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The behaviour of gases
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The behaviour of gases
2016
CHAPTER ONE
MATTER
1.0
Introduction
Long before the science of chemistry was established, materials
were described as existing in one of three physical states. There are
either rigid, solid objects, having a definite volume and a fixed shape,
nonrigid liquids, having no fixed shape other than that of their
containers but having definite volumes or gases, which have neither
fixed shape nor fixed volume.
The techniques used for handling various materials depend on
their physical states as well as their chemical properties. While it is
comparatively easy to handle liquids and solids, it is not as convenient
to measure out a quantity of a gas. Fortunately, except under rather
extreme conditions, all gases have similar physical properties, and the
chemical identity of the substance does not influence those properties.
For example, all gases expand when they are heated in a nonrigid
container and contract when they are cooled or subjected to increased
pressure. They readily diffuse through other gases. Any quantity of gas
will occupy the entire volume of its container, regardless of the size of
the container.
1.1.
States of Matter
Matter is anything that has mass and occupies space. All the
material things in the universe are composed of matter, including
anything we can touch as well as the planets in the solar system and all
the stars in the sky. It is composed of tiny particles such as atoms,
molecules, or ions and can exist in three physical states- solid, liquid
and gas.
Solid State
In the solid state, the individual particles of a substance are in
fixed positions with respect to each other because there is not enough
thermal energy to overcome the intermolecular interactions between
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the particles. As a result, solids have a definite shape, volume and are
incompressible. Most solids are hard, but some (like waxes) are
relatively soft. Some large crystals look the way they do because of the
regular arrangement of atoms (ions) in their crystal structure. Solids
usually have their constituent particles arranged in a regular, threedimensional array of alternating positive and negative ions called a
crystal. Some solids, especially those composed of large molecules,
cannot easily organize their particles in such regular crystals and exist
as amorphous (literally, ―without form‖) solids. Glass is one example
of an amorphous solid.
Liquid State
A liquid is a nearly incompressible fluid that conforms to the
shape of its container but retains a (nearly) constant volume
independent of pressure. The volume is definite if the temperature and
pressure are constant. The molecules have enough energy to move
relative to each other and the structure is mobile.
Gaseous State
Gases consist of tiny particles widely spaced (Figure 1.1). Under
typical conditions, the average distance between gas particles is about
ten times their diameter. Because of these large distances, the volume
occupied by the particles themselves is very small compared to the
volume of the empty space around them. For a gas at room
temperature and pressure, the gas particles themselves occupy about
0.1% of the total volume. The other 99.9% of the total volume is empty
space (whereas in liquids and solids, about 70% of the volume is
occupied by particles). Because of the large distances between gas
particles, the attractions or repulsions among them are weak.
The particles in a gas are in rapid and continuous motion. For
example, the average velocity of nitrogen molecules, N2, at 20 °C is
about 500 m/s. As the temperature of a gas increases, the particles‘
velocity increases. The average velocity of nitrogen molecules at 100 °C
is about 575 m/s.
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The particles in a gas are constantly colliding with the walls of the
container and with each other. Because of these collisions, the gas
particles are constantly changing their direction of motion and their
velocity. In a typical situation, a gas particle moves a very short
distance between collisions. For example, oxygen, O2, molecules at
normal temperatures and pressures move an average of 10-7 m between
collisions.
Fig.1.1. A Representation of the Solid, Liquid, and Gas States
The various characteristics or properties of the states of matter
discussed above are summarized in table 1.1 below.
Table 1.1. Characteristics of the Three States of Matter
Characteristic
Shape
Solid
Definite
Volume
Relative
intermolecular
interaction strength
Relative particle
positions
Definite
Strong
Liquid
conforms to the
shape of its
container
Definite
Moderate
in contact and
fixed in place
in contact but not
fixed
Compressibility
incompressible
incompressible
3
Gas
Indefinite
not in contact,
random
positions
Compressible
fluid
Indefinite
Weak
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1.2. Phase Transition
Phase transition is a term used to describe a state of change of
matter from one state to another. The state or phase of a given set of
matter can change depending on pressure and temperature conditions,
transitioning to other phases as these conditions change to favour their
existence; for example, solid transitions to liquid with an increase in
temperature. Near absolute zero, a substance exists as a solid. As heat
is added to this substance it melts into a liquid at its melting point,
boils into a gas at its boiling point, and if heated high enough would
enter a plasma state in which the electrons are so energized that they
leave their parent atoms.
1.2.1. Melting point
This is the temperature at which the solid and liquid forms of a
pure substance can exist at equilibrium. As heat is applied to a solid, its
temperature will increase until the melting point is reached. More heat
then will convert the solid into a liquid with no temperature change.
When the entire solid has melted, additional heat will raise the
temperature of the liquid. The melting temperature of crystalline solids
is a characteristic figure and is used to identify pure compounds and
elements. Most mixtures and amorphous solids melt over a range of
temperatures.
The melting temperature of a solid is generally considered to be
the same as the freezing point of the corresponding liquid; because a
liquid may freeze in different crystal systems and because impurities
lower the freezing point, however, the actual freezing point may not be
the same as the melting point. Thus, for characterizing a substance, the
melting point is preferred. A typical example is the change of solid ice
to liquid water as shown below.
H2O(s) → H2O(l)
Ice, snow
liquid water
(melting, fusion)
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1.2.2. Freezing point
This is the temperature at which a liquid becomes a solid. As
with the melting point, increased pressure usually raises the freezing
point. The freezing point is lower than the melting point in the case of
mixtures and for certain organic compounds such as fats. As a mixture
freezes, the solid that forms first usually has a composition different
from that of the liquid, and formation of the solid changes the
composition of the remaining liquid, usually in a way that steadily
lowers the freezing point. This principle is used in purifying mixtures,
successive melting and freezing gradually separating the components.
The heat of fusion (heat that must be applied to melt a solid), must be
removed from the liquid to freeze it. Some liquids can be supercooled
i.e., cooled below the freezing point without solid crystals forming.
Putting a seed crystal into a supercooled liquid triggers freezing,
whereupon the release of the heat of fusion raises the temperature
rapidly to the freezing point. Freezing of liquid water to ice is a
common example.
H2O(l)
→
liquid water
H2O(s)
Ice
(freezing)
1.2.3. Condensation
This is change of a gas to either liquid or solid state, generally
upon a surface that is cooler than the adjacent gas. The change of
vapour to solid is sometimes called deposition. A substance condenses
when the pressure exerted by its vapour exceeds the vapour pressure
of the liquid or solid phase of the substance at the temperature of the
surface where condensation occurs. Heat is released when a vapour
condenses. Unless this heat is removed, the surface temperature will
increase until it is equal to that of the surrounding vapour. In the
atmosphere, however, there is an abundant supply of aerosols, which
serve as nuclei, called condensation nuclei, on which water vapour
may condense. Some are hygroscopic (moisture-attracting), and
condensation begins on them when the relative humidity is less than
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100 percent, but other nuclei require some supersaturation before
condensation begins. Condensation accounts for the formation of dew
(liquid water formed by condensation of water vapour from the
atmosphere), and Frost (solid water formed by direct condensation of
water vapour from the atmosphere without first forming liquid water).
H2O(g)
→
Water vapour
H2O(g)
→
Water vapour
H2O(l)
dew
(condensation)
H2O(s) (condensation, deposition)
frost, snow
1.2.4. Vapourization
This refers to the conversion of a substance from the liquid or
solid phase into the gaseous (vapour) phase. Heat must be supplied to
a solid or liquid to effect vaporization. If the surroundings do not
supply enough heat, it may come from the system itself as a reduction
in temperature. The atoms or molecules of a liquid or solid are held
together by cohesive forces, and these forces must be overcome in
separating the atoms or molecules to form the vapour; the heat of
vaporization is a direct measure of these cohesive forces.
H2O(l)
→
(vaporization)
Liquid water
H2O(g)
water vapour
1.2.5. Sublimation
The change of a solid directly to the vapour without its becoming
liquid is specifically referred to as sublimation. Although the vapor
pressure of many solids is quite low, some (usually molecular solids)
have appreciable vapor pressure. Ice, for instance, has a vapour
pressure of 4.7 mmHg at 0oC. For this reason, a pile of snow slowly
disappears in winter even though the temperature is too low for it to
melt. The snow is being changed directly to water vapour.
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The behaviour of gases
H2O(s)
Ice, snow
→
2016
H2O(g)
(sublimation)
Water vapour
Sublimation can be used to purify solids such as impure iodine
that readily vaporize. Impure iodine is heated in a beaker so that it
vaporizes, leaving nonvolatile impurities behind. The vapour
crystallizes on the bottom surface of a dish containing ice that rests on
top of the beaker. Freeze-drying of foods is a commercial application of
sublimation. Brewed coffee, for example, is frozen and placed in a
vacuum to remove water vapour. The ice continues to sublime until it
is all gone, leaving freeze-dried coffee. Most freeze-dried foods are
easily reconstituted by adding water. The following diagram
summarizes these phase transitions.
Fig.1.2. Diagram showing the nomenclature for the different phase
transitions.
1.3. Heat of Phase Transition
Any change of state involves the addition or removal of energy
as heat to or from the substance. A simple experiment shows that this
is the case. Suppose you add heat at a constant rate to a beaker
containing ice at -20oC. In Figure 1.3 below, we have plotted the
temperature of the different phases of water as heat is added. The
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The behaviour of gases
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temperature of the ice begins to rise from -20oC, as you would expect;
the addition of heat normally raises the temperature of a substance. At
0oC, the ice begins to melt, so that you get a beaker of ice in water. Note
the flat region in the curve, labeled ice and water. Why is this region
flat? It means that heat is being added to the system without a change
in temperature; the temperature remains at 0oC. This temperature, of
course, is the melting point of ice. The heat being added is energy
required to melt ice to water at the same temperature. The
intermolecular forces binding water molecules to specific sites in the
solid phase must be partially broken to allow water molecules the
ability to slide over one another easily, as happens in the liquid state.
Note the flat regions for each of the phase transitions. Because heat is
being added at a constant rate, the length of each flat region is
proportional to the heat of phase transition.
Fig. 1.3. Heating curve for water: Heat is being added at a constant rate to a
system containing water. Note the flat regions of the curve. When heat is
added during a phase transition, the temperature does not change.
The heat needed for the melting of a solid is called the heat of
fusion (or enthalpy of fusion) and is denoted ∆Hfus. For ice, the heat of
fusion is 6.01 kJ per mole.
H2O(s) → H2O(l); ∆Hfus = 6.01 kJ/mol
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The heat needed for the vaporization of a liquid is called the heat of
vaporization (or enthalpy of vaporization) and is denoted ∆Hvap. At
100oC, the heat of vaporization of water is 40.7 kJ per mole.
H2O(l) → H2O(g); ∆Hvap = 40.7 kJ/mol
Note that much more heat is required for vaporization than for
melting. Melting needs only enough energy for the molecules to escape
from their sites in the solid. For vaporization, enough energy must be
supplied to break most of the intermolecular attractions. A refrigerator
relies on the cooling effect accompanying vaporization. The
mechanism contains an enclosed gas that can be liquefied under
pressure, such as ammonia or 1,1,1,2-tetrafluoroethane, CH2FCF3. As
the liquid is allowed to evaporate, it absorbs heat and thus cools its
surroundings (the interior space of the refrigerator). Gas from the
evaporation is recycled to a compressor and then to a condenser,
where it is liquefied again. Heat leaves the condenser, going into the
surrounding air.
1.4. Pressure of Gases
The molecules of a gas, being in continuous motion, frequently
strike the inner walls of their container. As they do so, they
immediately bounce off without loss of kinetic energy, but the reversal
of direction (acceleration) imparts a force to the container walls. This
force, divided by the total surface area on which it acts, is the pressure
of the gas.
The pressure of a gas is observed by measuring the pressure
that must be applied externally in order to keep the gas from
expanding or contracting. To visualize this, imagine some gas trapped
in a cylinder having one end enclosed by a freely moving piston. In
order to keep the gas in the container, a certain amount of weight
(more precisely, a force, f) must be placed on the piston so as to exactly
balance the force exerted by the gas on the bottom of the piston, and
tending to push it up. The pressure of the gas (P) is simply the quotient
f/A, where A is the cross-section area of the piston.
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Example 1.1. If a force of 16N is pressed against an area of 2.44 m2, what
is the pressure in pascals?
Solution
Given force, F = 16N, area, A = 2.44 m2
Apply the relationship,
𝐹
𝑃= 𝐴
𝑃=
16𝑁
2.44m 2
= 6.57𝑁𝑚−2
1.4.1. Pressure Units
The unit of pressure in the SI system is the pascal (Pa), defined
as a force of one newton per square metre (1 Nm–2 = 1 kg m–1 s–2 ). In
chemistry, it is more common to express pressures in units of
atmospheres or torr:
1 atm = 101325 Pa = 760 torr.
The older unit millimetre of mercury (mm Hg) is almost the
same as the torr; it is defined as one mm of level difference in a
mercury barometer at 0°C. In meteorology, the pressure unit most
commonly used is the bar:
1 bar = 106 N m–2 = 0.987 atm.
For conversion purposes,
1 atm = 760 torr =760 mmHg = 1.01325 × 105 Nm-2
Example 1.2. How many atmospheres are in 1547mmHg
Solution
Use the conversion factor;
1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔
∴ 𝑥 𝑎𝑡𝑚 = 1547 𝑚𝑚𝐻𝑔
Cross multiplying and making 𝑥 the subject gives
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The behaviour of gases
𝑥=
2016
1 𝑎𝑡𝑚 ×1547 𝑚𝑚𝐻𝑔
760 𝑚𝑚𝐻𝑔
𝑥 = 2.04 𝑎𝑡𝑚
Example 1.3. Write the conversion factor to determine how many
mmHg are in 9.65 atm.
Solution
Use the same conversion factor as in example 1.2 above
1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔
∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔
Cross multiplying and making 𝑥 the subject give
𝑥=
9.65 𝑎𝑡𝑚 ×760 𝑚𝑚𝐻𝑔
1 𝑎𝑡𝑚
𝑥 = 7334 𝑚𝑚𝐻𝑔
Example 1.4. How many torr are in 1.56 atm
Solution
Use the conversion factor;
1 𝑎𝑡𝑚 = 760 𝑡𝑜𝑟𝑟
∴ 1.56 𝑎𝑡𝑚 = 𝑥 𝑡𝑜𝑟𝑟
Cross multiplying and making 𝑥 the subject give
𝑥=
1.56𝑎𝑡𝑚 ×760 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
𝑥 = 1190 𝑡𝑜𝑟𝑟
Example 1.5. Blood pressures are expressed in mmHg. What would be
the blood pressure in atm if a patient‘s systolic and diastolic blood
pressures are 120 mmHg and 82 mmHg respectively? (In medicine,
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The behaviour of gases
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such a blood pressure would be reported as ―120/82‖, spoken as ―one
hundred twenty over eighty-two‖ ).
Solution
Use the same conversion factor as in example one above
1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔
∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔
Cross multiplying and making 𝑥 the subject give
Use the same conversion factor as in example one above
1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔
∴ 120 𝑚𝑚𝐻𝑔 =
120 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔
= 0.157 atm
82 𝑚𝑚𝐻𝑔 =
82 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔
= 0.107 atm
∴
120 𝑚𝑚𝐻𝑔
82 𝑚𝑚𝐻𝑔
= 0.157 𝑎𝑡𝑚: 0.107 𝑎𝑡𝑚
1.4.2. Atmospheric Pressure
This is defined as the force per unit area exerted against a
surface by the weight of the air above that surface. In most
circumstances atmospheric pressure is closely approximated by the
hydrostatic pressure caused by the weight of air above the
measurement point. On a given plane, low-pressure areas have less
atmospheric mass above their location, whereas high-pressure areas
have more atmospheric mass above their location. Likewise, as
elevation (altitude) increases, there is less overlying atmospheric mass,
so that atmospheric pressure decreases with increasing elevation.
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1.4.3. Measurement of Gas Pressure
A barometer is piece of lab equipment specifically designed to
measure the atmospheric pressure. Invented in the early 17th century
by the Italian EVANGELISTA TORRICELLI. The barometer consists of
a vertical glass tube closed at the top and evacuated, and open at the
bottom, where it is immersed in a dish of a liquid. The atmospheric
pressure acting on this liquid will force it up into the evacuated tube
until the weight of the liquid column exactly balances the atmospheric
pressure. If the liquid is mercury, the height supported will be about
760 cm; this height corresponds to standard atmospheric pressure.
Fig. 1.4. A simple barometer
The formula for this pressure in the atmosphere is derived as shown
below:
𝒇𝒐𝒓𝒄𝒆 = 𝒎𝒂𝒔𝒔 × 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
or
𝑭 = 𝒎𝒂 or mg
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The behaviour of gases
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Earth's acceleration of objects is based on its gravitational field and
equals approximately 9.80665 m s-2. Additionally, since pressure is the
force per the unit area being measured, then
𝑭
𝑷=
𝑨
𝒎𝒈
= 𝑨
Since mass (m) = density (d) × volume (v)
𝑷=
Since
𝑉𝑜𝑙𝑢𝑚𝑒 (𝑋 3 )
𝐴𝑟𝑒𝑎 (𝑋 2 )
𝒈 ×𝒅 ×𝑽
𝑨
= 𝑕𝑖𝑒𝑔𝑕𝑡 (𝑋)
𝑷 = 𝑔 ×𝑑 ×𝑕
Where d = density, g = gravity and h = height of the liquid or gas.
Example 1.6. Mercury has a density of 13.6 g/cm3 and water has a
density of 1.00 g/cm3. If a column of mercury has a height of 755 mm,
how high would a corresponding column of water be in feet?
Solution:
Let us begin by setting the pressures equal:
Pmercury = Pwater
Since
𝑷 = 𝑔 ×𝑑 ×𝑕
We can write:
𝑕𝑤𝑎𝑡𝑒𝑟 =
=
𝑑𝐻𝑔 × 𝑕𝐻𝑔
𝑑 𝑤𝑎𝑡𝑒𝑟
13.6𝑔𝑐𝑚 −3 ×755 𝑚𝑚
1.00𝑔𝑐𝑚 −3
= 10268 𝑚𝑚 = 33.7 𝑓𝑡
1.4.4. The Manometer
A modification of the barometer, the U-tube manometer,
provides a simple device for measuring the pressure of any gas in a
container. There are a variety of manometer designs. A simple,
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The behaviour of gases
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common design is to seal a length of glass tubing and bend the glass
tube into a U-shape. The glass tube is then filled with a liquid, typically
mercury, so that all trapped air is removed from the sealed end of the
tube. The glass tube is then positioned with the curved region at the
bottom. The mercury settles to the bottom.
After the mercury settles to the bottom of the manometer, a
vacuum is produced in the sealed tube. The open tube is connected to
the system whose pressure is being measured. In the sealed tube, there
is no gas to exert a force on the mercury (except for some mercury
vapor). In the tube connected to the system, the gas in the system
exerts a force on the mercury. The net result is that the column of
mercury in the sealed tube is higher than that in the unsealed tube. The
difference in the heights of the columns of mercury is a measure of the
pressure of gas in the system.
In the open-tube manometer, the pressure of the gas is given
by h (the difference in mercury levels) in units of torr or mmHg.
Atmospheric pressure pushes on the mercury from one direction, and
the gas in the container pushes from the other direction. In a
manometer, since the gas in the bulb is pushing more than the
atmospheric pressure, you add the atmospheric pressure to the height
difference:
Pgas > Patm
Gas pressure = atmospheric pressure + h (height of the mercury)
Pgas < Patm
Gas pressure = atmospheric pressure - h (height of the mercury)
The closed-tube manometer look similar to regular manometers except
that the end that is open to the atmospheric pressure in a regular
manometer is sealed and contains a vacuum. In these systems, the
difference in mercury levels (in mmHg) is equal to the pressure in torr.
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The behaviour of gases
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Fig.1.5. The Manometer
Example 1.7. Find the pressures using the manometer set up below.
since Pgas < Patm
Pgas= Patm ‒ h
Pgas= (763 ‒35)g
Pgas= 728 mmHg
Solution
since Pgas > Patm
Pgas= Patm + h
Pgas= (755 + 24 )mmHg
=779mmHg
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The behaviour of gases
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Example 1.8. Suppose you want to construct a closed-end manometer
to measure gas pressures in the range 0.000–0.200 atm. Because of the
toxicity of mercury, you decide to use water rather than mercury. How
tall a column of water do you need? (The density of water is 1.00
g/cm3; the density of mercury is 13.6 g/cm3).
Solution
Given: pressure range and densities of water and mercury, column
height unknown.
Strategy:
Step 1. Calculate the height of a column of mercury corresponding to
0.200 atm in millimeters of mercury. This is the height needed for a
mercury-filled column.
Step 2. From the given densities, use a proportion to compute the
height needed for a water-filled column.
In millimeters of mercury, a gas pressure of 0.200 atm
1atm = 760mmHg
0.200 𝑎𝑡𝑚 ×760𝑚𝑚𝐻𝑔
∴ 0.200 atm will be
1 𝑎𝑡𝑚
= 152 𝑚𝑚𝐻𝑔
Using a mercury manometer, you would need a mercury column of at
least 152 mm high.
Because water is less dense than mercury, you need a taller
column of water to achieve the same pressure as a given column of
mercury. The height needed for a water-filled column corresponding to
a pressure of 0.200 atm is proportional to the ratio of the density of
mercury to the density of water;
Using 𝑷 = 𝑔 × 𝑑 × 𝑕
Where d = density, g = gravity and h = height of the liquid or gas.
Let us begin by setting the pressures equal:
Pmercury = Pwater
We can then write:
𝑔 × 𝑑𝐻𝑔 × 𝑕𝐻𝑔 = 𝑔 × 𝑑𝑤𝑎𝑡𝑒𝑟 × 𝑕𝑤𝑎𝑡𝑒𝑟
17
The behaviour of gases
𝑕𝑤𝑎𝑡𝑒𝑟 =
=
2016
𝑑𝐻𝑔 × 𝑕𝐻𝑔
𝑑 𝑤𝑎𝑡𝑒𝑟
13.6𝑔𝑐𝑚 −3 × 152 𝑚𝑚
1.00𝑔𝑐𝑚 −3
= 2070 𝑚𝑚
Comment: it takes a taller column of a less dense liquid to achieve the
same pressure.
1.4.5. Effect of Pressure on the volume of gases
For a gas whose volume is not fixed, increasing the pressure
will cause the gas to contract (reducing the volume), and decreasing
the pressure will cause the gas to expand (increasing the volume). If
the volume is fixed, then increasing the pressure will increase the
temperature, and decreasing the pressure will decrease the
temperature.
1.4.6. Simple Pressure Related Applications
• Drinking straw: A drinking straw is used by creating a
suction with your mouth. Actually this causes a decrease in air
pressure on the inside of the straw. Since the atmospheric pressure is
greater on the outside of the straw, liquid is forced into and up the
straw.
• Siphon: With a siphon water can be made to flow "uphill". A
siphon can be started by filling the tube with water (perhaps by
suction). Once started, atmospheric pressure upon the surface of the
upper container forces water up the short tube to replace water flowing
out of the long tube.
1.5. Density of a Gas
This is defined as mass divided by the volume of a gas
𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑑 =
𝑚𝑎𝑠𝑠 (𝑔)
𝑣𝑜𝑙𝑢𝑚𝑒 (𝐿)
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The density of a gas in grams/L can be obtained from ideal gas
equation as follows:
𝑃𝑉 = 𝑛𝑅𝑇
𝑚𝑎𝑠𝑠 (𝑚)
Number of mole of a gas (n) =
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
Substituting ―n‖ into ideal gas equation above
𝑚
𝑃𝑉 = × 𝑅𝑇
𝑀
Cross multiplying we have
𝑀 × 𝑃𝑉 = 𝑚 × 𝑅𝑇
Divide both side by V gives
𝑚
𝑃 × 𝑀 = × 𝑅𝑇
𝑉
Lastly divide both by RT gives density
𝑚
𝑃×𝑀
= 𝑅𝑇
𝑉
𝑑 =
𝑃×𝑀
𝑅𝑇
Example 1.9. What is the density of oxygen at STP? [R= 0.8206L atm
mol-1K-1]
Solution
Data collection
S.t = 273K
S.p = 1 atm
R= 0.8206L atm mol-1K-1
Molecular weight, M of oxygen = 32.0gmol-1
𝑚
𝑃×𝑀
Using 𝑑 = 𝑉 = 𝑅𝑇
1 𝑎𝑡𝑚 × 32.0 𝑔𝑚𝑜𝑙 −1
=
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 273.15𝐾
= 1.428𝑔/𝐿
Example 1.10. A 0.0125g sample of a gas with an empirical formula of
CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at
22.5 °C. What is the molecular formula of the compound? [R= 0.8206L
atm mol-1K-1]
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Solution
Collect the available data and convert as necessary to agree with the given
unit of R then find the value of density from which the molecular weight of the
𝑃×𝑀
gas can be determined using the relation: 𝑑 = 𝑅𝑇
Mass of gas sample = 0.0125g
Volume = 165 mL = 0.156 L
Temperature, T = 22.5°C = 295.7K
Pressure, P = 13.7 mm Hg = 1 atm ×
𝑚
Now density, 𝑑= 𝑉𝑑 =
𝑚
𝑉
=
0.0125𝑔𝑔
0.156 𝐿𝐿
13.7 𝑚𝑚𝐻𝑔
= 0.0180
760 𝑚𝑚𝐻𝑔
𝑎𝑡𝑚𝑎𝑡𝑚
= 0.0758 𝑔𝑔
/𝐿𝐿
To find molecular weight of gas, we use
𝑃 × 𝑀
𝑅𝑇
Making molecular weight, M the subject and substituting
𝑑𝑅𝑇
M= 𝑃
0.075𝑔𝐿−1 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 273.15𝐾
=
0.0180 𝑎𝑡𝑚
𝑑=
M = 102𝑔𝑚𝑜𝑙 −1
The molecular formula is (CHF2)2 or C2H2F4.
Example 1.11. If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25
oC, what is the pressure of O2 & H2O?[0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 , H =
1, 0 = 16, ]
Solution
Step 1: Write the balanced chemical reaction.
Step 2: Calculate the moles of each product.
Step 3: Find the pressure of each via PV = nRT
Equation of reaction : 2H2O2(l) → 2H2O (g) + O2 (g)
From the equation of reaction, 2 mol of 2H2O2 produce 2 mol of H2O
and a mol of O2.
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Therefore mol of H2O2 =
2016
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
0.11𝑔
= 34𝑔/𝑚𝑜𝑙
1
2
=0.0032 mol
× 0.0032 mol of H2O2
mol of O2 =
=0.0016 mol
mol of H2O = 1 × 0.0032 mol of H2O2
=0.0032 mol
Using PV = nRT to calculate the pressure of the gases
𝑛𝑅𝑇
P(O2) = 𝑉
0.0016 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 298𝐾
=
2.5 𝐿
= 0.016 atm
𝑛𝑅𝑇
P(H2O) =
𝑉
0.0032 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 298𝐾
=
2.5 𝐿
= 0.032 atm
Example 1.12. A chemist has synthesized a greenish-yellow gaseous
compound of chlorine and oxygen and finds that its density is 8.14 g/L
at 47°C and 3.15 atm. Calculate the molar mass of the compound and
determine its molecular formula.
Solution
We can calculate the molar mass of a gas if we know its density,
temperature, and pressure. The molecular formula of the compound
must be consistent with its molar mass. What temperature unit should
we use?
Data provided
density = 8.14 g/L
T = 47°C = 320 K
P = 3.15 atm
𝑑𝑅𝑇
Using the relationship, 𝑀 = 𝑃 to solve for molar mass,
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=
2016
8.14 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 ×320 𝐾
3.15 𝑎𝑡𝑚
−1
= 67.9 𝑔𝑚𝑜𝑙
We can determine the molecular formula of the compound by trial and
error, using only the knowledge of the molar masses of chlorine (35.45
g) and oxygen (16.00 g). We know that a compound containing one Cl
atom and one O atom would have a molar mass of 51.45 g, which is too
low, while the molar mass of a compound made up of two Cl atoms
and one O atom is 86.90 g, which is too high. Thus, the compound
must contain one Cl atom and two O atoms and have the formula ClO2
, which has a molar mass of 67.45 g.
Example 1.13. The density of a gaseous organic compound is 3.38 g/L
at 40°C and 1.97 atm. What is its molar mass?
Solution
Data provided
d = 3.38 g/L
T = 40°C = 313 K
P = 1.97 atm
𝑑𝑅𝑇
Using the relationship 𝑀 = 𝑃
=
3.38 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 ×313 𝐾
1.97 𝑎𝑡𝑚
−1
= 44.0 𝑔𝑚𝑜𝑙
1.5.1. The effects of temperature on density
The density of a gas depends quite strongly on its temperature,
so hot air has a smaller density than does cold air; colder air is more
dense than hot air. From everyday experience, we know that
something is dense if it tries to drop, which is why a stone drops to the
bottom of a pond and a coin sinks to the bottom of a pan of water. This
relative motion occurs because both the stone and the coin have higher
densities than does water, so they drop. Similarly, we are more dense
than air and will drop if we fall off a roof. Just like the coin in water,
cold air sinks because it is denser than warmer air. We sometimes see
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this situation stated as warm air ‗displaces‘ the cold air, which
subsequently takes its place. Alternatively, we say ‗warm air rises‘,
which explains why we place our clothes above a radiator to dry them,
rather than below it.
Light entering the room above the radiator passes through
these pockets of warm air as they rise through colder air, and therefore
passes through regions of different density. The rays of light bend in
transit as they pass from region to region, much in the same way as
light twists when it passes through a glass of water. We say the light is
refracted. The eye responds to light, and interprets these refractions and
twists as different intensities.
So we see swirling eddy (or ‗convective‘) patterns above a
radiator because the density of air is a function of temperature. If all
the air had the same temperature, then no such difference in density
would exist, and hence we would see no refraction and no eddy
currents – which is the case in the summer when the radiator is
switched off. Then again, we can sometimes see a ‗heat haze‘ above a
hot road, which is caused by exactly the same phenomenon.
1.6. Temperature
`
This is the numerical measure of the degree of hotness or
coldness of a body. It is an important property of any gas. If two bodies
are at different temperatures, heat will flow from the warmer to the
cooler one until their temperatures are the same. This is the principle
on which thermometry is based; the temperature of an object is
measured indirectly by placing a calibrated device known as a
thermometer in contact with it. When thermal equilibrium is obtained,
the temperature of the thermometer is the same as the temperature of
the object.
1.6.1. Temperature Scale
A thermometer makes use of some temperature-dependent
quantity, such as the density of a liquid, to allow the temperature to be
found indirectly through some easily measured quantity such as the
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length of a mercury column. The resulting scale of temperature is
entirely arbitrary; it is defined by locating its zero point, and the size of
the degree unit.
Celsius temperature scale locates the zero point at the freezing
temperature of water; the Celsius degree (C °) is defined as 1/100 of
the difference between the freezing and boiling temperatures of water
at 1 atm pressure.
The older Fahrenheit scale placed the zero point at the coldest
temperature it was possible to obtain at the time (by mixing salt and
ice.) The 100° point was set with body temperature (later found to be
98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The
Fahrenheit scale is a finer one than the Celsius scale; there are 180
Fahrenheit degrees in the same temperature interval that contains 100
Celsius degrees, so 1F° = 9/5 C . Since the zero points are also different
by 32F, conversion between temperatures expressed on the two scales
requires the addition or subtraction of this offset, as well as
multiplication by the ratio of the degree size. These selections allow us
to write the following relations.
9
t(oF) = 5 t(oC) + 32
9
5
t(oC) = t(oF) – 32
Where
t(oF) is the temperature in degree Fahrenheit and
t(oC) is the temperature in degree Celsius.
1.6.2. Absolute temperature
In 1787 the French mathematician and physicist JACQUES
CHARLES discovered that for each Celsius degree that the
temperature of a gas is lowered, the volume of the gas will diminish by
1/273 of its volume at 0°C. The obvious implication of this is that if the
temperature could be reduced to –273°C, the volume of the gas would
contract to zero. Of course, all real gases condense to liquids before this
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The behaviour of gases
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happens, but at sufficiently low pressures their volumes are linear
functions of the temperature (Charles' Law), and extrapolation of a plot
of volume as a function of temperature predicts zero volume at -273°C.
This temperature, known as absolute zero, corresponds to the total
absence of thermal energy.
Because the Kelvin scale is based on an absolute, rather than on
an arbitrary zero of temperature, it plays a special significance in
scientific calculations; most fundamental physical relations involving
temperature are expressed mathematically in terms of absolute
temperature. The diagram below compares the different temperature
scales with respect to boiling and freezing point of water.
Fig. 1.6. Comparison of Temperature Scales (Schematic)
1.6.3. Conversion between Celsius and Kelvin Scale
In order to covert temperature in degree Celsius to temperature
in Kelvin, the expression below is used.
toC = (273 + t )K = T (K)
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Where t is the temperature on the Celsius scale, T is the temperature on
the Kelvin scale.
Example 1.14. Covert the following temperatures to Kelvin scale:
a. 27oC; b. -10oC.
Solutions
a. Using the relationship
toC = (273 + t )K = T (K)
27oC = (273 + 27)K = 300K
b. toC = (273 + t )K = T (K)
-10oC = (273- 10)K = 263K
In order to convert absolute temperature T K to degree Celsius,
273 is simply subtracted from the value.
Example 1.15. Covert the following temperatures to degree Celsius:
a. 298K
b. 25K
Solutions
a. Using the relationship
toC = (273 + t )K = T (K)
toC = (298 ‒ 273) oC = 25oC
b. Using the relationship
toC = (273 + t )K = T (K)
toC = (25 ‒ 273) oC = ‒ 248 oC
1.7. The Volume of Gas
The volume of a gas is simply the space in which the molecules
of the gas are free to move. If we have a mixture of gases, such as air,
the various gases will coexist within the same volume. In these
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The behaviour of gases
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respects, gases are very different from liquids and solids, the two
condensed states of matter. The volume of a gas can be measured by
trapping it above mercury in a calibrated tube known as a gas burette
(fig. 1.7). The SI unit of volume is the cubic meter, but in chemistry the
liter and the milliliter (mL) are commonly used.
Fig. 1.7. Gas burette
It is important to bear in mind, however, that the volume of a
gas varies with both the temperature and the pressure, so reporting the
volume alone is not very useful. A common practice is to measure the
volume of the gas under the ambient temperature and atmospheric
pressure, and then to correct the observed volume to what it would be
at standard atmospheric pressure and some fixed temperature, usually
0° C or 25°C. The table below shows some commonly used volume
measurement units and their conversion factor.
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1.8. Effect of Temperature on the volume of gases
If the volume of the container is not fixed, increasing the
temperature will cause a gas to expand (increase the volume), and
contract when cooled (decreasing the volume). This would be the case
for a gas inside a piston, or inside a rubber balloon. If the volume is
fixed, then increasing the temperature will increase the pressure, and
decreasing the temperature will decrease the pressure. This would be
the case for a gas in a closed solid container, like a canister or sealed
metal box.
