Download Gases Properties of Gases Kinetic Molecular Theory of Gases

Gases
Chemistry 120
Properties of Gases
Kinetic Molecular Theory of Gases
Pressure
Boyle’s and Charles’ Law
The Ideal Gas Law
Gas reactions
Partial pressures
Gases
Chemistry 120
Properties of Gases
All elements will form a gas at some temperature
Most small molecular compounds and elements are
either gases or have a significant vapor pressure.
1 H
1
2 Li
3
3 Na
11
4 K
19
5 Rb
37
6 Cs
55
He
2
Room Temperature Gases
Be
4
Mg
12
Ca
20
Sr
38
Ba
56
Sc
21
Y
39
Lu
71
Ti
22
Zr
40
Hf
72
V
23
Nb
41
Ta
73
Cr
24
Mo
42
W
74
Mn
25
Tc
43
Re
75
Fe
26
Ru
44
Os
76
Co
27
Rh
45
Ir
77
Ni
28
Pd
46
Pt
78
Cu
29
Ag
47
Au
79
Gases
Zn
30
Cd
48
Hg
80
B
5
Al
13
Ga
31
In
49
Tl
81
C
6
Si
14
Ge
32
Sn
50
Pb
82
N
7
P
15
As
33
Sb
51
Bi
83
O
8
S
16
Se
34
Te
52
Po
84
F
9
Cl
17
Br
35
I
53
At
85
Ne
10
Ar
18
Kr
36
Xe
54
Rn
86
Chemistry 120
Properties of Gases
As the temperature rises, all elements form a gas at
some point.
In the following diagram,
Blue
represents
solids
Green
represents
liquids
Red
represents
gases
At O K, all elements are solids
At 6000 K, all are gases
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Gases
Chemistry 120
Gases
Chemistry 120
Properties of Gases
Gases have no shape and no volume.
They take the volume and shape of the container
Their densities are low – usually measured in gL-1
The atoms or molecules of the gas are far further
apart than in a solid or a liquid.
Gases
Chemistry 120
Gases as an ensemble of particles
The attractive forces between liquids and solids are
very strong
LiF:
M.p.: 848°C
Solid
Liquid
B. p.: 1676°C
Liquid
Gas
In a gas, the forces between particles are negligible
and as there are no attractive forces, a gas will
occupy the volume of the container.
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Gases
Gases as an ensemble of particles
Chemistry 120
The structures of liquids and
solids are well ordered on a
microscopic level
CaCl2
Ethanol, C2H5OH
Gases
Gases as an ensemble of particles
Chemistry 120
In a gas, there is no order and all the properties of
the gas are isotropic – all the properties of the gas
are the same in all directions.
Gas particles are distributed uniformly throughout
the container.
They can move throughout the container in straight
line trajectories.
Gases
Gases as an ensemble of particles
Chemistry 120
The directions of the motions of the gas particles are
random
and
The velocities form a distribution – there is a range
of possible velocities around an average value.
The trajectories of the gas particles are straight lines
and there are two possible fates for a gas
molecule......
3
Gases
Gases as an ensemble of particles
Chemistry 120
A gas particle can collide with
– the walls of the container
Or
– another gas molecule
When this happens, the gas particle changes
direction.
Gases
Gases as an ensemble of particles
Chemistry 120
Kinetic energy can be transferred between the two
colliding particles
– one can slow down and the other speed up –
but the net change in kinetic energy is zero.
These collisions are termed elastic, meaning that
there is no overall change in kinetic energy.
Gases
Chemistry 120
The average kinetic energy for a given gas is
determined by the temperature alone and the width
and peak maximum is also determined by the
temperature.
The Maxwell-Boltzmann distribution for He
4
Gases
Gases as an ensemble of particles
Chemistry 120
The force exerted by the gas particles on the walls
of the container gives rise to the pressure of the gas.
We define pressure as the force exerted per unit
area:
P = Force = F
Area
A
The unit of pressure is the Pascal (Pa)
1 Pa = 1 Nm-2
In practice, the Pascal is too small - kPa or GPa
Gases
Pressure measurement
Chemistry 120
Pressure is also measured in several other non – SI
units:
In industry:
Pounds per square in (p.s.i.)
