrium.
meaning
of
the
term
dynamic
equilibrium
(b)
Explain
why
a
(c)
Suggest
two
reasons
than
5
MPa
high
would
yield
of
why
be
methanol
the
very
is
favoured
operation
of
this
by
high
process
at
(2
marks)
(2
marks)
pressure.
a
pressure
much
higher
expensive.
(2
(d)
State
and
the
effect
explain
of
your
an
increase
in
temperature
on
the
equilibrium
yield
of
answer.
(3
(e)
If
a
catalyst
have
to
be
marks)
methanol
were
not
greater
used
than
in
700
this
K.
process,
Suggest
the
why
operating
an
temperature
increased
marks)
would
temperature
would
berequired.
(1
mark)
AQA,
2
At
high
temperatures,
reversible
reaction
as
nitrogen
shown
is
in
oxidised
the
by
equation
oxygen
to
form
nitrogen
monoxide
in
2003
a
below.
–1
N
(g)
+
O
2
(a)
In
terms
(b)
State
of
(g)
ΔH
2NO(g)
=
+180
kJ
mol
2
electrons,
give
the
meaning
of
the
term
oxidation
(1
in
and
explain
temperature,
the
on
effect
the
of
yield
an
of
increase
nitrogen
in
pressure,
monoxide
in
and
the
the
effect
above
of
an
mark)
increase
equilibrium.
(6
marks)
AQA,
3
Hydrogen
is
produced
on
an
industrial
scale
from
methane
as
shown
by
the
2006
equation
below.
–1
CH
(g)
+
H
4
(a)
State
(b)
The
Le
O(g)
CO(g)
+
3H
2
ΔH
(g)
=
+205
kJ
mol
2
Châtelier’s
principle.
(1
following
what
Le
would
changes
happen
Châtelier’s
are
to
made
the
principle
(i)
The
overall
(ii)
The
concentration
to
pressure
yield
to
of
explain
is
of
this
reaction
hydrogen
your
at
from
equilibrium.
a
given
In
each
amount
of
case,
methane.
At
equilibrium,
typical
a
industrial
Suggest
two
high
increased.
steam
yield
process,
reasons
of
the
why
Use
answer.
in
the
reaction
mixture
is
increased.
(6
(c)
mark)
predict
hydrogen
operating
is
favoured
temperature
temperatures
higher
than
by
is
high
temperature.
usually
this
are
less
not
than
In
marks)
a
1200
K.
used.
(2
marks)
AQA,
4
The
equation
for
the
formation
N
of
ammonia
(g)
+
3H
2
Experiment
aconstant
Curve
A
Curves
3
128
mol
was
shows
B,
of
A
carried
temperature
C,
how
and
D
hydrogen.
the
refer
In
out
and
a
to
each
shown
2NH
2
starting
pressure
number
is
(g)
of
similar
1
20
moles
(g)
of
mol
of
nitrogen
and
3
mol
of
hydrogen
MPa.
ammonia
experiments,
experiment
below.
3
with
of
2004
different
present
starting
with
conditions
changed
1
mol
were
of
used.
with
time.
nitrogen
and
at
Chapter
6
Equilibria
B
C
moles
A
of
D
ammonia
time
(a)
On
is
a
copy
rst
(b)
State
(c)
Use
an
of
curve
reached.
Le
Le
A,
Label
Châtelier’s
Châtelier’s
experiment
pressure.
mark
this
point
that
represents
the
time
at
which
equilibrium
X
(1
mark)
(1
mark)
principle.
principle
carried
Explain
the
point
out
why
to
at
this
identify
the
which
same
curve
is
one
of
temperature
different
from
the
as
curves
B,
C,
experiment
A
curve
or
D
but
represents
at
a
higher
A
(4
(d)
Identify
which
conditions
reaction
are
one
the
mixture.
of
the
same
curves
as
Explain
in
or
D
represents
experiment
A
except
your
B,
C,
choice
of
an
that
experiment
a
catalyst
is
in
marks)
which
added
to
the
the
curve.
(3
marks)
AQA,
5
The
reaction
processes.
of
methane
Under
with
certain
steam
conditions
produces
the
hydrogen
following
for
reaction
use
in
many
2005
industrial
occurs.
–1
(g)
CH
+
2H
4
(a)
Initially,
with
0.25
(i)
a
1.0
mol
catalyst
mol
O(g)
CO
2
of
of
methane
until
carbon
Calculate
the
equilibrium
(g)
+
4H
2
and
equilibrium
ΔH
(g)
=
+165
kJ
mol
2
2.0
was
mol
of
steam
established.
were
The
placed
in
a
equilibrium
ask
and
mixture
heated
contained
dioxide.
amounts,
in
moles,
of
methane,
steam,
and
hydrogen
in
the
mixture.
(3
mar ks)
3
(ii)
The
volume
of
the
ask
was
5.0
dm
.
Calculate
the
concentration,
–3
in
mol
dm
,
of
methane
in
the
equilibrium
mixture.
(1
(b)
The
table
below
equilibrium
shows
mixture
the
in
equilibrium
the
same
CH
gas
ask
(g)
and
H
4
concentration
at
of
each
temperature
O(g)
CO
2
(g)
gas
in
a
mar k)
different
T
H
2
(g)
2
–3
0.10
concentration / mol dm
(i)
Write
an
expression
for
the
0.48
0.15
equilibrium
0.25
constant,
K
,
for
this
reaction.
c
(1
(ii)
Calculate
a
value
for
at
K
temperature
T
and
give
its
mar k)
units.
c
(c)
The
mixture
in
part
( b)
was
placed
in
a
ask
of
volume
greater
(3
mar ks)
(3
mar ks)
(2
marks)
than
3
5.0 dm
State
(d)
and
and
Explain
in
part
allowed
explain
why
(b)
the
to
the
reach
effect
amount
reaches
of
equilibrium
on
the
hydrogen
equilibrium
at
at
amount
a
temperature
of
decreases
lower
T.
hydrogen.
when
the
mixture
temperature.