Why does thunder accompany lightning?
Lightning is one of the most impressive and yet frightening
manifestations of nature. It reminds us just how powerful nature can
be. Lightning is quite a simple phenomenon. Just before a storm
breaks, perhaps following a period of hot, fine weather, we often note
how the air feels ‗tense‘. In fact, we are expressing an experiential
truth: the air contains a great number of ions – charged particles. The
existence of a large charge on the Earth is mirrored by a large charge in
the upper atmosphere. The only difference between these two charges
is that the Earth bears a positive charge and the atmosphere bears a
negative charge.
Accumulation of a charge difference between the Earth and the
upper atmosphere cannot proceed indefinitely. The charges must
eventually equalize somehow: in practice, negative charge in the upper
atmosphere passes through the air to neutralize the positive charge on
the Earth. The way we see this charge conducted between the Earth
and the sky is lightning: in effect, air is ionized to make it a conductor,
allowing electrons in the clouds and upper atmosphere to conduct
through the air to the Earth‘s surface. This movement of electrical
charge is a current, which we see as lightning. Incidentally, ionized air
emits light, which explains why we see lightning. Lightning comprises
a massive amount of energy, so the local air through which it conducts
tends to heat up to as much as a few thousand degrees centigrade. And
we have already seen how air expands when warmed, e.g. as described
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mathematically by Charles‘s law. In fact, the air through which the
lightning passes increases in volume to an almost unbelievable extent
because of its rise in temperature. And the expansion is very rapid.
1.9. Standard Temperature and Pressure, s.t.p.
Suppose two scientists work on the same research project, but
one resides in the far north of the Arctic Circle and the other lives near
the equator. Even if everything else is the same – such as the air
pressure, the source of the chemicals and the manufacturers of the
equipment – the difference between the temperatures in the two
laboratories will cause their results to differ widely. For example, the
‗room energy‘ RT will differ. One scientist will not be able to repeat the
experiments of the other, which is always bad science.
An experiment should always be performed at known
temperature. Furthermore, the temperature should be constant
throughout the course of the experiment, and should be noted in the
laboratory notebook. But to enable complete consistency, sets of
universally accepted arbitrary standards were devised and are called a
set of standard conditions. ‗Standard pressure‘ was set as 1 atm and
‗Standard temperature‘ has the value of 0oC (273 K). If both the
pressure and the temperature are maintained at these standard
conditions, then we say the measurement was performed at ‗standard
temperature and pressure‘, which is universally abbreviated to ‗s.t.p.‘
If the scientists at the equator and the Arctic Circle perform their work
in thermostatically controlled rooms, both at s.t.p., then the results of
their experiments will be identical. If we know the volume of a
sample of a gas at any condition, we can easily calculate the volume it
would have as an ideal gas at STP by employing the combined gas law.
1.10. Molar volume of a gas
The volume occupied by one mole of a gas under any
conditions of temperature and pressure is called the molar volume, Vm.
The molar volume of an ideal gas depends on the conditions of
temperature and pressure; at s.t.p. it is 22.4 L (or 22400 cm3).
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How did we arrive at this value?
It is simply the volume of 1.00 mol of gas at STP
At s.t.p, pressure (P) = 1atm, temperature (T) = 27K, for one mole of
gas, n = 1, R= 0.0821 L atm mol-1K-1
Using ideal gas equation to calculate the volume
PV = nRT
V=
=
𝑛𝑅𝑇
𝑃
1.00 𝑚𝑜𝑙 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 ×273 𝐾
1.00 𝑎𝑡𝑚
= 22.4 𝐿
1.11. Molecular weight and density of a gas
The molar volumes of all gases are the same when measured at
the same temperature and pressure. But the molar masses of different
gases will vary. This means that different gases will have different
densities (different masses per unit volume). If we know the molecular
weight of a gas, we can calculate its density.
More importantly, if we can measure the density of an
unknown gas, we have a convenient means of estimating its molecular
weight. This is one of many important examples of how a macroscopic
measurement (one made on bulk matter) can yield microscopic
information (that is, about molecular-scale objects).
Determination of the molecular weight of a gas from its density
is known as the Dumas method, after the French chemist JEAN DUMAS
(1800-1840) who developed it. One simply measures the weight of a
known volume of gas and converts this volume to its STP equivalent,
using Boyle's and Charles' laws. The weight of the gas divided by its
STP volume yields the density of the gas, and the density multiplied by
22.4Lmol–1 gives the molecular weight. Pay careful attention to the
examples of gas density calculations shown below.
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Example 1.16. Calculate the approximate molar mass of a gas whose
measured density is 3.33 g/L at 30oC and 780 torr.
Solution.
Data provided
Molar mass?
Density = 3.33 g/L
Volume = 1L
Temperature,T = 30oC = (30 +273)K
Pressure, P = 780 torr = (780/760) atm
From the ideal gas equation, the number of moles contained in one litre
of the gas is
𝑃𝑉
𝑛 = 𝑅𝑇
780
𝑎𝑡𝑚 × (1.00 𝐿)
760
=
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 393𝐾
= 0.413 𝑚𝑜𝑙
Now density, 𝑑 =
𝑚𝑎𝑠𝑠 (𝑚 )
𝑉𝑜𝑙𝑢𝑚𝑒 (𝑉)
Therefore, 𝑚 = 𝑑𝑣
But mass (m) = number of mole (n) × molar mass (M)
Therefore 𝑑𝑣 = 𝑛𝑚
M=
𝑑 ×𝑣
𝑛
Substituting gives
M=
33𝑔 𝐿−1 ×1.0 𝐿
0.0413 𝑚𝑜𝑙
= 80.6gmol-1
Example 1.17. The density of air at 15OC and 1.00 atm is 1.23g/L. What
is the molar mass of the air?
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The behaviour of gases
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Solution
First calculate the mole of air from which the molar mass can be gotten.
Data provided
Density = 1.23 g/L
Volume = 1L
Temperature,T = 15oC = (15 +273)K = 288K
Pressure, P = 1 atm
Molar mass?
From the ideal gas equation, the number of moles contained in one litre
of the air is
𝑃𝑉
𝑛=
𝑅𝑇
1 𝑎𝑡𝑚 × (1.00 𝐿)
=
0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾 −1 × 288𝐾
M=
𝑑 ×𝑣
= 0.0423 𝑚𝑜𝑙
𝑛
Substituting gives
M=
1.23𝑔 𝐿−1 ×1.0 𝐿
0.0423 𝑚𝑜𝑙
= 29.1gmol-1
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CHAPTER TWO
THE GAS LAWS
2.1. Introduction
Experience has shown that several properties of a gas can be
related to each other under certain conditions. The properties are
pressure (P), volume (V), temperature (T, in kelvins), and amount of
material expressed in moles (n). What we find is that a sample of gas
cannot have any random values for these properties. Instead, only
certain values, dictated by some simple mathematical relationships,
will occur. These properties and other variables such as rate of
diffusion of any gaseous substance bear a simple mathematical
relationship to each other. These are collectively called gas laws.
2.2. Pressure – Volume Relationship
Robert Boyle (1627–1691), an Irish physical scientist, discovered
that the volume of a given sample of a gas at a constant temperature is
inversely proportional to its pressure. This generalization, known as
Boyle’s law, applies approximately to any gas, no matter what its
composition. (It does not apply to liquids or solids.)
Inverse proportionality occurs when one variable gets larger by
the same factor as another gets smaller. For example, average speed and
the time required to travel a certain distance are inversely proportional.
If we double our speed, the time it takes us to complete the trip is
halved. Similarly, if the pressure on a given sample of gas at a given
temperature is doubled (increased by a factor of 2), its volume is
halved (decreased by a factor of 2).
Boyle might have observed the following data on volume and
pressure for a given sample of gas at a given temperature, under four
different sets of conditions:
Volume (L)
Pressure (atm)
1
4.00
1.00
2
2.00
2.00
3
1.00
4.00
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4
0.500
8.00
Note that tabulating data is very helpful when two or more variables
are being considered. The units are usually included in the column
headings in such a table. The data in the table show that the product of
the volume (V) and the pressure (P) is a constant. The table may be
expanded to show this relationship:
Volume (L)
Pressure (atm)
Volume × Pressure (L. atm)
1
4.00
1.00
4.00
2
2.00
2.00
4.00
3
1.00
4.00
4.00
4
0.500
8.00
4.00
Mathematically expression of the law;
1
[V α 𝑃 ]T
𝐾
[V = ]T
𝑃
PV = K
(Where K = constant of proportionality).
A more useful form of the law can be written as:
P1V1= P2V2
Where V1 and P1 refer to the original volume and pressure, V2 and P2
refer to the volume and pressure under the new or changed conditions.
If we place the values of P on the horizontal axis and the values
of V on the vertical axis, plot the preceding tabulated values for P and
V, and smoothly connect the points, we get a curve that can tell us
what the volume will be at any intermediate pressure (Figure 2.1a). We
can also plot 1/V versus P and get a straight line through the origin
(Figure 2.1b).
V (L)
1/V (1/L)
P (atm)
1
4.00
0.250
1.00
2
2.00
0.500
2.00
3
1.00
1.00
4.00
4
0.500
2.00
8.00
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(a)
(b)
Fig. 2.1. Graphical illustration of Boyle’s law: (a) Plot of V versus P. (b) Plot
of 1/V versus P.
2.3. Kinetic Theory and Boyle’s Law
The pressure of gas is due to continuous collision of the gaseous
molecules with the walls of the container. At constant temperature, the
average kinetic energy of the gas molecules is constant. If the size of
the container is reduced to a half (volume reduces), the frequency of
collision of the gas molecules with the walls of the container will be
doubled. This is due to the fact that the distance to the walls has been
reduced to a half. Therefore, the gas pressures will double the initial
value.
On the other hand if the volume of the container (size) is
doubled, the frequency of collision of the gas molecules with the walls
of the container will become reduced by a half, since the distance
between the molecules before colliding with the walls has been
doubled. Hence the pressure will be half of the initial value.
Example 2.1. A certain mass of a gas occupies 400cm3 at 1.0 × 105 Nm-2.
Calculate its volume when the pressure is 4.0 × 105 Nm-2 at constant
temperature.
36
The behaviour of gases
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Solution
In trying to solve this kind of problem, it is always good to collect the given
information together so as to easily detect the variable you are asked to find.
Data provided;
P1 = 1.0 × 105 Nm-2,
V1 =400cm3,
P2 = 2.0 × 105 Nm-2,
V2 = ?
According to boyle‘s law, P1V1= P2V2
Making V2 the subject,
P V
V2 = 1 1
P2
1 × 10 5 ×400
On substituting, V2 = 2 × 10 5 = 200cm3
Example 2.2. If 4 Liters of methane gas has a pressure of 1.0 atm, what
will be the pressure of the gas if we squish it down so it has a volume
of 2.5 L?
Solution
Data provided;
P1 = 1.0 atm
V1 = 4.0L
P2 = ?
V2 = 2.5 L
According to boyle‘s law, P1V1= P2V2
Making P2 the subject,
p v
P2 = v1 1
2
1.0 × 4
P2 = 2.5 = 1.6 𝑎𝑡𝑚
Example 2.3. A 3.50-L sample of gas has a pressure of 0.750 atm.
Calculate the volume after its pressure is increased to 1.50 atm at
constant temperature.
Solution
Alternatively, data collection can be in the form of table as shown
below
Pressure
Volume
1
0.750 atm
3.50 L
37
The behaviour of gases
2
1.50 atm
Using P1V1= P2V2
Making V2 the subject,
p v
V2 = 1 1
2016
?
P2
Substitution of the values into the equation yields
0.750 ×3.50
V2 = 1.50
= 1.75 𝐿
Note that multiplying the pressure by 2 causes the volume to be
reduced to half.
Example 2.4. A sample of gas initially occupies 35.0 mL at 1.50 atm.
Calculate the pressure required to reduce its volume to 20.5 mL at
constant temperature.
Solution
Data collection
Pressure
Volume
1
1.50 atm
35.0 mL
2
?
20.5 mL
Using P1V1= P2V2
Making P2 the subject,
p v
P2 = V1 1
2
Substitution of the values into the equation yields
1.50 ×35.0
P2 =
= 2.56 𝑎𝑡𝑚
20.5
Note that the units of pressure and volume must be the same on each side of
the equation P1V1= P2V2 . If the units given in a problem are not the same,
one or more of the units must be converted.
Example 2.5. A 1.45-L sample of gas has a pressure of 0.950 atm.
Calculate the volume after its pressure is increased to 787 torr at
constant temperature.
Solution
Because the pressures are given in two different units, one of them
must be changed.
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The behaviour of gases
1
Pressure
0.950 atm
2
787 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
760 𝑡𝑜𝑟𝑟
2016
Volume
1.45 L
= 1.036 𝑎𝑡𝑚
?
Using P1V1= P2V2
Making V2 the subject,
p v
V2 = 1 1
P2
Substitution of the values into the equation yields
V2 =
0.950 ×1.45
1.036
= 1.33 𝐿
Alternatively, we can change 0.950 atm to torr and still arrive at the
same answer.
(722 torr) (1.45 L) = (787 torr)V2
V2 = 1.33 𝐿
Note: 1 atm = 760 torr
Example 2.6. Calculate the initial volume of a sample of gas at 1.20 atm
if its volume is changed to 70.4 mL as its pressure is changed to 744
torr at constant temperature
Solution
Data collection
Pressure
1
1.20 atm
2
744 𝑡𝑜𝑟𝑟
1 𝑎𝑡𝑚
760 𝑡𝑜𝑟𝑟
Volume
?
= 0.979 𝑎𝑡𝑚
70.4 L
Using P1V1= P2V2
Making V1 the subject,
p v
V1 = 2 2
P1
Substitution of the values into the equation yields
0.979 ×70.4
V1 =
1.2
= 57.4 𝐿
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The behaviour of gases
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Example 2.7. Calculate the pressure required to change a 3.38-L sample
of gas initially at 1.15 atm to 925 mL, at constant temperature.
Solution
Collect the data and convert 925 mL to L (mL ≡ cm3, 1000mL = 1L)
Pressure
Volume
1
1.15 atm
3.38 L
2
?
925 mL = 0.925 L
Using P1V1= P2V2
Making P2 the subject,
p v
P2 = V1 1
2
Substitution of the values into the equation yields
1.15 ×3.38
P2 =
0.925
= 4.20 𝑎𝑡𝑚
The pressure must be raised to 4.20 atm.
Practice questions
1. State Boyle‘s law (i) in words (ii) mathematically
2. Explain Boyle‘s law in terms of kinetic theory.
3. Fill the following gaps: (Measurements
temperatures).
at
constant
Initial pressure
Initial volume
Final pressure
Final volume
1.0 × 105 Nm-2
1.0 × 105 Nm-2
800 mm Hg
300cm3
225cm3
3.50dm3
300cm3
1.5 × 105 Nm-2
760 mmHg
650 mmHg
900cm3
700 cm3
-
4. 30dm3 of oxygen at 10 atmospheres is placed in a 20dm3
container. Calculate the new pressure if temperature is kept
constant.
40
The behaviour of gases
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5. Calculate the initial pressure of a 485-mL sample of gas that has
been changed at constant temperature to 1.16 L and 1.18 atm.
2.4. Temperature – Volume Relationship
In 1787, 125 years after Boyle published the law that bears his
name, J. A. C. Charles (1746–1823) discovered a law relating the
volume of a given sample of gas to its absolute temperature. It took
more than a century to discover this law because of the requirement
that the temperature be absolute.
The volume of a sample of gas varies with the temperature, as
shown in Table 2.1 and plotted in Figure 2.2(a) for a particular sample.
Although the volume changes with the Celsius temperature, the
relationship is not a direct proportionality. That is, when the Celsius
temperature doubles, the volume does not double, all other factors
being held constant. On the graph, the plotted points form a straight
line, but the line does not pass through the origin. For a direct
proportionality to exist, the straight line must pass through the origin.
If the straight line corresponding to the points in Table 12.1 is extended
until the volume reaches 0 L, the Celsius temperature is -273K (Figure
2.2b). Charles defined a new temperature scale in which the lowest
possible temperature is absolute, corresponding to -273K. This
temperature is called absolute zero.
Table 2.1 Temperature and Volume Data for a Particular Sample of Gas at a
Given Pressure
Temperature(°C)
Volume(L)
1
0
0.400
2
100
0.548
3
200
0.692
4
300
0.840
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The behaviour of gases
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(a)
(b)
Fig. 2.2. Dependence of Volume on Temperature at Constant Pressure (a) Plot
of the data given in Table 2.1. (b) Extension of the line in part (a) to absolute
zero, with the Kelvin scale added to the horizontal axis.
We can state Charles‘ findings in simple terms:
At constant pressure, the volume of a fixed amount of gas is directly
proportional to its absolute temperature. This means an increase in the
temperature of a fixed mass of a gas leads to a corresponding increase
in the volume of the gas by the same proportion, and vice versa, with
the proviso that pressure remains the same.
Mathematically expression of the law;
[ V∝ T ]P
[ V= KT ] P
[ V/𝑇 = K ] P
(Where K = constant of proportionality).
A more useful form of the law can be express as:
𝑉1
𝑇1
=
𝑉2
𝑇2
Where V1 and T1 refer to the original volume and pressure, V2 and T2
refer to the volume and pressure under the new or changed conditions.
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The behaviour of gases
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2.5. Kinetic Theory and Charles’ Law
As the temperature of the gas molecules increase, the average
kinetic energy is equally raised, and hence, the average velocity of gas
molecules. The gas molecules move more rapidly colliding with one
another and more frequently with the walls of the container. For gas
pressure to remain constant, the volume of the container must be
increased with an increase in temperature.
Example 2.8. Assume that the volume of a balloon filled with H2 is 1.00
L at 25°C. Calculate the volume of the balloon when it is cooled to 78°C in a low-temperature bath made by adding dry ice to acetone.
Solution
Collect the given information and convert as necessary
Data provided;
V1 = 1.00L,
T1 = 250C = (25 + 273)K = 298K
T2 = -780C = (273 - 78)K =195K
V2 = ?
Applying Charles‘ law,
𝑉1
𝑇1
=
V2 =
𝑉2
𝑇2
195 ×1.00
293
= 0.65L
Example 2.9. The volume of a fixed mass of gas measured at
atmospheric pressure and 260C is 3.0 dm3. Calculate the volume at
1270C and at the same pressure.
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The behaviour of gases
2016
Solution
Data provided;
V1 = 3.0 dm3,
T1 = 260C = (25 + 273)K = 299K
T2 = 1270C = (273 + )K =400 K
V2 = ?
Applying Charles‘ law,
𝑉1
𝑇1
=
V2 =
𝑉2
𝑇2
400 ×3.00
299
= 4.0 dm3
Example 2.10. If 250cm3 of a gas at s.t.p. is heated to 270C at constant
pressure, calculate its new volume.
Solution
Data provided;
V1 = 250 cm3,
T1 = s.t = 273K
T2 = 270C = (273 +27 )K =300 K
V2 = ?
Applying Charles‘ law,
V1/ T1 = V2/ T2
V2 = V1 × T2/ T1
V2 = 250 × 300
273
44
The behaviour of gases
2016
= 274.7 cm3
Example 2.11. Show that the data in Table below prove (a) that the
Celsius temperature is not directly proportional to volume and (b) that
the Kelvin temperature is directly proportional to volume.
Temperature and Volume data for a particular Sample of gas at a given pressure
Solution
As the absolute temperature 273 K is increased to 373 K or 473 K, the
volume increases to 373/273 = 1.37 or 473/273 = 1.37 times the original
volume. The ratio of V to T is constant (see Table above). The volume is
directly proportional to absolute temperature.
Example 2.12. Calculate the Celsius temperature to which a 678-mL
sample of gas at 0oC must be heated at constant pressure for the
volume to change to 0.896 L.
Solution
Data provided
V1 = 678 mL = 0.678 L
T1 = 0oC = 273K
V2 = 0.896 L
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The behaviour of gases
2016
T2 = ?
Using the relationship
𝑉1
𝑇1
=
𝑉2
𝑇2
273 ×0.896
0.678
T2 =
= 361 𝐾
= (361 − 273)oC
= 88oC
Note: 1000 mL = 1L
Example 2.13. Calculate the original temperature of a 0.456-mL gas
sample if it is expanded at constant pressure to 1.75 L at 55°C.
Solution
Data provided
V1 = 0.456 mL = 0.000456 L
T1 = ?
V2 = 1.75 L
T2 = 55 OC = (273 + 55)K = 238K
Using the relationship below and making T1 the subject;
𝑉1
𝑇1
=
T1 =
𝑉2
𝑇2
238 ×0.000456
1.75
= 0.1 𝐾
K
= (0.1 − 273) oC
= −272.9 oC
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The behaviour of gases
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Example 2.14. A plastic bag of peanuts is laid on a windowsill in the
sun, where its temperature increases from 20OC to 30OC. If the original
volume is 100.0 cm3, what is the final volume after warming?
Solution
Data collection
V1 = 100 cm3
T1 = 20OC = 293 K
V2 = ?
T2 = 30 oC = 303 K
Using the relationship below and making
substituting;
𝑉1
𝑉
= 𝑇2
𝑇
1
V2 the subject and
2
303 ×100
293
V2=
= 103.4 cm3
Example 2.15. The temperature of a 4.00 L sample of gas is changed
from 10.0 °C to 20.0 °C. What will the volume of this gas be at the new
temperature if the pressure is held constant?
Solution
Data collection
V1 = 4.00L
T1 = 10OC = 283 K
V2 = ?
T2 = 20 OC = 293 K
47
The behaviour of gases
Using the relationship below and making
2016
V2 the subject and
substituting;
𝑉1
𝑇1
𝑉
= 𝑇2
2
𝑉2 =
=
𝑇2 𝑉1
𝑇1
293 ×4.00
283
= 4.1 𝐿
Example 2.16. Carbon dioxide is usually formed when gasoline is
burned. If 30.0 L of CO2 is produced at a temperature of 1.00 x103 °C
and allowed to reach room temperature (25.0 °C) without any pressure
changes, what is the new volume of the carbon dioxide?
Solution
Data collection
V1 = 30.0L
T1 = 1.00 x103 °C = (273 + 1000)K = 1273 K
V2 = ?
T2 = 25 OC = 298 K
Using the relationship below and making
substituting;
𝑉1
𝑇1
𝑉
= 𝑇2
2
𝑉2 =
=
𝑇2 𝑉1
𝑇1
298 × 30.00
1273
= 7.0 𝐿
48
V2 the subject and
The behaviour of gases
2016
Example 2.17. The volume of a gas syringe which contains 56.05
milliliters was raised to 67.7 milliliters at 107.5 oC. Determine the initial
temperature of the gas?
Solution
Data collection
V1 = 56.05 mm = 0.05605L
T1 =
V2 = 67.7 mm = 0.068L
T2 = 107.5 OC = 380.5 K
Using the relationship below and making T1 the subject and
substituting;
𝑉1
𝑇1
𝑉
= 𝑇2
2
𝑇1 =
=
𝑇2 𝑉1
𝑉2
380.5 × 0.05605
0.068
= 313.6 𝐾
= (313.6 − 273) = 40.6 oC
Example 2.18. If 15.0 liters of neon at 25.0 °C is allowed to expand to
45.0 liters, what is the new temperature?
Solution
Data provided
V1 = 15.0L
49
The behaviour of gases
2016
T1 = 25 °C = (273 + 25)K = 298 K
V2 = 45.0 L
T2 = ?
Using the relationship below and making
T2 the subject and
substituting;
𝑉1
𝑇1
𝑉
= 𝑇2
2
𝑇1 𝑉2
𝑉1
𝑇2 =
=
298 × 45.00
15
= 294 𝐾
Example 2.19. A balloon has a volume of 2500.0 mL on a day when the
temperature is 30.0 °C. If the temperature at night falls to 10.0 °C, what
will be the volume of the balloon if the pressure remains constant?
Solution
Data collection
V1 = 2500 mL
T1 = 30OC = 303 K
V2 = ?
T2 = 10 OC = 283 K
Using the relationship below and making
substituting;
𝑉1
𝑇1
𝑉
= 𝑇2
2
50
V2 the subject and
The behaviour of gases
𝑉2 =
=
2016
𝑇2 𝑉1
𝑇1
283 ×2500.00
303
= 2335 𝑚𝐿
2.6. Temperature-Pressure Relationship
Boyle‘s Law is the relationship between Pressure and Volume
but does not address temperature. How does temperature change
affect the properties of a sample of gas? Recall that temperature is a
measure of the average kinetic energy of particles. As the particles of a
substance move faster, the substance‘s temperature increases. The
particles bump into each other and the sides of the container more
often.
How would this affect a system where the volume is closed and
constant? This observation was first made by Gay-Lussac. He observed
that pressure has a direct proportional link with temperature of a
sample of gas in a closed container (volume constant). Properly put,
this law states that at constant volume, the pressure of a fixed mass of a
gas is directly proportional to its absolute temperature. The law is
expressed mathematically as follows:
𝑃 ∝ 𝑇 (Constant volume)
𝑃
𝑇
= 𝑘
A more useful form of the law can be express as:
𝑃1
𝑃
= 2
𝑇1
𝑇2
Where P1 and T1 refer to the original pressure and temperature, P2 and
T2 refer to the pressure and temperature under the new or changed
conditions.
Note: in solving or addressing mathematical problems with this law,
the temperature must be expressed in Kelvin and the pressure in a
standard uint.
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The behaviour of gases
2016
Example 2.20. 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What
would be the required temperature (in Celsius) to change the pressure
to standard pressure?
Solution
Data provided
P1 = 97.0 kPa
T1 = 25.0°C = ( 25 + 273)K = 298.0 K
P2 = s.p. = 101.325kPa
T2 = ?
𝑃
𝑃
Applying 𝑇1 = 𝑇2 and making T2 the subject
1
T2 =
=
2
𝑇1 𝑃2
𝑃1
298.0 𝐾 ×101.325 𝑘𝑃𝑎
97.0 𝑘𝑃𝑎
= 311K
Converting to degree in Celsius;
311K = (311 ‒ 273) °C
= 38°C
Example 2.21. If a gas in a closed container is pressurized from 15.0
atmospheres to 16.0 atmospheres and its original temperature was 25.0
°C, what would the final temperature of the gas be?
Solution
Data provided
P1 = 15 atm
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The behaviour of gases
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T1 = 25.0°C = ( 25 + 273)K = 298.0 K
P2 = 16 atm
T2 = ?
𝑃
𝑃
Applying 𝑇1 = 𝑇2 and making T2 the subject
1
T2 =
=
2
𝑇1 𝑃2
𝑃1
298.0 𝐾 ×16 𝑎𝑡𝑚
15 𝑎𝑡𝑚
= 317 K
Example 2.22. A 30.0 L sample of nitrogen inside a metal container at
20.0 °C was placed inside an oven whose temperature is 50.0 °C. The
pressure inside the container at 20.0 °C was 3.00 atm. What is the
pressure of the nitrogen after its temperature was increased?
Solution
Collect data and convert temperatures to Kevin
P1 = 3.00 atm
T1 = 25.0°C = ( 20 + 273)K = 293.0 K
P2 = ?
T2 = 50.0°C = ( 50 + 273)K = 323.0 K
Applying
P2 =
𝑃1
𝑇1
=
𝑃2
𝑇2
and making P2 the subject
𝑇2 𝑃1
𝑇1
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The behaviour of gases
=
2016
323.0 𝐾 ×3.00 𝑎𝑡𝑚
293 𝑎𝑡𝑚
= 3.3 atm
Example 2.23. The temperature of a sample of gas in a steel container at
30.0 kPa is increased from ‒100.0 °C to 1.00 x 103 °C. What is the final
pressure inside the tank?
Solution
Collect data and convert temperatures to Kevin
P1 = 30 kPa
T1 = ‒100.0 °C = (‒100.0 + 273)K = 173.0 K
P2 = ?
T2 = 1.00 x 103 °C = (1.00 x 103 + 273)K = 1273.0 K
Applying
P2 =
=
𝑃1
𝑇1
=
𝑃2
𝑇2
and making P2 the subject
𝑇2 𝑃1
𝑇1
1273 𝐾 ×30 𝑘𝑃𝑎
173 𝐾
= 220 kPa
2.7. The Combined Gas Law
Boyle‘s and Charles‘ laws may be merged into one law, called the
combined gas law, expressed in equation form as derived below:
From Boyle‘ law: V∝ 1/𝑃 (T constant)
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The behaviour of gases
2016
From Charles‘ law: V ∝ 𝑇 (P constant)
V ∝ 1/𝑃 ∝ T
𝑉 = 𝑘𝑇/𝑃
𝑃𝑉
=k
𝑇
That is, for a given sample of a gas, PV/T remains constant, and
therefore
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
(a given sample of a gas)
𝑇2
This expression is a mathematical statement of the combined (or
general) gas law. In words, the volume of given sample of a gas is
inversely proportional to its pressure and directly proportional to its
absolute temperature.
Note that if the temperature is constant, T1 = T2, then the
expression reduces to the equation for Boyle‘s law, P1V1 = P2V2.
Alternatively, if the pressure is constant, P1 = P2, the expression is
equivalent to Charles‘ law, V1/T1 = V2/T2.
When the initial volume V1 of a gas at temperature T1 and pressure P1
is subjected to changes in temperature to T2 and pressure to P2, its new
volume V2 is obtained from the equation.
To apply this gas law, the amount of gas should remain
constant. As with the other gas laws, the temperature must be
expressed in kelvins, and the units on the similar quantities should be
the same. Because of the dependence on three quantities at the same
time, it is difficult to tell in advance what will happen to one property
of a gas sample as two other properties change. The best way to know
is to work it out mathematically.
Example 2.24. A certain mass of a gas occupies 330 cm3 at 27oC and 9.0
× 104 Nm-2 pressure. Calculate its volume at s.t.p. (s.p = 1.0 × 105 Nm-2).
Solution
Write the given data down, convert as variable to appropriate units
and substitute into the form to find the unknown.
55
The behaviour of gases
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Data provided:
V1 = 330 cm3
P1 = 9.0 × 104 Nm-2
T1 = 27oC = (27 + 273)K = 300K
T2 = s.t. = 273K
P2 = s.p. = 1.0 × 105Nm-2
V2 = ?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
9.0 × 10 4 × 330 × 273
1.0 × 10 5 × 300
= 270 cm3
Example 2.25. Calculate the volume of a sample of gas originally
occupying 908 mL at 717 torr and 20OC after its temperature and
pressure are changed to 72OC and 1.07 atm.
Solution
In attempting this problem, the volume can be stated in millilitres in
both states. The pressure can be stated in atmospheres in both but the
temperature must be in kelvins in both states.
Data provided
V1 = 908 mL
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The behaviour of gases
717
760
P1 =
2016
𝑎𝑡𝑚 = 0.94 𝑎𝑡𝑚
T1 = 20oC = (20 + 273)K = 293K
T2 = 72 oC = 345K
P2 = 1.07 atm
V2 = ?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
0.94 × 908 × 345
1.07 × 293
= 943 mL
Example 2.27. Calculate the original volume of a sample of gas that is
at 700 torr and 22 oC before its volume, temperature, and pressure are
changed to 998 mL, 82°C, and 2.07 atm
Solution
Data provided
V1 = ?
P1 =
700
760
𝑎𝑡𝑚 = 0.92 𝑎𝑡𝑚
T1 = 22oC = (22 + 273)K = 295K
T2 = 82 oC = 355K
P2 = 2.07 atm
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The behaviour of gases
2016
V2 = 998 mL
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V1 the subject of the formula:
V1 =
=
𝑃2 𝑉2 𝑇1
𝑃1 𝑇2
2.07 × 998 × 298
0.92 × 355
= 1884 mL
Example 2.28. 17.3-mL sample of gas originally at standard
temperature and pressure is changed to 10.9 mL at 678 torr. Calculate
its final temperature in degrees
Celsius.
Solution
Data provided
V1 = 17.3 mL
P1 = s.p. = 760 torr
T1 = s.t. = 273 K
T2 = ?
P2 = 678 torr
V2 =10.9 mL
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making T2 the subject of the formula:
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The behaviour of gases
𝑃2 𝑉2 𝑇1
𝑃1 𝑉1
T2 =
=
2016
678 × 10.9 × 273
760 × 17.3
= 153 K
Converting to degree Celsius
= (153 − 273) oC
= 120 oC
Example 2.29. Calculate the volume at standard temperature and
pressure of a sample of gas that has a volume of 49.7 mL at 52°C and
811 torr.
Solution
Data provided
V1 = 49.7 mL
811
760
P1 =
𝑎𝑡𝑚 = 1.07 𝑎𝑡𝑚
T1 = 52°C = 325 K
T2 = s.t. = 273 K
P2 = s.p. = 1 atm
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making T2 the subject of the formula:
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
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The behaviour of gases
=
2016
1.07 × 49.7 × 273
1.0 × 325
= 45 mL
Example 2.30. Calculate the new volume after a 2.00-L sample of gas at
1.50 atm and 127oC is changed to 27oC at 3.50 atm.
Solution
Data provided
V1 = 2.00 L
P1 =1.50 𝑎𝑡𝑚
T1 = 127°C = 400 K
T2 = 27oC =300 K
P2 = 3.50 atm
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making T2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
1.5 × 2.00 × 300
3.50 × 400
= 0.64 L
Example 2.31. 500.0 liters of a gas are prepared at 700.0 mmHg and
200.0 °C. The gas is placed into a tank under high pressure. When the
tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the
volume of the gas?
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Solution
Data provided
V1 = 500 L
P1 = 700.0 𝑚𝑚𝐻𝑔 = 700 760 𝑎𝑡𝑚 = 0.92 𝑎𝑡𝑚
T1 = 200°C = 473 K
T2 = 20oC =293 K
P2 = 30.0 atm
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
0.92 × 500 × 293
30 × 473
= 9.5 L
Example 2.32. A gas balloon has a volume of 106.0 liters when the
temperature is 45.0 °C and the pressure is 740.0 mm of mercury. What
will its volume be at 20.0 °C and 780 .0 mm of mercury pressure?
Solution
Data provided
V1 = 106 L
P1 = 740.0 𝑚𝑚𝐻𝑔
T1 = 45°C = 318 K
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T2 = 20oC =293 K
P2 = 780.0 𝑚𝑚𝐻𝑔
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
740 × 106 × 293
780 × 318
= 92.7 L
Example 2.33. The volume of a gas originally at standard temperature
and pressure was recorded as 488.8 mL. What volume would the same
gas occupy when subjected to a pressure of 100.0 atm and temperature
of -245.0 °C?
Solution
Data provided
V1 = 488.8 L
P1 = 𝑠. 𝑝. = 1.0 𝑎𝑡𝑚
T1 = s.t. = 273 K
T2 = ‒245oC =28 K
P2 = 100 𝑎𝑡𝑚
V2 =?