In research:
Pascal, atmosphere, bar, Torr
Gases
Pressure conversion factors
Chemistry 120
Atmospheric pressure = 101,325 Pa
1 Atmosphere = 101,325 Pa = 1 bar
1 Atmosphere = 101,325 Pa
= 1 bar
= 760 Torr
= 760 mmHg
= 14.7 p.s.i.
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Gases
Pressure Measurement
Chemistry 120
Pressure is measured using a manometer or
barometer
– either one containing Hg or an electronic gauge
A mercury manometer is a U–tube connected to the
gas vessel, with the other end either evacuated or
open to the atmosphere.
The measurement of the height difference between
the mercury levels on both sides of the ‘U’ gives the
pressure........
Gases
Pressure Measurement
Chemistry 120
Let the height difference between the two Hg levels
be ∆h
Then the gas pressure is given by
Pgas = P0 + ∆h
As
P = Force = F = mg where g = 9.81 ms-2
Area
A A
Gases
Pressure Measurement
Chemistry 120
How is the height difference related to the pressure?
As density, ρ = m
V
Then
m=ρV
The volume of the column of mercury is
V = A.∆h
And so
m = ρ V = ρ A.∆h
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Gases
Pressure Measurement
Chemistry 120
The pressure above the baseline pressure P0 is
therefore
Pgas = mg =ρgA.∆h = ρg∆h
A
A
Gases
Gases as an ensemble of particles
Chemistry 120
Kinetic energy can be transferred between the two
colliding particles
– one can slow down and the other speed up –
but the net change in kinetic energy is zero.
These collisions are termed elastic, meaning that
there is no overall change in kinetic energy.
The Gas Laws
Chemistry 120
The factors that control the behavior of a gas are
•
The nature of the gas
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The Gas Laws
Chemistry 120
The factors that control the behavior of a gas are
•
The nature of the gas
•
The quantity of the gas - n
The Gas Laws
Chemistry 120
The factors that control the behavior of a gas are
•
The nature of the gas
•
The quantity of the gas - n
•
The pressure - P
The Gas Laws
Chemistry 120
The factors that control the behavior of a gas are
•
The nature of the gas
•
The quantity of the gas - n
•
The pressure - P
•
The temperature - T
8
The Gas Laws
Chemistry 120
The factors that control the behavior of a gas are
•
The nature of the gas
•
The quantity of the gas - n
•
The pressure - P
•
The temperature - T
•
The volume of a gas -V
The Gas Laws
Chemistry 120
These laws apply to a perfect gas or and ideal gas.
All gases behave as ideal gases at ordinary
temperatures and pressures.
The qualities of an ideal gas are:
•
Zero size to the gas particles
We assume that the volume of the container is
very much larger than the total volume of the gas
molecules
•
No attractive forces between atoms
The Gas Laws
Chemistry 120
These laws apply to a perfect gas or and ideal gas.
All gases behave as ideal gases at ordinary
temperatures and pressures.
At low temperatures and high pressures gases
deviate from ideality.
The ideal gas laws are based on three
interdependent laws – Boyle’s Law, Charles’ Law
and Avogadro’s Law.
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The Gas Laws
Chemistry 120
Boyle’s Law
Robert Boyle experimented with
gases in Oxford in 1660.
He discovered that the product of
the volume and the pressure of a
gas is a constant, so long as the
quantity of gas and the
temperature are constant.
The Gas Laws
Chemistry 120
Boyle’s Law
Mathematically,
PV = a constant
as long as n and T are constant
The Gas Laws
Chemistry 120
Boyle’s Law
Mathematically,
PV = a constant, k
or
P=k
V
as long as n and T are constant.
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The Gas Laws
Chemistry 120
Boyle’s Law
A graph of Boyle’s data
shows this relationship:
PV = k
The Gas Laws
Chemistry 120
Boyle’s Law
A graph of 1/P as the
abscissa and V as the
ordinate.
V=k
P
The graph shows a
straight line of slope k
The Gas Laws
Chemistry 120
Boyle’s Law
As the pressure rises, 1/P becomes smaller and the
graph passes through the origin.