AQA,
2010
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
129
7
Oxidation,
redox
7.1
Learning
➔
reductio n,
reactions
Oxidation
objective s:
and
Redox
and
re d uc t ion
reactions
Dene a redox reaction in
The
terms of oxygen or hydrogen
oxidation
word
redox
was
is
short
used
for
for
reduction–oxidation.
reactions
in
which
Historically,
oxygen
was
added.
transfer.
In
➔
Dene a redox reaction in
this
called
reaction
an
copper
oxidising
has
been
oxidised
terms of electron transfer.
copper
oxide.
Oxygen
is
1
Cu(s)
+
O
2
➔
to
agent
(g)
CuO(s)
➝
2
Dene a half equation.
Reduction
described
a
reaction
in
which
oxygen
was
removed.
Specication reference: 3.1.7
In
this
reaction
reducing
copper
oxide
has
been
reduced
and
+
H
(g)
➝
Cu(s)
+
H
2
hydrogen
hydrogen
In
this
added
was
was
often
called
reaction
to
used
to
remove
reverse,
chlorine
where
and
describing
a
much
electrons,
redox
also
has
been
(g)
+
what
more
and
called
show
reduced
H
and
at
because
addition
of
hydrogen
has
been
a
1:
–
electrons
When
is
in
it
above
gains
You
into
can
is
reactions,
oxidised
electrons.
of
two
oxidation.
reactions
the
movement
reaction
called
something
reduced
the
was
redox
reactions.
redox
of
2HCl(g)
➝
removed,
the
involve
transfer
loss
the
to
something
example
again
was
picture.
always
separating
gain
Worked
copper
happens
when
(g)
electrons
general
electron
by
the
Look
the
2
hydrogen
losing
reactions
electrons
oxygen,
it.
Gaining
get
O(l)
2
2
By
the
reduction.
Cl
The
is
agent
CuO(s)
As
hydrogen
half
the
you
loses
Since
electrons
see
it
they
transfer
equations
are
of
that
electrons.
Half
reaction
equations
between
copper
and
oxygen
to
form
oxide:
1
Cu
+
O
Copper
oxide
is
an
ionic
CuO
➝
2
2
compound
so
you
2+
balanced
show
symbol
the
ions
equation
present
in
using
copper
(Cu
+
Next
look
at
the
copper.
It
+
O
2+
O
2
➝
write
the
)
(instead
of
CuO)
to
oxide:
1
Cu
can
2–
(Cu
2–
+
O
)
2
has
lost
two
electrons
so
it
has
been
oxidised.
2+
Cu
130
–
2e
➝
Cu
2+
or
Cu
➝
Cu
+
2e
➔
Oxidation,
reductio n ,
and
red ox
re act io ns
7
This
is
a
half
equation.
plus
electrons
rather
It
than
is
usual
minus
to
write
half
equations
electrons,
that
is:
with
2+
Cu
➝
Cu
+
2e
rather
than
2+
Cu
Next
look
been
reduced:
at
the
oxygen.
It
–
2e
has
Cu
➝
gained
two
electrons
so
it
has
1
2–
O
2
If
you
add
original
the
two
equation.
half
(g)
+
2e
➝
O
2
equations
Notice
that
the
together,
you
numbers
of
end
up
with
electrons
the
cancel
out.
2+
Cu
Cu
➝
1
2–
O
2
O
➝
2
1
Cu(s)
+
2+
O
Worked
When
example
copper
magnesium
oxide
oxide
2:
Half
reacts
are
(g)
(Cu
➝
2–
+
O
)(s)
2
2
equations
with
magnesium,
copper
produced:
CuO(s)
+
Mg(s)
MgO(s)
➝
+
Cu(s)
2+
Write
the
equation
with
copper
2+
magnesium
oxide
as
Look
at
the
as
O
It
2–
(Cu
+
O
)
to
show
+
the
2–
+
copper.
oxide
O
ions
2+
)
has
)
and
2–
(Mg
2+
(Cu
and
+
Mg
Cu
➝
gained
two
+
present.
2–
(Mg
+
electrons
O
so
)
it
has
been
reduced.
2+
Cu
Look
at
the
magnesium.
It
+
has
2e
lost
Cu
➝
electrons
so
it
has
been
oxidised.
2+
Mg
Mg
➝
+
2e
2–
Notice
that
spectator
If
you
add
the
O
ion
takes
no
part
in
the
reaction.
It
is
called
a
ion
these
half
equations
you
get:
2+
Cu
2+
+
Mg
➝
Cu
+
Mg
Hint
This
is
the
ionic
equation
for
the
redox
reaction.
The phrase OIL RIG makes the
The
denition
of
oxidation
and
reduction
now
used
is:
denition of oxidation and reduction
easy to remember.
Oxidation
Is
Loss
Reduction
Is
Gain
of
of
electrons.
Oxidation
electrons.
Is
By
this
denition,
magnesium
is
oxidised
by
anything
that
removes
Loss (of electrons)
electrons
from
it
(not
just
oxygen)
leaving
a
positive
ion.
For
Reduction
example,
chlorine
oxidises
magnesium:
Is
2+
Mg(s)
+
Cl
(g)
2
➝
(Mg
+
2Cl
)(s)
➔
Gain (of electrons)
131
7.1
Oxidation
and
re d uc t ion
Look
been
at
the
magnesium.
It
has
lost
electrons
and
has
therefore
oxidised.
2+
Mg
Look
been
at
the
chlorine.
It
has
Mg
➝
+
gained
2e
electrons
and
has
therefore
reduced.
Cl
+
2e
2Cl
➝
2
And
adding
cancel
the
two
half
equations
together,
the
electrons
out:
2+
Mg(s)
+
Cl
(g)
➝
(Mg
+
2Cl
)(s)
2
You
the
may
nd
transfer
Figure
that
of
adding
electrons,
arrows
helps
to
the
keep
equation,
track
of
them,
of
2
+
2
+
O
2
2+
(Cu
(g)
2–
(Mg
+
O
) (s)
O
) (s)
gain
+
H
(g)
Cu (s)
+
H
of
2
must
Oxidising
reaction,
be
and
from
•
reducing
•
oxidising
O (l)
electrons
ions
if
reduced
reducing
the
agents
agents
above
give
one
species
(gains
is
oxidised
electrons),
them).
agents
that:
away
accept
electrons
–
they
are
electron
electrons.