Using the gas equation:
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The behaviour of gases
𝑃1 𝑉 1
𝑇1
=
2016
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
V2 =
=
1.0 × 488.8 × 28
100 × 273
= 0.5 L
Example 2.34. A gas is heated from 263.0 K to 298.0 K and the volume
is increased from 24.0 liters to 35.0 liters by moving a large piston
within a cylinder. If the original pressure was 1.00 atm, what would the
final pressure be?
Solution
Data provided
V1 = 24.0 L
P1 = 1.0 𝑎𝑡𝑚
T1 = 263.0 K
T2 =298.0 K
P2 = ?
V2 = 35.0 L
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making P2 the subject of the formula:
P2 =
𝑃1 𝑉1 𝑇2
𝑉2 𝑇1
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The behaviour of gases
=
2016
1.0 × 24.0 × 298
35 × 263
= 0.78 atm
Example 2.35. The pressure of a gas is reduced from 1200.0 mmHg to
850.0 mmHg as the volume of its container is increased by moving a
piston from 85.0 mL to 350.0 mL. What would the final temperature be
if the original temperature was 90.0 °C?
Solution
Data provided
V1 = 85.0 mL
P1 = 1200 𝑚𝑚𝐻𝑔
T1 = 90.0 °C = 363 K
T2 =?
P2 = 850 𝑚𝑚𝐻𝑔
V2 = 350.0 mL
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making T2 the subject of the formula:
T2 =
=
𝑃2 𝑉2 𝑇1
𝑉1 𝑃1
850 × 350 × 363
85 × 1200
= 1059 K
Converting to degree Celsius
= (1059 − 273) oC
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2016
= 786 oC
Example 2.36. If a gas is heated from 298.0 K to 398.0 K and the
pressure is increased from 2.230 x 103 mmHg to 4.560 x 103 mmHg
what final volume would result if the volume is allowed to change
from an initial volume of 60.0 liters?
Solution
Data provided
V1 = 60.0 L
P1 = 2.230 × 103 𝑚𝑚𝐻𝑔
T1 = 298.0 K
T2 =398.0 K
P2 = 4.560 × 103 𝑚𝑚𝐻𝑔
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
2.230 × 10 3 × 60.0 × 398
4.560 × 10 3 × 298
= 39.2 L
Example 2.37. A balloon containing a sample of gas has a temperature
of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The
balloon has a volume of 1,070 mL. The balloon is transported by plane
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2016
to Denver, where the temperature is 11°C and the pressure is 655 torr.
What is the new volume of the balloon?
Solution
Data provided
V1 = 1070 mL
P1 = 1.09 𝑎𝑡𝑚
T1 = 22°C = 295 K
T2 =11°C = 284 K
P2 = 655 𝑡𝑜𝑟𝑟 = 655 760 𝑎𝑡𝑚 = 0.86 𝑎𝑡𝑚
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
1.09 × 1070 × 284
0.86 × 295
= 1306 mL
Example 2.38. A balloon used to lift weather instruments into the
atmosphere contains gas having a volume of 1,150 L on the ground,
where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this
gas has a pressure of 6.88 torr and a temperature of −15°C. What is the
new volume of the gas?
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Solution
Data provided
V1 = 1150 L
P1 = 0.977 𝑎𝑡𝑚
T1 = 18°C = 291 K
T2 = −15°C = 258 K
P2 = 6.88 𝑡𝑜𝑟𝑟 = 6.88 760 𝑎𝑡𝑚 = 0.0091 𝑎𝑡𝑚
V2 =?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
=
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
0.977 × 1150 × 258
0.0091 × 291
= 109465 L
2.8. Relationship between Amount and Volume
2.8.1. Gay-Lussac's Law of Combining Volumes
In the same 1808 article in which Gay-Lussac published his
observations on the thermal expansion of gases, he pointed out that
when two gases react, they do so in volume ratios that can always be
expressed as small whole numbers. This came to be known as the Law
of combining volumes.
Example 2.39. Ammonium carbonate decomposes when heated to
yield carbon dioxide, ammonia, and water vapour. Calculate the ratio
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The behaviour of gases
2016
of the (separate) volume of ammonia to that of water vapour, each at
450°C and 1.00 atm.
Solution
The mole ratio of the gases, given in the balanced equation, is
2 mol NH3: 1 mol CO2: 1 molH2O
The ammonia and water vapour are separated and measured at the
given temperature and pressure. The ratio of their volumes can be
calculated as follows:
𝑉 𝑁𝐻 3
𝑉𝐻 2 𝑂
=
𝑛 𝑁𝐻 3 𝑅𝑇 /𝑃
𝑛 𝐻 2 𝑂 𝑅𝑇 /𝑃
Because R is a constant and both T and P are the same for the two
gases, this equation reduces to
𝑉 𝑁𝐻 3
𝑉𝐻 2 𝑂
=
𝑛 𝑁𝐻 3
𝑛 𝐻 2𝑂
The right side of this equation is the ratio of the numbers of moles—the
ratio given by the balanced chemical equation. The left side of the
equation is the ratio of the volumes, so the ratio given by the balanced
chemical equation is equal to the volume ratio under these conditions.
The ratio is 2: 1.
Example 2.40. If 2.00 L H2 of and 1.00 L of both at standard temperature
and pressure, are allowed to react, will the water vapor they form at
250°C and 1.00 atm occupy 2.00 L?
Solution
2H2 (g) + O2 (g) → 2H2O (g)
The volumes of H2 and O2 that react are in the ratio given in the
balanced equation because the two gases have the same temperature and
pressure. The volume of water vapour formed is not in that ratio,
however, because its temperature is different. Its volume will be much
greater than 2 L.
2.8.2. Avogadro's Law
The work of the Italian scientist Amedeo Avogadro
complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811
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The behaviour of gases
2016
he published a hypothesis stating that at the same temperature and
pressure, equal volumes of different gases contain the same number of
molecules (or atoms if the gas is monatomic). This law states that equal
volumes of all gases, under the same conditions of temperature and
pressure, contain the same number of molecule.
Mathematically: V ∝ 𝑛 (at constant T and P)
𝑉 = 𝑘𝑛
𝑉
=𝑘
𝑛
Where V is the volume of gas, n is the number of molecules and 𝑘 is
the proportionality constant.
This law relates the volume of a fixed mass of a gas to the
number of molecules it contains. It shows that the volume occupied by
a gas depends on the number of molecules it contains, at a given
temperature and pressure. An increase in the number of gas molecules
leads to an increase in gas volume, and vice versa.
According to Avogadro‘s law we see that when two gases react
with each other, their reacting volumes have a simple ratio to each
other. If the product is a gas, its volume is related to the volume of the
reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac).
For example, consider the synthesis of ammonia from molecular
hydrogen and molecular nitrogen:
3H2(g) + N2(g)
→ 2NH3(g)
3 mol
1 mol
2 mol
Because, at the same temperature and pressure, the volumes of gases
are directly proportional to the number of moles of the gases present,
we can now write
3H2(g) + N2(g)
→ 2NH3(g)
3 volume 1 volume 2 volume
The volume ratio of molecular hydrogen to molecular nitrogen is 3:1,
and that of ammonia (the product) to molecular hydrogen and
molecular nitrogen combined (the reactants) is 2:4, or 1:2.
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The behaviour of gases
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Example 2.41. 50 cm3 of sulphur (IV) oxide were produced at s.t.p.
when some quantity of powdered sulphur was burnt in excess oxygen.
(a) Write a balanced chemical equation for the reaction. (b) Calculate
the volume of oxygen used up during the reaction. (c) Which of the
laws is applicable? State the law.
Solution
(a). S(g) + O2(g) → SO2(g)
(b). From the balanced chemical equation in (a) above;
At s.t.p: 22400 cm3 of SO2 required 22400 cm3 of O2
Hence 1 cm3 of SO2 will require 1 cm3 of O2
∴ 50 cm3 of SO2 will use 50 cm3 of O2
(c). Avogadro‘s law is applicable in (b) above and it state that at the
same temperature and pressure equal volume of gases contain the
same number of molecules.
2.9. The Ideal Gas Law
So far, the gas laws we have used have focused on changing one or
more properties of the gas, such as its volume, pressure, or
temperature. There is one gas law that relates all the independent
properties of a gas under any particular condition, rather than a change
in conditions. This gas law is called the ideal gas law. The general
ideal gas equation is a combination of Boyle‘s, Charles‘ and
Avogadro‘s laws involving the four gas variables: pressure (P), volume
(V), number of mole of gas (n), and temperature (T).
From Boyle‘ law: V∝ 1/𝑃 (T constant)
From Charles‘ law: V ∝ 𝑇 (P constant)
From Avogadro‘s law: V ∝ 𝑛 (P,T constant)
V ∝ 1/𝑃 ∝ T ∝ 𝑛
V = R × 1/𝑃 × T × 𝑛
PV = nRT
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The behaviour of gases
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In this equation, P is pressure, V is volume, n is amount of moles,
and T is temperature. R is called the ideal gas law constant and is a
proportionality constant that relates the values of pressure, volume,
amount, and temperature of a gas sample. The variables in this
equation do not have the subscripts i and f to indicate an initial
condition and a final condition. The ideal gas law relates the four
independent properties of a gas under any conditions.
2.10. Evaluation of the Gas Constant, R
The gas constant can be expressed in various units, all having
the dimension of energy per degree per mol.
From the general equation PV = nRT we get:
𝑃𝑉
𝑅=
𝑛𝑇
Where P is pressure, V is volume, n is amount, and T is temperature.
R is most easily calculated from the fact that the hypothetical volume
of an ideal gas is 22.4L at STP (273.K and 1 atm).
i. If volume is expressed in liters and pressure in atmospheres,
then the proper value of R is as follows:
𝑃𝑉
𝑅=
𝑛𝑇
1.0 𝑎𝑡𝑚 ×22.414 𝐿
= 1.0 𝑚𝑜𝑙 ×273.15 𝐾
ii. if pressure is in
thus:
Nm-2
R= 0.08206 atm L mol-1K-1
and volume in m3 then the proper value of R is
𝑃𝑉
𝑛𝑇
Where P = 101325 Nm-2, V = 22.4/1000 = 0.0224 m3
𝑅=
101325 𝑁𝑚 −2 ×0.0224 𝑚 3
R=
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 8.314N m mol-1K-1
iii. if pressure is in atm and volume in cm3 then
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The behaviour of gases
𝑅=
2016
𝑃𝑉
𝑛𝑇
=
1.0 𝑎𝑡𝑚 ×22414 𝑐𝑚 3
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 82.06 atm cm3 mol-1K-1
[22.414 L =22400 cm3]
iv. if pressure is in Pa and volume in liter
𝑅=
𝑃𝑉
𝑛𝑇
[1 atm = 1.01325× 105 Pa; 1 L= 10−3m3]
=
1.01325 ×10 5 𝑃𝑎 ×22.414 × 10 −3 𝑚 3
1.0 𝑚𝑜𝑙 ×273.15 𝐾
= 8.314 Pa m3 mol-1K-1
v. In JK−1 mol−1,
R = 8.314 kgm2s−2
= 8.314 JK−1 mol−1 [1 Pa = 1 kgm−1 s−2]
vi. In cal K−1 mol−1 (1 cal = 4.184 J),
R = 1.987 calK−1mol−1
Example 2.42. 50.0 g of N2 (M = 28.0 g) occupies a volume of 750mL at
298.15 K. Assuming the gas behaves ideally, calculate the pressure of
the gas in kPa.
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The behaviour of gases
2016
Solution
Collect the data, convert volume to liter and find the number of mole
of nitrogen before substituting into the ideal gas equation to find
pressure
Data given
Mass of N2 = 50g
Molar mass of N2 = 28g/mol
Volume , V = 750mL =
750
1000
𝐿 = 0.750 𝐿
Temperature, T = 298.15 K
𝑚𝑎𝑠𝑠
𝑚𝑎𝑠𝑠
Number of mole of nitrogen gas (n) = 𝑚𝑜𝑙𝑎𝑟
50 𝑔
= 28𝑔𝑚𝑜𝑙 −1
=1.79 mol
Using PV = nRT
𝑃=
=
𝑛𝑅𝑇
𝑉
1.79 𝑚𝑜𝑙 ×0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙 −1 𝐾 −1 ×298.15 𝐾
0.750 𝐿
= 58.39 atm
Converting to kPa
=
(58.39 × 101325 )
1000
= 5.916 × 103 𝑘𝑃𝑎
Example 2.43. Calculate the volume occupied by 2.5 moles of an ideal
gas at -23 oC, and 4.0 atmospheres. [R = 0.082 atm dm3 K-1 mol-1]
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The behaviour of gases
2016
Solution
Data provided:
P = 4.0 atm
T = -23 + 273 = 250K
n = 2.5 moles
R = 0.082 atm dm3 K-1 mol-1
V=?
Applying PV = nRT
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
2.5 ×0.082 ×250
4
=12.8 dm3
Example 2.44. Calculate the volume of 1.63 mol of carbon dioxide gas
at 295 K and 1.14 atm.
Solution
Data provided:
P = 1.14 atm
T = 295K
n = 1.63 moles
R = 0.082 atm L K-1 mol-1
V=?
Applying PV = nRT
Making V the subject, and substituting:
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The behaviour of gases
V=
=
2016
𝑛𝑅𝑇
𝑃
1.63 ×0.082 ×295
1.14
=34.6 L
Example 2.45. Calculate the volume of 0.898 mol of methane gas, CH4,
at 292 K and 1.06 atm.
Solution
Data provided:
P = 1.06 atm
T = 292 K
n =0 .898 moles
R = 0.082 atm L K-1 mol-1
V=?
Applying PV = nRT
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
0.898 mol ×0.082 atm L mol −1 K −1 ×292 K
1.06 atm
= 20.3 L
Example 2.46. Calculate the volume of 42.6 g of oxygen gas at 35oC and
792 torr
Solution
1. First convert temperature to Kelvin and pressure to atm.
2. Find the number of mole of oxygen
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2016
3. Plug data into idea gas equation to find the volume
Data provided:
P=
792
760
𝑎𝑡𝑚 = 1.04 atm
T = 35oC = 308 K
R = 0.082 atm L K-1 mol-1
n = mass/molar mass = 42.6g/32gmol-1 = 1.33 mol
V=?
Applying PV = nRT
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
1.33 mol ×0.082 atm L mol −1 K −1 ×308 K
1.04 atm
=32.3 L
Students sometimes wonder “How do I decide when to use the combined gas
law and when to use the ideal gas law?” The answer depends on the problem,
naturally. If moles are involved, the combined gas law cannot be used.
Example 2.47. Decide which gas law should be used to solve each of
the following:
(a) Calculate the final volume of a sample of gas that has an initial
volume of 7.10 L at STP if the temperature and pressure are changed to
33oC and 696 torr.
(b) Calculate the volume of 0.977 mol of gas at 33oC and 792 torr.
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The behaviour of gases
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Solution
(a) The combined gas law can be used because it does not involve
number of moles and initial and final conditions are involved.
(b) This problem involves moles and must be solved with the ideal gas
law.
Example 2.48. Calculate the pressure of 0.0789 mol of chlorine gas that
occupies 891 mL at ‒15°C.
Solution
The quantities given are converted to the units generally used with the
ideal gas law equation. Note that the nature of the gas is immaterial as
long as the number of moles is known.
Data provided:
T = ‒15°C = (‒15 +273) K = 258 K
n =0 .0789 moles
R = 0.082 atm L K-1 mol-1
V = 891 mL = (891/1000) L = 0.891 L
P=?
Applying PV = nRT
Making P the subject, and substituting:
P=
=
𝑛𝑅𝑇
𝑉
0.0789 mol ×0.082 atm L mol −1 K −1 ×258 K
0.891 L
=1.87 atm
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The behaviour of gases
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Example 2.49. Calculate the pressure of 0.0855 mol of neon gas that
occupies 66.1 mL at 25°C.
Solution
Data provided:
T = 25°C = (25 +273) K = 298 K
n =0 .0855 moles
R = 0.082 atm L K-1 mol-1
V = 66.1 mL = (66.1/1000) L = 0.0661 L
P=?
Applying PV = nRT
Making P the subject, and substituting:
P=
=
𝑛𝑅𝑇
𝑉
0.0855 mol × 0.082 atm L mol −1 K −1 ×298 K
0.0661 L
= 31.6 atm
Example 2.50. Calculate the number of moles of oxygen gas in a 2.60-L
container at 19°C and
755 torr.
Solution
Data provided:
T = 19°C = (19 +273) K = 292 K
R = 0.082 atm L K-1 mol-1
V = 2.60 L
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The behaviour of gases
P = 755 torr =
755
760
2016
𝑎𝑡𝑚 = 0.99 atm
n =?
Applying PV = nRT
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
0.99 atm ×2.60 L
0.082 atm L mol −1 K −1 ×292 K
= 0.12 mol
5.0g of neon is at 256 mm Hg and at a temperature of 35º C. What is the
volume?
Solution
Step 1: Write down your given information:
P = 256 mmHg
V=?
m = 5.0 g
R = 0.082 L atm mol-1K-1
T = 35oC
Step 2: Convert as necessary:
T = 35oC = (35 + 273)K = 308 K
P = 256 mmHg = (256/760) atm = 0.34 atm
n = mass/molar mass = 5.0g/20.1797 gmol-1 = 2.5 mol
Applying PV = nRT
Making V the subject, and substituting:
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The behaviour of gases
V=
=
2016
𝑛𝑅𝑇
𝑃
2.5 mol ×0.082 atm L mol −1 K −1 ×308 K
0.34 atm
=186 L
Example 2.51. What is a gas‘s temperature in Celsius when it has a
volume of 25 L, 203 mol, 143.5 atm?
Solution
Data provided:
R = 0.082 atm L K-1 mol-1
V = 25 L
P = 143.5 atm
n = 203 mol
T =?
Applying PV = nRT
Making T the subject, and substituting:
T=
=
𝑃𝑉
𝑛𝑅
143.5 atm × 25 L
0.082 atm L mol −1 K −1 ×203 mol
= 215.5 K
Converting to degree Celsius
= (215.5 − 273) oC
= −57.5 oC
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The behaviour of gases
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Example 2.52. Sodium azide (NaN3) is used in some automobile air
bags. The impact of a collision triggers the decomposition of NaN3 as
follows:
2NaN3(s) →2Na(s) + 3N2(g)
The nitrogen gas produced quickly inflates the bag between the driver
and the windshield and dashboard. Calculate the volume of N2
generated at 85°C and 812 mmHg by the decomposition of 50.0 g of
NaN3.
Strategy From the balanced equation we see that 2 mol NaN3 gives 3
mol N2 so the conversion factor between NaN3 and N2 is
3 mol N2
2 𝑚𝑜𝑙 𝑁𝑎𝑁3
Because the mass of NaN3 is given, we can calculate the number of
moles of NaN3 and hence the number of moles of N2 produced. Finally,
we can calculate the volume of N2 using the ideal gas equation.
Solution
The sequence of conversions is as follows:
grams of NaN3 → moles of NaN3 →moles of N2 →volume of N2
First, we calculate the number of moles of N2 produced by 50.0 g of
NaN3:
mole of NaN3 = mass/ molar mass
= 50g/65.02gmol-1
= 0.769 mol
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The behaviour of gases
Mole of N2 = 0.769 mol ×
2016
3 mol N 2
2 𝑚𝑜𝑙 𝑁𝑎𝑁 3
= 1.15 mol N2
The volume of 1.15 mol of N2 can be obtained by using the ideal gas
equation:
PV = nRT
Making V the subject, and substituting:
V=
=
=
𝑛𝑅𝑇
𝑃
1.15 mol ×0.082 atm L mol −1 K −1 ×(85+273) K
812
) atm
760
(
33.7594 𝐿
1.068
=31.6 L
Example 2.53. The equation for the metabolic breakdown of glucose
(C6H12O6) is the same as the equation for the combustion of glucose in
air:
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
Calculate the volume of CO2 produced at 37°C and 1.00 atm when 5.60
g of glucose is used up in the reaction. [C=12, O=16,H=1]
Solution
Collect given data and convert as necessary
R = 0.082 atm L K-1 mol-1
V=?
P = 1.00 atm
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n=?
T = 37°C = 310 K
Mass of glucose = 5.6g
Strategy
1. First calculate the molar mass of glucose
2. Calculate the mole of C6H12O6 from which the mole of CO2 can
be gotten since 1 mole of C6H12O6 produced 6 moles of CO2
from the equation.
3. Use ideal gas equation to find the volume of CO2 produced.
C6H12O6 =[(12× 6 )+ (1 × 12) + (16× 6)] = 179gmol-1
Mole of C6H12O6 = 5.6g/179gmol-1
= 0.03 mol
From the balanced equation we see that 1 mol C6H12O6 gives 6 mol CO2
so the conversion factor between C6H12O6 and CO2 is
6 mol CO2
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
Mole of CO2 = 0.03 mol ×
6 mol CO 2
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
=0.18 mol CO2
The volume of 0.18 mol of CO2 can be obtained by using the ideal gas
equation:
PV = nRT
Making V the subject, and substituting:
V=
𝑛𝑅𝑇
𝑃
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The behaviour of gases
=
2016
0.18 mol ×0.082 atm L mol −1 K −1 ×310 K
1.0 atm
= 4.6 𝐿
Example 2.54. Assuming ideal behaviour, which of the following
samples of gases will have the greatest volume at STP? Which of these
gases will have the greatest density at STP? (a) 0.82 mole of He. (b) 24 g
of N2. (c) 5.0 × 1023 molecules of Cl2
Solution
At STP, [T = 273K, P=1.0atm, R= 0.082 atm L mol−1 K −1 ]
(a)
n = 0.82 mole He
Using PV = nRT
Making V the subject, and substituting:
V=
𝑛𝑅𝑇
𝑃
=
0.82 mol ×0.082 atm L mol −1 K −1 × 273 K
1.0 atm
= 18.4 L
𝑚
𝑉
Density, d of He =
But mass, 𝑚 = 𝑛𝑚 = 0.82 × 4.003
= 0.33𝑔
∴ dendity, d =
0.88 𝑔
18.4 𝐿
=0.018g/L
(b)
24 g of N2
No of mole of nitrogen 𝑛 =
84
𝑚𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
The behaviour of gases
=
2016
24g
14𝑔𝑚𝑜𝑙 −1
= 1.71 𝑚𝑜𝑙
Using PV = nRT to calculate the volume of N2 at STP
Making V the subject, and substituting:
V=
𝑛𝑅𝑇
𝑃
=
1.71 mol ×0.082 atm L mol −1 K −1 × 273 K
1.0 atm
= 38.3 L
Density, d of N2 =
𝑚
𝑉
=
24 𝑔
38.3 𝐿
= 0.627 g/L
(c)
5.0 × 1023 molecules of Cl2
Using PV = nRT to calculate the volume of chlorine molecule at STP,
Make V the subject, and substite:
V=
𝑛𝑅𝑇
𝑃
=
5.0 × 10 23 mol ×0.082 atm L mol −1 K −1 × 273 K
1.0 atm
= 1.1193 × 1025 L
Density, d of Cl2 =
=
𝑚
𝑉
35.5 𝑔
1.1193 ×10 25 𝐿
= 3.17 × 10−24 g/L
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The behaviour of gases
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Summary of the results
He gas
N2 gas
Cl2 gas
Volume at STP (L)
18.4
38.3
1.1193 × 1025
Density at STP (g/L)
0.018
0.627
3.17 × 10−24
Results from the calculations showed chlorine gas has the highest
volume at STP 1.1193 × 1025 L while nitrogen gas has the greatest
density of 0.627 g/L.
Example 2.55. How many moles of O2 are present in a 0.500-L sample
at 25oC and 1.09 atm?
Solution
Collect the given data and convert as necessary
T = 25°C = (25 +273) K = 298 K
R = 0.082 atm L K-1 mol-1
V = 0.500 L
P = 1.09 atm
n =?
Applying PV = nRT to find n of O2;
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
1.09 atm ×0.500 L
0.082 atm L mol −1 K −1 × 298 K
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The behaviour of gases
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=0.022 mol of O2
Example 2.56. What is the volume of 1.00 mol of gas at STP?
Solution
Data provided:
P = s.p. = 1.0 atm
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
n = 1.0 mol
V=?
Applying PV = nRT
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
1.0mol ×0.082 atm L mol −1 K −1 ×273 K
1.0 atm
= 22.4 L
Note that the volume of 1.00 mol of gas at STP is called the molar volume of a
gas. This value should be memorized.
Example 2.57. How many moles of SO2 are present in a 765-mL sample
at 37oC and 775 torr?
Solution
Since R is defined in terms of liters and atmospheres, the pressure and
volume are first converted to those units.
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The behaviour of gases
2016
Collect the given data
T = 37°C = (37 +273) K = 310 K
R = 0.082 atm L K-1 mol-1
V = 765 mL = (765/1000)L = 0.765 L
P = 775 torr = (775/760) atm = 1.02 atm
n =?
Applying PV = nRT to find n of SO2;
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
1.02 atm ×0.765 L
0.082 atm L mol −1 K −1 × 310 K
= 0.03 mol of SO2
Example 2.57. At what temperature will 0.0750 mol of CO2 occupy 2.75
L at 1.11 atm?
Solution
Collect the given data
V = 2.75 L
P = 1.11 atm
n = 0.0750 mol
T =?
R = 0.082 atm L K-1 mol-1
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The behaviour of gases
2016
Applying PV = nRT and Making T the subject, and substituting:
T=
=
𝑃𝑉
𝑛𝑅
1.11 atm × 2.75 L
0.0750 mol × 0.082 atm L mol −1 K −1
= 496 K
Example 2.58. What volume will 7.00 g of Cl2 occupy at STP?
Solution
The value of n is not given explicitly in the problem, but the mass is
given, from which we can calculate the number of moles:
Data provided
V=?
P = s.p. = 1.0 atm
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
Number of mole of Cl2 =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑕𝑙𝑜𝑟𝑖𝑛𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
=
7.00 𝑔
35.5 × 2 𝑔/𝑚𝑜𝑙
= 0.0986 mol of Cl2
Applying PV = nRT to find the volume of Cl2
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
0.0986 mol × 0.082 atm L mol −1 K −1 273 K
1.0 atm
= 2.23 L
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The behaviour of gases
2016
Example 2.59. If 4.58 g of a gas occupies 3.33 L at 27oC and 808 torr,
what is the molar mass of the gas?
Solution
If you do not see at first how to solve this problem to completion, at
least you can recognize that P, V, and T data are given. First calculate
the number of moles of gas present from which you can get the molar
mass.
Collect the given data
T = 27°C = (27 +273) K = 300 K
R = 0.082 atm L K-1 mol-1
V = 3.33 L
P = 808 torr = (808/760) atm = 1.06 atm
Mass (m) = 4.58g
n =?
Applying PV = nRT to find n of gas;
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
1.06 atm × 3.33 L
0.082 atm L mol −1 K −1 × 300 K
= 0.143 mol of gas
𝑚𝑎𝑠𝑠 (𝑚)
𝑚𝑎𝑠𝑠 (𝑀)
Recall, number of mole (n) = 𝑚𝑜𝑙𝑎𝑟
𝑚𝑎𝑠𝑠 (𝑚 )
𝑜𝑓 𝑚𝑜𝑙𝑒 (𝑛)
Therefore molar mass of gas (M) = 𝑛𝑢𝑚𝑏𝑒𝑟
4.58 𝑔
= 0.143 𝑚𝑜𝑙
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The behaviour of gases
2016
= 32.0g/mol
Example 2.60. What volume is occupied by the oxygen liberated by
heating 0.250 g of KClO3 until it completely decomposes to KCl and
oxygen? The gas is collected at STP.
Solution
From the balanced equation below, we see that 2 mol KClO3 gives 3
mol O2 so the conversion factor between KClO3 and O2 is
3 mol O2
2 𝑚𝑜𝑙 KClO3
2 KClO3 → 2KCl + 3O2
Because the mass of KClO3 is given, we can calculate the number of
moles of KClO3 and hence the number of moles of O2 produced.
Finally, we can calculate the volume of O2 using the ideal gas equation.
Collect the given data
T = s.t. = 273 K
R = 0.082 atm L K-1 mol-1
V= ?
P = 𝑠. 𝑝. = 1.0 atm
Mass (m) = 0.250 g of KClO3
n =?
Molar mass of KClO3 [39.10 +35.5 +(16× 3)] = 122.6g/mol
mole of KClO3 = mass/ molar mass
= 0.250g/122.6gmol-1
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The behaviour of gases
2016
= 0.002 mol KClO3
Mole of O2 = 0.002 mol ×
3 mol O 2
2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 3
= 0.003 mol O2
The volume of 0.003 mol O2 can be obtained by using the ideal gas
equation:
PV = nRT
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
0.003 mol ×0.082 atm L mol −1 K −1 × 273 K
1.0 atm
= 0.067 L
5.0 moles of oxygen gas are contained in a 1.13 dm3 container at 127oC.
wha is the pressure of the system in Nm-2? [R = 8.314NmK-1mol-1]
Solution
Collect information provided and convert as necessary
T = 127oC = 400 K
n = 0.32 mol
R = 8.314NmK-1mol-1
V = 1.13 dm3 = (1.13/1000) m3 = 1.13 × 10−3 𝑚3
[1000 dm3 = 1 m3]
P =?
Use PV = nRT to find P
Making P the subject, and substituting:
92
The behaviour of gases
P=
=
2016
𝑛𝑅𝑇
𝑉
0.32 mol ×8.314 Nm mol −1 K −1 ×400 K
0.00113 m 3
= 941762.8 𝑁𝑚−2
= 9.43 × 105 𝑁𝑚−2
Example 2.61. A vessel contains 2.5 dm3 of oxygen gas at 29oC under
2.1 atmospheres. Estimate the amount of the gas at STP.
Solution
We will use the general gas equation to get the volume of oxygen gas
at STP then apply ideal gas law to get the amount of the gas.
Collect information provided and convert as necessary
T1 = 29oC = (29 + 273) K = 302 K
P1 = 2.1 atm
V1 = 2.5 dm3
T2 = s.t. = 273 K
P2 = s.p. = 1.0 atm
V2 = ?
n=?
Using the gas equation:
𝑃1 𝑉 1
𝑇1
=
𝑃2 𝑉 2
𝑇2
Making V2 the subject of the formula:
V2 =
𝑃1 𝑉1 𝑇2
𝑃2 𝑇1
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The behaviour of gases
=
2016
2.1 atm × 2.5 dm 3 × 273 K
1.0 atm × 302 K
= 4.75 dm3
Now we can apply ideal gas equation in order to find the amount of
oxygen gas at s.t.p;
PV = nRT
Making n the subject, and substituting:
n=
𝑃𝑉
𝑅𝑇
But P = 1 atm = 101325 𝑁𝑚−2
Volume, V = 4.75 dm3 = (4.75/1000) m3 = 4.75 × 10−3 𝑚3
Temperature, T = 273 K
R = 8.314NmK-1mol-1
n=
101325 Nm −2 × 4.75 × 10 −3 m 3
8.314 NmK −1 mol −1 × 273 K
= 0.212 mol of O2 gas
Example 2.62.Calculate the volume occupied by 40g of carbon dioxide
(CO) at 4.58 × 104 𝑁𝑚−2 and 50oC , assuming ideal gas law is obeyed.
[O = 16, C =12]
Solution
First we calculate the molar mass of CO and find its number f mole
then apply ideal gas equation to get the volume.
Collect the given data and convert as necessary
T = 50°C = (50 +273) K = 323 K
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The behaviour of gases
2016
R = 8.314NmK-1mol-1
V= ?
P = 4.58 × 104 𝑁𝑚−2
Mass of CO(m) = 40g
n =?
Molar mass of CO = 12 +16 = 28g/mol
Recall, number of mole (n) =
𝑚𝑎𝑠𝑠 (𝑚)
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
40 𝑔
= 28𝑔/ 𝑚𝑜𝑙
= 1.43 mol
Applying PV = nRT to find V of CO gas;
Making V the subject, and substituting:
V=
=
𝑛𝑅𝑇
𝑃
1.43 mol 8.314 Nm mol −1 K −1 × 323 K
4.58 × 10 4 𝑁𝑚 −2
= 0.082 m3 of CO gas
= 82.0 dm3 of CO gas
[1000dm3 = 1m3]
Example 2.63. How many moles of a gas are contained in 890.0 mL at
21.0 °C and 750.0 mmHg pressure?
Solution
Collect the given data and convert to appropriate units
T = 21°C = (21 +273) K = 294 K
R = 0.082 atm L K-1 mol-1
95
The behaviour of gases
2016
V = 890 mL = (890/1000)L = 0.890 L
P = 750.0 mmHg = (750/760) atm = 0.99 atm
n =?
Applying PV = nRT to find n of gas;
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
0.99 atm × 0.890 L
0.082 atm L mol −1 K −1 × 294 K
= 0.037 mol of gas
Example 2.64. Calculate the molecular weight of a gas if 35.44 g of the
gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a constant
temperature of 35.5 °C.
Solution
We first find the number of mole of the gas using ideal gas equation
and then relate the mole and molar mass.
Collect the given data and convert as necessary
T = 27°C = (35.5 +273) K = 308.5 K
R = 0.082 atm L K-1 mol-1
V = 7.50 L
P = 60.0 atm
Mass (m) = 20.44g
n =?
96
The behaviour of gases
2016
Applying PV = nRT to find n of gas;
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
60.0 atm × 8.5 L
0.082 atm L mol −1 K −1 × 308.5 K
= 20.1 mol of gas
Recall, number of mole (n) =
𝑚𝑎𝑠𝑠 (𝑚)
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
𝑚𝑎𝑠𝑠 (𝑚 )
𝑜𝑓 𝑚𝑜𝑙𝑒 (𝑛)
Therefore molar mass of gas (M) = 𝑛𝑢𝑚𝑏𝑒𝑟
=
20.44 𝑔
20.1 𝑚𝑜𝑙
= 1.0 g/mol
2.11. Dalton’s Law of Partial Pressure
Thus far we have concentrated on the behaviour of pure
gaseous substances, but experimental studies very often involve
mixtures of gases. For example, for a study of air pollution, we may be
interested in the pressure-volume-temperature relationship of a sample
of air, which contains several gases. In this case, and all cases involving
mixtures of gases, the total gas pressure is related to partial pressures,
that is, the pressures of individual gas components in the mixture. In 1801
Dalton formulated a law, now known as Dalton’s law of partial
pressures, which states that the total pressure of a mixture of gases is just
the sum of the pressures that each gas would exert if it were present alone.
Figure 2.3 illustrates Dalton‘s law.
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The behaviour of gases
2016
Figure 2.3. Schematic illustration of Dalton’s law of partial pressures.
Consider a case in which two gases, A and B, are in a container of
volume V.