This implies that at infinitely large pressure, the
volume of a gas is zero.
We know that molecules and
and atoms have a definite
volume, so Boyle’s law must
fail at very high pressures.
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The Gas Laws
Chemistry 120
Charles’ Law
Jacques Charles was a Feench
scientist and aeronaut who
discovered (1787) that all gases
expand by the same amount when
the temperature of the gas rises by
the same amount.
The Gas Laws
Chemistry 120
Charles’ Law
Mathematically, we express this as
V = k’T
And a graph of Charles’ Law
is a straight line:
The Gas Laws
Chemistry 120
Charles’ Law
As the temperature is lowered,
the volume decreases…….
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The Gas Laws
Chemistry 120
Charles’ Law
The graph intersect
the temperature axis
and the zero volume
occurs at –273.16oC,
the lowest possible
temperature.
Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
Combining Boyle’s Law, Charles’ Law and
Avogadro’s Law,
V=k
P
and
V = k’T
and V = k”n
we can say that
V ∝ nT
P
Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
V ∝ nT
P
Or
V =K nT
P
Rearranging we find
PV = a constant
nT
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Chemistry 120
The Gas Laws
The Combined Gas Law for a Perfect Gas
The constant is termed the Universal Gas Constant,
R, and takes the value
R = 8.314 Jmol-1K-1
So the Universal Gas Law is written as
PV = nRT
This relationship applies to all gases as long as they
fulfill the conditions for near ideal behavior – not at
high pressure and not at low temperature
The Gas Laws
Using the Combined Gas Law
Chemistry 120
If the quantity of gas is the same, then changes in
pressure, temperature or volume can be calculated
easily as
P1V1 = n = P2V2
RT1
RT2
Or
P1V1 = P2V2
T1
T2
The Gas Laws
Using the Combined Gas Law
Chemistry 120
The advantage of this expression is that the units do
not matter; the units used for P1 ,V1, and T1 will be
returned in the calculation for P2 ,V2, and T2.
However, if you have to use PV = nRT, you must use
the correct units which are consistent with R.
The easiest way is to convert all temperatures to K,
all pressures to Pa and all volumes to m3; the value
for R is then 8.314 Jmol-1K-1
14
The Gas Laws
The absolute temperature scale
Chemistry 120
From Charles’ Law, the decrease in volume per unit
temperature is always the same and therefore there
must be a minimum temperature that can be reached.
This is absolute zero O K, and is the zero point for
the absolute temperature scale.
The temperature in K is related to the temperature in
oC through
T/K = T/oC + 273.16
Chemistry 120
The Gas Laws
Example: Molecular Mass determinations
If we know the mass of gas in a sample of known
volume, pressure and temperature, then we can
calculate the relative molecular mass as we can
calculate n.
As n =
m
then, PV = mRT , so RMM = mRT
RMM
RMM
PV
RMM = mRT
PV
The Gas Laws
Example: Molar volumes
Chemistry 120
From Avogadro’s Law, equal quantities of gas
occupy equal volumes.
The volume of one mole of gas is therefore
independent of the nature of the gas, as long as the
gas behaves as ideal.
One mole of a perfect gas at 0oC and 1 atm pressure
occupies
22.4 L
15
The Gas Laws
Example: Volumes and moles
Chemistry 120
When we react solids or liquids, the easiest way is to
measure the mass of the sample and then convert to
moles by dividing by the relative molecular mass.
For gases, the easiest way is to measure the pressure
or the volume, as the densities of gases are so low.
For these calculations, you must use the same
temperatures and pressures for each gas.
The Gas Laws
Partial pressures
Chemistry 120
In a mixture of gases, we can measure the total
pressure of the mixture – PTotal and therefore we can
use PV = nRT to determine the total number of
moles of gas present.