The following questions are about the reaction:
2+
Ca(s) + Br
(g) ➝ (Ca
+ 2Br
)(s)
2
132
reduced
(loses
Summary questions
1
are
Writing the electrons that are transferred helps to keep track of them
chemical
another
oxidised
2
copper
▲ Figure 1
is
2–
+
2
follows
in
2
magnesium
It
shown
electrons
1
Mg (s)
a
as
show
1.
loss
In
which
a
Which element has gained electrons?
b
Which element has lost electrons?
c
Which element has been oxidised?
d
Which element has been reduced?
e
Write the half equations for these redox reactions.
f
What is the oxidising agent?
g
What is the reducing agent?
donors
7.2
Oxidation
Oxidation
states
states
Learning
Oxidation
has
been
states
reduced
oxidation
are
in
a
used
to
redox
see
what
reaction.
has
been
Oxidation
oxidised
states
are
and
also
objective s:
what
➔
Dene an oxidation state.
➔
Describe how oxidation states
called
numbers
are worked out.
The
idea of oxidation
states
Specication reference: 3.1.7
Each
element
compound
it
has
state.
of
lost
In
the
or
a
in
oxidation
molecule,
electrons
Every
compound
gained,
the
state
in
its
given
with
oxidation
is
an
simply
elements
element
element
is
compared
between
electronegative
•
a
of
given
oxidation
tells
the
state
us
tells
different
the
how
element
us
state
many
in
its
about
In
an
electrons
the
oxidation
has
an
ionic
uncombined
distribution
electronegativity.
negative
uncombined
state.
The
more
state.
oxidation
state
ofzero.
•
A
positive
number
shows
that
the
element
has
lost
electrons
and
2+
has
therefore
state
•
A
of
has
state
•
The
of
oxidised.
For
example,
Mg
has
an
oxidation
+2.
negative
and
been
number
therefore
shows
been
that
the
reduced.
element
For
has
example,
gained
Cl
has
electrons
an
oxidation
–1.
more
oxidised.
positive
The
the
more
number,
negative
the
the
more
number,
the
element
the
more
it
has
has
been
been
reduced.
•
The
numbers
always
have
a
Rules for nding oxidation
The
following
rules
1
Uncombined
2
Some
gives
▼ Table 1
the
allow
elements
elements
compounds.
will
always
Others
oxidation
+
–
sign
unless
they
are
zero.
states
you
have
to
work
oxidation
have
usually
states
or
the
same
have
of
the
these
out
oxidation
state
of
0.
oxidation
same
states:
state
oxidation
in
all
state.
their
Table
1
elements.
The usual oxidation states of some elements
Element
Oxidation state in compound
Example
+1 (except in metal hydrides, e.g., NaH,
HCl
hydrogen, H
where it is –1)
Group 1
always +1
NaCl
Group 2
always +2
CaCl
aluminium, Al
always +3
AlCl
2
3
–2 (except in peroxides where it is –1, and
Na
oxygen, O
O
2
the compound OF
, where it is +2)
2
uorine, F
Always –1
NaF
–1 (except in compounds with F and O,
NaCl
chlorine, Cl
where it has positive values)
3
The
all
sum
of
all
compounds
the
are
oxidation
states
electrically
in
a
compound
equals
0,
since
neutral.
133
7.2
Oxidation
stat e s
+
4
The
sum
of
the
oxidation
states
of
a
complex
ion,
such
as
NH
or
4
2–
SO
,
equals
the
charge
on
the
ion.
4
5
In
a
compound
negative
the
oxidation
most
electronegative
Working out oxidation
in
element
always
has
a
state.
states of
elements
compounds
Start
with
states
any
you
other
the
correct
know
formula.
from
element.
the
Some
Look
rules.
for
Then
examples
the
elements
deduce
are
the
shown
whose
oxidation
oxidation
states
of
below.
Phosphorus pentachloride, PCl
5
Study tip
Chlorine
You should know the rules for
to
make
has
the
an
oxidation
sum
of
the
state
of
oxidation
–1,
so
states
the
phosphorus
must
be
+5,
zero.
nding oxidation states.
Ammonia, NH
3
Hydrogen
to
make
has
the
an
electronegative
oxidation
oxidation
sum
of
the
than
state
of
oxidation
hydrogen,
+1,
so
states
so
the
zero.
hydrogen
nitrogen
Also,
must
must
be
nitrogen
have
a
is
–3,
more
positive
state.
Nitric acid, HNO
3
Each
oxygen
Hydrogen
So
the
Notice
has
has
an
nitrogen
that
oxidation
oxidation
must
be
nitrogen
compounds.
combined
an
Here
with
a
to
of
make
have
nitrogen
more
state
+5,
may
state
of
–2,
a
the
sum
of
the
in
total.
oxidation
oxidation
positive
electronegative
Hydrogen sulde, H
–6
+1.
different
has
making
states
oxidation
element,
state
in
states
zero.
different
because
it
is
oxygen.
S
2
Hydrogen
make
the
has
sum
an
of
oxidation
the
state
oxidation
of
+1,
states
so
the
sulfur
must
be
–2,
to
zero.
2–
Sulfate ion, SO
4
Each
So
oxygen
the
equal
sulfur
to
Notice
the
that
has
an
must
be
charge
sulfur
oxidation
+6,
on
may
to
the
state
make
of
the
–2,
making
sum
of
the
–8
in
total.
oxidation
states
ion.
have
different
oxidation
states
in
different
compounds.
Black copper oxide, CuO
Oxygen
make
has
the
an
sum
oxidation
of
the
state
of
oxidation
Red copper oxide, Cu
–2,
states
so
the
copper
must
be
+2,
to
zero.