The pressure exerted by gas A, according to the ideal gas equation, is
𝑛 𝑅𝑇
𝑃𝐴 = 𝐴
𝑉
Where nA is the number of moles of A present. Similarly, the pressure
exerted by gas B is
𝑛 𝑅𝑇
𝑃𝐵 = 𝐵𝑉
In a mixture of gases A and B, the total pressure PT is the result of the
collisions of both types of molecules, A and B, with the walls of the
container. Thus, according to Dalton‘s law,
𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵
=
𝑛 𝐴 𝑅𝑇
𝑉
=
𝑅𝑇
𝑉
=
𝑛𝑅𝑇
𝑉
+
𝑛 𝐵 𝑅𝑇
𝑉
(𝑛𝐴 + 𝑛𝐵 )
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The behaviour of gases
2016
Where n, the total number of moles of gases present, is given by n = nA
+ nB, and PA and PB are the partial pressures of gases A and B,
respectively. For a mixture of gases, then, PT depends only on the total
number of moles of gas present, not on the nature of the gas molecules.
In general, the total pressure of a mixture of gases is given by
𝑃𝑇 = 𝑃1 + 𝑃2 + 𝑃3 + − − − −
Where P1, P2, P3, . . . are the partial pressures of components 1, 2, 3, . . . .
To see how each partial pressure is related to the total pressure,
consider again the case of a mixture of two gases A and B. Dividing PA
by PT, we obtain
𝑃𝐴
𝑃𝑇
=
𝑛 𝐴 𝑅𝑇
𝑉
(𝑛 𝐴 +𝑛 𝐵 ) 𝑅𝑇 𝑉
=
𝑛𝐴
𝑛 𝐴 +𝑛 𝐵
= 𝑋𝐴
Where XA is called the mole fraction of A. The mole fraction is a
dimensionless quantity that expresses the ratio of the number of moles of one
component to the number of moles of all components present. In general, the
mole fraction of component i in a mixture is given by
𝑛
𝑋𝑖 = 𝑛 𝑖
𝑇
Where ni and nT are the number of moles of component i and the total
number of moles present, respectively. The mole fraction is always
smaller than 1. We can now express the partial pressure of A as
𝑃𝐴 = 𝑋𝐴 𝑃𝑇
Similarly,
𝑃𝐵 = 𝑋𝐵 𝑃𝑇
Note that the sum of the mole fractions for a mixture of gases must be
unity. If only two components are present, then
𝑛
𝑛
𝑋𝐴 + 𝑋𝐵 = 𝑛 +𝐴𝑛 + 𝑛 +𝐵𝑛 = 1
𝐴
𝐵
𝐴
𝐵
If a system contains more than two gases, then the partial pressure of
the ith component is related to the total pressure by
𝑃𝑖 = 𝑋𝑖 𝑃𝑇
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The behaviour of gases
2016
How are partial pressures determined? A manometer can measure
only the total pressure of a gaseous mixture. To obtain the partial
pressures, we need to know the mole fractions of the components,
which would involve elaborate chemical analyses. The most direct
method of measuring partial pressures is using a mass spectrometer.
The relative intensities of the peaks in a mass spectrum are directly
proportional to the amounts, and hence to the mole fractions, of the
gases present.
Example 2.65. A mixture of gases contains 3.85 moles of neon (Ne), 0.92
mole of argon (Ar), and 2.59 moles of xenon (Xe). Calculate the partial
pressures of the gases if the total pressure is 2.50 atm at a certain
temperature.
Solution
The partial pressure of a gas is equal to the product of its mole fraction
and the total pressure (PT)
Given data
Mole of Ne = 3.85 moles
Mole of Ar = 0.92 moles
Mole of Xe = 2.59 moles
Total pressure, PT = 2.5
𝑛 𝑁𝑒
Mole fraction of neon (𝑋𝑁𝑒 ) =
𝑛 𝑁𝑒 +𝑛 𝐴𝑟 +𝑛 𝑋𝑒
3.85 𝑚𝑜𝑙
= 3.85 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙
= 0.523
Therefore
𝑃𝑁𝑒 = 𝑋𝑁𝑒 𝑃𝑇
= 0.523 × 2.50
= 1.31 𝑎𝑡𝑚
Similarly, we can calculate the mole fraction of argon and its partial
pressure:
𝑛
Mole fraction of Argon (𝑋𝐴𝑟 ) = 𝑛 + 𝑛𝐴𝑟 +𝑛
𝐴𝑟
=
𝑁𝑒
𝑋𝑒
0.92 𝑚𝑜𝑙
0.92 𝑚𝑜𝑙 + 3.85 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙
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= 0.125
Therefore
𝑃𝐴𝑟 = 𝑋𝐴𝑟 𝑃𝑇
= 0.125 × 2.50
= 0.313 𝑎𝑡𝑚
Finally, we calculate the mole fraction of xenon and its partial pressure:
𝑛
Mole fraction of Xenon (𝑋𝑋𝑒 ) = 𝑛 +𝑛 𝑋𝑒 +𝑛
𝑋𝑒
𝐴𝑟
𝑁𝑒
2.59 𝑚𝑜𝑙
= 2.59 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙
= 0.352
Therefore
𝑃𝑋𝑒 = 𝑋𝑋𝑒 𝑃𝑇
= 0.352 × 2.50
= 0.88 𝑎𝑡𝑚
Check: The individual partial pressures must be less than the total pressure
and make sure that the sum of the partial pressures is equal to the total
pressure; that is, (1.31 + 0.313 + 0.880) atm = 2.50 atm.
Example 2.66. A sample of natural gas contains 8.24 moles of methane
(CH4), 0.421 mole of ethane (C2H6), and 0.116 mole of propane (C3H8).
If the total pressure of the gases is 1.37 atm, what are the partial
pressures of the gases?
Solution
Data provided
Mole of methane = 8.24 moles
Moles of ethane = 0.421 moles
Mole of propane = 0.116 moles
Total pressure = 1.37 atm
Mole fraction of neon (𝑋𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 ) = 𝑛
=
𝑛 𝑚𝑒𝑡 𝑕 𝑎𝑛𝑒
+𝑛
𝑚𝑒𝑡 𝑕 𝑎𝑛𝑒
𝑒𝑡 𝑕 𝑎𝑛𝑒 +𝑛 𝑝𝑟𝑜𝑝𝑎𝑛𝑒
8.24 𝑚𝑜𝑙
8.24 𝑚𝑜𝑙 +0.421 𝑚𝑜𝑙 +0.116 𝑚𝑜𝑙
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= 0.94
Therefore
𝑃𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 = 𝑋𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 𝑃𝑇
= 0.94 × 1.37
= 1.29 𝑎𝑡𝑚
Similarly, we can calculate the mole fraction of ethane and its partial
pressure:
𝑛 𝑒𝑡 𝑕 𝑎𝑛𝑒
𝑛 𝑒𝑡 𝑕 𝑎𝑛𝑒 + 𝑛 𝑚𝑒𝑡 𝑕 𝑎𝑛𝑒 +𝑛 𝑝𝑟𝑜𝑝𝑎𝑛𝑒
Mole fraction of Ethane (𝑋𝑒𝑡 𝑕𝑎𝑛𝑒 ) =
=
0.421 𝑚𝑜𝑙
0.92 𝑚𝑜𝑙 + 8.24 𝑚𝑜𝑙 +0.116 𝑚𝑜𝑙
= 0.05
Therefore
𝑃𝑒𝑡 𝑕𝑎𝑛𝑒 = 𝑋𝑒𝑡 𝑕𝑎𝑛𝑒 𝑃𝑇
= 0.05 × 1.37
= 0.069 𝑎𝑡𝑚
Finally, we calculate the mole fraction of propane and its partial
pressure:
Mole fraction of Propane (𝑋𝑝𝑟𝑜𝑝𝑎𝑛𝑒 )
=
=
𝑛 𝑝𝑟𝑜𝑝𝑎𝑛𝑒
𝑛 𝑝𝑟𝑜𝑝𝑎𝑛𝑒 +𝑛 𝑚𝑒𝑡 𝑕 𝑎𝑛𝑒 +𝑛 𝑒𝑡 𝑕 𝑎𝑛𝑒
0.116 𝑚𝑜𝑙
0.116 𝑚𝑜𝑙 +8.24 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙
= 0.01
Therefore
𝑃𝑝𝑟𝑜𝑝𝑎𝑛𝑒 = 𝑋𝑝𝑟𝑜𝑝𝑎𝑛𝑒 𝑃𝑇
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The behaviour of gases
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= 0.01 × 1.37
= 0.014 𝑎𝑡𝑚
2.12. Gases Collected Over Water
Whenever a gas is collected over water, it becomes wet by
water vapour. Since water vapour is a gas, it exerts its own pressure,
and a mixture of gases is obtained. The pressure exerted is, therefore,
the sum of the partial pressures of the gas and that of the water vapour
at that temperature, i.e.
PTotal = Pgas +Pwater vapour
Pgas = PTotal − Pwater vapour
Dalton‘s law of partial pressures is useful for calculating volumes of
gases collected over water. For example, when potassium chlorate
(KClO3) is heated, it decomposes to KCl and O2:
2KClO3(s) → 2KCl(s) + 3O2(g)
The oxygen gas can be collected over water, as shown in Figure 2.4.
Initially, the inverted bottle is completely filled with water. As oxygen
gas is generated, the gas bubbles rise to the top and displace water
from the bottle. This method of collecting a gas is based on the
assumptions that the gas does not react with water and that it is not
appreciably soluble in it. These assumptions are valid for oxygen gas,
but not for gases such as NH3, which dissolves readily in water. The
oxygen gas collected in this way is not pure, however, because water
vapour is also present in the bottle. The total gas pressure is equal to
the sum of the pressures exerted by the oxygen gas and the water
vapour:
PT = P 𝑜2 + P𝐻2 O
Consequently, we must allow for the pressure caused by the presence
of water vapour when we calculate the amount of O2 generated. Table
2.2 shows the pressure of water vapour at various temperatures.
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The behaviour of gases
Table 2.2. pressure of water vapour at various temperatures
Temperature
Water vapour pressure
(oC )
(mmHg)
0
4.58
10
9.21
30
31.82
50
92.51
70
233.7
80
355.1
90
525.76
100
760.00
104
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Figure 2.4: An apparatus for collecting gas over water. The oxygen generated
by heating potassium chlorate (KClO3) in the presence of a small amount of
manganese dioxide (MnO2), which speeds up the reaction, is bubbled through
water and collected in a bottle as shown. Water originally present in the bottle
is pushed into the trough by the oxygen gas.
Example 2.67. Oxygen gas generated by the decomposition of
potassium chlorate is collected as shown in Figure 2.4. The volume of
oxygen collected at 26°C and atmospheric pressure of 771 mmHg is 141
mL. Calculate the mass (in grams) of oxygen gas obtained. The
pressure of the water vapour at 26°C is 25.2 mmHg. [O = 16]
Solution
To solve for the mass of O2 generated, we must first calculate the
partial
pressure of O2 in the mixture. What gas law do we need? How do we
convert pressure of O2 gas to mass of O2 in grams?
Data provided
PT = 771 mmHg
P𝐻2 O = 25.2 mmHg
P 𝑜2 = ?
V = 141mL = (141/1000)L = 0.141 L
Molar mass (M) = 32.0 g/mol
T = 26°C = (26 + 273)K = 299 K
R = 0.082 atm L K-1 mol-1
From Dalton‘s law of partial pressures we know that
PT = P 𝑜2 + P𝐻2 O
P 𝑜2 = PT ‒ P𝐻2 O
= 771 mmHg ‒ 25.2 mmHg
= 746 mmHg
= (746/760) atm
= 0.98 atm
From the ideal gas equation we write
𝑃𝑉 = 𝑛𝑅𝑇
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The behaviour of gases
But
𝑛=
𝑚
𝑀
2016
𝑚
Therefore 𝑃𝑉 = 𝑀 . 𝑅𝑇
and 𝑃𝑉𝑀 = 𝑚𝑅𝑇
𝑃𝑉𝑀
𝑚 = 𝑅𝑇
0.98 𝑎𝑡𝑚 × 0.141 𝐿 × 32.0 𝑔/𝑚𝑜𝑙
𝐿 𝐾 −1 𝑚𝑜𝑙 −1 ×299 𝐾
= 0.0821 𝑎𝑡𝑚
= 0.180 g
Example 2.68. Hydrogen gas generated when calcium metal reacts with
water is collected as shown in Figure 5.14. The volume of gas collected
at 30°C and pressure of 988 mmHg is 641 mL. What is the mass (in
grams) of the hydrogen gas obtained? The pressure of water vapor at
30°C is 31.82 mmHg. [H =1].
Solution
Data provided
PT = 988 mmHg
P𝐻2 O = 31.82 mmHg
P 𝐻2 = ?
V = 641mL = (641/1000)L = 0.641 L
Molar mass (M) = 2.0 g/mol
T = 30°C = (30 + 273)K = 303 K
R = 0.082 atm L K-1 mol-1
From Dalton‘s law of partial pressures we know that
PT = P 𝑜2 + P𝐻2 O
P 𝐻2 = PT ‒ P𝐻2 O
= 988 mmHg ‒ 31.82 mmHg
= 956.18 mmHg
= (956.18/760) atm
= 1.26 atm
From the ideal gas equation we write
𝑃𝑉 = 𝑛𝑅𝑇
𝑚
But 𝑛 =
𝑀
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The behaviour of gases
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𝑚
Therefore 𝑃𝑉 = . 𝑅𝑇
𝑀
and 𝑃𝑉𝑀 = 𝑚𝑅𝑇
𝑃𝑉𝑀
𝑚 = 𝑅𝑇
1.26 𝑎𝑡𝑚 × 0.641 𝐿 × 2.0 𝑔/𝑚𝑜𝑙
=
0.0821 𝑎𝑡𝑚 𝐿 𝐾 −1 𝑚𝑜𝑙 −1 ×303 𝐾
= 0.065 g
Example 2.69. 20 dm3 of hydrogen were collected over water at 17oC
and 79.7KNm-2 pressure. Calculate (i) pressure of dry hydrogen at this
temperature. (ii) volume of dry hydrogen gas at s.t.p. (vapour pressure
of water is 1.9 KNm-2 at 17oC; 1atm = 101.3 KNm-2 ).
Solution
Data given:
PTotal = 79.7KNm-2
Pwater vapour = 1.9 KNm-2 at 17oC
Pgas = ?
(i)
According to Dalton‘s law:
Pgas = PTotal − Pwater vapour
p(H2) = (79.7 − 1.9 ) KNm-2
= 77.8 KNm-2
(ii)
To find the volume of dry hydrogen gas at s.t.p.
Data provided.
P1= 77.8 KNm-2
V1= 20 dm3
T1 = 17oC = (17 + 273)K = 290K
P2 = s.p. = 101.3 KNm-2
T2 = s.t. = 273K
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The behaviour of gases
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V2 = ?
Using the general gas equation and substituting,
P1V1/T1 = P2V2/T2
V2 =
77.8 ×20 ×273
101.3 ×290
= 14.5 dm3
Example 2.70. (i). If I try to put a 1.00-L sample of O2 at 300 K and 1.00
atm plus a 1.00-L sample of N2 at 300 K and 1.00 atm into a rigid 1.00-L
container at 300 K, will they fit? (ii) If so, what will be their total
volume and total pressure?
Solution
(i) The gases will fit; gases expand or contract to fill their containers.
(b) The total volume is the volume of the container—1.00 L. The
temperature is 300 K, given in the problem. The total pressure is the
sum of the two partial pressures. Partial pressure is the pressure of each
gas (as if the other were not present). The oxygen pressure is 1.00 atm.
The oxygen has been moved from a 1.00-L container at 300 K to
another 1.00-L container at 300 K, and so its pressure does not change.
The nitrogen pressure is 1.00 atm for the same reason. The total
pressure is 1.00 atm + 1.00 atm = 2.00 atm.
Example 2.71. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L
container at 300 K. What will be their total volume, temperature, and
total pressure?
Solution
The total volume is the volume of the container—1.00 L. The
temperature is 300 K, given in the problem. The total pressure is the
sum of the two partial pressures. The oxygen pressure is 1.00 atm. The
nitrogen pressure is 0.500 atm, since it was moved from 0.500 L at 1.00
atm to 1.00 L at the same temperature (Boyle‘s law).
The total pressure is 1.00 atm + 0.500 atm = 1.50 atm
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The behaviour of gases
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Example 272. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L
container at 300 K. What will be their total volume, temperature, and
total pressure? Calculate the number of moles of O2 both before and
after mixing.
Solution
Data provided
P = 1.0 atm
T = 300 K
V = 1.0 L
R = 0.082 atm L K-1 mol-1
n=?
Applying the ideal gas equation to find n before mixing
𝑃𝑉 = 𝑛𝑅𝑇
Making n the subject, and substituting:
n=
=
𝑃𝑉
𝑅𝑇
1 atm × 1.0 L
0.082 atm L mol −1 K −1 × 300 K
= 0.0406 mol of O2 gas
After mixing
n𝑜2 =
=
𝑃𝑜2 𝑉
𝑅𝑇
1 atm × 1.0 L
0.082 atm L mol −1 K −1 × 300 K
= 0.0406 mol of O2 gas
There is no change in the number of moles of oxygen gas before and
after mixing.
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The behaviour of gases
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Example 2.73. Calculate the volume of 1.00 mol of H2O at 1.00-atm
pressure and a temperature of 25oC.
Solution
Water (H2O) is not a gas under these conditions, and so the equation
PV = nRT does not apply. (The ideal gas law can be used for water
vapour, e.g., water over 100 oC at 1 atm or water at lower temperatures
mixed with air). At 1-atm pressure and 25 oC, water is a liquid with a
density of about 1.00 g/mL.
1 mol of H2O contain 18g
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑉 =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
=
18 𝑔
1.00 𝑔/𝑚𝐿
= 18 mL
Example 2.74. A container holds three gases: oxygen, carbon dioxide,
and helium. The partial pressures of the three gases are 2.00 atm, 3.00
atm, and 4.00 atm, respectively. What is the total pressure inside the
container?
Solution
Data provided
Poxygen = 2.00 atm
Pcarbon dioxide = 3.00 atm
Phelium = 4.00 atm
PT = ?
From Dalton‘s law of partial pressure,
PT = sum of the individual partial pressures in the reaction vessel
= Poxygen + Pcarbon dioxide + Phelium
= (2.0 + 3.0 +4.0) atm = 9.0 atm
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The behaviour of gases
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Example 2.75. A tank contains 480.0 grams of oxygen and 80.00 grams
of helium at a total pressure of 7.00 atmospheres. Calculate the
following.
a) How many moles of O2 are in the tank?
b) How many moles of He are in the tank?
c) Total moles of gas in tank.
d) Mole fraction of O2.
e) Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.
Solution
Let us first find the mole of the gases with the simple relation (n =
m/M)
Data given:
Mass of O2 = 480.0 g
Mass of helium = 80g
Total pressure (PT) = 7.00 atm
a). Mole of O2 in the tank (n) =
𝑚𝑎𝑠𝑠 (𝑚)
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)
480 𝑔
= 32 𝑔/𝑚𝑜𝑙
= 15 mol of O2
𝑚𝑎𝑠𝑠 (𝑚 )
𝑚𝑎𝑠𝑠 (𝑀)
b). Mole of He in the tank (n) = 𝑚𝑜𝑙𝑎𝑟
=
111
80 𝑔
4.0 𝑔/𝑚𝑜𝑙
The behaviour of gases
2016
= 20 mol of He
c). Total mole of gas in the tank = mole of O2 + mole of He
= (15 + 20)mole
= 35 moles
d). Mole fraction of O2 (𝑋𝑂2 ) =
𝑛𝑂2
𝑛 𝑂 2 + 𝑛 𝐻𝑒
15 𝑚𝑜𝑙
= 15 𝑚𝑜𝑙 +20 𝑚𝑜𝑙
= 0.4286
e). Mole fraction of He (𝑋𝐻𝑒 ) =
𝑛 𝐻𝑒
𝑛 𝑂 2 + 𝑛 𝐻𝑒
20 𝑚𝑜𝑙
= 15 𝑚𝑜𝑙 +20 𝑚𝑜𝑙
= 0.5714
f). Partial pressure of O2 (𝑃𝑂2 ) = 𝑋𝑂2 𝑃𝑇
= 0.4286 × 7.00
= 3.0 atm
g). Partial pressure of He (𝑃𝐻𝑒 ) = 𝑋𝐻𝑒 𝑃𝑇
= 0.5714× 7.00
= 3.99 atm
Example 2.76. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00
moles of H2S, and 4.00 moles of argon at a total pressure of 1620.0
mmHg. Calculate the following.
a) Total moles of gas in tank
b) Mole fraction of gases
c) Partial pressure of gases
d) Pressure fraction of gases
Solution
Data given:
Mole of O2 = 5.00 moles
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The behaviour of gases
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Mole of Ne = 3.00 moles
Mole of H2S = 6.00 moles
Mole of Ar = 4.00 moles
Total pressure (PT) = 1620.0 mmHg = (1620/760) atm = 2.1 atm
a). Total mole of gas in the tank = mole of O2 + mole of Ne + mole of
H2S + mole of Ar
= (5+3+6 + 4) mole
= 18 moles
bi). Mole fraction of O2 (𝑋𝑂2 ) =
𝑛𝑂2
𝑛 𝑂 2 + 𝑛 𝑁𝑒 + 𝑛 𝐻 2 𝑆 + 𝑛 𝐴𝑟
5 𝑚𝑜𝑙
= 5 𝑚𝑜𝑙 +3 𝑚𝑜𝑙 +6 𝑚𝑜𝑙 +4 𝑚𝑜𝑙
= 0.28
bii). Mole fraction of Ne (𝑋𝑁𝑒 ) = 𝑛 + 𝑛
𝑂2
𝑛 𝑁𝑒
𝑁𝑒 + 𝑛 𝐻 2 𝑆 + 𝑛 𝐴𝑟
3 𝑚𝑜𝑙
= 5 𝑚𝑜𝑙 +3 𝑚𝑜𝑙 +6 𝑚𝑜𝑙 +4 𝑚𝑜𝑙
= 0.17
𝑛𝐻 𝑆
biii). Mole fraction of 𝐻2 𝑆 (𝑋𝐻2 𝑆 ) = 𝑛 + 𝑛 + 2𝑛 + 𝑛
𝑂2
𝑁𝑒
𝐻 2𝑆
𝐴𝑟
6 𝑚𝑜𝑙
5 𝑚𝑜𝑙 +3 𝑚𝑜𝑙 +6 𝑚𝑜𝑙 +4 𝑚𝑜𝑙
=
= 0.33
biv). Mole fraction of Ar (𝑋𝐴𝑟 ) = 𝑛 + 𝑛
𝑂2
𝑛 𝐴𝑟
𝑁𝑒 + 𝑛 𝐻 2 𝑆 + 𝑛 𝐴𝑟
4 𝑚𝑜𝑙
=
5 𝑚𝑜𝑙 +3 𝑚𝑜𝑙 +6 𝑚𝑜𝑙 +4 𝑚𝑜𝑙
= 0.22
ci). Partial pressure of O2 (𝑃𝑂2 ) = 𝑋𝑂2 𝑃𝑇
= 0.28 × 2.1
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The behaviour of gases
2016
= 0.588 atm
cii). Partial pressure of Ne (𝑃𝑁𝑒 ) = 𝑋𝑁𝑒 𝑃𝑇
= 0.17 × 2.1
= 0.357 atm
ciii). Partial pressure of 𝐻2 𝑆(𝑃𝐻2 𝑆 ) = 𝑋𝐻2 𝑆 𝑃𝑇
= 0.33 × 2.1
= 0.693 atm
civ). Partial pressure of Ar (𝑃𝐴𝑟 ) = 𝑋𝐴𝑟 𝑃𝑇
= 0.22 × 2.1
= 0.462 atm
di). pressure fraction of O2 (𝑃𝑂2 ) =
=
𝑃𝑂 2
𝑃𝑂 2 + 𝑃𝑁𝑒 + 𝑃𝐻 2 𝑆 + 𝑃𝐴𝑟
0.588 𝑎𝑡𝑚
0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚
= 0.28 atm
dii). Pressure fraction of Ne (𝑃𝑁𝑒 ) =
=
𝑃𝑁𝑒
𝑃𝑂 2 + 𝑃𝑁𝑒 + 𝑃𝐻 2 𝑆 + 𝑃𝐴𝑟
0.357 𝑎𝑡𝑚
0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚
= 0.17 atm
diii). Pressure fraction of 𝐻2 𝑆 (𝑃𝐻2 𝑆 ) =
=
𝑃𝐻 2 𝑆
𝑃 𝑂 2 + 𝑃𝑁𝑒 + 𝑃𝐻 2 𝑆 + 𝑃𝐴𝑟
0.693 𝑎𝑡𝑚
0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚
= 0.33 atm
div). Pressure fraction of Ar (𝑃𝐴𝑟 ) =
=
𝑃𝐴𝑟
𝑃𝑂 2 + 𝑃𝑁𝑒 + 𝑃𝐻 2 𝑆 + 𝑃𝐴𝑟
0.462 𝑎𝑡𝑚
0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚
= 0.22 atm
Example 2.77. A cylinder of compressed natural gas has a volume of
20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate
the partial pressure of each gas at 22.0°C and the total pressure in the
cylinder. [H = 1, C = 12].
Solution
Let us first find the mole of the gases with the simple relation (n =
m/M) then apply ideal gas equation to find the partial pressure of the
gases.
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Data given:
Mass of methane = 1813 g
Mass of ethane = 336g
T = 22.0°C = 295 K
V = 20.0 L
𝑚𝑎𝑠𝑠 (𝑚)
𝑚𝑎𝑠𝑠 (𝑀)
Mole of methane in the cylinder (n) = 𝑚𝑜𝑙𝑎𝑟
But molar mass of methane (CH4) = [12 + (1× 4)] = 16 g/mol
n
=
1813 𝑔
16 𝑔/𝑚𝑜𝑙
= 113 mol of methane
𝑚𝑎𝑠𝑠 (𝑚 )
𝑚𝑎𝑠𝑠 (𝑀)
Mole of ethane in the cylinder (n) = 𝑚𝑜𝑙𝑎𝑟
But molar mass of methane (C2H6) = [(12 ×2) + (1× 6)] = 30 g/mol
n
338 𝑔
= 30 𝑔/𝑚𝑜𝑙
= 11.3 mol of ethane
Partial pressure of methane (𝑃) =
=
𝑛 𝑚𝑒𝑡 𝑕 𝑎𝑛𝑒 × 𝑅𝑇
𝑉
113 𝑚𝑜𝑙 × 0.0821 𝑎𝑡𝑚 𝐿 𝐾 −1 𝑚𝑜𝑙 −1 ×295 𝐾
20 𝐿
= 136.8 atm
Partial pressure of methane (𝑃) =
𝑛 𝑒𝑡 𝑕 𝑎𝑛𝑒 × 𝑅𝑇
𝑉
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=
2016
11.3 𝑚𝑜𝑙 × 0.0821 𝑎𝑡𝑚 𝐿 𝐾 −1 𝑚𝑜𝑙 −1 ×295 𝐾
20 𝐿
= 13.6 atm
Total pressure in the cylinder (PT) = Pmethane + Pethane
= (136 + 13.6)atm
= 149.4 𝑎𝑡𝑚
2.13. Gas Diffusion and Effusion
2.13.1. Gas Diffusion
A direct demonstration of random motion is provided by
diffusion, the gradual mixing of molecules of one gas with molecules of
another by virtue of their kinetic properties. Despite the fact that molecular
speeds are very great, the diffusion process takes a relatively long time
to complete. For example, when a bottle of concentrated ammonia
solution is opened at one end of a laboratory bench, it takes some time
before a person at the other end of the bench can smell it. The reason is
that a molecule experiences numerous collisions while moving from
one end of the bench to the other. Thus, diffusion of gases always
happens gradually, and not instantly as molecular speeds seem to
suggest. Furthermore, because the root-mean square speed of a light
gas is greater than that of a heavier gas, a lighter gas will diffuse
through a certain space more quickly than will a heavier gas. Figure 2.5
illustrates gaseous diffusion.
In 1832 the Scottish chemist Thomas Graham found that under
the same conditions of temperature and pressure, rates of diffusion for gases
are inversely proportional to the square roots of their molar masses. This
statement, now known as Graham’s law of diffusion, is expressed
mathematically as
𝑟1
𝑟2
=
𝑀2
𝑀1
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Where r1 and r2 are the diffusion rates of gases 1 and 2, and M1 and M2
are their molar masses, respectively.
Figure 2.5: A demonstration of gas diffusion. NH3 gas (from a bottle
containing aqueous ammonia) combines with HCl gas (from a bottle
containing hydrochloric acid) to form solid NH4Cl. Because NH 3 is lighter
and therefore diffuses faster, solid NH4Cl first appears nearer the HCl bottle
(on the right).
2.13.2. Gas Effusion
Whereas diffusion is a process by which one gas gradually
mixes with another, effusion is the process by which a gas under pressure
escapes from one compartment of a container to another by passing through a
small opening. Figure 2.6 shows the effusion of a gas into a vacuum.
Although effusion differs from diffusion in nature, the rate of effusion
of a gas has the same form as Graham‘s law of diffusion [see Equation
for diffusion of a gas]. A helium-filled rubber balloon deflates faster
than an air-filled one because the rate of effusion through the pores of
the rubber is faster for the lighter helium atoms than for the air
molecules. Industrially, gas effusion is used to separate uranium
isotopes in the forms of gaseous 235UF6 and 238UF6. By subjecting the
gases to many stages of effusion, scientists were able to obtain highly
enriched 235U isotope, which was used in the construction of atomic
bombs during World War II.
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Figure 2.6: Gas effusion. Gas molecules move from a high-pressure region
(left) to a low-pressure one through a pinhole.
Example 2.78. A flammable gas made up only of carbon and hydrogen
is found to effuse through a porous barrier in 3.50 min. Under the same
conditions of temperature and pressure, it takes an equal volume of
chlorine gas 7.34 min to effuse through the same barrier. Calculate the
molar mass of the unknown gas, and suggest what this gas might be.
Solution
Data given
Time for effusion of chlorine = 7.34 min
Time of effusion of unknown gas X = 3.5 min
Molar mass of chlorine gas = 70.90 g/mol
Molar mass of X = ?
Strategy
We find the rate of effusion of both gases and the use graham‘s law to
find the molar mass of gas X
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑐𝑕𝑙𝑜𝑟𝑖𝑛𝑒
7.34 𝑚𝑖𝑛
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋
3.5 𝑚𝑖𝑛
Rate of effusion of Cl2 =
Rate of effusion of X =
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𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐶𝑙2
=
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑋
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙 2
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑋
7.34 𝑚𝑖𝑛
3.5 𝑚𝑖𝑛
=
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙 2
7.34 𝑚𝑖𝑛
×
2016
𝑀𝑋
𝑀𝐶𝑙 2
3.5 𝑚𝑖𝑛
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋
3.5 𝑚𝑖𝑛
= 7.34 𝑚𝑖𝑛
From the molar mass of Cl2, we write
3.5 𝑚𝑖𝑛
=
7.34 𝑚𝑖𝑛
𝑀𝑋
70.90 𝑔/𝑚𝑜𝑙
3.5 𝑚𝑖𝑛
𝑀𝑋 = (
)2 × 70.90 g/mol
7.34 𝑚𝑖𝑛
𝑀𝑋 = 16.1 𝑔/𝑚𝑜𝑙
Because the molar mass of carbon is 12.01 g and that of hydrogen is
1.008 g, the gas is methane (CH4).
Note: Because lighter gases effuse faster than heavier gases, the molar mass of
the unknown gas must be smaller than that of chlorine gas. Indeed, the molar
mass of methane (16.04 g) is less than the molar mass of chlorine gas (70.90
g).
Example 2.79. Methane effuses through a small opening in the
side of a container at rate of 1.3 mols -1 . An unknown gas X
effuses through the same opening at the rate of 5.42 mols-1 when
maintained at the same temperature and pressure as methane.
Determine the molar mass of the unknown gas. [H= 1, C= 12]
Solution
Data provided
Rate of effusion of methane = 1.3 mols -1
Rate of effusion of gas X = 5.42 mols-1
Molar mass of methane = 16.0g/mol
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Molar mass of gas X = ?
Use graham‘s law to find molar mass of gas X
𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐶𝐻4
=
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑋
1.3 mols −1
=
5.42 mols−1
𝑀𝑋
𝑀𝐶𝐻4
𝑀𝑋
16.0
𝑀
(2.3985)2 = 16𝑋
𝑀𝑋 = 16 × 5.7528
= 92. 045
Therefore molar mass of gas X is 92.045g/mol
Example 2.80. The time required for a volume of O2 to diffuse through
opening is 40seconds. Calculate the molar mass of gas which requires
50seconds for the same volume to diffuse through the same opening
under the same conditions [O = 16].
Solution
Data provided
Rate of diffusion of oxygen = 40s
Rate of diffusion of gas X = 50s
Molar mass of oxygen = 32.0g/mol
Molar mass of gas X = ?
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
35𝑠
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋
50𝑠
Rate of diffusion of O2 =
Rate of diffusion of X =
𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑂2
=
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑋
120
𝑀𝑋
𝑀𝑂2
The behaviour of gases
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑋
40𝑠
50𝑠
=
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2
40𝑠
×
50𝑠
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋
=
2016
50𝑠
40𝑠
From the molar mass of O2, we write
50𝑠
=
40𝑠
𝑀𝑋
32.0 𝑔/𝑚𝑜𝑙
50𝑠
𝑀𝑋 = (40𝑠 )2 × 32.0 g/mol
𝑀𝑋 = (1.25)2 × 32.0 𝑔/𝑚𝑜𝑙
𝑀𝑋 = 1.5625 × 32.0 𝑔/𝑚𝑜𝑙
𝑀𝑋 = 50.0 𝑔/𝑚𝑜𝑙
Example 2.81. If equal amounts of hydrogen and argon are placed in a
porous container and allowed to escape, which gas will escape faster
and why?
Solution
If equal amounts of hydrogen and argon are placed in a porous
container and allowed to escape, hydrogen gas will effuse faster than
argon because hydrogen is a lighter gas (with molecular mass of
2.0g/mol) than argon (with molecular mass of 39.95g/mol) and the
rate of effusion varies inversely with molecular weight of a gas.
Example 2.82. The time required for a volume of gas y to effuse
through a small hole was 112.2seconds. The time required for the same
volume of oxygen was 84.7seconds. Calculate the molecular weight of
gas y.
Solution
Data provided
Rate of effusion of oxygen = 48.7s
Rate of effusion of gas y = 112.2s
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Molar mass of oxygen = 32.0g/mol
Molar mass of gas y = ?
Rate of effusion of O2 =
Rate of effusion of y =
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛
48.5𝑠
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑌
112.2𝑠
𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑂2
=
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑌
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑌
48.5𝑠
112.2𝑠
=
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2
48.5𝑠
×
𝑀𝑌
𝑀𝑂2
112.2𝑠
𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑌
=
112.2𝑠
48.5𝑠
From the molar mass of O2, we write
112.2𝑠
=
48.5𝑠
𝑀𝑌
32.0 𝑔/𝑚𝑜𝑙
112.2𝑠
𝑀𝑌 = (
)2 × 32.0 g/mol
48.5𝑠
= (2.31)2 × 32.0 𝑔/𝑚𝑜𝑙
= 5.35 × 32.0 𝑔/𝑚𝑜𝑙
= 171.2 𝑔/𝑚𝑜𝑙
Example 2.83. Consider the reaction represented by the following
equation:
State what would happen to the vapour density of N2O4 as the
temperature of the system is increased. If the system is cooled, would
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The behaviour of gases
2016
the gases become lighter or darker in colour? Explain your answer in
each case.