As the mixture contains more than one gas, we can
write the contribution of the pressure of each gas to
the total pressure
The Gas Laws
Partial pressures
Chemistry 120
So the total pressure Ptotal is written as the sum of all
the individual pressures of the components of the gas
mixture:
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The Gas Laws
Partial pressures
Chemistry 120
So the total pressure Ptotal is written as the sum of all
the individual pressures of the components of the gas
mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
The Gas Laws
Partial pressures
Chemistry 120
So the total pressure Ptotal is written as the sum of all
the individual pressures of the components of the gas
mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
As PV = nRT then
The Gas Laws
Partial pressures
Chemistry 120
So the total pressure Ptotal is written as the sum of all
the individual pressures of the components of the gas
mixture:
PTotal = P1 + P2 + P3 + P4 + ...........
As PV = nRT then
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
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The Gas Laws
Partial pressures
Chemistry 120
So
PTotal = P1 + P2 + P3 + P4 + ...........
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
The Gas Laws
Partial pressures
Chemistry 120
So
PTotal = P1 + P2 + P3 + P4 + ...........
nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........
nTotal = n1 + n2 + n3 + n4 + ...........
The Gas Laws
Partial pressures
Chemistry 120
So the pressures of each component of the gas
mixture correlate with the number of moles of the
gas component of the mixture – a simple extension
of Avogadro’s Law.
18
The Gas Laws
Partial pressures
Chemistry 120
We can also write the fraction of the total pressure
that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
The Gas Laws
Partial pressures
Chemistry 120
We can also write the fraction of the total pressure
that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
The Gas Laws
Partial pressures
Chemistry 120
We can also write the fraction of the total pressure
that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
P1 = n1RT
19
The Gas Laws
Partial pressures
Chemistry 120
We can also write the fraction of the total pressure
that is due to one of the component:
PTotal = P1 + P2 + P3 + P4 + ...........
nTotal = n1 + n2 + n3 + n4 + ..........
P1 = n1RT
=
n1
So, P1
PTotal
n1 + n2 + n3 + n4 + ..........
The Gas Laws
Partial pressures
P1 =
PTotal
Chemistry 120
n1
n1 + n2 + n3 + n4 + ..........
The fraction on the RHS is called the mole fraction
and is written as x1 so we can write
P1 =
Or
n1 PTotal
n1 + n2 + n3 + n4 + ..........
P1 = x1 PTotal
The Gas Laws
Partial pressures
Chemistry 120
The sum of the partial pressures for all the
components and for the mole fractions is 1
ΣiPi = Ptotal
and
Σixi = 1
20
Thermochemistry
Energy
Chemistry 120
Energy is defined as the ability to do work.
There are several forms of energy
Kinetic energy – energy due to motion
EK = 1/2mv2
Potential energy – the energy due to the position of a
particle in a field
e.g. Gravitational, electrical, magnetic etc.
Thermochemistry
Energy
Chemistry 120
The unit of energy is the Joule (J) and
1 J = 1 kgm2s-2
Thermochemistry is the study of chemical energy
and of the conversion of chemical energy into other
forms of energy.
It is part of thermodynamics – the study of the flow
of heat.
Chemistry 120
Thermochemistry
Thermochemically, we define the system as the part
of the universe under study and the surroundings as
everything else.
Systems come in three forms:
Open
The system can exchange matter and
energy with the surroundings
Closed
The system can exchange energy only
with the surroundings
Isolated
There is no exchange of matter or of
energy with the surroundings
21
Chemistry 120
Thermochemistry
Matter is continually in motion and has an internal
energy that is composed of several different types
There is
Translation
Rotation
Vibration Potential
between molecules and inside molecules.
The internal energy is written as U
Chemistry 120
Thermochemistry
Matter is continually in motion and has an internal
energy that is composed of several different types
There is
Translation
Rotation
Vibration Potential
between molecules and inside molecules.
The internal energy is written as U
The internal energy is directly connected to heat and
the transfer of heat.
Chemistry 120
Thermochemistry
Heat is the transfer of internal energy between the
surroundings and the system or between systems.
The direction of the heat flow is indicated by the
temperature
– heat flows along a Temperature gradient
from high temperature to low temperature.
When the temperature of the system and that of the
surroundings are equal, the system is said to be
in thermal equilibrium
22
Thermochemistry
Energy is the capacity to do work
Chemistry 120
but what is work?
Thermochemistry
Energy is the capacity to do work
Chemistry 120
but what is work?