O
2
Oxygen
make
has
the
Oxidation
similar
▲ Figure 1
The two oxides of copper –
So,
an
sum
oxidation
of
states
the
are
compounds
black
copper
state
of
oxidation
written
in
in
which
oxide
is
–2,
states
so
Roman
the
metal
copper(II)
each
copper
must
numerals
has
oxide
a
to
oxide (right)
134
oxide.
These
compounds
+1,
distinguish
different
to
are
and
shown
red
in
between
oxidation
copper
oxide
copper(II) oxide (left) and copper(I)
copper(I)
be
zero.
Figure
1.
state.
is
Oxidation,
reductio n ,
and
red ox
re act io ns
7
+
Chlorine – an element with many oxidation states
The element chlorine, Cl
, has an oxidation state of zero by denition. When
2
combined with other elements, it can exhibit several dierent oxidation
states from −1 to +7
.
When combined with a metal, chlorine forms ionic compounds which
contain the Cl
ion whose oxidation state is −1. However, chlorine forms
a number of compounds which also contain oxygen. Oxygen is more
electronegative than chlorine and so the chlorine forms positive oxidation
states. For example:
Formula
Oxidation state
Name
of chlorine
0
Cl
Chlorine
2
NaCl
−1
Sodium chloride
NaClO
+1
Sodium hypochlorite (sodium chlorate(I))
+3
Sodium chlorite (sodium chlorate(III))
+4
Chlorine dioxide
+5
Sodium chlorate (sodium chlorate(V))
+6
Dichlorine hexoxide
+7
Sodium perchlorate (sodium chlorate(VII))
NaClO
2
ClO
2
NaClO
3
Cl
O
2
6
NaClO
4
1
Suggest why uorine does not form any compounds in which it has a
positive oxidation state.
.sdnuopmoc sti lla ni etats
noitadixo evitagen a sah ti os tnemele evitagenortcele tsom eht si eniruolF
Summary questions
1
Work out the oxidation states of each element in the following compounds:
a
PbCl
b
CCl
c
NaNO
2
4
3
2
In the reaction: CuO + Mg
➝ Cu + MgO, what are the oxidation states of
oxygen before and after the reaction?
3
In the reaction: 2Cu + O
➝ 2CuO, what are the oxidation states of
2
oxygen before and after the reaction?
1
4
In the reaction: FeCl
+
2
2
Cl
➝ FeCl
2
,
3
what are the oxidation states of iron before and after the reaction?
5
Give the oxidation state of the following:
3–
a
P in PO
4
b
N in NO
3
+
c
N in NH
4
135
7.3
Learning
➔
➔
Redox
objective s:
equations
Using oxidation
Explain how half equations
You
are used to balance an
been
equation.
considering
Deduce half equations from a
redox equation.
saw
in
Topic
oxidised
Remember
gain
of
7.1
and
that
you
which
electron
that
states
has
redox
can
work
been
equations
out
reduced
which
in
a
element
redox
has
reaction
by
transfer.
oxidation
electrons
in
is
loss
of
electrons
(OIL)
and
reduction
is
(RIG).
Specication reference: 3.1.7
You
can
also
use
oxidation
states
to
help
you
to
understand
redox
reactions.
When
an
element
is
reduced,
it
gains
electrons
and
its
oxidation
state
Hint
goes
down.
state
has
In
the
gone
reaction
down
from
below,
+3
to
iron
+2,
is
reduced
whilst
because
iodide
is
its
oxidation
oxidised:
Another way of working is to
remember that when an element is
+3
–1
+2
0
1
reduced it gains electrons and its
3+
2+
Fe
+
I
Fe
➝
+
I
2
2
oxidation state is reduced.
Even
3+
For example, in:
M
in
complicated
reactions,
you
can
see
which
element
has
been
2+
➝
M
oxidised
the number of plusses has been
and
which
has
been
reduced
when
you
put
in
the
oxidation
states:
reduced so M has been reduced.
+5
2+
It follows that for:
M
➝
–2
+1
+4
–2
0
+6
–2
+1
2–
3+
2IO
M
+
5HSO
3
I
➝
3
+
2
+1
–2
+
5SO
+
3H
+
H
4
O
2
the number of plusses has been
Iodine
in
IO
is
reduced
(+5
to
0)
and
sulfur
in
HSO
3
is
oxidised
3
increased so M has been oxidised.
(+4
to
+6).
The
Balancing
You
can
redox
For
•
redox
the
idea
states
of
all
the
other
atoms
have
not
changed.
reactions
of
oxidation
states
to
help
balance
equations
for
reactions.
an
equation
to
the
numbers
of
must
•
use
oxidation
the
be
total
the
be
balanced:
atoms
of
each
element
on
each
side
of
the
equation
same
charge
on
each
side
of
the
equation
must
be
the
same.
Example 1: the thermite reaction
This
is
a
iron(III)
The
strongly
oxide
exothermic
to
unbalanced
produce
equation
Fe
O
2
Write
the
oxidation
+3
2
If
you
have
Each
can
look
at
changed
iron
write
the
(s)
+
Al(s)
above
+
Fe(l)
➝
each
0
0
Fe(l)
+
Al
O
➝
+3
+
been
weld
reacts
with
railway
lines.
you
can
state.
reduced
see
The
by
that
only
oxygen
gaining
equation:
3+
+
3e
(s)
3
–2
Al
O
2
oxidation
has
to
element:
Al(s)
Fe
136
aluminium
used
2
equation
half
was
3
their
atom
the
(s)
which
It
3
states
O
in
iron.
is:
–2
Fe
reaction
molten
➝
Fe
(s)
3
the
is
iron
and
aluminium
unchanged.
three
electrons
so
you
Oxidation,
reductio n ,
and
red ox
re act io ns
7
Each
aluminium
atom
has
been
oxidised
by
losing
three
electrons:
3+
Al
In
of
the
reaction,
electrons
the
lost.
aluminium
atoms
started
with
atoms.
The
number
This
two
as
iron
iron
balanced
Fe
of
means
O
there
atoms.
atoms,
(s)
+
+
3e
electrons
that
so
equation
2
Al
➝
(The
you
is
gained
must
oxygen
must
must
be
the
is
also
a
equal
same
the
spectator
have
two
number
number
ion.)
of
You
aluminium
therefore:
2Al(s)
2Fe(l)
➝
+
Al
3
O
2
(s)
3
Example 2: aqueous solutions
Sometimes
in
are
oxidised
neither
aqueous
solutions,
nor
species
reduced.