Solution
The density becomes lighter as the temperature of the system increases.
If the system is cooled, the product becomes lighter in colour. At low
temperature, dinitrogen (IV) oxide (N2O4) predominates. Hence, the
gases become lighter while at high temperature, N2O4 dissolves to
nitrogen (IV) oxide (NO2) molecules and the gases becomes darker in
colour and lighter in density.
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CHAPTER THREE
THE KINETIC MOLECULAR THEORY OF GASES
3.1 Introduction
The gas laws help us to predict the behaviour of gases, but they
do not explain what happens at the molecular level to cause the
changes we observe in the macroscopic world. For example, why does
a gas expand on heating?
This section introduces the kinetic molecular theory of gases,
which explains the gas laws and when extended, also explains some
properties of liquids and solids.
Five postulates explain why gases behave as they do:
1. A gas is composed of molecules that are separated from each other
by distances far greater than their own dimensions.
2. The volume of the molecules is insignificant compared with the
volume occupied by the gas.
3. Forces between the molecules are negligible, except when the
molecules collide with one another.
4. Molecular collisions are perfectly elastic; that is, no energy is lost
when the molecules collide.
5. The average kinetic energy of the molecules is proportional to the
temperature of the gas in kelvins. Any two gases at the same
temperature will have the same average kinetic energy. The average
kinetic energy of a molecule is given by
𝐾𝐸 = 1 2 𝑚𝑢2
Where m is the mass of the molecule and u is its speed. The horizontal
bar denotes an average value. The quantity 𝑢2 is called mean square
speed; it is the average of the square of the speeds of all the molecules:
2
𝑢2 = 𝑢12 + 𝑢22 + − − − 𝑢𝑁
Where N is the number of molecules.
Assumption 5 enables us to write
𝐾𝐸 𝛼 𝑇
1
2
2 𝑚𝑢 𝛼 𝑇
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𝐾𝐸 = 1 2 𝑚𝑢2 = 𝐶𝑇
Where C is the proportionality constant and T is the absolute
temperature.
According to the kinetic molecular theory, gas pressure is the
result of collisions between molecules and the walls of their container.
It depends on the frequency of collision per unit area and on how
―hard‖ the molecules strike the wall. The theory also provides a
molecular interpretation of temperature. According to Equation above,
the absolute temperature of a gas is a measure of the average kinetic
energy of the molecules. In other words, the absolute temperature is a
measure of the random motion of the molecules—the higher the
temperature, the more energetic the molecules.
Because it is related to the temperature of the gas sample, random
molecular motion is sometimes referred to as thermal motion.
3.2. Application to the Gas Laws
Although the kinetic theory of gases is based on a rather simple
model, the mathematical details involved are very complex. However,
on a qualitative basis, it is possible to use the theory to account for the
general properties of substances in the gaseous state. The following
examples illustrate the range of its utility:
• Compressibility of Gases. Because molecules in the gas phase are
separated by large distances (assumption 1), gases can be compressed
easily to occupy less volume.
• Boyle’s Law. The pressure exerted by a gas results from the impact
of its molecules on the walls of the container. The collision rate, or the
number of molecular collisions with the walls per second, is
proportional to the number density (that is, number of molecules per
unit volume) of the gas. Decreasing the volume of a given amount of
gas increases its number density and hence its collision rate. For this
reason, the pressure of a gas is inversely proportional to the volume it
occupies; as volume decreases, pressure increases and vice versa.
• Charles’s Law. Because the average kinetic energy of gas molecules
is proportional to the sample‘s absolute temperature (assumption 5),
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The behaviour of gases
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raising the temperature increases the average kinetic energy.
Consequently, molecules will collide with the walls of the container
more frequently and with greater impact if the gas is heated, and thus
the pressure increases. The volume of gas will expand until the gas
pressure is balanced by the constant external pressure.
• Avogadro’s Law. We have shown that the pressure of a gas is
directly proportional to both the density and the temperature of the
gas. Because the mass of the gas is directly proportional to the number
of moles (n) of the gas, we can represent density by n/V. Therefore,
𝑛
𝑃 ∝ 𝑉 𝑇
For two gases, 1 and 2, we write
𝑛 𝑇
𝑛 𝑇
𝑃1 ∝ 𝑉1 1 = 𝐶 𝑉1 1
1
𝑃2 ∝
𝑛 2 𝑇2
𝑉2
1
=𝐶
𝑛 2 𝑇2
𝑉2
Where C is the proportionality constant. Thus, for two gases under the
same conditions of pressure, volume, and temperature (that is, when P1
= P2, T1 = T2, and V1 = V2), it follows that n1 = n2, which is a
mathematical expression of Avogadro‘s law.
• Dalton’s Law of Partial Pressures. If molecules do not attract or
repel one another
(assumption 3), then the pressure exerted by one type of molecule is
unaffected by the presence of another gas. Consequently, the total
pressure is given by the sum of individual gas pressures.
3.3. Distribution of Molecular Speeds
The kinetic theory of gases enables us to investigate molecular
motion in more detail. Suppose we have a large number of gas
molecules, say, 1 mole, in a container. As long as we hold the
temperature constant, the average kinetic energy and the mean square
speed will remain unchanged as time passes. As you might expect, the
motion of the molecules is totally random and unpredictable. At a
given instant, how many molecules are moving at a particular speed?
To answer this question Maxwell analyzed the behaviour of gas
molecules at different temperatures.
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Figure 3.1(a) shows typical Maxwell speed distribution curves for
nitrogen gas at three different temperatures. At a given temperature,
the distribution curve tells us the number of molecules moving at a
certain speed. The peak of each curve represents the most probable speed,
that is, the speed of the largest number of molecules. Note that the
most probable speed increases as temperature increases (the peak
shifts toward the right). Furthermore, the curve also begins to flatten
out with increasing temperature, indicating that larger numbers of
molecules are moving at greater speed. Figure 3.1(b) shows the speed
distributions of three gases at the same temperature. The difference in
the curves can be explained by noting that lighter molecules move
faster, on average, than heavier ones.
(a)
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(b)
Figure 3.1 (a) The distribution of speeds for nitrogen gas at three different
temperatures. At the higher temperatures, more molecules are moving at faster
speeds. (b) The distribution of speeds for three gases at 300 K. At a given
temperature, the lighter molecules are moving faster, on the average.
3.4. Root-Mean-Square Speed
How fast does a molecule move, on the average, at any
temperature T? One way to estimate molecular speed is to calculate the
root-mean-square (rms) speed (urms), which is an average molecular speed.
One of the results of the kinetic theory of gases is that the total kinetic
3
energy of a mole of any gas equals 2 𝑅𝑇. Earlier we saw that the
average kinetic energy of one molecule is 1 2 𝑚𝑢2 and so we can write
3
𝑁𝐴 1 2 𝑚𝑢2 = 𝑅𝑇
2
Where NA is Avogadro‘s number and m is the mass of a single
molecule. Because NAm = M, where M is the molar mass, this equation
can be rearranged to give
3𝑅𝑇
𝑢2 = 𝑀
Taking the square root of both sides gives
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The behaviour of gases
𝑢2 = 𝑢𝑟𝑚𝑠 =
2016
3𝑅𝑇
𝑀
(3.1)
Equation (3.1) shows that the root-mean-square speed of a gas
increases with the square root of its temperature (in kelvins). Because
M appears in the denominator, it follows that the heavier the gas, the
more slowly its molecules move. If we substitute 8.314 J/K-1 mol-1 for R
and convert the molar mass to kg/mol, then urms will be calculated in
meters per second (m/s).
Example 3.1. Calculate the root-mean-square speeds of helium atoms
and nitrogen molecules in m/s at 25°C.
Strategy
To calculate the root-mean-square speed we use
𝑢𝑟𝑚𝑠 =
3𝑅𝑇
𝑀
expressed in m/s and convert temperature to kelvin
Solution
To calculate urms, the units of R should be 8.314 J/K-1 mol-1 and,
because 1 J= 1 kg m2s-2, the molar mass must be in kg/mol.
The molar mass of He is 4.003 g/mol, or 4.003 × 10-23 kg/mol.
25°C = (25 + 273)K = 298 K
3𝑅𝑇
𝑀
From 𝑢𝑟𝑚𝑠 =
=
3 8.314 𝐽 𝐾 −1 𝑚𝑜𝑙 −1 × 298 𝐾
4.003 × 10 −3 𝑘𝑔/𝑚𝑜𝑙
= 1.86 × 106 𝐽/𝑘𝑔
Use the conversion factor 1 J= 1 kg m2s-2 we get
= 1.86 × 106 m2 /s2
= 1.36 × 103 m/s
The procedure is the same for N2, the molar mass of which is 28.02
g/mol, or 2.802 × 10-2 kg/mol so that we write
𝑢𝑟𝑚𝑠 =
3 8.314 𝐽 𝐾 −1 𝑚𝑜𝑙 −1 × 298 𝐾
2.802 × 10 −2 𝑘𝑔/𝑚𝑜𝑙
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= 2.65 × 105 m2 /s2
= 515 m/s
Example 3.2. Calculate the root-mean-square speed of molecular
chlorine in m/s at 20°C.
Solution
To calculate the root-mean-square speed we use
𝑢𝑟𝑚𝑠 =
3𝑅𝑇
𝑀
expressed in m/s and convert temperature to kelvin
20°C = (20 + 273)K = 293 K
Molar mass of Cl2 = 71g/mol = 7.1 × 10-2 kg/mol
3𝑅𝑇
𝑀
From 𝑢𝑟𝑚𝑠 =
=
3 8.314 𝐽 𝐾 −1 𝑚𝑜𝑙 −1 × 293 𝐾
7.1 × 10 −2 𝑘𝑔 /𝑚𝑜𝑙
= 1.03 × 105 m2 /s2
= 320.8 m/s
The calculation in Example 3.1 has an interesting relationship to
the composition of Earth‘s atmosphere. Unlike Jupiter, Earth does not
have appreciable amounts of hydrogen or helium in its atmosphere.
Why is this the case? A smaller planet than Jupiter, Earth has a weaker
gravitational attraction for these lighter molecules. A fairly
straightforward calculation shows that to escape Earth‘s gravitational
field, a molecule must possess an escape velocity equal to or greater
than 1.1 × 104 m/s. Because the average speed of helium is
considerably greater than that of molecular nitrogen or molecular
oxygen, more helium atoms escape from Earth‘s atmosphere into outer
space. Consequently, only a trace amount of helium is present in our
atmosphere. On the other hand, Jupiter, with a mass about 320 times
greater than that of Earth, retains both heavy and light gases in its
atmosphere.
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3.5. Ideal and Real Gases
Any gas that obeys the gas laws at all temperatures and pressures is
called an ideal or perfect gas. Hence, the gas the equation: PV = nRT is
applicable to ideal gases only. Real gases only obey gas laws under
normal conditions of temperature and pressure.
3.6. Differences between ideal gas and real gases
1. An ideal gas obeys the gas laws at all temperatures and
pressure, while a real gas obeys the gas laws under normal
conditions of temperature and pressure.
2. The actual volume of the molecules of an ideal gas is negligible
compared with the volume of the container, while the actual
volume of the molecules of a real gas is not negligible i.e.
molecules of a real gas occupy space.
3. In an ideal gas there are no intermolecular attractions at all
temperatures and pressures, while intermolecular attraction is
strong and appreciable in a real gas at high pressure and low
temperature.
3.7. Deviation from Ideal Behaviour
The gas laws and the kinetic molecular theory assume that
molecules in the gaseous state do not exert any force, either attractive
or repulsive, on one another. The other assumption is that the volume
of the molecules is negligibly small compared with that of the
container.
A gas that satisfies these two conditions is said to
exhibit ideal behaviour.
Although we can assume that real gases behave like an ideal
gas, we cannot expect them to do so under all conditions. For example,
without intermolecular forces, gases could not condense to form
liquids. The important question is: Under what conditions will gases
most likely exhibit nonideal behaviour?
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Figure 3.2 shows PV/RT plotted against P for three real gases and an
ideal gas
at a given temperature. This graph provides a test of ideal gas
behaviour. According to the ideal gas equation (for 1 mole of gas),
PV/RT equals 1, regardless of the actual gas pressure. (When n = 1, PV
= nRT becomes PV = RT, or PV/RT = 1.) For real gases, this is true only
at moderately low pressures (≤ 5 atm); significant deviations occur as
pressure increases. Attractive forces operate among molecules at
relatively short distances. At atmospheric pressure, the molecules in a
gas are far apart and the attractive forces are negligible. At high
pressures, the density of the gas increases; the molecules are much
closer to one another. Intermolecular forces can then be significant
enough to affect the motion of the molecules, and the gas will not
behave ideally.
Another way to observe the nonideal behaviour of gases is to
lower the temperature. Cooling a gas decreases the molecules‘ average
kinetic energy, which in a sense deprives molecules of the drive they
need to break from their mutual attraction.
To study real gases accurately, then, we need to modify the
ideal gas equation, taking into account intermolecular forces and finite
molecular volumes. Such an analysis was first made by the Dutch
physicist J. D. van der Waals in 1873. Besides being mathematically
simple, van der Waals‘s treatment provides us with an interpretation
of real gas behaviour at the molecular level.
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Figure 3.2: Plot of PV/RT versus P of 1 mole of a gas at 0°C. For 1 mole of an
ideal gas, PV/RT is equal to 1, no matter what the pressure of the gas is. For
real gases, we observe various deviations from ideality at high pressures. At
very low pressures, all gases exhibit ideal behavior; that is, their PV/RT values
all converge to 1 as P approaches zero
Consider the approach of a particular molecule toward the wall
of a container. The intermolecular attractions exerted by its neighbours
tend to soften the impact made by this molecule against the wall. The
overall effect is a lower gas pressure exerted by an ideal gas, Pideal, is
related to the experimentally measured; that is, observed pressure, Pobs,
by the equation
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Where a is a constant and n and V are the number of moles and
volume of the gas, respectively. The correction term for pressure
(an2/V2) can be understood as follows. The intermolecular interaction
that gives rise to nonideal behavior depends on how frequently any
two molecules approach each other closely. The number of such
―encounters‖ increases with the square of the number of molecules per
unit volume, (n/V)2, because the presence of each of the two molecules
in a particular region is proportional to n/V and so a is just a
proportionality constant. The quantity Pideal is the pressure we would
measure if there were no intermolecular attractions.
Another correction concerns the volume occupied by the gas
molecules. In the ideal gas equation, V represents the volume of the
container. However, each molecule does occupy a finite, although
small, intrinsic volume, so the effective volume of the gas becomes (V ‒
nb), where n is the number of moles of the gas and b is a constant. The
term nb represents the volume occupied by n moles of the gas.
Having taken into account the corrections for pressure and
volume, we can rewrite the ideal gas equation as follows:
Equation (3.2), relating P, V, T, and n for a nonideal gas, is known as the
van der Waals equation. The van der Waals constants a and b are
selected to give the best possible agreement between Equation (3.2)
and observed behaviour of a particular gas. Table 3.1 lists the values of
a and b for a number of gases. The value of a indicates how strongly
molecules of a given type of gas attract one another. We see that
helium atoms have the weakest attraction for one another, because
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helium has the smallest a value. There is also a rough correlation
between molecular size and b. Generally, the larger the molecule (or
atom), the greater b is, but the relationship between b and molecular (or
atomic) size is not a simple one.
Example 3.3. Given that 2.75 moles of CO2 occupy 4.70 L at 53°C,
calculate the pressure of the gas (in atm) using (a) the ideal gas
equation and (b) the van der Waals equation.
Solution
Strategy
To calculate the pressure of CO2 using the ideal gas equation and van
der Waals equation, we proceed by applying the ideal gas equation
and then find the correction terms in van der Waals equation and
substitute in the equation to get the pressure.
(a).
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Data provided:
V = 4.70 L
T = 53°C = (53 + 273) K = 326 K
n = 2.75 mol
R = 0.0821 L atm/K -1 mol-1
Substituting these values in the ideal gas equation, we write
𝑃𝑉 = 𝑛𝑅𝑇
𝑛𝑅𝑇
𝑃=
𝑉
2.75 𝑚𝑜𝑙 ×0.0821 𝐿 𝑎𝑡𝑚 𝐾 −1 𝑚𝑜𝑙 −1 ×326 𝐾
=
4.70 𝐿
= 15.7 atm
(b). Applying equation 3.2, It is convenient to first calculate the
correction terms in Equation separately. From Table 3.1, we have
a = 3.59 atm L2/mol2
b = 0.0427 L/mol
so that the correction terms for pressure and volume are
𝑎𝑛2
3.59 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2 × (2.75 𝑚𝑜𝑙)2
=
𝑉2
(4.70 𝐿)2
= 1.23 atm
𝑛𝑏 = 2.75 𝑚𝑜𝑙 × (0.0427 L/mol)
= 0.117 L
Finally, substituting these values in the van der Waals equation, we
have
(P + 1.23 atm) (4.70 L ‒ 0.117 L)= (2.75 mol) (0.0821 L atm/K
(326 K)
(P + 1.23 atm) (4.688 L) =73.6 L
4.688P + 5.77 = 73.6
4.688P = 73.6 ‒ 5.77
4.688P = 67.83
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P = 14.5 atm
Example 3.4. Using the data shown in Table 3.1, calculate the pressure
exerted by 4.37 moles of nitrogen gas confined in a volume of 2.45 L at
38°C using van der Waals equation.
Solution
Data provided
V = 2.45 L
T =38°C = (38 + 273) K = 311 K
n = 4.37 mol
R = 0.0821 L atm/K -1 mol-1
Now let us apply equation 3.2, to find the pressure. From Table 3.1, we
have
a = 1.39 atm L2/mol2
b = 0.0913 L/mol
so that the correction terms for pressure and volume are
𝑎𝑛2
1.39 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2 × (4.37 𝑚𝑜𝑙)2
=
𝑉2
(2.45 𝐿)2
= 4.4 atm
𝑛𝑏 = 4.37 𝑚𝑜𝑙 × (0.0913 L/mol)
= 0.399 L
Finally, substituting these values in the van der Waals equation, we
have
(P + 4.4 atm) (2.45 L ‒ 0.399 L)= (4.37 mol) (0.0821 L atm/K
(311 K)
P = 50. 0 atm
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Example 3.5. Using (a) the ideal gas law equation and (b) Van der
Waal‘s eqauation, cal culate the pressure exerted by 50.0g of carbon
(IV) oxide in 1.00 L vessel at 25oC. [Find values of a and b in table 3.1]
Solution
Convert the mass of CO2 to mole and temperature to Kelvin.
Data provided
a = 3.592 atm L2/mol2
b = 0.043 L/mol
V = 1.00 L
T =25°C = (25 + 273) K = 298 K
R = 0.0821 L atm/K -1 mol-1
Mass of CO2 = 50.0g
𝑚𝑎𝑠𝑠
Number of mole of CO2 (n) =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
But molar mass of CO2 = 12 + (16 × 2) = 44 g/mol
50 𝑔
∴ (n) = 44 𝑔/𝑚𝑜𝑙 = 1.136 𝑚𝑜𝑙
(a). Let us substitute these values in the ideal gas equation, we write
𝑃𝑉 = 𝑛𝑅𝑇
𝑛𝑅𝑇
𝑃= 𝑉
1.136 𝑚𝑜𝑙 ×0.0821𝐿 𝑎𝑡𝑚 𝐾 −1 𝑚𝑜𝑙 −1 ×298 𝐾
=
1.00 𝐿
= 27.78 atm
(b). Let us apply equation 3.2, to find the pressure.
so that the correction terms for pressure and volume are
𝑎𝑛2
3.592 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2 × (1.136 𝑚𝑜𝑙)2
=
𝑉2
(1.00 𝐿)2
= 4.635 atm
𝑛𝑏 = 1.136 𝑚𝑜𝑙 × (0.043 L/mol)
= 0.0488 L
Finally, substituting these values in the van der Waals equation, we
have
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(P + 4.635 atm) (1.00 L ‒ 0.0488 L)
= (1.136 mol) (0.0821 L atm/K -1 mol-1) (298 K)
P = 24.61 atm
Example 3.6. Oxygen is supplied to hospital and chemical laboratories
in large steel cylinders. Typically, such a cylinder has an internal
volume of 28.0 litres and contains 6.80kg of oxygen. Use Van der
Waal‘s equation to estimate the pressure inside such cylinder at 20oC.
Solution
Data provided
a = 1.36 atm L2/mol2
b = 0.032 L/mol
V = 28.0 L
T =20°C = (20 + 273) K = 293 K
R = 0.0821 L atm/K -1 mol-1
Mass of O2 = 6.8kg = 6800g
𝑚𝑎𝑠𝑠
Number of mole of O2 (n) = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
But molar mass of O2 = (16 × 2) = 32 g/mol
6800 𝑔
∴ (n) =
= 212.5 𝑚𝑜𝑙
32 𝑔/𝑚𝑜𝑙
Let us apply Van der Wall‘s equation to find the pressure.
so that the correction terms for pressure and volume are
𝑎𝑛2
1.36 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2 × (212.5 𝑚𝑜𝑙)2
=
𝑉2
(28.00 𝐿)2
= 78.3 atm
𝑛𝑏 = 212.5 𝑚𝑜𝑙 × (0.032 L/mol)
= 6.8 L
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Finally, substituting these values in the van der Waals equation, we
have
(P + 78.3 atm) (28 L ‒ 6.8 L)= (212.5 mol) (0.0821 L atm/K -1 mol-1) (293
K)
P = 162.82 atm
Example 3.7. Use Van der Waal‘s equation, calculate the pressure
exerted by 1 mol of ammonia at 0OC in a volume of
(a)
1 .0 litre and
(b)
0.05 litre
Solution
Use Van der Waal‘s equation. Convert temperatures to kelvins.
Data provided
a = 4.17 atm L2/mol2
b = 0.037 L/mol
R = 0.0821 L atm/K -1 mol-1
(a)
V = 1.0 L
T =0°C = (0 + 273) K = 273 K
n = 1 mol
Applying Van der Waal‘s equation to find the pressure;
But for 1 mol, the equation is reduced to
𝑎
𝑃 + 𝑉 2 𝑉 − 𝑏 = 𝑅𝑇
Substituting data into above equation;
𝑃+
4.17 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2
(1.00 𝐿)2
−1
−1
0.0821 L atmK
22.4 L atm
mol
1.00 𝐿 − 0.037 𝐿/𝑚𝑜𝑙 =
× 273 K
𝑃 + 4.17 𝑎𝑡𝑚𝑚𝑜𝑙 −1 0.963 𝐿/𝑚𝑜𝑙 =
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0.963𝑃 𝐿/𝑚𝑜𝑙 + 4. 𝐿 𝑎𝑡𝑚 = 22.4 L atm
0.963𝑃𝐿/𝑚𝑜𝑙 = (22.4 − 4 )𝐿𝑎𝑡𝑚
0.963𝑃𝐿/𝑚𝑜𝑙 = 18.4 𝐿𝑎𝑡𝑚
18 𝐿𝑎𝑡𝑚
𝑃 =
0.963 𝐿/𝑚𝑜𝑙
= 19 𝑎𝑡𝑚 𝑚𝑜𝑙 −1
(b)
V = 0.05 L
T =0°C = (0 + 273) K = 273 K
n = 1.0 mol
Applying Van der Waal‘s equation to find the pressure;
But for 1 mol, the equation is reduced to
𝑎
𝑃 + 𝑉 2 𝑉 − 𝑏 = 𝑅𝑇
Substituting data into above equation;
𝑃+
4.17 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2
(0.05 𝐿)2
−1
−1
0.05 𝐿 − 0.037 𝐿/𝑚𝑜𝑙 =
0.0821 L atmK mol × 273 K 𝑃 + 1668 𝑎𝑡𝑚𝑚𝑜𝑙 −1 0.013 𝐿/𝑚𝑜𝑙 =
22.4 L atm
0.013𝑃 𝐿/𝑚𝑜𝑙 + 21.68 𝐿 𝑎𝑡𝑚 = 22.4 L atm
0.013𝑃 𝐿/𝑚𝑜𝑙 = (22.4 − 21.68 )𝐿𝑎𝑡𝑚
0.013𝑃 𝐿/𝑚𝑜𝑙 = 0.716 𝐿𝑎𝑡𝑚
0.716 𝐿𝑎𝑡𝑚
𝑃 = 0.013𝑃 𝐿/𝑚𝑜𝑙𝑙
= 55.0 𝑎𝑡𝑚 𝑚𝑜𝑙 −1
3.8. Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intermolecular forces are responsible for the nonideal behaviour of
gases. They exert even more influence in the condensed phases of
matter—liquids and solids. As the temperature of a gas drops, the
average kinetic energy of its molecules decreases. Eventually, at a
sufficiently low temperature, the molecules no longer have enough
energy to break away from the attraction of neighbouring molecules.
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At this point, the molecules aggregate to form small drops of liquid.
This transition from the gaseous to the liquid phase is known as
condensation.
In contrast to intermolecular forces, intramolecular forces hold
atoms together in a molecule. Intramolecular forces stabilize individual
molecules, whereas intermolecular forces are primarily responsible for
the bulk properties of matter (for example, melting point and boiling
point).
Generally, intermolecular forces are much weaker than
intramolecular forces. Much less energy is usually required to
evaporate a liquid than to break the bonds in the molecules of the
liquid. For example, it takes about 41 kJ of energy to vaporize 1 mole of
water at its boiling point; but about 930 kJ of energy are necessary to
break the two O‒H bonds in 1 mole of water molecules. The boiling
points of substances often reflect the strength of the intermolecular
forces operating among the molecules.
At the boiling point, enough energy must be supplied to
overcome the attractive forces among molecules before they can enter
the vapour phase. If it takes more energy to separate molecules of
substance A than of substance B because A molecules are held together
by stronger intermolecular forces, then the boiling point of A is higher
than that of B. The same principle applies also to the melting points of
the substances. In general, the melting points of substances increase
with the strength of the intermolecular forces.
To discuss the properties of condensed matter, we must
understand the different types of intermolecular forces. Dipole-dipole,
dipole-induced dipole, and dispersion forces make up what chemists
commonly refer to as van der Waals forces, after the Dutch physicist
Johannes van der Waals. Ions and dipoles are attracted to one another
by electrostatic forces called ion-dipole forces, which are not van der
Waals forces. Hydrogen bonding is a particularly strong type of dipoledipole interaction. Because only a few elements can participate in
hydrogen bond formation, it is treated as a separate category.
Depending on the phase of a substance, the nature of chemical bonds,
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and the types of elements present, more than one type of interaction
may contribute to the total attraction between molecules, as we will see
below.
3.8.1. Dipole-Dipole Forces
Dipole-dipole forces are attractive forces between polar molecules,
that is, between molecules that possess dipole moments. Their origin is
electrostatic, and they can be understood in terms of Coulomb‘s law.
The larger the dipole moment, the greater the force.
3.8.2. Ion-Dipole Forces
Coulomb‘s law also explains ion-dipole forces, which attract an
ion (either a cation or an anion) and a polar molecule to each other. The
strength of this interaction depends on the charge and size of the ion
and on the magnitude of the dipole moment and size of the molecule.
The charges on cations are generally more concentrated, because
cations are usually smaller than anions. Therefore, a cation interacts
more strongly with dipoles than does an anion having a charge of the
same magnitude.
3.8.3. Dispersion Forces
What attractive interaction occurs in nonpolar substances? To answer
this question, let us consider the arrangement shown in Figure 3.2
below.
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Figure 3.2: (a) Spherical charge distribution in a helium atom. (b) Distortion
caused by the approach of a cation. (c) Distortion caused by the approach of a
dipole.
If we place an ion or a polar molecule near an atom (or a nonpolar
molecule), the electron distribution of the atom (or molecule) is
distorted by the force exerted by the ion or the polar molecule,
resulting in a kind of dipole. The dipole in the atom (or nonpolar
molecule) is said to be an induced dipole because the separation of
positive and negative charges in the atom (or nonpolar molecule) is due to the
proximity of an ion or a polar molecule. The attractive interaction between
an ion and the induced dipole is called ion-induced dipole interaction,
and the attractive interaction between a polar molecule and the
induced dipole is called dipole-induced dipole interaction.
The likelihood of a dipole moment being induced depends not only on
the charge on the ion or the strength of the dipole but also on the
polarizability of the atom or molecule—that is, the ease with which the
electron distribution in the atom (or molecule) can be distorted. Generally,
the larger the number of electrons and the more diffuse the electron
cloud in the atom or molecule, the greater its polarizability. By diffuse
cloud we mean an electron cloud that is spread over an appreciable
volume, so that the electrons are not held tightly by the nucleus.
Figure 3.3: Induced dipoles interacting with each other. Such patterns exist
only momentarily; new arrangements are formed in the next instant. This
type of interaction is responsible for the condensation of nonpolar gases.
Polarizability allows gases containing atoms or nonpolar
molecules (for example, He and N2) to condense. In a helium atom, the
electrons are moving at some distance from the nucleus. At any instant
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it is likely that the atom has a dipole moment created by the specific
positions of the electrons. This dipole moment is called an instantaneous
dipole because it lasts for just a tiny fraction of a second. In the next
instant, the electrons are in different locations and the atom has a new
instantaneous dipole, and so on. Averaged over time (that is, the time
it takes to make a dipole moment measurement), however, the atom
has no dipole moment because the instantaneous dipoles all cancel one
another. In a collection of He atoms, an instantaneous dipole of one He
atom can induce a dipole in each of its nearest neighbors (Figure 3.3).
At the next moment, a different instantaneous dipole can create
temporary dipoles in the surrounding He atoms. The important point
is that this kind of interaction produces dispersion forces, attractive
forces that arise as a result of temporary dipoles induced in atoms or
molecules. At very low temperatures (and reduced atomic speeds),
dispersion forces are strong enough to hold He atoms together, causing
the gas to condense. The attraction between nonpolar molecules can be
explained similarly.
A quantum mechanical interpretation of temporary dipoles was
provided by the German physicist Fritz London in 1930. London
showed that the magnitude of this attractive interaction is directly
proportional to the polarizability of the atom or molecule. As we might
expect, dispersion forces may be quite weak. This is certainly true for
helium, which has a boiling point of only 4.2 K, or 2269°C. (Note that
helium has only two electrons, which are tightly held in the 1s orbital.
Therefore, the helium atom has a low polarizability.)
Dispersion forces, which are also called London forces, usually
increase with molar mass because molecules with larger molar mass
tend to have more electrons, and dispersion forces increase in strength
with the number of electrons. Furthermore, larger molar mass often
means a bigger atom whose electron distribution is more easily
disturbed because the outer electrons are less tightly held by the nuclei.
Table 3.2 compares the melting points of similar substances that consist
of nonpolar molecules. As expected, the melting point increases as the
number of electrons in the molecule increases. Because these are all
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nonpolar molecules, the only attractive intermolecular forces present
are the dispersion forces.
In many cases, dispersion forces are comparable to or even
greater than the dipole-dipole forces between polar molecules. For a
dramatic illustration, let us compare the boiling points of CH3F
(278.4°C) and CCl4 (76.5°C). Although CH3F has a dipole moment of
1.8 D, it boils at a much lower temperature than CCl4, a nonpolar
molecule. CCl4 boils at a higher temperature simply because it contains
more electrons. As a result, the dispersion forces between CCl4
molecules are stronger than the dispersion forces plus the dipoledipole forces between CH3F molecules. (Keep in mind that dispersion
forces exist among species of all types, whether they are neutral or bear
a net charge and whether they are polar or nonpolar.)
Table 3.2 melting points of similar Nonpolar compounds
Compound
Melting point (OC)
CH4
‒182.5
CF4
‒150.0
CCl4
‒23.0
CBr4
90.0
Cl4
171.0
Example 3.8. What type(s) of intermolecular forces exist between the
following pairs: (a) HBr and H2S, (b) Cl2 and CBr4, (c) I2 and NO3‒, (d)
NH3 and C6H6?
Strategy
Classify the species into three categories: ionic, polar (possessing a
dipole moment), and nonpolar. Keep in mind that dispersion forces
exist between all species.
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Solution
(a)
Both HBr and H2S are polar molecules.
Therefore, the intermolecular forces present are dipole-dipole forces, as
well as dispersion forces.
(b)
Both Cl2 and CBr4 are nonpolar, so there are only dispersion
forces between these molecules.
(c)
I2 is a homonuclear diatomic molecule and therefore
nonpolar, so the forces between it and the ion NO3‒ are ioninduced dipole forces and dispersion forces.
(d)
NH3 is polar, and C6H6 is nonpolar. The forces are dipoleinduced dipole forces and dispersion forces.
3.9. The Hydrogen Bond
Normally, the boiling points of a series of similar compounds
containing elements in the same periodic group increase with
increasing molar mass. This increase in boiling point is due to the
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increase in dispersion forces for molecules with more electrons.
Hydrogen compounds of Group 4A follow this trend, as Figure 3.4
shows. The lightest compound, CH4, has the lowest boiling point, and
the heaviest compound, SnH4, has the highest boiling point. However,
hydrogen compounds of the elements in Groups 5A, 6A, and 7A do
not follow this trend. In each of these series, the lightest compound
(NH3, H2O, and HF) has the highest boiling point, contrary to our
expectations based on molar mass. This observation must mean that
there are stronger intermolecular attractions in NH3, H2O, and HF,
compared to other molecules in the same groups. In fact, this
particularly strong type of intermolecular attraction is called the
hydrogen bond, which is a special type of dipole-dipole interaction between
the hydrogen atom in a polar bond, such as N‒ H, O‒ H, or F‒ H, and an
electronegative O, N, or F atom. The interaction is written
A‒H •••• B or A‒H ••• A
A and B represent O, N, or F; A‒H is one molecule or part of a
molecule and B is a part of another molecule; and the dotted line
represents the hydrogen bond. The three atoms usually lie in a straight
line, but the angle AHB (or AHA) can deviate as much as 30° from
linearity. Note that the O, N, and F atoms all possess at least one lone
pair that can interact with the hydrogen atom in hydrogen bonding.
The average energy of a hydrogen bond is quite large for a
dipole-dipole interaction (up to 40 kJ/mol). Thus, hydrogen bonds
have a powerful effect on the structures and properties of many
compounds. Figure 3.5 shows several examples of hydrogen bonding.
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Figure 3.4: Boiling points of the hydrogen compounds of Groups 4A, 5A, 6A,
and 7A elements. Although normally we expect the boiling point to increase as
we move down a group, we see that three compounds (NH3, H2O, and HF)
behave differently. The anomaly
can be explained in terms of intermolecular hydrogen bonding.
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Figure 3.5: Hydrogen bonding in water, ammonia, and hydrogen fluoride.
Solid lines represent covalent bonds, and dotted lines represent hydrogen
bonds.