Work is the action of a force over a distance. To be
able to do work, we must be able to exert a force
over a distance. During this process, energy is
expended.
Thermochemistry
Energy is the capacity to do work
Chemistry 120
but what is work?
Work is the action of a force over a distance. To be
able to do work, we must be able to exert a force
over a distance. During this process, energy is
expended.
w=Fxd
where w is the work, F is the force and d is the
distance. Work is measured in Joules.
23
Thermochemistry
PV work
Chemistry 120
When a gas expands against an external pressure, for
example in a cylinder, against a constant weight
(weight being a force.....) the work done can be
written as
w=Fxd
As
Thus
P=F
A
w = PAd
and as
Ad = Vfinal – Vinitial = ∆V
Then
w = P∆V
then
F = PA
Thermochemistry
PV work
Chemistry 120
By convention, the work done when a gas expands is
negative,
Thus
w = - P∆V
for an expanding gas
Thermochemistry
State Functions
Chemistry 120
The state of a system is defined by the precise
conditions of the system:
The quantity and type of matter present
The temperature and pressure
The molecular structure of the system
As 1 mole = 6.02 x 1023 particles, defining the state
of a system uniquely is experimentally impossible in
an absolute sense.
24
Thermochemistry
State Functions and U
Chemistry 120
The internal energy, U, of a system is a function of
the state of the system.
Although we cannot measure the absolute state of a
system, we can measure changes in the state of the
system in a relative way, by measuring the work and
the heat that takes place during a chemical change.
As U is a function of the state of the system, it does
not depend on the way the state of the system is
prepared – it is independent of the path.
Thermochemistry
State Functions and U
Chemistry 120
U is therefore a state function of the system. It
depends only on the present state of the system and
not on the previous history or the path by which the
system was prepared.
Because we have no measure of the state of a
system, or of the internal energy, we can only
measure the change in the state, through the
observation of work and transfers of heat into and
out of the system.
Thermochemistry
Internal Energy, U and State Functions
Chemistry 120
Energy, and therefore the capacity to do work is
present in all matter.
This internal energy is stored in translational,
rotational, vibrational and potential forms or modes
in the material.
The exact distribution of energy defines the state of
the system, together with external variables such as
pressure, temperature.
25
Thermochemistry
Internal Energy, U and State Functions
Chemistry 120
U is a function of the state of the material only, not
of the history of the sample or the path taken to
prepare the state of the sample.
Heat is the transfer of energy between the
surroundings and the sample
- the symbol for heat is q
Work is the result of a force acting over a distance
- the symbol for work is w
Thermochemistry
Internal Energy, U and State Functions
Chemistry 120
Heat and work are the only two ways of changing
the internal energy of a system.
Temperature is defined by the direction of the flow
of heat, which is always from high temperature to
low temperature.
When the the temperature of the system and the
surroundings are the same, the system is at thermal
equilibrium with it’s surroundings.
Chemistry 120
Thermochemistry
The sign conventions of thermochemistry
When the internal energy of the system rises, this
energy change has a positive sign.
- The energy of the system rises when heat is
absorbed
- The energy of the system rises when work is
done on the system e.g. a gas is compressed
- in these cases, q is positive
w is positive
26
Chemistry 120
Thermochemistry
The sign conventions of thermochemistry
When the internal energy of the system lowers, this
energy change has a negative sign.
- The energy of the system lowers when heat is
leaves the system
- The energy of the system rises when the
system does work e.g. a gas expands against an
external pressure
- in these cases, q is negative
w is negative
Thermochemistry
Internal energy rises:
Chemistry 120
q>0
w>0
Internal energy drops:
q<0
w<0
Thermochemistry
The First Law of Thermodynamics
Chemistry 120
Energy can be exchanged but cannot be
created or destroyed.
It is a statement of the Law of Conservation of
Energy
∆U = Ufinal – Uinitial = q + w
27
Thermochemistry
Chemical applications of the 1st Law
Chemistry 120
Any chemical change can be characterized as an
Endothermic change
or an
Exothermic change.
In an exothermic reaction, internal chemical energy
is converted into heat, which leaves the system if the
system is not isolated or causes the temperature to
rise if the system in isolated.