You
take
must
part
in
balance
redox
them
reactions
but
separately.
+
These
include
water
ions(in
alkaline
species
that
Suppose
are
you
molecules,
solution).
oxidised
want
to
H
ions
Oxidation
or
(in
states
acid
solution),
only
help
us
and
to
OH
balance
the
reduced.
balance
the
following
equation,
where
dark
purple
2+
manganate(VII)
ions
react
in
acid
solution
with
Fe
ions
to
produce
pale
▲ Figure 1
2+
pink
Mn
A demonstration of the
3+
ions
and
Fe
ions.
thermite reaction
The
unbalanced
equation
is:
2+
MnO
+
+
Fe
+
H
2+
➝
3+
Mn
+
Fe
+
H
4
1
Write
+7
the
oxidation
–2
+2
state
+
above
+1
2+
MnO
+
each
+2
+
Fe
+3
2+
H
Mn
➝
element.
+1
–2
H
O
3+
+
Fe
+
4
2
Identify
been
O
2
2
the
species
that
has
been
oxidised
and
the
species
that
has
reduced.
+7
+2
2+
MnO
Mn
➝
Manganese
has
been
reduced
from
+7
to
+2
4
therefore
ve
electrons
must
be
gained.
2+
MnO
+
5e
➝
Mn
(this
equation
is
not
chemically
balanced)
4
+2
+3
2+
3+
Fe
Fe
➝
must
be
Fe
has
been
oxidised
2+
order
this
+2
to
+3
so
one
electron
3+
Fe
In
from
lost.
to
step
balance
must
be
the
Fe
➝
number
multiplied
by
of
e
electrons
that
are
transferred,
5:
2+
5Fe
+
3+
➝
5Fe
+
5e
2+
So,
you
know
that
there
are
5Fe
ions
to
every
MnO
ion.
4
3
Include
the
this
redox
information
in
the
unbalanced
equation,
2+
+
5Fe
+
+
H
2+
➝
Mn
3+
+
5Fe
+
H
4
4
equation
Balance
nor
the
is
In
side,
O
2
still
not
remaining
reduced.
left-hand
balance
process.
MnO
(this
to
order
you
chemically
atoms,
to
‘use
need
4H
those
up’
O
balanced)
that
the
on
are
four
the
neither
oxygen
right-hand
oxidised
atoms
side,
on
the
which
will
2
+
in
turn
require
8H
MnO
+
on
the
left-hand
2+
5Fe
+
+
8H
side.
2+
➝
Mn
3+
+
5Fe
+
4H
4
Notice
that
this
O
2
equation
is
balanced
for
both
atoms
and
charge.
137
7.3
Redox
+
equations
Dispropor tionation
In some chemical reactions, atoms of the same element
1
Suggest why hairdressers, who use hydrogen
can be both oxidised and reduced. For example, hydrogen
peroxide as a bleach, store it in the fridge and in
peroxide decomposes to oxygen and water.
bottles with a small hole in the cap.
−2
−1
0
2
2H
O
2
➝
2H
2
Here is another dispropor tionation reaction.
O + O
2
2
Cu
O ➝ Cu +CuO
2
Work out the oxidation states of each atom
Check that you can work out the oxidation state of each
using the rules in Topic 7
.2. Which element
oxygen (shown in red) using the rules in Topic 7
.2.
dispropor tionates?
Two
of
have
the
oxygen
increased
reduced
atoms
their
in
the
oxidation
hyd rogen
state
a nd
peroxide
two
)2+ dna 0 ot 1+( uC
have
2
.eloh llams hguorht epacse nac noitisopmoced morf
it.
2
decudorp sesaG .
Half
equations from the
2
O
H fo noitisopmoced nwod swolS
balanced
1
equation
Example 1
The
reaction
nitrogen
between
monoxide.
copper
The
and
cold
balanced
dilute
symbol
+
3Cu
+
8H
nitric
acid
equation
is
produces
work
out
elements
1
Put
the
have
in
the
half
been
2NO
➝
equations,
oxidised
numbers
0
+1
+5
3Cu
+
2NO
+
4H
and
you
and
look
–2
+
8H
rst
which
for
a
need
have
change
+2
+2
to
2
Now
Each
six
has
work
of
in
−2
the
+1
which
reduced.
oxidation
states:
–2
2+
+
2NO
been
out
the
know
been
3Cu
➝
+
2NO
+
H
3
Copper
O
2
+
3Cu
gas
2+
+
3
To
the
shown:
oxidised
the
three
O
2
half
and
nitrogen
has
been
reduced.
equations.
copper
atoms
loses
two
electrons,
a
total
of
electrons:
2+
3Cu
The
two
nitrogen
atoms
3Cu
➝
NO
+
have
6e
each
gained
three
electrons
so
3
the
half
equation
must
be
based
2NO
+
on:
6e
2NO
➝
3
This
half
oxygen
side.
no
equation
atoms
The
on
total
is
not
the
charge
charge.
Look
ions
the
at
balanced
left-hand
on
the
the
left
original
for
side
is
atoms
and
–8
or
only
whereas
equation.
charge.
two
You
on
the
There
the
are
right-hand
need
to
six
right-hand
include
side
the
has
eight
+
H
four
on
oxygen
left-hand
atoms
that
side
are
of
our
half
equation
unaccounted
for)
and
(to
use
also
up
the
the
four
extra
H
O
on
2
the
right-hand
half
equation
side.
This
also
accounts
for
the
charge,
so
the
complete
is:
+
2NO
+
8H
+
6e
➝
2NO
+
4H
3
This
equation
is
balanced
O
2
in
both
atoms
and
charge.