The strength of a hydrogen bond is determined by the
coulombic interaction between the lone-pair electrons of the
electronegative atom and the hydrogen nucleus. For example, fluorine
is more electronegative than oxygen, and so we would expect a
stronger hydrogen bond to exist in liquid HF than in H 2O. In the liquid
phase, the HF molecules form zigzag chains:
The boiling point of HF is lower than that of water because each H 2O
takes part in four intermolecular hydrogen bonds. Therefore, the forces
holding the molecules together are stronger in H2O than in HF.
Example 3.9. Which of the following can form hydrogen bonds with
water? CH3OCH3, CH4, F‒, HCOOH, Na+.
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Strategy
A species can form hydrogen bonds with water if it contains one of the
three electronegative elements (F, O, or N) or it has an H atom bonded
to one of these three elements.
Solution
There are no electronegative elements (F, O, or N) in either CH4 or
Na+. Therefore, only CH3OCH3, F‒ and, HCOOH can form hydrogen
bonds with water.
The intermolecular forces discussed so far are all attractive in
nature. Keep in mind, though, that molecules also exert repulsive
forces on one another. Thus, when two molecules approach each other,
the repulsion between the electrons and between the nuclei in the
molecules comes into play. The magnitude of the repulsive force rises
very steeply as the distance separating the molecules in a condensed
phase decreases. This is the reason that liquids and solids are so hard
to compress. In these phases, the
molecules are already in close contact with one another, and so they
greatly resist being compressed further.
Example 3.10. What intermolecular forces besides dispersion forces, if
any, exist in each substance? Are any of these substances solids at room
temperature?
1. potassium chloride (KCl)
2. ethanol (C2H5OH)
3. bromine (Br2)
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Solution
1. Potassium chloride is composed of ions, so the intermolecular
interaction in potassium chloride is ionic forces. Because ionic
interactions are strong, it might be expected that potassium chloride is
a solid at room temperature.
2. Ethanol has a hydrogen atom attached to an oxygen atom, so it
would experience hydrogen bonding. If the hydrogen bonding is
strong enough, ethanol might be a solid at room temperature, but it is
difficult to know for certain. (Ethanol is actually a liquid at room
temperature.)
3. Elemental bromine has two bromine atoms covalently bonded to
each other. Because the atoms on either side of the covalent bond are
the same, the electrons in the covalent bond are shared equally, and the
bond is a nonpolar covalent bond. Thus, diatomic bromine does not
have any intermolecular forces other than dispersion forces. It is
unlikely to be a solid at room temperature unless the dispersion forces
are strong enough. Bromine is a liquid at room temperature.
3.10. Intermolecular Forces at Low Temperature, High Molecular
Weight, and High Pressure
At low temperatures where the gas molecules have lower
kinetic energies, the contribution for attractive forces increases, which
the ideal gas law does not account for.
The universal attractive force, or London dispersion force, also
generally increases with molecular weight. The London dispersion
force is caused by correlated movements of the electrons in interacting
molecules. Electrons that belong to different molecules start "fleeing"
and avoiding each other at the short intermolecular distances, which is
frequently described as formation of "instantaneous dipoles" that
attract each other.
Finally, as a gas is compressed and pressure increases,
repulsive forces from the gas molecules oppose the decrease in
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volume. The frequency of collisions also increases at higher pressure,
thereby increasing the contribution of these intermolecular forces.
3.11. The Mean Free Path of Gas Molecules
The motion of a molecule in a gas is complicated. Besides
colliding with the walls of the confinement vessel, the molecules
collide with each other. A useful parameter to describe this motion is
the mean free path. The mean free path is the average distance
traversed by a molecule between collisions. The mean free path of a
molecule is related to its size; the larger its size the shorter its mean
free path.
Suppose the gas molecules are spherical and have a diameter d.
Two gas molecules will collide if their centers are separated by less
than 2d. Suppose the average time between collisions is ∆t. During
this time, the molecule travels a distance v . ∆t, and sweeps a volume
equal to see diagrams below
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If on average it experiences one collision, the number of molecules in
the volume V must be 1. If N is the number of molecules per unit
volume, this means that
or
The time interval ∆t defined in this manner is the mean time between
collisions, and the mean free path is given by
Here we have assumed that only one molecule is moving while all
others are stationary. If we carry out the calculation correctly (all
molecules moving), the following relation is obtained for the mean free
path:
The Number Density
For N1 stationary particles, the number of molecules per unit volume
 N 1  N
Nd  

 V  V
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Collision Frequency
Z1 = 𝜋𝑑2 𝑁𝑑
The Mean Collision Time
The mean collision time is average time elapsed between successive
collisions.
coll
1
𝑍1
1
= 𝜋𝑑 2 𝑁
𝑑
Factors affecting mean free path
1. Density: As gas density increases, the molecules become closer
to each other. Therefore, they are more likely to run into each
other, so the mean free path decreases.
2. Radius of molecule: increasing the radius of the molecules will
decrease the space between them, causing them to run into each
other thereby decreasing the mean free path.
3. Pressure, temperature, and other factors that affect density can
indirectly affect mean free path.
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CHAPTER FOUR
THE MOLE CONCEPT
4.1 Introduction
Most chemical experiments involve enormous numbers of
atoms or molecules. In order to estimate the quantities of various
chemical substances, Chemists adopted a convenient concept involving
collection of elementary units such as atoms, molecules, ions, or
electrons of a chemical substance. This is concept is ―the mole‖.
This quantity is sometimes referred to as the chemical amount. In
Latin mole means a "massive heap" of material. It is convenient to think
of a chemical mole as such. Visualizing a mole as a pile of particles,
however, is just one way to understand this concept. A sample of a
substance has a mass, volume (generally used with gases), and number
of particles that is proportional to the chemical amount (measured in
moles) of the sample. For example, one mole of oxygen gas (O2 )
occupies a volume of 22.4 L at standard temperature and pressure
(STP; 0°C and 1 atm), has a mass of 31.998 grams, and contains about
6.022 × 10 23 molecules of oxygen. Measuring one of these quantities
allows the calculation of the others and this is frequently done in
stoichiometry.
The mole is the amount of a chemical substance which contains as
many elementary entities as there are atoms in 0.012 kilogram of carbon-12;
its symbol is "mol." When the mole is used, the elementary entities
must be specified and may be atoms, molecules, ions, electrons, other
particles, or specified groups of such particles.
The term "mole" commonly is used to represent the number of
molecules (or atoms) in a quantity of material; that is, one mole of
molecules = Avogadro's number of molecules. In this sense, a mole is a
dimensionless number, just as a dozen means 12. For example, you
could talk about a mole of caterpillars, meaning 6.023 x 1023
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caterpillars. However, the number is seldom useful except in talking
about molecules or atoms. The metric prefixes commonly are used to
give such units as millimoles, nanomoles, or picomoles.
4.2. Avogadro's Number
We have noted that one mole of a substance always contains a
certain number of molecules (or atoms), regardless of the substance
involved. This number is called Avogadro’s number (in honor of the
scientist who first suggested the concept, long before the value of the
number could be determined). The number represented as NA is 6.023
x 1023. We know that the number of particles (atoms or molecules) in 1
mole of a substance is 6.022×1023 atoms or molecules. We can therefore
say:
1 mole of carbon atoms weighs 12.0 g and contains 6.023 x 1023 atoms
1 mole of sodium atoms weighs 23.0 g and contains 6.023 x 1023 atoms
4.3. Calculations Using Mole Concept
Example 4.1. How many molecules are there in 20.0 g of benzene,
C6H6?
Solution:
First find how many moles of C6H6 there are in 20.0 g, then use
Avogadro's number to find the number of molecules.
Molar mass of C6H6 = (6 x 12.0) + (6 x 1.0)
= 78.0 g/mol
Moles of C6H6 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6 𝐻6
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6 𝐻6
=
20 𝑔
78 𝑔/𝑚𝑜𝑙
= 0.256 𝑚𝑜𝑙𝑒𝑠
Number of molecules = moles × Avogadro‘s number
=(0.256 𝑚𝑜𝑙𝑒𝑠)(6.023 x 1023 molecules/mole)
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= 1.54 × 1023 molecules
Example 4.2. How many atoms are represented by 3.00 moles of
calcium (Ca) ?
Solution:
3.00 mol Ca ×
6.023 x 10 23
1 .00 mol Ca
= 1.81 x 1024 atoms Ca
Example 4.3. How many atoms are represented by 3 g of calcium (Ca) ?
[Ca = 40, Avogadro constant = 6.023 x 1023]
Solution
40 g (1 mol) of calcium contains 6.023 x 1023 atoms
1 g of calcium will contain
∴ 3 g will contain
6.023 x 10 23
40
6.023 x 10 23 × 3
40
atoms
atoms
= 4.5 × 1022 atoms
Example 4.4. What is the mass of 6.02 × 1024 atoms of magnesium many
atoms are represented by 3 g of calcium (Ca) ? [Mg = 24, Avogadro
constant = 6.023 x 1023]
Solution
6.023 x 1023 atoms of magnesium weigh 24 g (1 mol)
∴ 6.02 × 1024 atoms weigh
6.02 x 10 24 ×24
6.023 x 10 23
= (24 × 10)g
= 240 g
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Example 4.5. How many moles are in 9.03 × 1023 atoms of sodium? [Na
= 23, Avogadro constant = 6.023 x 1023]
Solution
Let y represent the amount of Na
6.023 x 1023 atoms ≡ 1 mole
9.03 × 1023 atoms ≡ y mole
9.03 x 10 23 × 1.0
y=
6.023 x 10 23
= 1.50 mol
Example 4.6. Calculate the mass of sodium, which would contain the
same number of atoms as 9.0 g of carbon? [Na = 23, C = 12]
Solution
Since equal amount of two or more elements contain the same number
of atoms, then:
1 mole (12.0 g) of carbon contains the same number of atoms as 1 mole
(23.0 g) of sodium.
i.e. 12.0 g of carbon ≡ 23.0 g of sodium
9.0 × 23.0
∴ 9.0 g of carbon ≡
12
= 30.0 g of sodium
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CHAPTER FIVE
GASES IN CHEMICAL REACTION
5.1. Introduction
Gases that are involved in chemical reactions obey the same
laws of stoichiometry that apply to substances in any other state
therefore, the ideal gas law can be used to calculate the quantities of
gaseous substances involved in a reaction and then those results used
to find the quantities of other substances. Figure 5.1 presents the
conversions allowed by the ideal gas law to determine the number of
moles of a gaseous reactant or product.
Figure 5.1: Mole Conversions, Including Application of the Ideal Gas Law to
Determine the Number of Moles of a Gaseous Reactant or Product.
Example 5.1. How many liters of oxygen gas at 21oC and 1.13 atm can
be prepared by thermal decomposition of 0.950 g of KClO3? [K = 39, Cl
= 35.5, O = 16]
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Solution
We first find the number of mole of KClO3 from which the number of
mole of oxygen can be determine through the balanced equation and
finally apply ideal gas equation to find the volume of oxygen that can
be prepared.
Solution
Data provided and necessary conversion
T = 21oC = (21 +273)K = 294K
P = 1.13 atm
Mass of KClO3 = 0.950 g
V=?
Molar mass of KClO3 = [39 + 35.5 + (16× 3)] = 122.5 g/mol
𝑚𝑎𝑠𝑠
Mole of KClO3 (n) = 𝑚𝑜𝑙𝑎𝑟
=
𝑚𝑎𝑠𝑠
0.950 g
122.5 𝑔/𝑚𝑜𝑙
= 0.007787 mol of KClO3
The number of moles of O2 produced is
3 𝑚𝑜𝑙 𝑂2
0.007787 mol of KClO3 (
) = 0.01168 mol O2
2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂 3
We can now use the ideal gas law equation:
𝑉=
𝑛𝑅𝑇
𝑃
=
0.0116 8 mol
0.0821 𝐿 𝑎𝑡𝑚 𝐾 −1 𝑚𝑜𝑙 −1 (294 𝐾)
1.13 𝑎𝑡𝑚
= 0.249 L
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Example 5.2. A chemist decomposes 1.06 g of Hg2O in a sealed system.
The oxygen produced has a pressure of 0.514 atm and a volume of 62.5
mL at 35°C Calculate the value of R from these data.
Solution
Data provided and necessary conversion
T = 35oC = (35 +273)K = 308K
P = 0.514 atm
Mass of Hg2O = 1.06 g
V = 62.5 mL = (62.5/1000) L = 0.0625 L
Molar mass of Hg2O = [ (200.5 × 2 )+ 16] = 417 g/mol
𝑚𝑎𝑠𝑠
Mole of Hg2O (n) =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
1.06 g
= 417 𝑔/𝑚𝑜𝑙
= 0.00254 mol of Hg2O
The number of moles of O2 produced is
1 𝑚𝑜𝑙 𝑂2
0.00254 mol of Hg2O (2 𝑚𝑜𝑙
𝑅=
𝑃𝑉
𝑛𝑇
=
0.514 atm
𝐻𝑔2 𝑂
) = 0.00127 mol O2
(0.0625 L)
0.00127 𝑚𝑜𝑙 (308 𝐾)
= 0.0821 L atm K-1mol-1
Example 5.3. As N2 and H2 react to form NH3 in a large cylinder at
500°C, what happens to (a) the total number of atoms? (b) the total
number of molecules? (c) the total pressure?
Solution
(a) The number of atoms stays the same, as is true for all reactions.
That is the basis for the balanced chemical equation.
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(b) N2 (g) + 3H2 (g) → 2NH3 (g)
The total number of moles of gas decreases as this reaction proceeds, so
the number of molecules also decreases.
(c) The total pressure decreases as the total number of moles of gas
decreases.
Example 5.4. How many litres of CO2 at STP can be prepared by the
complete thermal decomposition of 0.150 mol of Ca(HCO3)2. The
products are CaO, CO2 and H2O.
Solution
Data provided
T = 273K
P = 1.0 atm
R = = 0.0821 L atm K-1mol-1
Mole of Ca(HCO3)2 = 0.150 mol
V=?
The number of moles of CO2 produced is
2 𝑚𝑜𝑙 𝑂2
)
𝐶𝑎(𝐻𝐶𝑂3 )2
0.150 mol of Ca(HCO3)2 (1 𝑚𝑜𝑙
= 0.30 mol CO2
We can now use the ideal gas law equation:
𝑉=
𝑛𝑅𝑇
𝑃
=
0.30 mol
0.0821 𝐿 𝑎𝑡𝑚 𝐾 −1 𝑚𝑜𝑙 −1 (273 𝐾)
1.0 𝑎𝑡𝑚
= 6.7249 L of CO2
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CHAPTER SIX
CHEMICAL EQUILIBRIUM IN GASES
6.1. Introduction
Few chemical reactions proceed in only one direction. Most are,
at least to some extent, reversible. At the start of a reversible process,
the reaction proceeds toward the formation of products. As soon as
some product molecules are formed, the reverse process—that is, the
formation of reactant molecules from product molecules—begins to
take place. When the rates of the forward and reverse reactions are equal and
the concentrations of the reactants and products no longer change with time,
chemical equilibrium is reached.
Chemical equilibrium is a dynamic process. As such, the rate of
product formation and conversion back to reactant molecules is
constant.
Note that a chemical equilibrium reaction involves different
substances as reactants and products. Equilibrium between two phases
of the same substance is called physical equilibrium because the changes
that occur are physical processes. The vaporization of water in a closed
container at a given temperature is an example of physical equilibrium.
In this instance, the number of H2O molecules leaving and the number
returning to the liquid phase are equal:
H2O(l) ⇌ H2O(g)
(The double arrow means that the reaction is reversible.)
The study of physical equilibrium yields useful information,
such as the equilibrium vapour pressure. However, chemists are
particularly interested in chemical equilibrium processes, such as the
reversible reaction involving nitrogen dioxide (NO2) and dinitrogen
tetroxide (N2O4). The progress of the reaction
N2O4(g) ⇌ 2NO2(g)
can be monitored easily because N2O4 is a colorless gas, whereas NO2
has a darkbrown colour that makes it sometimes visible in polluted air.
Suppose that a known amount of N2O4 is injected into an evacuated
flask. Some brown colour appears immediately, indicating the
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formation of NO2 molecules. The colour intensifies as the dissociation
of N2O4 continues until eventually equilibrium is reached. Beyond that
point, no further change in colour is observed. By experiment we find
that we can also reach the equilibrium state by starting with pure NO2
or with a mixture of NO2 and N2O4. In each case, we observe an initial
change in colour, caused either by the formation of NO2 (if the colour
intensifies) or by the depletion of NO2 (if the colour fades), and then
the final state in which the colour of NO2 no longer changes.
Depending on the temperature of the reacting system and on the initial
amounts of NO2 and N2O4, the concentrations of NO2 and N2O4 at
equilibrium differ from system to system (Figure 6.1).
Figure 6.1: Change in the concentrations of NO2 and N2O4 with time, in three
situations. (a) Initially only NO2 is present. (b) Initially only N2O4 is present.
(c) Initially a mixture of NO2 and N2O4 is present. In each case, equilibrium is
established to the right of the vertical line.
The NO2–N2O4 System at
6.2. The Equilibrium Constant
Let us consider the following reversible reaction:
aA + bB ⇌cC + d D
in which a, b, c, and d are the stoichiometric coefficients for the reacting
species A, B, C, and D. The equilibrium constant for the reaction at a
particular temperature is
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𝐾=
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
2016
( 6.1)
Equation (6.1) is the mathematical form of the law of mass action. It
relates the concentrations of reactants and products at equilibrium in terms
of a quantity called the equilibrium constant. The equilibrium constant
is defined by a quotient. The numerator is obtained by multiplying
together the equilibrium concentrations of the products, each raised to a
power equal to its stoichiometric coefficient in the balanced equation.
The same procedure is applied to the equilibrium concentrations of
reactants to obtain the denominator. This formulation is based on
purely empirical evidence, such as the study of reactions like NO2–
N2O4.
The equilibrium constant has its origin in thermodynamics,
however, we can gain some insight into K by considering the kinetics
of chemical reactions. Let us suppose that this reversible reaction
occurs via a mechanism of a single elementary step in both the forward
and reverse directions:
in which kf and kr are the rate constants for the forward and reverse
directions, respectively. At equilibrium, when no net changes occur,
the two rates must be equal:
ratef = rater
Or
kf[A][B]2 = kr[AB2]
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𝐾𝑓
𝐾𝑟
=
2016
[𝐴𝐵2 ]
𝐴 [𝐵]2
Because both kf and kr are constants at a given temperature, their ratio
is also a constant, which is equal to the equilibrium constant Kc.
𝐾𝑓
𝐾𝑟
= 𝐾𝑐 =
[𝐴𝐵2 ]
𝐴 [𝐵]2
So Kc is a constant regardless of the equilibrium concentrations of the
reacting species because it is always equal to kf/kr, the quotient of two
quantities that are themselves constant at a given temperature. Because
rate constants are temperature-dependent, it follows that the
equilibrium constant must also change with temperature.
Finally, we note that if the equilibrium constant is much greater
than 1 (that is, K > 1), the equilibrium will lie to the right of the reaction
arrows and favour the products. Conversely, if the equilibrium
constant is much smaller than 1 (that is, K ˂ 1), the equilibrium will lie
to the left and favour the reactants (Figure 6.2).
Figure 6.2: (a) At equilibrium, there are more products than reactants, and
the equilibrium is said to lie to the right. (b) In the opposite situation, when
there are more reactants than products, the equilibrium is said to lie to the left.
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6.3. Ways of Expressing Equilibrium Constants
To use equilibrium constants, we must express them in terms of
the reactant and product concentrations. Our only guidance is the law
of mass action [Equation (6.1)]. However, because the concentrations of
the reactants and products can be expressed in different units and
because the reacting species are not always in the same phase, there
may be more than one way to express the equilibrium constant for the
same reaction. To begin with, we will consider reactions in which the
reactants and products are in the same phase.
6.4. Homogeneous Equilibria
The term homogeneous equilibrium applies to reactions in
which all reacting species are in the same phase. An example of
homogeneous gas-phase equilibrium is the dissociation of N2O4. The
equilibrium constant is
𝐾𝑐 =
[𝑁𝑂2 ]2
[𝑁2 𝑂4 ]
Note that the subscript in Kc denotes that the concentrations of the
reacting species are expressed in moles per liter. The concentrations of
reactants and products in gaseous reactions can also be expressed in
terms of their partial pressures. At constant temperature the pressure P
of a gas is directly related to the concentration in moles per liter of the
gas; that is, P = (n/V)RT. Thus, for the equilibrium process
N2O4(g) ⇌ 2NO2(g)
We can write
𝐾𝑃 =
𝑃 2 𝑁𝑂2
𝑃𝑁2 𝑂4
in which P𝑁𝑂2 and 𝑃𝑁2 𝑂4 are the equilibrium partial pressures (in
atmospheres) of NO2 and N2O4, respectively. The subscript in KP tells
us that equilibrium concentrations are expressed in terms of pressure.
In general, Kc is not equal to Kp, because the partial pressures of
reactants and products are not equal to their concentrations expressed
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in moles per liter. A simple relationship between KP and Kc can be
derived as follows. Let us consider this equilibrium in the gas phase:
aA(g) ⇌ bB(g)
in which a and b are stoichiometric coefficients. The equilibrium
constant Kc is
𝐾𝑐 =
and the expression for KP is
[𝐵]𝑏
[𝐴]𝑎
𝑃𝑏 𝐵
𝐾𝑝 = 𝑃 𝑎 𝐴
in which PA and PB are the partial pressures of A and B. Assuming
ideal gas behaviour,
𝑃𝐴 𝑉 = 𝑛𝐴 𝑅𝑇
𝑛 𝑅𝑇
𝑃𝐴 = 𝐴𝑉
in which V is the volume of the container in liters. Also,
𝑃𝐵 𝑉 = 𝑛𝐵 𝑅𝑇
𝑛 𝑅𝑇
𝑃𝐵 = 𝐵
𝑉
Substituting these relations into the expression for KP, we obtain
Now both nA/V and nB/V have the units of moles per liter and can be
replaced by [A] and [B], so that
in which
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∆𝑛 = 𝑏 − 𝑎
= moles of gaseous products ‒ moles of gaseous reactants
Because pressure is usually expressed in atmospheres, the gas constant
R is given by 0.0821 L atm/K mol, and we can write the relationship
between KP and Kc as
𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛
(6.2)
In general, KP ≠ Kc except in the special case when ∆n = 0. In that case,
Equation (5.2) can be written as
𝐾𝑃 = 𝐾𝐶 (0.0821T)0
𝐾𝑃 = 𝐾𝐶
6.5. Equilibrium Constant and Units
Note that it is general practice not to include units for the
equilibrium constant. In thermodynamics, the equilibrium constant is
defined in terms of activities rather than concentrations. For an ideal
system, the activity of a substance is the ratio of its concentration (or
partial pressure) to a standard value, which is 1 M (or 1 atm). This
procedure eliminates all units but does not alter the numerical parts of
the concentration or pressure. Consequently, K has no units.
Example 6.1. Write expressions for KP if applicable, for the following
reversible reactions at equilibrium:
(a).
HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)
(b).
2NO (g) + O2(g) ⇌ 2NO2 (g)
Strategy
Keep in mind the following facts: (1) the KP expression applies only to
gaseous reactions and (2) the concentration of solvent (usually water)
does not appear in the equilibrium constant expression.
Solution
(a)
Because there are no gases present, KP does not apply.
(b)
𝐾𝑃 =
𝑃 2 𝑁𝑂2
𝑃 2 𝑁𝑂 𝑃𝑂2
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Example 6.2
The equilibrium constant KP for the decomposition of phosphorus
pentachloride to phosphorus trichloride and molecular chlorine
PCl5(g) ⇌
PCl3(g) + Cl2(g)
is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5
and PCl3 are 0.973 atm and 0.548 atm, respectively, what is the
equilibrium partial pressure of Cl2 at 250°C.
Strategy
The concentrations of the reacting gases are given in atm, so we can
express the equilibrium constant in KP. From the known KP value and
the equilibrium pressures of PCl3 and PCl5, we can solve for P𝐶𝑙2 .
Solution
Data given
Partial pressures of PCl5 = 0.973 atm
Partial pressures of PCl3 = 0.548 atm
𝐾𝑃 = 1.05
Partial pressures of Cl2 = ?
First, we write KP in terms of the partial pressures of the reacting
species
𝐾𝑃 =
𝑃𝑃𝐶𝑙 3 𝑃 𝐶𝑙
2
𝑃𝐶𝑙 5
Knowing the partial pressures, we write
1.05 =
𝑃𝐶𝑙 2 =
0.548 ( 𝑃𝐶𝑙 2 )
(0.973)
1.05 ×0.973
0.548
= 1.86 atm
Example 6.3
The equilibrium constant KP for the reaction
2NO2 ⇌ 2NO + O2 (g)
is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270
atm.
Solution
Data given
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Partial pressures of NO2 = 0.400 atm
Partial pressures of NO = 0.270 atm
𝐾𝑃 = 158
Partial pressures of O2 =?
𝑃 2 𝑁𝑂 𝑃 𝑂
2
𝑃 2 𝑁𝑂 2
(0.270)2 × 𝑃𝑂 2
158 =
(0.400)2
0.16 (158)
𝑃𝑂2 = 0.073
𝐾𝑃 =
= 346 atm
Example 6.4
Methanol (CH3OH) is manufactured industrially by the reaction
CO(g) + 2H2(g) ⇌ CH3OH(g)
The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is
the value of KP at this temperature?
Strategy
Apply the relationship between Kc and KP is given by Equation (6.2).
What is the change in the number of moles of gases from reactants to
product? Recall that ∆n = moles of gaseous products ‒ moles of
gaseous reactants. Convert temperature to kelvins
Solution
Data given
Kc = 10.5
T = 220oC = (220 + 273)K = 493 K
∆n = (1‒3) = ‒2
Applying
𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛
𝐾𝑃 = (10.5)(0.0821 × 493)-2
= 6.41× 10-3
Example 6.5
For the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g)
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KP is 4.3 × 10-4 at 375°C. Calculate Kc for the reaction.
Solution
Data given
Kp = 4.3 × 10-4
T = 375oC = (375 + 273)K = 648 K
∆n = (2‒4) = ‒2
Applying
𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛
= 𝐾𝐶 (0.0821 × 648)-2
4.3 × 10-4 = 𝐾𝐶 (3.5 × 10-4)
𝐾𝐶 =
4.3 × 10 −4
3.5 × 10 −4
= 1.2
6.6. Heterogeneous Equilibria
A reversible reaction involving reactants and products that are in
different phases leads to a heterogeneous equilibrium. For example,
when calcium carbonate is heated in a closed vessel, this equilibrium is
attained:
CaCO3(s) ⇌ CaO(s) + CO2(g)
The two solids and one gas constitute three separate phases. At
equilibrium, we might write the equilibrium constant in terms of
partial pressure for gases as
𝐾𝑃 = 𝑃𝐶𝑂2
The equilibrium constant in this case is numerically equal to the
pressure of CO2 gas, an easily measurable quantity.
Example 6.6
Consider the following heterogeneous equilibrium:
CaCO3(s) ⇌ CaO(s) + CO2(g)
At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc
for the reaction at this temperature.
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Strategy
Remember that pure solids do not appear in the equilibrium constant
expression. The relationship between KP and Kc is given by Equation
(6.2).
Solution
(a) Using 𝐾𝑃 = 𝑃𝐶𝑂2 we write
𝐾𝑃 = 𝑃𝐶𝑂2
= 0.236
Applying 𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛
Where T = 800°C = (800 + 273)K = 1073 K
∆n = 1
𝐾𝑃 = 0.236
0.236 = 𝐾𝐶 (0.0821 × 1073)1
0.236 = 𝐾𝐶 (88.09)
0.236
𝐾𝐶 = 88.09
= 2.68 × 10-3
Example 6.7
Consider the following equilibrium at 395 K:
NH4HS(s) ⇌ NH3(g) + H2S(g)
The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for
the reaction.
Solution
Data given
T = 395oC = (395 + 273)K = 670 K
∆n = 2
𝑃𝑁𝐻3 = 0.265
𝑃𝐻2 𝑆 = 0.265
Kp = ?
Kc = ?
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𝐾𝑃 = 𝑃𝑁𝐻3 𝑃𝐻2 𝑆
= 0.265 (0.265)
= 0.07
Applying
𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛
= 𝐾𝐶 (0.0821 × 670)2
0.07
= 𝐾𝐶 (3025)
𝐾𝐶 =
0.07
3025
= 2.3 × 10-5
Example 6.8
Starting with a 3: 1 mixture of H2 and N2 at 450.0°C, the equilibrium
mixture is found to be 9.6% NH3, 22.6% N2, and 67.8% H2 by volume.
The total pressure is 50.0 atm. Calculate KP and Kc.
The reaction is N2 + 3H2 ⇌ 2NH3.
Solution:
According to Dalton's law of partial pressures, the partial pressure of a
gas in a mixture is given by the product of its volume fraction and the
total pressure. Therefore the equilibrium pressure of each gas is
𝑃𝑁𝐻3 = (0.096)(50.0 atm) = 4.8 atm
𝑃𝑁2 = (0.226)(50.0 atm) =11.3 atm
𝑃𝐻2 = (0.678)(50.0 atm) = 33.9 atm
Total pressure = 50.0 atm
T = 450.0°C = (450 + 273)K = 723 K
∆n = (2‒4) = ‒2
The equilibrium expression for the reaction is
𝐾𝑃 =
𝑃 2 𝑁𝐻 3
𝑃 3 𝐻 2 𝑃𝑁 2
Substituting the data gives
=
=
(4.80)2
(33.9)3 ×(11.3)
23.04
(440227 )
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= 5.2 × 10-5
𝐾𝑃 = 𝐾𝐶 (0.0821T)∆𝑛 to find KC
= 𝐾𝐶 (0.0821 × 723)-2
5.2 × 10-5 = 𝐾𝐶 (2.8 × 10-4)
𝐾𝐶 =
5.2 × 10 −5
2.8 × 10 −4
= 0.184
6.7. The Effect of Change in Partial Pressure of One Gas on Kp and
Equilibrium Position
It often is important to know the yield of a chemical reaction—
that is, the percentage of reactants converted to products. The
following example shows how this yield may be calculated, and how
conditions may be altered to increase the yield.
Example 6.9
Kp = 54.4 at 355.0°C for the reaction H2 + I2 ⇌ 2HI. What percentage of
I2 will be converted to HI if 0.20 mole each of H2 and I2 are mixed and
allowed to come to equilibrium at 355.0°C and a total pressure of 0.50
atm?
Solution:
Assume that X moles each of H2 and I2 are used up in reaching
equilibrium
to give 2X moles of HI, in accordance with the chemical equation,
leaving 0.20 - X moles each of H2 and I2. The partial pressure of each
gas is given by the product of its mole fraction and the total pressure.
𝑃𝐻𝐼 =
2𝑋
0.4
(0.50 𝑎𝑡𝑚)
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𝑃𝐻2 = 𝑃𝐼2 =
𝐾𝑃 =
𝑃 2 𝐻𝐼
𝑃𝐻 2 𝑃𝐼 2
=
0.2 −𝑋
0.4
2016
(0.50 𝑎𝑡𝑚)
2𝑋
(0.50 𝑎𝑡𝑚 )]2
0.4
0.2 −𝑋
(0.50 𝑎𝑡𝑚 )]2
0.4
[
[
2𝑋
54.4 = (0.2 −𝑋 )2
Taking the square root of each side, we obtain
2𝑋
7.4 = 0.2 −𝑋
2X = 7.4 (0.2 ‒ X)
2X = 1.48 ‒ 7.4X
9.4X = 1.48
X = 0.157 = moles of H2 and I2 used up
Percentage conversion (yield) =
0.157
0.200
× 100 = 78.5%
Example 6.10
What percentage of I2 will be converted to HI at equilibrium at 355.0°C,
if 0.200 mole of I2 is mixed with 2.00 moles of H2 at total pressure of
0.50 atm?
Solution:
In this problem, it is advantageous first to assume that the large excess
of H2 will use almost the entire amount of I2, leaving only X moles of it
unused. In general, it is always advantageous to let X represent the
smallest unknown entity because it often simplifies the mathematical
solution. If X moles of I2 are not used, then 0.20 ‒ X moles are used. For
every mole of I2 used up, one of H2 is used up, and two of HI are
formed. Proceeding as in the last problem, the number of moles of each
component at equilibrium is
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The partial pressure of each component will be the mole fraction of
each times the total pressure, as follows.
𝑃𝐻2 =
𝑃𝐼2 =
𝑃𝐻𝐼 =
1.80+ 𝑋
2.20
(0.50 𝑎𝑡𝑚)
𝑋
(0.50 𝑎𝑡𝑚)
2.20
0.40 − 2𝑋
(0.50 𝑎𝑡𝑚)
2.20
When we substitute these partial pressures into the expression for Kp,
we get an expression that will be tedious to solve unless we make a
reasonable approximation: we assume that X is negligible in
comparison with 0.20 and 1.80.
𝐾𝑃 = 54.4 =
X=
0.40 − 2𝑋
(0.50 )]2
2.20
1.80+ 𝑋
𝑋
(0.50 ) [
(0.50
2.20
2.20
[
)]
≅
(0.40)2
𝐼.80 (𝑋)
(0.40)2
𝐼.80 (54.4)
= 0.0016 moles of I2 not used
0.200 ‒ 0.0016 = 0.1984 moles of I2 used
Percentage of I2 used =
0.1984
0.200
× 100 = 99.2%
Note that the wise decision to let X = the amount of I2 not used instead
of the amount of I2 that was used really did simplify the solution by
making it possible to neglect X when added to or subtracted from
larger numbers. If we had solved the quadratic equation instead, we
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would have found that 99.197% of the I2 had been used up. This is a
common method of simplifying a mathematical problem, and at the end
you can always check to see whether your answer really is negligible
compared to what you said it was. Many chemists say that if X is less
than 10.0% of what it is added to or subtracted from, it is okay to
neglect it.
The preceding problem illustrates the fact that, although the
value of Kp does not change with changes in concentration, the
equilibrium position will change to use up part of the excess of any one
reagent that has been added. In this problem, the large excess of H 2
shifts the equilibrium position to the right, causing more of the I2 to be
used up (99.2% compared to 78.5%) than when H2 and I2 are mixed in
equal proportions. Advantage may be taken of this principle by using a
large excess of a cheap chemical to convert the maximum amount of an
expensive chemical to a desired product. In this case I2, the more
expensive chemical, is made to yield more HI by using more of the
cheaper H2.
6.8. The Percentage Decomposition of Gases
Many gases decompose into simpler ones at elevated
temperatures, and it often is important to know the extent to which
decomposition takes place.
Example 6.11
Kp = 1.78 atm at 250.0°C for the decomposition reaction
PCl5(g) ⇌
PCl3(g) + Cl2(g). Calculate the percentage of PC15 that
dissociates if 0.0500 mole of PC15 is placed in a closed vessel at 250.0°C
and 2.00 atm pressure.