Example 2
The
the
138
reaction
gas
between
nitrogen
copper
dioxide.
and
hot
concentrated
nitric
acid
produces
Oxidation,
reductio n ,
and
red ox
re act io ns
7
1
The
balanced
symbol
equation
is
shown
with
the
oxidation
states
included:
0
+1
+5
–2
+2
+1
+
Cu
+
−2
+
2NO
Cu
➝
+
2H
3
Copper
2
Now
has
work
Copper
been
out
has
+4
–2
2+
4H
lost
half
two
+
2NO
2
oxidised
the
O
and
2
nitrogen
has
been
reduced.
equations.
electrons
so
the
half
equation
is:
2+
Cu
Nitrogen
in
NO
has
Cu
➝
gained
an
+
2e
electron
so
the
half
are
an
equation
must
3
be
based
on:
2NO
+
2e
2NO
➝
3
This
is
not
oxygens
balanced
on
the
for
charge
left-hand
side
2
or
atoms.
and
a
There
total
charge
of
extra
−4
two
whereas
the
+
right-hand
left-hand
charge.
side
side
You
is
to
neutral.
use
then
up
need
You
the
to
need
extra
add
to
add
oxygen.
two
H
O
to
the
four
These
the
H
will
ions
also
right-hand
to
the
balance
the
side.
2
The
half
equation
is:
+
2NO
+
4H
+
2e
2H
➝
3
Note
out
that
and
if
you
you
get
add
the
back
to
O
+
2NO
2
half
the
equations
original
2
together,
balanced
the
electrons
cancel
equation.
Summary questions
1
The following questions are about the equation:
1
2+
3+
+
Fe
Cl
2
2
➝
Fe
+ Cl
2
a
Write the oxidation states for each element.
b
Which element has been oxidised? Explain your answer.
c
Which element has been reduced? Explain your answer.
d
Write the half equations for the reaction.
a
Use oxidation states to balance the following equations:
i
Cl
+ NaOH ➝ NaClO
2
ii
Sn + HNO
➝
3
b
+ NaCl + H
3
SnO
+ NO
2
Write the half equations for
O
2
+ H
2
O
2
par t a i and ii
139
Practice
questions
1
(a)
In
terms
of
electron
transfer,
what
does
the
reducing
(b)
What
is
the
oxidation
state
of
an
atom
in
an
(c)
Deduce
the
oxidation
state
of
nitrogen
in
each
(i)
NCl
(ii)
Mg
agent
uncombined
of
the
do
in
a
redox
reaction?
(1
mark)
(1
mark)
element?
following
compounds.
3
N
3
(iii)
NH
2
OH
2
(3
(d)
Lead(IV)
oxide,
PbO
,
reacts
with
concentrated
hydrochloric
acid
to
produce
marks)
chlorine,
2
2+
lead(II)
ions,
,
Pb
and
water.
2+
(i)
Write
a
half
equation
for
the
formation
of
and
Pb
water
from
PbO
in
the
2
+
presence
(ii)
Write
(iii)
Hence
a
of
ions.
H
half
equation
deduce
hydrochloric
an
for
the
equation
acid
is
added
formation
for
to
the
of
chlorine
reaction
lead(IV)
which
oxide,
from
chloride
occurs
when
ions.
concentrated
PbO
2
(3
marks)
AQA,
2
Chlorine
and
(a)
Dene
(b)
In
bromine
an
are
oxidising
both
agent
oxidising
in
terms
of
2002
agents.
electrons.
(1
mark)
2–
aqueous
solution,
bromine
oxidises
sulfur
dioxide,
SO
,
to
sulfate
ions,
SO
2
4
2–
(i)
Deduce
the
oxidation
state
of
sulfur
in
SO
and
in
SO
2
4
(ii)
Deduce
a
half
equation
for
the
reduction
of
bromine
(iii)
Deduce
a
half
equation
for
the
oxidation
of
SO
in
in
aqueous
aqueous
solution.
solution
forming
2
2–
SO
+
and
H
ions.
4
(iv)
Use
these
between
two
half
aqueous
equations
bromine
to
and
construct
sulfur
an
overall
equation
for
the
reaction
dioxide.
(5
marks)
AQA,
3
(a)
By
referring
to
electrons,
explain
the
meaning
of
the
term
oxidising
(1
(b)
For
the
element
oxidation
X
in
the
ionic
compound
MX,
explain
the
meaning
of
the
mark)
term
state
(1
(c)
2004
agent
Complete
elements
the
in
table
the
below
given
ion
by
or
deducing
the
oxidation
state
of
each
of
the
mark)
stated
compound.
Oxidation state
2–
carbon in CO
3
+
phosphorus in PCl
4
nitrogen in Mg
N
3
2
(3
(d)
In
acidied
aqueous
solution,
nitrate
ions,
NO
react
with
copper
metal
marks)
forming
3
nitrogen
monoxide,
NO,
and
copper(
ii)
ions.
(i)
Write
a
half
equation
for
the
oxidation
(ii)
Write
a
half
equation
for
the
reduction,
nitrogen
(iii)
Write
an
of
copper
in
an
to
copper(
acidied
ii)
ions.
solution,
of
nitrate
ions
overall
equation
for
this
reaction.
(3
marks)
AQA,
4
(a)
Nitrogen
to
monoxide.
monoxide,
NO,
is
formed
when
silver
metal
reduces
nitrate
ions,
2005
NO
3
in
acid
solution.
Deduce
the
oxidation
state
of
nitrogen
in
NO
and
in
NO
3
(b)
Write
a
half
equation
for
the
reduction
of
ions
NO
3
monoxide
140
and
water.
in
acid
solution
to
form
nitrogen
Chapter
7
Oxidation,
reductio n ,
a nd
redox
re a ct i ons
+
(c)
Write
a
(d)
Hence,
deduce
ions
acid
in
half
equation
an
for
the
overall
oxidation
equation
for
of
silver
the
metal
reaction
to
Ag
(aq)
between
ions.
silver
metal
and
nitrate
solution.
(5
marks)
AQA,
5
Iodine
reacts
with
concentrated
nitric
acid
to
produce
nitrogen
dioxide,
2006
NO
2
(a)
(i)
Give
I
the
state
of
each
of
the
following.