Solution:
Although you are told that you are starting with 0.0500 mole PCl5, this
piece of information is not needed to find the percentage dissociation
at the given pressure and temperature. If you were asked for the
volume of the reaction vessel, then you would need to know the actual
number of moles; otherwise not. To answer the question that is asked,
it is simpler to just start with one mole (don't worry about the volume)
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and assume that X moles of PCl5 dissociate to give X moles each of
PCl3 and Cl2 and 1 - X moles of PCl5 at equilibrium.
The partial pressures are given by the mole fractions times the total
pressure, and are substituted into the Kp expression, to give
𝑋
1+𝑋
𝐾𝑝 = 1.78 =
=
1.78 =
2.0 𝑎𝑡𝑚
1 −𝑋
1+𝑋
𝑋
1+𝑋
2.0 𝑎𝑡𝑚
0.50 𝑎𝑡𝑚
2𝑋 2
1−𝑋 (1+𝑋)
2𝑋 2
1− 𝑋 2
1.78 ‒ 1.78X2 = 2X2
𝑋2 =
1.78
𝑋=
0.478
3.78
= 0.478
= 0.686 moles PCl5 dissociate
Percentage of PCl5 dissociated =
0.686
1.00
× 100 = 68.6%
This was not a difficult quadratic equation to solve but, even if it had
been, it would not be possible to neglect X compared to 1.00; it is too
large. If we had neglected X, we would have obtained the extremely
erroneous answer of 94.3% dissociated. If Kp is very large (or very
small), it means that the equilibrium position lies far to the right (or to
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the left). In either of these cases it is possible to choose X so that it will
be very small and amenable to a simplified math solution. The value of
Kp for the PCl5 equilibrium is neither very large nor very small, and
hence it never will be possible to neglect X.
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CHAPTER SEVEN
WORK OF EXPANSION AND COMPRESSION OF GAS
7.1. Introduction
We have seen that work can be defined as force F multiplied by
distance d:
w = Fd
In thermodynamics, work has a broader meaning that includes
mechanical work (for example, a crane lifting a steel beam), electrical
work (a battery supplying electrons to light the bulb of a flashlight),
and surface work (blowing up a soap bubble). In this section we will
concentrate on mechanical work.
One way to illustrate mechanical work is to study the
expansion or compression of a gas. Many chemical and biological
processes involve gas volume changes. Breathing and exhaling air
involves the expansion and contraction of the tiny sacs called alveoli in
the lungs. Another example is the internal combustion engine of the
automobile. The successive expansion and compression of the
cylinders due to the combustion of the gasoline-air mixture provide
power to the vehicle. Figure 6.1 shows a gas in a cylinder fitted with a
weightless, frictionless movable piston at a certain temperature,
pressure, and volume. As it expands, the gas pushes the piston upward
against a constant opposing external atmospheric pressure P. The work
done by the gas on the surroundings is
Work done = 𝑃 × 𝐴 × 𝑕
But 𝐴 × 𝑕 = 𝑉
Therefore 𝑤 = 𝑃∆𝑉
w = ‒P∆V
(7.1)
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Where ∆V, the change in volume, is given by Vf ‒ Vi. The minus sign in
Equation (7.1) takes care of the sign convention for w. For gas
expansion (work done by the system), ∆V > 0, so ‒P∆V is a negative
quantity. For gas compression (work done on the system), ∆V ˂ 0, and
‒P∆V is a positive quantity.
Note that ―‒P∆V‖ is often referred to as ―P-V‖ work.
Figure 7.1: The expansion of a gas against a constant external pressure (such
as atmospheric pressure). The gas is in a cylinder fitted with a weightless
movable piston. The work done is given by ‒P∆V.
According to Equation (6.1), the units for work done by or on a
gas are liters atmospheres. To express the work done in the more
familiar unit of joules, we use the conversion factor
1 L . atm = 101.3 J
Example 7.1. A certain gas expands in volume from 2.0 L to 6.0 L at
constant temperature. Calculate the work done by the gas if it expands
(a) against a vacuum and (b) against a constant pressure of 1.2 atm.
Strategy A simple sketch of the situation is helpful here:
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The work done in gas expansion is equal to the product of the external,
opposing pressure and the change in volume.
Solution
(a) Because the external pressure is zero, no work is done in the
expansion:
w = ‒P∆V
= ‒ (0)(6.0 ‒ 2.0) L
=0
(b) The external, opposing pressure is 1.2 atm, so
w = ‒P∆V
= ‒ (1.2 atm) (6.0‒ 2.0) L
= ‒ 4.8 L . atm
To convert the answer to joules, we write
101.3𝐽
w = ‒ 4.8 L . atm × 1 𝐿.𝑎𝑡𝑚
= ‒ 4.9 × 102 J
Because this is gas expansion (work is done by the system on the
surroundings), the work done has a negative sign.
Example 7.2. A gas expands from 264 mL to 971 mL at constant
temperature. Calculate the work done (in joules) by the gas if it
expands (a) against a vacuum and (b) against a constant pressure of
4.00 atm.
Solution
This is similar to above example. Convert volume to litre.
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(a) Because the external pressure is zero, no work is done in the
expansion:
w = ‒P∆V
= ‒ (0)(971 ‒ 264) mL
= ‒ (0)(527/1000) L
=0
(b) The external, opposing pressure is 4.0 atm, so
w = ‒P∆V
= ‒ (4.0 atm) (971 ‒ 264) Ml
=‒ (4.0 atm) (971 ‒ 264)
=‒ (4.0 atm)(527/1000) L
= ‒ (4.0 atm)(0.527) L
= ‒ 2.108 L. atm
To convert the answer to joules, we write
101.3𝐽
w = ‒ 2.108 L . atm × 1 𝐿.𝑎𝑡𝑚
= ‒ 2.13 × 102 J
Example 7.3. The work done when a gas is compressed in a cylinder is
387 J. During this process, there is a heat transfer of 152 J from the gas
to the surroundings. Calculate the energy change for this process.
Strategy
Compression is work done on the gas, so what is the sign for w? Heat
is released by the gas to the surroundings. Is this an endothermic or
exothermic process? What is the sign for q?
Solution
To calculate the energy change of the gas, we need ∆U = q + w. Work of
compression is positive and because heat is released by the gas, q is
negative. Therefore, we have
∆U = q + w
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= ‒152 J + 387 J
= 235 J
As a result, the energy of the gas increases by 235 J.
Example 7.4. A gas expands and does P-V work on the surroundings
equal to 279 J. At the same time, it absorbs 216 J of heat from the
surroundings. What is the change in energy of the system?
Expansion is work done by system on surrounding, so what is the sign
for w? Heat is absorbed from the surroundings. Is this an endothermic
or exothermic process? What is the sign for q?
Solution
To calculate the energy change of the gas, we use ∆U = q + w. Work of
expansion is negative and because heat is absorbed from the
surroundings, q is positive. Therefore, we have
∆U = q + w
= 216 J ‒ 279 J
= ‒ 63 J
7.2. Enthalpy
Internal energy of a system is a state function, and this property
is used in the discussion of any change in the heat content of the
chemical reaction. Consider the reaction:
2CO(g) +O2(g)
→ 2CO2(g)
The reactants CO and oxygen are the initial states of the atoms
and the product CO2 gives the final state. Since internal energy is a
state function, the energy change ∆E associated with the reaction
depends only on the initial and final states but not on the path taken by
the reaction.
The energy change, ∆E of each chemical reaction run at
constant temperature is a measure of the relative bond strength of
reactants and products. The expression for pressure – volume (P-V)
work shows than ∆E can be measured using the equation
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∆𝐸 = 𝑞 – 𝑤
=q–
𝑣2
𝑣1
𝑝𝑒𝑥 𝑑𝑣
For reaction at constant volume
V1=V2,
∆V = 0
Therefore ∆E = q – 0 (constant volume)
So ∆E = 𝑞𝑉
(Change in internal energy is equal to the heat absorbed by the system
when the process occurs at constants volume).
Change in internal energy can easily be measured using closed
vessel such as bomb calorimeter but most often than not, chemical
reactions are run at constant pressure like using open vessels, instead
of constant volume, such that heat absorbed by the system is neither
equal to 𝑞𝑣 or ∆E.
We shall therefore develop another state function called the
enthalpy which takes care of both the internal energy and work done
by the system.
𝜕𝑈 = 𝜕𝑞 − 𝑃𝜕𝑉
∆𝑈 = 𝑞𝑃 – 𝑃∆𝑉
∆𝑈 = 𝑈2 ‒ 𝑈1
∆𝑉 = 𝑉2 ‒ 𝑉1
Hence 𝑈2 ‒ 𝑈1 = 𝑞𝑝 – 𝑃 (𝑉2 ‒ 𝑉1 )
𝑈2 ‒ 𝑈1 = 𝑞𝑝 – 𝑃𝑉2 + 𝑃𝑉1
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𝑈2 + 𝑃𝑉2 = 𝑞𝑝 + 𝑈1 + 𝑃𝑉1
𝑈, 𝑃 𝑎𝑛𝑑 𝑉 are all state functions therefore 𝑈 + 𝑃𝑉 (internal energy
plus work of expansion) will therefore give us another state function
represented by 𝐻 (𝑈 + 𝑃𝑉) - enthalpy.
Therefore at constant pressure,
𝜕𝑞 = 𝑈2 + 𝑃𝑉2 − (𝑈1 + 𝑃𝑉1 )
= 𝐻2
−
𝐻1
𝑞𝑝 = ∆𝐻
Addition of heat at constant pressure result to increase in enthalpy.
The enthalpy change is equal to the heat absorbed only when the
process is carried out at constant pressure. Because of equality between
∆H and 𝑞𝑝 , the enthalpy is often called heat content of a system.
For a process at constant pressure in which heat is evolved by
the system to the surroundings, ∆H and 𝑞𝑝 are both negative. This is
because enthalpy of the final state is lower than the enthalpy of the
initial state.
∆H = 𝐻𝑓 − 𝐻𝑖 , negative
This process is exothermic. There is rise in temperature, on the other
hand when the system absorbs heat from the surroundings both ∆H
and 𝑞𝑝 are the positive, the process is endothermic. There is fall in
temperature.
Our results so far show that
∆E = 𝑞𝑉
∆H = 𝑞𝑝
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
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= ∆𝐸 + 𝑃2 𝑉2 − 𝑃1 𝑉1
∆H and ∆E differ only by the difference in the PV products of the final
and initial states.
If gases are produced or consumed in a chemical reaction,
∆(𝑃𝑉) may be quite appreciable and ∆H and ∆E differ significantly.
For ideal gases reacting at constant temperature producing
gases as the only products,
aA (g) + bB (g) → cC (g) + dD (g)
𝑃𝑉 (products) = (𝑐 + 𝑑)𝑅𝑇
𝑃𝑉 (reactants) = (𝑎 + 𝑏)𝑅𝑇
Therefore ∆(𝑃𝑉) is given by the expression
∆(𝑃𝑉) = [(𝑐 + 𝑑)𝑅𝑇 − (𝑎 + 𝑏)𝑅𝑇 ]
= ∆𝑛𝑅𝑇
The total number of moles of gaseous products minus the total number
of moles of gaseous reactants is defined as ∆𝑛, change in number of
moles
In general then we can say
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
= ∆𝐸 + ∆𝑛𝑅𝑇
Example 6.7. For decomposition of MgCO3(s) by the reaction,
MgCO3(s)
→
MgO(s) + CO2(g)
∆H = 108.78kJ at 900K and 1atm pressure. If the molar volume of
MgCO3(s) is 0.028dm3 and that of MgO(g) is 0.011dm3, find ∆E
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Solution:
The ∆(𝑃𝑉) is divided into two terms, that due to volume change in
solids and that due to production of gas.
∆(𝑃𝑉) = ∆(𝑃𝑉) solids + ∆(𝑃𝑉)gas
∆(𝑃𝑉)s = p (0.011− 0.028) = − p × 0.017 negligible
Since 1mole of gas appear in the products, ∆n = 1
∆(𝑃𝑉)g = ∆nRT = 1×8.314×900 = 7.483kJ
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)
= (108.78 – 7.483) kJ
= 101.297 kJ
= 101.30 kJ
Example 7.8. Liquid water is vaporized at 100oC and 1.013bar. The heat
of vaporization is 40.69 kJ mol-1. What are the values of
(i)
(ii)
(iii)
(iv)
Wrev per mole
q per mole
∆E and
∆H?
Solution
(i)
Assuming that water vapour is an ideal gas and that the
volume of liquid water is negligible compared to steam.
𝑊𝑟𝑒𝑣 = + 𝑃∆𝑉 = + 𝑅𝑇
= + (8.314×10-3 kJ mol-1K-1)(373.15K)
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= + 3.10 kJ mol-1
(ii)
(iii)
(iv)
The heat of vapourization is 40.69kJmol-1 and since heat is
absorbed, q has a positive value,
q = 40.69 kJ mol-1
∆𝐸 = 𝑞 − 𝑤
= (40.69 – 3.10) kJ mol-1
= 37.59 kJ mol-1
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
= (37.59+3.10) kJ mol-1
=40.69 kJ mol-1
Example 7.9. Calculate the work done on the surrounding when one
mole of water is vaporized at 100oC and 1atm. Giving molar volume of
liquid water as 18cm3 mol-1 and molar volume of steam as 24dm3 mol-1
at 1atm
Solution:
Work done on the surrounding is = 𝑃∆𝑉
Molar volume of liquid water is 18cm3 mol-1
Molar volume of steam = 24dm3 mol-1
Therefore the volume of 1 mole of liquid water is negligible
1 atm = 1.01325× 105 𝑁𝑚−2
24dm3 = 24× 10−3 m3
Work done = 1.01325× 105 𝑁𝑚−2 × 24× 10−3 m3 mol-1
= 24.319 × 102 Nm mol-1
= 2.43 kJ mol-1
Example 7.10. If the enthalpy change for the process in question 3
above is 40.70 kJ mol-1, what is the change in internal energy ∆E?
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Solution
From
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)
Where ∆H = 40.70 kJ mol-1
∆E = 40.70 kJ mol-1 – 2. 43 kJ mol-1
= 38.27 kJ mol-1
Example 7.11. For the combustion of benzene according to the equation
1
C6H6 (l ) + 72 O2 (g) → 6CO2 (g) + 3H2O (l)
If the heat of reaction at constant pressure is
∆H25oC = −3267.62 kJ mol-1, find ∆E
Solution:
There is contraction of gaseous volume from 7.5 to 6 moles of gas.
Hence ∆n = 6 − 7.5 = −1.5
∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)
= ∆𝐸 + ∆𝑛𝑅𝑇
∆𝐸 = ∆𝐻 − ∆𝑛𝑅𝑇
= ∆H – (– 1.5× 8.314 × 298.2)J
= (– 3267.62 + 3.72) kJ
= – 3263.90 kJ
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Example 7.12. To vaporize 100.0g of CCl4 at its normal boiling point,
349.0K and 1atm; 19.5kJ of heat is required. Calculate ∆Hvap for CCl4
and compare it with ∆E for the same process.
Solution:
Molar mass of CCl4 = 154.0g mol-1
100𝑔
Number of moles of CCl4 in 100g = 154𝑔 𝑚𝑜𝑙 −1
= 0.65 mol
For 0.65mol, enthalpy change
= 19.5 kJ
For 1mol, enthalpy change will be
1 𝑚𝑜𝑙 𝐶𝐶𝑙 4
0.65 𝑚𝑜𝑙
× 19.5 kJ
= 30.0 kJ
∴ ∆𝐻vap = 30.0K
∆𝐻 = ∆𝐸 + ∆ 𝑃𝑉
∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)
= ∆𝐻vap − ∆𝑛𝑅𝑇
∆n = 1 because CCl4 (l) → CCl4 (g)
Volume of CCl4 (l) is negligible compared to CCl4 (g)
For ∆n = 1 at 349.0K,
∆E = 30.0KJ – 1mol × 8.314JK-1mol-1) (349K)
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= 30.0 kJ – 2.90 kJ
= 27.1 kJ
Thus of the 30.0KJ of energy transferred from the surroundings in the
form of heat, 27.1KJ is used to increase the internal energy of the
molecules ∆E and 2.9KJ is used to expand the resulting vapour, ∆(𝑃𝑉).
7.3. Heat Capacities of Gases
The amount of heat expressed in joules, necessary to produce a
standard change of 1oC in one gram of material is called specific heat.
The product of the specific heat and the molar mass of a substance is
the heat required to raise the temperature of one mole of that substance
by 1oC. This is called the molar heat capacity and is a positive number
which has the unit of joules per mole – degree (J.mol-1 K-1).
Since the heat capacity C is the amount of heat needed to produce a
temperature change of 1oC, it would appear that the quantity of heat
required to produce a total temperature change, ∆T, is
q = C∆T = C(T2 – T1).
Heat is not a state function therefore it does not depend on the
initial and final states only but on how the process is carried out. The
equation above says nothing about the process of producing the
temperature change. We can remove this vagueness by defining two
molar heat capacities for gases, one Cp for processes at constant
pressure the other Cv for processes at constant volume.
Thus Cp =
𝑞𝑝
∆𝑇 and Cv =
𝑞𝑣
∆𝑇
The equivalent definitions in terms of infinitesimal changes are:
Cp =
𝑑𝑞 𝑝
𝑑𝑇
=
𝑑𝐻
𝑑𝑇
and
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The behaviour of gases
Cv =
𝑑𝑞 𝑣
𝑑𝑇
2016
𝑑𝐸
= 𝑑𝑇
From the above, we can find that the heat required to change the
temperature of n moles of material from T1 to T2 is:
qp =
qv =
𝑇2
𝑛𝐶𝑝 𝑑𝑇
𝑇1
𝑇2
𝑛𝐶𝑣 𝑑𝑇
𝑇1
for process at constant pressure and
for process at constant volume
Near room temperature, 𝐶𝑝 and 𝐶𝑣 are constants independent of
temperature therefore:
qp = nCp (T2 − T1)
qv = nCv (T2 − T1)
For an ideal gas, there is a relationship between Cp and Cv. This
relationship is shown easily by combing the definition of enthalpy with
definitions of Cp and Cv. Thus for one mole of an ideal gas,
𝐻 = 𝐸 + 𝑃𝑉
Differentiating the above equation gives
𝑑𝐻 = 𝑑𝐸 + 𝑑(𝑃𝑉)
Dividing through by 𝑑𝑇
𝑑𝐻
𝑑𝑇
= 𝑑𝑇 +
𝑑𝐸
𝑑(𝑃𝑉)
𝑑𝑇
Cp = Cv +
𝑑(𝑃𝑉)
𝑑𝑇
Cp = Cv +
𝑑(𝑅𝑇)
𝑑𝑇
(PV = RT)
Cp = Cv +
𝑅𝑑𝑇
𝑑𝑇
(R is gas constant)
𝑑𝐻
𝑑𝐸
(Cp = 𝑑𝑇 , Cv = 𝑑𝑇 )
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Cp = Cv + R
The heat capacity at constant pressure Cp is always larger than the heat
capacity at constant volume Cv because PV – work is done when gas is
heated at constant pressure. The relationship is visualized quite easily
for an ideal gas. When one mole of ideal gas is heated at constant
pressure, the work done in pushing back the piston is P∆V = R∆T. For
a 1oC change in temperature, the amount of work done is equal to R,
and this is just the extra energy required to heat a mole of ideal gas at
constant pressure over that required to heat it through 1oC at constant
volume.
Example 6.13. How much heat is required to raise the temperature of
10g of argon through 10OC at (a) constant volume (b) constant
pressure?
[ Cv = 12.468J, Cp = 20.794J, Ar = 40]
Solution:
Atomic mass of Ar, a mono atomic gas = 40gmol-1
10𝑔
No. of moles in 10g = 40𝑔𝑚𝑜𝑙 −1 = 0.25mol
(a).
qv = nCv∆T
= 0.25 mol× 12.468𝐽𝐾 −1 mol−1 × 10𝐾
= 31.17J = 31J
(b).
qp = nCp∆T
= 0.25mol× 20.794JK-1 mol-1× 10𝐾
= 51.985J
= 51.99J = 52J
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Example 6.14. Suppose that 1.00kJ of heat is transferred to 2.00mol of
argon at 298.0K, 1atm. What will the final temperature Tf be if the heat
is transferred at (a) Constant volume (b) Constant pressure? Calculate
the energy change ∆E in each case. [ Cv = 12.468 JK-1 mol-1, Cp = 20.794
JK-1 mol-1]
Solution:
(a). At constant volume
First calculate the rise in temperature
qv = nCv∆T
𝑞
∆T = 𝑛𝐶𝑣
𝑣
=
1000 𝐽
2𝑚𝑜𝑙 (12.468𝐽 𝐾 −1 𝑚𝑜 𝑙 −1)
= 40.10K
∴ 𝑇𝑓 = 298K + 40.10K = 338.1K
Use the rise in temperature to calculate ∆E
∆E = nCv∆T
= (2 mol)( 12.468 JK-1 mol-1)(40.10K)
=1000J
(b). At constant pressure
First calculate the rise in temperature
qp = nCp∆T
𝑞
∆T = 𝑛𝐶𝑝
𝑝
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The behaviour of gases
=
2016
1000 𝐽
2𝑚𝑜𝑙 (20.794𝐽𝐾 −1 𝑚𝑜 𝑙 −1)
= 24.05K
∴ 𝑇𝑓 = 298K + 24.05K = 322.05K
Use this rise in temperature to calculate ∆E
∆E = nCp∆T
= (2 mol)( 12.468 JK-1 mol-1)(24.05K)
=600J
Note: the expression for ∆E involves Cv even though the process is
carried out at constant pressure. The difference of 400J between the
input qp and ∆E is the work done by the gas as it expands.
7.4. Reversible Isothermal and Adiabatic Processes
Before discussing adiabatic processes, let us look at reversible
isothermal expansion of gases. Isothermal expansion of gas is the
expansion carried out at constant temperature. The maximum work
that is obtainable from the isothermal expansion of an ideal gas is
easily calculated using the ideal equation of state,
𝑃𝑉 = 𝑛𝑅𝑇
P = 𝑛𝑅𝑇 𝑉
In reversible isothermal expansion,
W=−
𝑉2
𝑃𝑑𝑉
𝑉1
Substituting for p,
Wrev = −
𝑉2 𝑛𝑅𝑇
𝑉1 𝑉
𝑑𝑉
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The behaviour of gases
= − nRT
2016
𝑣2 𝑑𝑉
𝑣1 𝑉
𝑉
= − 𝑛𝑅𝑇𝐼𝑛 𝑉2
1
𝑉
= − 2.303𝑛𝑅𝑇 𝑙𝑜𝑔 𝑉2
1
Since V2 > V1, and the logarithm is positive then Wrev < 0. A negative
W indicates that work is done by the system on the surroundings.
In compression, the final volume V2 is less than V1 so
Wrev
positive. The positive value means that work is done on the gas.
is
Example 7.15. A mole of CH4 expands reversibly and isothermally
from 1dm3 to 50 dm3 at 250C. Calculate the work by the gas in joules
assuming the gas is ideal.
Solution:
𝑊𝑟𝑒𝑣 = − 𝑛𝑅𝑇𝑙𝑛
𝑉2
𝑉1
𝑉
= −2.303nRTlog 𝑉2
1
= − 2.303× 1𝑚𝑜𝑙 8.314𝐽𝐾 −1 𝑚𝑜𝑙 −1 (298.1𝐾 log
50
)
1
= − 2.303× 1 × 8.314𝐽 × 298.15 log 50
= −2.303× 1 × 8.314 × 298.15 × 1.6990
= − 9698.9J
= − 9.70KJ
7.5. Expression of Wrev in terms of Pressure
Let us consider a situation in which a change in pressure causes an
infinitesimal change in volume of the gas. This means there are
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changes in both pressure and volume, so work done by the system will
be W = ‒ 𝑑(𝑃𝑉). Since it is assumed that the change in pressure is
more pronounced, above equation can be expressed as W = ‒ 𝑉𝑑𝑃.
But V is not constant, rather a function of pressure, so using general
gas equation of state
PV = nRT
V=
𝑛𝑅𝑇
𝑃
Substituting for V in the above equation W =‒
𝑛𝑅𝑇
𝑝
𝑑𝑝
For isothermal process, T is constant
∴ 𝑤 = 𝑛𝑅𝑇
𝑃2 𝑑𝑃
𝑃1 𝑃
𝑃
= ‒ 𝑛𝑅𝑇𝑙𝑛 𝑃2
1
𝑃
∴ 𝑊𝑟𝑒𝑣 =‒ 𝑛𝑅𝑇𝑙𝑛 𝑃2
1
Assuming Boyles‘ law P1V1 = P2V2
𝑃2
𝑃1
=
𝑉1
𝑉2
𝑜𝑟
𝑉2
𝑉1
=
𝑃1
𝑉2
𝑉
𝑃
From the equation Wrev = ‒nRTln𝑉2 = nRTln𝑃1
1
𝑉
2
𝑃
‒ nRTln𝑉2 = 𝑛𝑅𝑇𝑙𝑛 𝑃1
1
𝑊𝑟𝑒𝑣 = +𝑛𝑅𝑇𝑙𝑛
2
𝑃2
𝑃1
When P2 is greater than P1, work is done ON the system by the
surroundings.
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The behaviour of gases
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Example 7.16. One mole of nitrogen at 250C and 1.01325× 105 Nm-2 is
expanded reversibly and isothermally to a pressure of 1.32× 104 Nm-2
(a) what is the value of Wrev? (b) What is the value of Wrev if the gas
is expanded against a constant pressure of 1.32× 104 Nm-2 ?
Solution:
(a). For one mole of gas
𝑃
Wrev = RTln𝑃2
1
= (8.314JK-1mol-1)(298.15K)ln
1.32×10 4
1.01325 ×10 5
= − 5.05kJmol-1
(b). P1V1 = RT,
V1 =
=
𝑅𝑇
𝑃1
8.314𝐽𝐾 −1 𝑚𝑜 𝑙 −1 (298.15𝐾)
1.01325 × 10 5 𝑁𝑚 −2
= 2.446× 10−2 𝑚3 𝑚𝑜𝑙 −1
V2 =
(8.314𝑁𝑚 𝐾 −1 𝑚𝑜 𝑙 −1 )(298.15𝐾)
1.32×104 𝑁𝑚 −2
= 0.1878m3 mol-1
𝑤 = 𝑃∆𝑉
= 1.32× 104 (0.1878 − 0.02446)
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The behaviour of gases
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= 1.32× 104 × 0.1633𝐽
= 2.156KJ
Example 7.17. An ideal gas expanded reversibly and isothermally from
10 bar to 1 bar at 250C. What are the values of
(a) W per mole?
(b) q per mole
(c) ∆E and
(d) ∆H
Solution:
(a). For one mole,
𝑃
W = RTln𝑃2
1
1
= (8.314JK-1mol-1) (298.15K)ln10
= - 5707.67J
= - 5.71kJ
(b). For isothermal expansion,
q=−w
= − (− 5.71kJ)
= 5.71kJ
(c). ∆E = 0, since this is an ideal gas
(d) 𝑑𝐻 = 𝑑𝐸 + 𝑑(𝑃𝑉) = 0 (Expansion is not at constant pressure).
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7.6. Adiabatic Expansion
An adiabatic process is one in which there is neither gain nor loss of
heat, that is one in which the system under investigation is thermally
isolated from its environment such that q = 0. In an adiabatic
expansion, work is done at the expense of the internal energy of the
gas, resulting in drop of temperature. When a gas expands
adiabatically to a larger volume and a lower pressure, the volume is
usually smaller than it would be for an isothermal expansion to the
same pressure.
The work done by isothermal reversible expansion of a gas is always
larger than the work done by the adiabatic expansion. The energy for
doing the additional work in isothermal expansion is provided by heat
absorbed from constant temperature reservoir. The energy for doing
work in the adiabatic expansion comes only from the cooling of the gas
itself.
Now let us consider the reversible adiabatic expansion of one mole
of ideal gas. Since for an adiabatic process, dq = 0, then by the first law
of thermodynamics,
𝑑𝐸 = − 𝑑𝑤 (𝑑𝐸 = 𝑑𝑞 – 𝑑𝑤)
dw = − dE
From the equation dE = CvdT
dw = − CvdT
∴ 𝑤𝑟𝑒𝑣 = −
𝑇2
𝐶𝑣𝑑𝑇
𝑇1
Since work, PV, is done at the expense of internal energy
dE = − PdV
CvdT = − PdV
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The behaviour of gases
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For one mole of gas,
PV = RT
P=
𝑅𝑇
𝑉
CvdT = −
𝐶𝑣 𝑑𝑇
𝑇
𝑅𝑇
𝑉
=−R.
dV
𝑑𝑉
𝑉
If V1 is the volume of the gas at T1 and V2 at T2 and Cv is independent
of temperature,
𝑇2
𝑑𝑇
𝐶
𝑇1 𝑣 𝑇
=−
𝑉2 𝑑𝑉
𝑅 𝑉
𝑉1
Within these limits, on integration,
𝑇
𝑇1
Cvln 2 = − Rln
𝑉2
𝑉1
For ideal gas during expansion V2>V1 so T2<T1, there is temperature
drop in adiabatic expansion. In other words the gas cools. Conversely,
adiabatic compression of gas produces an increase in temperature.
Example 7.18. Calculate the temperature increase of helium if a mole of
it is compressed adiabatically and reversibly from 44.8dm3 at 10C to
22.4dm3
[Cv for He = 12.55JK-1 mol-1, R = 8.314JK-1 mol-1]
Solution:
V1 = 44.8dm3, V2 = 22.4dm3
T1 = 273.15K , T2 = ?
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The behaviour of gases
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Cv = 12.55K-1 mol-1 R = 8.314JK-1 mol-1
Use equation
𝑇
𝑇1
Cvln 2 = − Rln
𝑉2
𝑉1
𝑇
22.4
2
=12.55JK-1 mol-1 ln(273.15
) = − 8.314JK-1mol-1ln 44.8
= ln
𝑇2
273.15𝐾
=−
8.314
12.55
ln
22.4
44.8
1
2
lnT2 − ln273.15 = − 0.6625ln( )
lnT2 = − 0.6625 × − 0.6931 + 𝑙𝑛273.15
lnT2 = 0.4592 + 5.6100
= 6.0692
T2 = 432.34K
Temperature increase = 432.34 – 273.15
= 159.19oC
≃ 159oC
Above calculation shows that the compression was carried out so
rapidly that there was no heat transfer to the container but sufficiently
slow to make it reversible.
To find the change in pressure, we use the equation,
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The behaviour of gases
𝑇
𝑇1
Cv ln 2 = − Rln
𝑇
𝑅
2016
𝑉2
𝑉1
𝑉
ln𝑇2 = − 𝐶 In 𝑉2
1
𝑣
𝑅
1
𝑉
= 𝐶 In 𝑉1
𝑣
2
Taking the antilog of both sides reduces the equation to
𝑇2
𝑇1
𝑉
= (𝑉1 ) R/Cv
2
From Cp = Cv + R,
R = Cp − Cv
∴
𝑇2
𝑇1
𝐶𝑝 − 𝐶𝑉
=
𝑉1
𝑉2
=
𝑉1 γ −1
𝑉2
𝐶𝑉
𝐶
Where γ = 𝐶𝑝
𝑣
Example 7.19. A mole of argon is allowed to expand adiabatically and
reversibly from a pressure of 10 bar (106 Pa) and 298.15K to 1 bar (105
Pa). What is the final temperature and how much work is done by the
5
2
3
2
gas? [Cp = R, Cv = R ]
Solution:
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The behaviour of gases
From Cp ln
𝑇2
𝑇1
= 𝑅 𝑙𝑛
𝑃2
𝑃1
5
2
2.5 × log
log
𝑇
1
𝑇2
𝑃2
= log
𝑇1
𝑃1
𝑇2
1
=
𝑙𝑜𝑔 (0.1)
𝑇1
2.5
log 𝑇2 − log 𝑇1 =
1
log 0.1
2.5
log 𝑇2 = log(298.15 ) − 0.4
= 2.4744 – 0.4
= 2.0744
T2 = 102.0744
∴ 𝑇2 = 118.70𝐾
Alternatively,
5 𝑇2
𝑃2
𝐼𝑛
= 𝐼𝑛
2 𝑇1
𝑃1
207
𝑃
R ln𝑇2 = 𝑅 𝑙𝑛 𝑃2
1
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The behaviour of gases
2. 5 𝐼𝑛
= 𝐼𝑛
2016
𝑇2
105 𝑝𝑎
= 𝐼𝑛 ( 6 )
298.15
10 𝑝𝑎
𝑇2
1
1
=
𝐼𝑛
298.15
2.5
10
ln T2 – In 298.15 = –0.9210
𝑙𝑛 T2 = 𝐼𝑛 298.15 – 0.9210
= 5.6979 – 0.9210
= 4.7766
= 118.70K
(b). W =
𝑇2
𝐶 𝑑𝑇
𝑇1 𝑉
= 𝐶𝑉 𝑇2 − 𝑇1
3
2
= R 𝑇2− 𝑇1
3
= 2 ( 8.31455JK-1mol-1)(118.7−298.15𝐾)
=−2.238kJmol-1
The maximum work that can be done by the gas is 2.24kJmol-1
Example 7.20. An ideal monatomic gas at 298.15K and 105Nm-2 is
expanded in reversible adiabatic process to a final pressure of 5×
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104 𝑁𝑚−2 . Calculate: (a) final temperature (b) q per mole (c) W per
5
3
mole, and (d) ∆E. [Cp = 2 𝑅, 𝑅 = 8.3145𝐽𝐾 −1 𝑚𝑜𝑙 −1 , 𝐶𝑉 = 2R]
Solution:
From Cp ln
5
2
𝑇2
𝑇1
= 𝑅 𝑙𝑛
𝑇
𝑇1
R ln 2 = 𝑅 𝑙𝑛
𝑇
𝑃2
𝑃1
𝑃2
𝑃1
𝑃
2.5 ln𝑇2 = 𝑙𝑛 𝑃2
1
𝑇
1
2
ln 298.15𝐾
= 2.5 In
1
5×10 4 𝑁𝑚 −2
10 5 𝑁𝑚 −2
ln 𝑇2 − 𝑙𝑛 298.15 =
1
𝑙𝑛
2.5
ln 𝑇2 = 𝑙𝑛 298.15 +
1
𝑙𝑛
2.5
0.5
0.5
= 5.6976 – 0.2773
= 5.4203
T2 = e5.4203
= 225.95K
= 226.0K
(b). For adiabatic process, no heat gain or less
∴𝑞=0
(C). W = − Cv (𝑇2 − 𝑇1 )
3
= − 2 𝑅 226.0 − 298.15 𝐽𝑚𝑜𝑙 −1
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= − 1.5(8.3145JK-1)(-72.15K)
= + 899.84Jmol-1
PdV = W = −∆E
∆E = q – W
=0–W
= −W
= − 899.84Jmol-1
Note: for infinitesimal increase in volume dV at the pressure P, work
done by the gas is PdV. Since this work is accomplished at the expense
of the internal energy of the gas, the internal energy must decrease by
an amount dE.
Example 7.21. A tank contains 20dm3 of compressed N2 at 10 bar (106
Pa) and 250C. Calculate w when the gas is allowed to expand reversibly
to 1 bar (105Pa) pressure (a) isothermally (b) adiabatically.
Solution:
For the isothermal expansion of ideal gas,
𝑃
𝑊 rev = RT ln 𝑃2
1
10 5 𝑝𝑎
= (8.3145JK-1mol-1)(298.15K)(ln 10 6 𝑝𝑎 )
= (2478.9582) (−2.3026)Jmol-1
= − 5708.0122 = −5.71kJmol-1
However there are n moles present
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From 𝑃𝑉 = 𝑛𝑅𝑇
n=
=
𝑃𝑉
𝑅𝑇
10 6 𝑁𝑚 −2 20×10 −3 𝑚 3
8.3145 𝑁𝑚𝐾 −1 𝑚𝑜 𝑙 −1 298.15𝐾
= 8.068
= 8.07moles
𝑊 = 𝑊 rev per mole × 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
= − 5.71× 8.07𝑘𝐽
= − 46.08 𝑘𝐽
= − 46.1 𝑘𝐽
(b). For adiabatic expansion, Cp = 29.125JKmol-1, Cv = 20.811JK-1mol-1, R
= 8.314JK-1mol-1.
𝑇
𝑃
Cp ln 𝑇2 = 𝑅 𝑙𝑛 𝑃2
1
29.125 ln
1
T2
298.15K
= 8.314ln
8.314
29.125
8.314
10 5 Nm −2
)
10 6 Nm −2
ln (
1
ln T2 − ln 298.15 = 29.125 ln (10 )
= − 0.6573
ln T2 = ln 298.15 – 0.6573
= 5.6976 – 0.6573
= 5.0403
∴ 𝑇2 = 154.5𝐾
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The behaviour of gases
W=
𝑇2
𝐶 𝑑𝑇
𝑇1 𝑉
2016
= Cv (T2-T1)
= (20.811JK-1mol-1)(154.5 – 298.15K)
= (20.811)(−143.65)Jmol-1
= −2.99kJmol-1
For 8.07moles
W = −2.99kJmol-1× 8.07𝑚𝑜𝑙𝑒𝑠
= − 24.12kJ
7.7. Degrees of freedom and equipartition of energy
For each atom in a solid or gas phase, three coordinates have to
be specified to describe the atom‘s position – a single atom has 3
degrees of freedom for its motion. A solid or a molecule composed of
N atoms has 3N degrees of freedom. We can also think about the
number of degrees of freedom as the number of ways to absorb energy.
The theorem of equipartition of energy (classical mechanics) states
that in thermal equilibrium the same average energy is associated with
each independent degree of freedom and that the energy is ½kBT. For
the interacting atoms, e.g. liquid or solid, for each atom we have
½ kBT for kinetic energy and ½ kBT for potential energy equality of kinetic and potential energy in harmonic approximation is
addressed by the virial theorem of classical mechanics.
Based on equipartition principle, we can calculate heat capacity of the
ideal gas of atoms - each atom has 3 degrees of freedom and internal
energy of 3/2kBT. The molar internal energy U=3/2NAkBT=3/2RT
and the molar heat capacity under conditions of constant volume is Cv
=[dU/dT]V=3/2R
In an ideal gas of molecules only internal vibrational degrees of
freedom have potential energy associated with them. For example, a
diatomic molecule has 3 translational + 2 rotational + 1 vibrational = 6
total degrees of freedom. Potential energy contributes ½ kBT only to
the energy of the vibrational degree of freedom, and Umolecule =
7/2kBT if all degrees of freedom are “fully” excited.
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CHAPTER EIGHT
CHEMICAL KINETICS
We have learnt on several occasions that a balanced chemical equation
is a chemical statement that gives the mole ratios of reactants and
products as well as the ratios of formula units. A balanced chemical
equation as ordinarily written provides valuable chemical information
as to the masses, or volumes (if gases are involved) and is therefore an
essential quantitative tool for calculating product yields from amounts
of reacting substances. However, a balanced chemical equation tells us
nothing about how fast or quickly chemical changes occur, or what
energy changes are associated with the molecular interaction in a given
chemical reaction. Knowing how quickly a chemical reaction occurs is
a crucial factor in how the reaction affects its surroundings. Therefore,
knowing the rate of a chemical reaction and the energy changes
associated with the molecular interaction during the reaction are
integral to understanding the reaction.
The questions of ―how fast does the reaction go‘? and ‗what
conditions or factors bring about variations in speed‘ in a given
chemical reaction are the subject of this chapter.
The concept of rate applies to a number of phenomenon in our
daily life. For example the change in distance by an athlete over time is
the running rate of the athlete. The number of soap bars that are
produced in a given time is the rate of production of soap etc. We
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apply the same principle in chemical reaction. This time as products
are formed reactants are used up and rate (speed) of a chemical
reaction can be expressed as the ratio of the change in the
concentration of a reactant (or product) to a change in time. The study
that deals with the movement/motion-the speeds, or rates of chemical
reactions is known as chemical kinetics.
We know that any reaction can be represented by the general equation
Reactants → products
This equation tells us that, during the course of a reaction, reactant
molecules are consumed while product molecules are formed. As a
result, we can follow the progress
of a reaction by monitoring either the decrease in concentration of the
reactants or the increase in concentration of the products.
Let us consider a simple reaction in which A molecules are converted
to B molecules (for example, the conversion of cis-1,2-dichloroethylene
to trans-1,2-dichloroethylene):
A → B
The decrease in the number of A molecules leads to increase in the
number of B molecules with time. In general, it is more convenient to
express the rate in terms of change in concentration with time. Thus,
for the preceding reaction we can express the rate as
𝑅𝑎𝑡𝑒 = −
∆[𝐴]
∆𝑡
𝑜𝑟 𝑅𝑎𝑡𝑒 =
∆[𝐵]
∆𝑡
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in which Δ[A] and Δ[B] are the changes in concentration (in molarity)
over a period Δt. Because the concentration of A decreases during the
time interval, Δ[A] is a negative quantity. The rate of a reaction is a
positive quantity, so a minus sign is needed in the rate expression to
make the rate positive. On the other hand, the rate of product
formation does not require a minus sign because Δ[B] is a positive
quantity (the concentration of B increases with time).
For more complex reactions, we must be careful in writing the rate
expression.
Consider, for example, the reaction
2A → B
Two moles of A disappear for each mole of B that forms—that is, the
rate at which B forms is one half the rate at which A disappears. We
write the rate as either
𝑟𝑎𝑡𝑒 = −
1 ∆[𝐴]
2 ∆𝑡
𝑜𝑟 𝑟𝑎𝑡𝑒 =
∆[𝐵]
∆𝑡
Consider the following hypothetical reaction between reactants
A and B to form
products C and D
aA + bB → cC + dD
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The behaviour of gases
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( a, b, c and d are the stoichiometric coefficients of A, B, C, and D
respectively). The rate of this reaction is the speed at which A or B is
consumed or, alternatively, the speed at which C or D is formed.
Mathematically this is given by
𝑟𝑎𝑡𝑒 =
−1 ∆[𝐴]
𝑎 ∆𝑡
=
−1 ∆[𝐵]
𝑏 ∆𝑡
=
1 ∆[𝐶]
𝑐 ∆𝑡
=
1 ∆[𝐷]
𝑑 ∆𝑡
Figure 8.1: The rate of reaction A →B, represented as the decrease of A
molecules with time and as the increase of B molecules with time.
Example 8.1
Write the rate expressions for the following reactions in terms of the
disappearance of the reactants and the appearance of the products:
(a). I-(aq) + OCl-(aq) → Cl-(aq) + OI- (aq)
(b). 3O2 (g) → 2O3 (g)
(c). 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O (g)
Solution
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(a). Because each of the stoichiometric coefficients equals 1
𝑟𝑎𝑡𝑒 = −
∆[𝐼 − ]
∆𝑡
∆[𝑂𝐶𝐼 − ]
∆𝑡
=
=
∆[𝐶𝑙 − ]
∆𝑡
=
∆[𝑂𝐼 − ]
∆𝑡
(b). Here the coefficients are 3 and 2, so
𝑟𝑎𝑡𝑒 = −
1 ∆[𝑂2 ]
3
∆𝑡
=
1 ∆[𝑂3 ]
2
∆𝑡
(c) In this reaction
𝑟𝑎𝑡𝑒 = −
1 ∆[𝑁𝐻3 ]
4
∆𝑡
1 ∆[𝐶𝑂2 ]
∆𝑡
= −5
=
1 ∆[𝑁𝑂]
4 ∆𝑡
=
1 ∆[𝐻2 𝑂]
6
∆𝑡
Practice Exercise
Write the rate expression for the following reaction:
Example 8.2
Consider the reaction
4NO2 (g) + O2 (g) → 2N2O5 (g)
Suppose that, at a particular moment during the reaction, molecular
oxygen is reacting at the rate of 0.037 M/s. (a) At what rate is N2O5
being formed? (b) At what rate is NO2 reacting?
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Strategy
To calculate the rate of formation of N2O5 and disappearance of NO2,
we need to express the rate of the reaction in terms of the
stoichiometric coefficients as in Example 8.1:
𝑟𝑎𝑡𝑒 = −
1 ∆[𝑁𝑂2 ]
4
∆𝑡
= −
∆[𝑂2 ]
∆𝑡
=
1 ∆[𝑁2 𝑂5 ]
2
∆𝑡
We are given
∆[𝑂2 ]
∆𝑡
= 0.037 M/s
Where the minus sign shows that the concentration of O2 is decreasing
with time.
Solution
(a) From the preceding rate expression, we have
−
∆[𝑂2 ]
∆𝑡
=
1 ∆[𝑁2 𝑂5 ]
2
∆𝑡
Therefore,
∆[𝑁2 𝑂5 ]
∆𝑡
= −2(− 0.037 𝑀𝑠 −1 ) = 0.074 M/s
(b) Here we have
=−
1 ∆[𝑁𝑂2 ]
4
∆𝑡
= −
∆[𝑂2 ]
∆𝑡
So
∆[𝑁𝑂2 ]
∆𝑡
= 4 − 0.037 𝑀𝑠 −1 = −0.15𝑀/𝑠
Practice Exercise
Consider the reaction
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Suppose that, at a particular moment during the reaction, molecular
hydrogen is being
formed at the rate of 0.078 M/s. (a) At what rate is P4 being formed? (b)
At what rate is
PH3 reacting?
Example 8.3
Hydrogen gas is used for fuel aboard the Space Shuttle and may be
used in
Earth-bound engines of the future:
a) Express the rate of this reaction in terms of changes in [H 2], [O2], and
[H2O] with time.
b) When [O2] is decreasing at 0.23 mol/L.s, at what rate is [H2O]
increasing?
Solution
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Example 8.4
Consider the reaction of nitrogen dioxide with fluorine to give
nitrylfluoride, NO2F.
How is the rate of formation of NO2F related to the rate of reaction of
fluorine?
Problem Strategy
We need to express the rate of this reaction in terms of concentration
changes with time of the product, NO2F, and reactant, F2, and then
relate these two rates. The rate of disappearance of reactants is
expressed as a negative quantity of concentration change
per some time interval as we all know. The rate of formation of
products is expressed as a positive quantity of concentration change
per some time interval. In order to equate rate expressions, we need to
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divide each by the coefficient of the corresponding substance in the
chemical equation.
Solution
You write
Rate of formation of NO2F =
∆[𝑁𝑂2 𝐹]
∆𝑡
and
Rate of reaction of F2 = −
∆[𝐹2 ]
∆𝑡
You divide each rate by the corresponding coefficient (if applicable)
and then equate them:
1 ∆[𝑁𝑂2 𝐹]
2
∆𝑡
= −
∆[𝐹2 ]
∆𝑡
Example 8.5
Calculate the average rate of decomposition of N2O5, - ∆[N2O5]/∆t, by
the reaction
during the time interval from t = 600s to t =1200s (regard all time
figures as significant).
Use the following data:
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Problem Strategy
An average reaction rate is the change in concentration of a reactant or
product over a time interval; in this case it‘s the rate of decomposition
of the reactant N2O5, (- ∆[N2O5]/∆t). The ∆[ N2O5] in the equation is the
change in concentration of N2O5 (final value minus initial value). The
∆t is the time interval (final minus initial) over which the concentration
change occurred.
Solution
Average rate of decomposition of N2O5 = −
∆[𝑁2 𝑂5 ]
∆𝑡
=
Practice Exercise
For the reaction given in Example 8.3, how is the rate of formation of
NO2F related to the rate of reaction of NO2?
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Example 8.6
Would a wet piece of iron metal rust faster in air or pure oxygen?
Explain (Hint: consider the effect of concentration on reaction rate)
Solution
Iron exposed to moist air will react with oxygen to form iron oxide.
This oxidation process is called rusting.
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Oxygen is more concentrated in its pure form than mixed with other
gases in air. And we know that, as concentration increases the rate of a
reaction will also increase. Therefore, a wet piece of iron metal will rust
faster in pure oxygen than in air.
Factors Affecting Reaction Rates
At the beginning of this unit it was pointed out that various reactions
or changes that occur in nature or otherwise take place at a variety of
speeds depending on the conditions. Why do these times for different
changes turn out as they do? As the speed of an athlete depends on
several factors such as temperature surroundings, wind direction,
health condition etc. so is the speed of a chemical reaction affected by
several factors. In general, the rate at which a given chemical reaction
takes place depends upon a number of factors. The rates of chemical
reactions can be affected by:
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Reactant concentrations
Often the rate of reaction increases when the concentration of a
reactant is increased. For a chemical reaction to have noticeable rate,
there should be noticeable number of molecules with the energy equal
or greater than the activation energy. With an increase in concentration
the number of molecules with the minimum required energy for a
reaction to take place will increase and thereby the rate of the reaction
increases. Suppose that at any one time 1 in a million particles have
enough energy to equal or exceed the activation energy. If you had 100
million particles, 100 of them would react. If you had 200 million
particles in the same volume, 200 of them would now react. The rate of
reaction has doubled by doubling the concentration. A piece of steel
wool burns with some difficulty in air (20% O2) but bursts into a
dazzling white flame in pure oxygen. The rate of burning increases
with the concentration of O2. In some reactions, however, the rate is
unaffected by the concentration of a particular reactant, as long as it is
present at some concentration. Another example is the reaction of zinc
and hydrochloric acid in the lab, zinc granules react fairly slowly with
dilute hydrochloric acid, but much faster if the acid is concentrated.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
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Concentration of catalyst.
A catalyst is a substance that increases the rate of reaction without being
consumed in the overall reaction. Because the catalyst is not consumed by
the reaction, it does not appear in the balanced chemical equation
(although its presence may be indicated by writing its formula over the
arrow). A solution of pure hydrogen peroxide, H2O2, is stable, but
when hydrobromic acid, HBr(aq), is added, H2O2 decomposes rapidly
into H2O and O2.
Here HBr acts as a catalyst to speed decomposition.
Temperature at which the reaction occurs
Usually reactions speed up when the temperature increases. It takes
less time to boil an egg at sea level than on a mountaintop, where
water boils at a lower temperature. Reactions during cooking go faster
at higher temperature.
Surface area of a solid reactant
If a reaction involves a solid with a gas or liquid, the surface area of
the solid affects the reaction rate. Because the reaction occurs at the
surface of the solid, the rate increases with increasing surface area. A
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wood fire burns faster if the logs are chopped into smaller pieces.
Similarly, the surface area of a solid catalyst is important to the rate of
reaction. The greater the surface area per unit volume, the faster the
reaction.
Pressure
Increasing the pressure on a reaction involving reacting gases increases
the rate of reaction. Increasing the pressure of a gas is exactly the same
as increasing its concentration. If you have a given mass of gas, the
way you increase its pressure is to squeeze it into a smaller volume. If
you have the same mass in a smaller volume, then its concentration is
higher. Thus the effect is the same as the concentration effect. In the
manufacture of ammonia by the Haber Process, the rate of reaction
between the hydrogen and the nitrogen is increased by the use of very
high pressures.
N2(g) + 3H2(g) ↔ 2NH3(g) + heat (ΔH = -92kJmol-1)
In fact, the main reason for using high pressures is to improve the
percentage of ammonia in the equilibrium mixture, but there is a useful
effect on rate of reaction as well. Changing the pressure on a reaction
which involves only solids or liquids has no effect on the rate.
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Nature of reactants
Substances differ markedly in the rates at which they undergo
chemical change. For example, hydrogen and fluorine molecules react
explosively, even at room temperature, producing hydrogen fluoride
molecules:
H2 + F2 → 2HF (very fast at room temp.)
Under similar conditions, hydrogen and oxygen molecules react so
slowly that no chemical change is apparent:
2H2 + O2 → 2H2O (Very slow at room temp.)
The differences in reactivity between reactions may be attributed to the
different structures of the atoms and molecules of the reacting
materials (for instance whether the substances are in solution or in the
solid state-include) If a reaction involves two species of molecules with
atoms that are already joined by strong covalent bonds (for example
quartz (SiO2) and water (H2O)) collisions between these molecules at
ordinary temperatures may not provide enough energy to break these
bonds unlike the collisions which take place between molecules whose
atoms are joined by weak covalent bonds. Therefore, reactions between
molecules whose atoms are bound by weak covalent bonds take place
at a faster rate than reactions between molecules whose atoms are
bound by strong covalent bonds. For example, when methane gas is
mixed with chlorine gas and exposed to sunlight an explosive reaction
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takes place in which chlorinated methane products are produced along
with hydrogen chloride.
CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
Concentration dependence of Rate: Order of a reaction
We just discussed how the rate of a chemical reaction depends on
several factors such as temperature, catalyst, surface area of reactants,
presence or absence of a catalyst, nature of reactants and concentration.
We have seen qualitatively that the rates of most chemical reactions
increase when the concentrations of the reactants increase. In this
section, you will explore the quantitative relationships between the
rate of a reaction and the concentrations of the reactants.
Now consider the general reaction
aA + bB → products (occuring at a constant temperature)
where A and B represent the reactant formulae and, a and b represent
the stoichiometric coefficients. In this section, you will study reaction
rates that are not affected by the concentrations of the products.
Therefore, you do not need to use symbols for the products.
In general, the rate of a reaction is proportional to the
concentration of each reactant raised to some power, where the power
on a given reactant is called, the order of the reaction with respect to
that reactant. The overall order of a reaction is the sum of all the
exponents of the concentration terms in the rate equation.
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The rate for the above reaction can be given as
Rate α [A]m[B]n ………………………………….………………8.1
where m and n are the rate law exponents and indicate the order of the
reaction with respect to the corresponding reactants. The values of m
and n for a given reaction must be determined experimentally and do
not change with temperature.
This relationship given by equation 8.1 can be expressed in a general
equation given below, called the rate law equation.
Rate = k [A]m [B]n ………………………………….………….8.2
The rate law equation expresses the relationship between the
concentrations of the reactants and the rate of the reaction. The letter k
represents a proportionality constant called the rate constant and it
indicates how fast or slow a reaction is proceeding. A small rate
constant indicates a slow reaction and a large rate constant indicates a
fast reaction. The value of k for a given reaction is temperature
dependent and is constant under constant temperature and pressure
conditions.
The exponents m and n do not necessarily correspond to the
stoichiometric coefficients of their reactants. Usually the value of a rate
law exponent is 1 or 2. But, seldom values of 0, 3; and even fractions
can occur. If the exponent for a given reactant is 1, then the reaction is
said to be first order with respect to that reactant. Similarly, if the
exponent of a reactant is 2, the reaction is said to be second order in
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this reactant. For example, the rate law equation below represents a
reaction that is first order in A, second order in B, and third order (1 +
2) overall.
Rate = k[A]1[B]2
For example, the reaction between nitric oxide and ozone
NO(g) + O3(g) → NO2(g) + O2(g)
is first order in nitric oxide and first order in ozone. The rate law
equation for this reaction is:
Rate = k{NO]1[O3]1
The overall order of the reaction is 1 + 1 = 2.
Determining reaction Orders
The order of a reaction with respect to its reactants can be determined
by running a series of experiments each of which starts with a different
set of reactant concentrations and the initial rate is obtained. The
experiments are designed to change one reactant concentration while
keeping the other constant. This method of determining order of a
reaction is known as the initial rate method.
Example 8.7
For the reaction A + B → products, the following rate data were
obtained in three separate experiments:
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a) What is the order of the reaction with respect to A and B?
b) What is the rate constant, k?
Solution
a) The general rate expression for this reaction is Rate = k[A] x [B]y and
the values of x and y must be determined from the above data.
The data obtained shows that the rate doubles in experiment number 2
than it was in experiment number 1 when the concentration of B is
doubled keeping A constant. In such a condition, when the rate of a
reaction doubles by doubling the concentration of a given reactant,
order of the reaction with respect to that reactant is 1.
In experiment number 3 the rate is found not to change, than it was in
experiment number 2, when the concentration of A is doubled. This
indicates that the rate does not depend on the concentration of A. This
means, order of the reaction with respect to A is zero.
The experimentally determined rate equation will then be Rate =
k[A]o[B] = k[B] and the overall order = 1
The order with respect to each reactant can also be determined by
calculation as follows:
Rate1 = 2 x 10-5 M min-1 = k[2]x [1]y
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Rate2 = 4 x 10-5 M min-1 = k[2]x [2]y
Taking the ratio of the two rates we have
The order of the reaction with respect to the reactant A can be
calculated by taking the ratio of the rate expressions of experiment
number 2 and 3.
Thus, the experimentally determined rate equation is given by
Rate = k[A]x[B]y = K[A]o[B] = k[B]
b) Once we get the order of the reaction the rate constant can be
calculated by taking the data obtained in any of the three experiments.
Consider experiment number 1
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Example 8.8
For the reaction X + 2Y + 2Z → products, the following rate data were
obtained:
a) What is the order of each reactant in the system?
b) What is the value of the rate constant?
c) What is the rate of disappearance of X in experiment 4?
Solution
a) In the first experiment the rate of the reaction was determined to be
1.0 x 10-6 M/min when the concentrations of all reactants were kept at
0.1 M.
In the second experiment the rate was found to be tripled when the
concentrations of X and Y were tripled and that of Z was kept constant.
This indicates that the rate of the reaction depends on the
concentration on either X or Y or on both of them.
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In the third experiment, the rate remained the same as in experiment
number 1 when the concentration of X was quadrupled and those of Y
and Z were kept constant. This fact indicates that, the rate of the
reaction does not depend on the concentration of X.
Therefore, Rate α [X]0
Going back to experiment number 2, the rate was tripled when the
concentrations of X and Y were each tripled. But we have decided that
the rate does not depend on the concentration of X. Therefore, the rate
was tripled when the concentration of Y was tripled.
Thus, Rate α [Y]1
When we compare experiments 2 and 4, neglecting X; the
concentration of Y is kept constant while that of Z is tripled. As a result
the rate was found to increase by a factor of 9.
Thus, Rate α [Z]2 since 9 = 32
b) k can be evaluated from any one of the 4 data sets, once orders are
known; e.g., expt. 1.
Rate = k[X]0[Y]1[Z]2
c) When [Z] = 0.15, [Y] = 0.15, and [X] = 0.125, from balanced equation,
Rate = k[Y] [Z]2
= 10-3 M-2 / min (0.15) (0.15)2 = 3.4 x 10-6 M/min
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Relation between Reactant Concentrations and Time
Rate laws enable us to calculate the rate of a reaction from the rate
constant and reactant concentrations. They can also be converted into
equations that enable us to determine the concentrations of reactants at
any time during the course of a reaction. We will illustrate this
application by considering first one of the simplest kind of rate
laws—that applying to reactions that are first order overall.
First-Order Reactions
A first-order reaction is a reaction whose rate depends on the reactant
concentration raised to the first power. In a first-order reaction of the type
A→ product
the rate is
From the rate law, we also know that
Thus,
We can determine the units of the first-order rate constant k by
transposing:
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Because the unit for ∆[A] and [A] is M and that for ∆t is s, the unit for k
is
Derivation of first order reaction
The preceding first order rate in differential form becomes
Half-Life
The half-life of a reaction, t1/2, is the time required for the concentration of
a reactant to decrease to half of its initial concentration.
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By the definition of half-life,
The equation above tells us that the half-life of a first-order reaction is
independent of the initial concentration of the reactant. Measuring the
half-life of a reaction is one way to determine the rate constant of a
first-order reaction.
Example 8.9
The decomposition of ethane (C2H6) to methyl radicals is a first-order
reaction with a rate constant of 5.36 × 10-4 s-1 at 700°C:
Calculate the half-life of the reaction in minutes.
Strategy
To calculate the half-life of a first-order reaction, we use half life
equation above. A
conversion is needed to express the half-life in minutes.
Solution
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For a first-order reaction, we only need the rate constant to calculate
the half-life of the reaction.
Using
Second-Order Reactions
A second-order reaction is a reaction whose rate depends on the
concentration of one reactant raised to the second power or on the
concentrations of two different reactants, each raised to the first power. The
simpler type involves only one kind of reactant molecule:
A → product
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Another type of second-order reaction is
A + B → product
and the rate law is given by
rate = k[A][B]
The reaction is first order in A and first order in B, so it has an overall
reaction order of 2.
Using calculus, we can obtain the following expressions for ―A→
product‖ second-order reactions:
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The equation above is a result of
Zero-Order Reactions
First- and second-order reactions are the most common reaction types.
Reactions whose order is zero are rare. For a zero-order reaction
A → product
the rate law is given by
rate = k[A]0
=k
Thus, the rate of a zero-order reaction is a constant, independent of
reactant concentration.
Using calculus, we can show that
The Equation above has the form of a linear equation. A plot of [A] t
versus t gives a straight line with slope = -k and y intercept = [A]0. To
calculate the half-life of a zero-order reaction, we set [A]t = [A]0/2 in
the equation above equation and obtain
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Many of the known zero-order reactions take place on a metal surface.
An example is the decomposition of nitrous oxide (N2O) to nitrogen
and oxygen in the presence of platinum (Pt):
2N2O(g) → 2N2(g) + O2(g)
When all the binding sites on Pt are occupied, the rate becomes
constant regardless of the amount of N2O present in the gas phase.
Third-order and higher order reactions are quite complex; they are not
presented in this book. Table 8.1 summarizes the kinetics of zero-order,
first-order, and second order reactions.
Table 8.1: Summary of the Kinetics of zero-order, first order, and second-order
reactions
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Activation Energy and Temperature Dependence of Rate Constants
With very few exceptions, reaction rates increase with increasing
temperature. For example, much less time is required to hard-boil an
egg at 100°C (about 10 min) than at 80°C (about 30 min). Conversely,
an effective way to preserve foods is to store them at subzero
temperatures, thereby slowing the rate of bacterial decay. Figure 8.2
shows a typical example of the relationship between the rate constant
of a reaction and temperature. To explain this behavior, we must ask
how reactions get started in the first place.
Figure 8.2: Dependence of rate constant on temperature. The rate constants of
most reactions increase with increasing temperature.
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The Collision Theory of Chemical Kinetics
The kinetic molecular theory of gases states that gas molecules
frequently collide with one another. Therefore it seems logical to
assume—and it is generally true—that chemical reactions occur as a
result of collisions between reacting molecules.
In terms of the collision theory of chemical kinetics, then, we
expect the rate of a reaction to be directly proportional to the number
of molecular collisions per second, or to the frequency of molecular
collisions:
This simple relationship explains the dependence of reaction rate on
concentration.
Consider the reaction of A molecules with B molecules to form
some product. Suppose that each product molecule is formed by the
direct combination of an A molecule and a B molecule. If we doubled
the concentration of A, say, then the number of A-B collisions would
also double, because, in any given volume, there would be twice as
many A molecules that could collide with B molecules. Consequently,
the rate would increase by a factor of 2. Similarly, doubling the
concentration of B molecules would increase the rate twofold. Thus, we
can express the rate law as
Rate = k[A][B]
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The reaction is first order in both A and B and obeys second-order
kinetics.
The
collision
theory
is
intuitively
appealing,
but
the
relationship between rate and molecular collisions is more complicated
than you might expect. The implication of the collision theory is that a
reaction always occurs when an A and a B molecule collide. However,
not all collisions lead to reactions. Calculations based on the kinetic
molecular theory show that, at ordinary pressures (say, 1 atm) and
temperatures (say, 298 K), there are about 1 × 1027 binary collisions
(collisions between two molecules) in 1 mL of volume every second, in
the gas phase. Even more collisions per second occur in liquids. If
every binary collision led to a product, then most reactions would be
complete almost instantaneously. In practice, we find that the rates of
reactions differ greatly. This means that, in many cases, collisions alone
do not guarantee that a reaction will take place.
Any molecule in motion possesses kinetic energy; the faster it moves,
the greater the kinetic energy. When molecules collide, part of their
kinetic energy is converted to vibrational energy. If the initial kinetic
energies are large, then the colliding molecules
will vibrate so strongly as to break some of the chemical bonds. This
bond fracture is the first step toward product formation. If the initial
kinetic energies are small, the molecules will merely bounce off each
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other intact. Energetically speaking, there is some minimum collision
energy below which no reaction occurs.
We postulate that, to react, the colliding molecules must have a
total kinetic energy equal to or greater than the activation energy (Ea),
which is the minimum amount of energy required to initiate a chemical
reaction. Lacking this energy, the molecules remain intact, and no
change results from the collision. The species temporarily formed by the
reactant molecules as a result of the collision before they form the product is
called the activated complex (also called the transition state).
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Figure 8.3: Potential energy profiles for (a) exothermic and (b) endothermic
reactions. These plots show the change in potential energy as reactants A and
B are converted to products C and D. The transition state is a highly unstable
species with a high potential energy. The activation energy is defined for the
forward reaction in both (a) and (b). Note that the products C and D are more
stable than the reactants in (a) and less stable than those in (b).
Figure 8.3
shows two different potential energy profiles for the
reaction
A+B→C+ D
If the products are more stable than the reactants, then the reaction will
be accompanied
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The behaviour of gases
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by a release of heat; that is, the reaction is exothermic [Figure 8.3(a)].
On the other hand, if the products are less stable than the reactants,
then heat will be absorbed by the reacting mixture from the
surroundings and we have an endothermic reaction [Figure 8.3(b)]. In
both cases, we plot the potential energy of the reacting system versus
the progress of the reaction. Qualitatively, these plots show the
potential energy changes as reactants are converted to products.
We can think of activation energy as a barrier that prevents less
energetic molecules from reacting. Because the number of reactant
molecules in an ordinary reaction is very large, the speeds, and hence
also the kinetic energies of the molecules, vary greatly. Normally, only
a small fraction of the colliding molecules—the fastest-moving ones—
have enough kinetic energy to exceed the activation energy. These
molecules can therefore take part in the reaction. The increase in the
rate (or the rate constant) with temperature can now be explained: The
speeds of the molecules obey the Maxwell distributions shown in
earlier chapters. Compare the speed distributions at two different
temperatures. Because more high-energy molecules are present at the
higher temperature, the rate of product formation is also greater at the
higher temperature.
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The Arrhenius Equation
The dependence of the rate constant of a reaction on temperature can
be expressed by this equation, now known as the Arrhenius equation:
in which Ea is the activation energy of the reaction (in kilojoules per
mole), R is the gas constant (8.314 J/K . mol), T is the absolute
temperature, and e is the base of the natural logarithm scale. The
quantity A represents the collision frequency and is called the frequency
factor. It can be treated as a constant for a given reacting system over a
fairly wide temperature range. The equation above shows that the rate
constant is directly proportional to A and, therefore, to the collision
frequency. Further, because of the minus sign associated with the
exponent Ea/RT, the rate constant decreases with increasing activation
energy and increases with increasing temperature. This equation can
be expressed in a more useful form by taking the natural logarithm of
both sides:
Equation above can take the form of a linear equation:
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Thus, a plot of ln k versus 1/T gives a straight line whose slope m is
equal to –Ea/R and whose intercept b with the ordinate (the y-axis) is
ln A.
Example 8.10
The rate constants for the decomposition of acetaldehyde
were measured at five different temperatures. The data are shown in
the table. Plot ln k versus 1/T, and determine the activation energy (in
kJ/mol) for the reaction. This reaction has been experimentally shown
to be ―3/2‖ order in CH3CHO, so k has the units of 1/M1 /2 . s.
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Strategy
Consider the Arrhenius equation written as a linear equation
A plot of ln k versus 1/T (y versus x) will produce a straight line with a
slope equal to –Ea/R. Thus, the activation energy can be determined
from the slope of the plot.
Solution
First, we convert the data to the following table:
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Figure 8.4: Plot of ln k versus 1/T. The slope of the line is equal to –Ea/R
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A plot of these data yields the graph in Figure 8.4. The slope of the line
is calculated from two pairs of coordinates:
From the linear form of Equation
An equation relating the rate constants k1 and k2 at temperatures T1 and
T2 can be used to calculate the activation energy or to find the rate
constant at another temperature if the activation energy is known. To
derive such an equation we start with equation below:
Subtracting ln k2 from ln k1 gives
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The rate constant of a first-order reaction is 4.68 ×10-2 s-1 at 298 K. What
is the rate constant at 375 K if the activation energy for the reaction is
33.1 kJ/mol?
Strategy
A modified form of the Arrhenius equation relates two rate constants
at two different temperatures [see above]. Make sure the units of R and
Ea are consistent.
Solution
The data are
Substituting in Equation
gives
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We convert Ea to units of J/mol to match the units of R. Solving the
equation gives
Reaction Mechanisms
As we mentioned earlier, an overall balanced chemical equation does
not tell us much about how a reaction actually takes place. In many
cases, it merely represents the sum of several elementary steps, or
elementary reactions, a series of simple reactions that represent the progress of
the overall reaction at the molecular level. The term for the sequence of
elementary steps that leads to product formation is reaction mechanism.
The reaction mechanism is comparable to the route of travel followed
during a trip; the overall chemical equation specifies only the origin
and destination.
As an example of a reaction mechanism, let us consider the
reaction between nitric oxide and oxygen:
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2NO(g) + O2(g) → 2NO2(g)
We know that the products are not formed directly from the collision
of two NO molecules with an O2 molecule because N2O2 is detected
during the course of the reaction. Let us assume that the reaction
actually takes place via two elementary steps as follows:
In the first elementary step, two NO molecules collide to form a N2O2
molecule. This event is followed by the reaction between N2O2 and O2
to give two molecules of NO2. The net chemical equation, which
represents the overall change, is given by the sum of the elementary
steps:
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Species such as N2O2 are called intermediates because they appear in the
mechanism of the reaction (that is, the elementary steps) but not in the overall
balanced equation. Keep in mind that an intermediate is always formed
in an early elementary step and consumed in a later elementary step.
The molecularity of a reaction is the number of molecules reacting
in an elementary step. These molecules may be of the same or different
types. Each of the elementary steps just discussed is called a
bimolecular reaction, an elementary step that involves two molecules. An
example of a unimolecular reaction, an elementary step in which only one
reacting molecule participates, is the conversion of cyclopropane to
propene is an example. Very few termolecular reactions, reactions that
involve the participation of three molecules in one elementary step, are
known, because the simultaneous encounter of three molecules is a far
less likely event than a bimolecular collision.
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