(2
Complete
+
the
balancing
10HNO
of
industry,
iodine
is
the
HIO
➝
3
In
in
marks)
3
I2
(b)
iodine
HIO
2,
(ii)
oxidation
following
+
NO
3
produced
equation.
+
H
2
from
the
O
(1
mark)
2
NaIO
that
remains
after
sodium
nitrate
has
3
been
The
crystallised
nal
stage
from
the
involves
mineral
the
Chile
reaction
saltpetre.
between
and
NaIO
NaI
in
acidic
solution.
3
Half
equations
for
the
redox
processes
are
given
below.
1
+
+
IO
5e
+
6H
3H
➝
3
O
+
I
2
2
2
1
I
I
➝
Use
these
half
equations
iodine
by
this
process.
When
concentrated
to
deduce
Identify
the
an
overall
oxidising
+
e
2
2
ionic
equation
for
the
production
(2
(c)
and
(d)
a
black
Identify
(ii)
Deduce
the
sulfuric
acid.
When
iodide
Identify
is
this
yellow
acid,
is
a
half
of
(ii)
State
added
to
potassium
iodide,
solid
to
the
marks)
sulfur
with
sulfur
gas
for
the
formation
concentrated
changes
that
has
from
an
of
sulfuric
+6
to
sulfur
acid
–2.
unpleasant
from
in
The
a
(1
mark)
(1
mark)
concentrated
different
reduction
redox
product
reaction,
of
this
smell.
gas.
(1
precipitate
Write
equation
poisonous
added
(i)
is
solid.
react
state
acid
formed.
black
ions
oxidation
reaction
A
the
are
(i)
the
(e)
solid
sulfuric
of
agent.
an
is
formed
aqueous
simplest
when
solution
ionic
silver
nitrate
containing
equation
for
the
solution,
iodide
acidied
with
dilute
nitric
ions.
formation
of
the
yellow
precipitate.
(1
what
is
observed
when
concentrated
mark)
ammonia
solution
is
added
to
mark)
this
precipitate.
(iii)
(f)
State
Consider
why
the
the
silver
following
Cl
nitrate
reaction
(aq)
+
is
in
2I
acidied
which
(aq)
2
(i)
(ii)
(iii)
In
terms
Write
a
Explain
of
equation
why
iodide
iodide
I
testing
ions
(aq)
+
for
iodide
behave
2Cl
as
mark)
(1
mark)
ions.
reducing
agents.
(aq)
2
electrons,
half
➝
when
(1
state
for
ions
the
the
meaning
of
conversion
react
differently
of
the
term
chlorine
from
reducing
into
chloride
agent
chloride
(1
mark)
(1
mark)
ions.
ions.
(3
marks)
AQA,
2012
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
141
Section
1
practice
1
Antimony
is
extraction
of
(a)
q ue stio ns
a
solid
element
antimony
Antimony
antimony
can
be
Sb
Write
an
(ii)
Write
this
(b)
In
the
a
half
the
in
industry.
grade
of
the
by
reacting
scrap
the
reaction
of
The
method
used
for
the
ore.
iron
with
low-grade
ores
that
contain
3
for
iron(II)
equation
iron
with
antimony
sulde
to
form
sulde.
to
show
what
happens
to
the
iron
atoms
(1
mark)
(1
mark)
in
reaction.
rst
sulde
and
used
S
equation
antimony
is
on
extracted
sulde,
2
(i)
that
depends
is
stage
of
roasted
the
in
air
extraction
of
to
it
convert
antimony
into
from
a
high-grade
antimony(III)
oxide
(Sb
ore,
O
2
)
antimony
and
sulfur
3
dioxide.
(i)
Write
(ii)
an
Identify
dioxide
(c)
In
the
equation
one
antimony(III)
(i)
Use
the
given
for
stage
in
of
oxide
this
the
is
standard
below
this
this
substance
formed
second
for
reaction.
that
is
extraction
reacted
1
directly
from
the
mark)
(1
mark)
sulfur
reaction.
with
enthalpies
Table
manufactured
(1
to
of
antimony
carbon
of
monoxide
formation
calculate
a
from
value
in
at
Table
for
the
a
high-grade
high
1
and
ore,
temperature.
the
standard
equation
enthalpy
change
reaction.
▼ Table 1
Sb
O
2
(s)
CO(g)
Sb(I)
CO
3
(g)
2
–1
– 705
H / kJ mol
Δ
– 111
+ 20
– 394
f
Sb
O
2
(s)
+
3CO(g)
➝
2Sb(l)
+
3CO
3
(g)
2
(3
(ii)
Suggest
why
antimony,
(iii)
State
this
(d)
Deduce
a
the
given
type
value
in
of
for
Table
reaction
the
1,
standard
is
that
not
enthalpy
of
formation
of
marks)
liquid
zero.
antimony(III)
oxide
has
undergone
(1
mark)
(1
mark)
(1
mark)
in
reaction.
one
low-grade
include
the
the
reason
ore,
cost
why
the
described
of
the
in
method
part
of
(a),
extraction
is
a
of
low-cost
antimony
process.
Do
from
not
ore.
AQA,
2
(a)
Complete
the
following
2014
table.
Relative mass
Relative charge
proton
electron
(2
(b)
An
atom
has
twice
as
many
protons
and
twice
as
many
neutrons
as
mark)
an
19
atom
F
.
of
Deduce
the
symbol,
including
3+
(c)
The
the
(i)
Give
(ii)
Explain
and
the
the
Na
electron
why
ion
have
the
arrangement
more
energy
3+
Al
mass
number,
of
this
atom.
(2
marks)
(3
marks)
+
ion
Al
is
of
same
these
needed
to
electron
arrangement.
ions.
remove
an
electron
from
the
+
ion
than
from
the
Na
ion.
AQA,
142
2007
Section
3
Molecules
of
molecules
are
NH
,
H
3
(a)
O,
and
HF
contain
covalent
bonds.
The
bonds
in
1
AS
Physical
chemist r y
1
these
2
polar.
(i)
Explain
(ii)
State
(iii)
Explain
(iv)
Explain
why
which
the
of
H–F
the
bond
is
polar.
molecules
NH
,
H
3
polar
why
than
the
the
why
bond
bonds
H
O
in
your
found
has
a
in
bond
O,
the
or
HF
contains
the
least
polar
bond.
2
chosen
molecule
other
angle
of
two
from
part
( b)(ii)
is
less
molecules.
104.5°.
(4
marks)
(2
marks)
2
(b)
The
boiling
points
of
NH
,
H
3
This
is
due
to
(i)
Identify
(ii)
Draw
to
a
all
the
type
type
diagram
each
and
the
other
lone
by
of
of
to
and
HF
intermolecular
show
this
pairs
O,
of
are
all
high
for
how
type
of
two
force
force
present
intermolecular
electrons
in
your
When
an
each
of
their
size.
case.
of
ammonia
force.
are
Include
attracted
partial
charges
diagram.
(4
marks)
+
ion
H
in
responsible.
molecules
+
(c)
molecules
2
intermolecular
reacts
with
an
NH
molecule,
an
NH
3
ion
is
formed.
4
+
(i)
Give
an
the
name
of
the
type
of
bond
formed
when
an
ion
H
reacts
with
molecule.
NH
3
(ii)
Draw
the
shape,
including
any
lone
pairs
of
electrons,
of
an
molecule
NH
3
+
and
of
an
ion.
NH
4
(iii)
Name
the
shape
produced
by
the
arrangement
of
atoms
in
the
molecule.
NH
3
+
(iv)
Give
the
bond
angle
in
the
NH
ion.
(7
marks)
4
AQA,
2007
Answers to the Practice Questions and Section Questions are available at
www.oxfordsecondary.com/oxfordaqaexams-alevel-chemistry
143
Section
1
summary
catalysts
ionisation
with
energy
E
bB
cC
K
+
dD
activation
d
energy
(D)
with
=
c
a
catalyst
b
(B)
activation
without
energy
catalyst
enthalpy
combustion
Δ
tsrif
of
noitasinoi
(A)
standard
ygrene
c
(C)
/
+
equilibrium
H
c
E
organic
trends
1200
chemistry
in
Cl
ionisation
S
1000
energy
P
Si
800
Mg
mass
spectrometry
600
Al
Na
400
E
cat
moves
in
1400
K
c
aA
Ar
Jk
constant
spectrometry
1600
l om
particles
equilibrium
energy
mass
of
1–
fraction
11
12
13
14
15
16
17
18
a
to
energy
E
atomic
number
reduce
isotopes
disturbance
atomic
orbitals
•
Maxwell–Boltzmann
•
s-orbital:
distribution
•
p-orbital:
6e
•
d-orbital:
10e
same
2e
number
of
protons
equilibrium
standard
of
•
different
number
enthalpy
formation
Δ
of
neutrons
H
f
collision
theory
enthalpy
H
sub-atomic
factors
Hess’
measuring
affect
law
that
rate
•
protons
•
neutrons
•
electrons
particles
(+1)
(neutral)
enthalpy
change
•
temperature
•
pressure
•
concentration
•
surface
(–1)
ΔH
Group
the
2
and
halogens
redox
area
Oxidation
oxidation
electrons
Is
states
Section
1:
Loss
Physical
chemistry
Reduction
fundamental
Is
particles
Gain
Group
2
bonding
periodicity
the
and
halogens
metallic
ionic
covalent
bonding
•
metal
and
•
transfer
non-metal
metals
•
non-metal
•
delocalised
•
shared
amount
electrons
of
substance
around
cations
non-metal
to
e
form
electrons
and
metal
of
bonding
•
bonding
ions
pair
e
e
e
of
2+
2+
Mg
moles
2+
Mg
Mg
electrons
×
chlorine
×
e
e
atoms
e
‘sea’
of
one
mole
contains
electrons
×
e
e
23
×
×
O
×
Mg
6.022
Cl
%
10
particles
Cl
2+
2+
Mg
2+
Mg
Mg
×
×
e
magnesium
atom
oxygen
e
atom
Cl
×
2+
e
×
Cl
2–
number
of
moles
×
×
×
O
×
Mg
a
chlorine
ideal
molecule
×
•
solid:
•
solution:
n
magnesium
O
ion
M
oxygen
equation
=
2–
2+
Mg
gas
m
×
PV
ion
electron
=
nRT
pair
n
repulsion
C
=
theory
V
forces
van
•
permanent
der
Waals
dipole-dipole
•
hydrogen
bonding
gnisaercni
•
htgnerts
intermolecular
covalent
bonding
properties
actual
percentage
yield
of
product
=
•
•
•
electronegativity
electrical
melting
yield
%
theoretical
yield
of
100
product
conductivity
and
boiling
points
mass
atom
of
product
=
economy
144
%
total
mass
of
reactants
100
Section
1
AS
Physical
chemist r y
Practical skills
Maths skills
In this section you have met the following
In this section you have met the following
ideas:
maths skills:
•
Finding the concentration of a
solution by titration.
•
Finding the yield of a reaction.
•
Finding ΔH of reactions using
•
•
Using standard form in calculations.
•
Carrying out calculations with
the Avogradro constant.
•
Carrying out calculations using Hess’s law.
calorimetry and Hess’s law.
•
Using appropriate signicant gures.
Investigating the eect of temperature,
•
Calculating weighted means.
•
Interpreting mass spectra.
•
Working out the shape of molecules
concentration and a catalyst on the
rate of reactions.
•
of a reaction.
Finding out K
using ideas about electron pair repulsion.
c
Extension
Produce a timeline detailing how our understanding of atoms,
atomic structure and chemical bonding has developed.
Suggested resources:
•
Atkins, P
. (2014), Physical Chemistry: A very shor t Introduction.
Oxford University Press, UK . ISBN:
•
978-0-19-968909-5
Dunmar, D., Sluckin, T., (2014), Soap, Science and Flat-Screen TVs.
Oxford University Press, UK . ISBN: 978-0-19-870083-8
•
Scerri, E. (2013), The Tale of 7 Elements. Oxford University Press, UK .
ISBN: 978-0-19-539131-2.
145
1