Thermochemistry
Chemical applications of the 1st Law
Chemistry 120
In an endothermic reaction, heat is required to drive
the chemical reaction and in an isolated system, the
temperature will fall. In an non-isolated system, heat
is absorbed from the surroundings.
Exothermic
T rises (isolated)
q negative (non-isolated)
Endothermic
T falls (isolated)
q positive (non-isolated)
Chemistry 120
Thermochemistry
Reactions at constant pressure and constant volume
At constant volume, ∆V = 0 and so
∆UV = qV - P∆V
∆UV = qV + 0 = qV
When the system can do PV work, i.e. a system at
constant pressure,
∆UP = qP - P∆V
where w = - P∆V
28
Chemistry 120
Thermochemistry
Most reactions take place at constant pressure and
therefore we define a new function, which is a state
function in the same way that U is a state function
Rearranging
∆UP = qP - P∆V
∆UP + P∆V = qP
We term qP the enthalpy of the reaction
qP = ∆H = ∆UP + P∆V
Chemistry 120
Thermochemistry
Enthalpy is an extensive property – one that depends
on the quantity of the material present in the
reaction.
This follows directly from the fact that the enthalpy
is the heat generated by a reaction
– there is more energy released from 1000 kg
of methane when it burns than from 1 g.
Chemistry 120
Thermochemistry
Enthalpies and internal energies are measured in kJ
mol-1 and the stoichiometry of a reaction is directly
applicable to the enthalpy – half the quantity of the
reaction results in half the enthalpy change taking
place.
29
Chemistry 120
Thermochemistry
We can characterize reactions as endothermic or
exothermic using the enthalpy, ∆H.
If the enthalpy change is
negative, the reaction is
exothermic and heat is
given out by the system
Reactants
∆H < 0, negative
Products
H
Chemistry 120
Thermochemistry
We can characterize reactions as endothermic or
exothermic using the enthalpy, ∆H.
If the enthalpy change is
negative, the reaction is
endothermic and heat is
absorbed by the system
Products
∆H >0, positive
Reactants
H
Chemistry 120
Thermochemistry
Using the enthalpy, we can account for the heat
entering a reaction at constant pressure – in the same
way that we account for the products and reactants
in a reaction.
In an endothermic reaction, the energy absorbed by
the system can be considered as a reactant.
Conversely, an exothermic reaction, one which
evolves heat, has the energy as a product.
30
Chemistry 120
Thermochemistry
Enthalpies and internal energies are measured in kJ
mol-1 and the stoichiometry of a reaction is directly
applicable to the enthalpy – half the quantity of the
reaction results in half the enthalpy change taking
place.
DO EXCERCISES AND EXAMPLES 6.1 – 6.5 (pp
235 – 245)
BRING THEM TO THE NEXT DISCUSSION
PERIOD
Thermochemistry
Heat Capacities
Chemistry 120
When a definite quantity of energy is absorbed by
materials, the temperature rises.With different
materials, the temperature rise, ∆T, is different.
The quantity of energy required to raise a quantity of
material by 1 K is termed the heat capacity.
Mathematically,
C=q
∆T
where C is the heat capacity, q is the heat.
Thermochemistry
Heat Capacities
Chemistry 120
The specific heat is the heat per gram of sample and
the molar heat capacity is the heat capacity per
mole.
31
Chemistry 120
Thermochemistry
Specific Heats, Molar Heats and Calorimetry
The heat capacity is the quantity of heat required to
raise a given quantity of a substance by 1 K
The specific heat
1 gram though 1 K
The molar heat
1 mole through 1 K
The units of heat capacity are
Jg-1K-1 (specific heat) or Jmol-1K-1 (molar heat)
Chemistry 120
Thermochemistry
Specific Heats, Molar Heats and Calorimetry
To calculate the heat transferred to a sample we use
q = quantity x heat capacity x ∆T
For the specific heat
q = mCs∆T
where m = mass
For the molar heat
q = nCm∆T
where n = no. of moles
Make sure that the units of the heat capacity matches
the units of quantity that is in the heat equation
Chemistry 120
Thermochemistry
Specific Heats, Molar Heats and Calorimetry
To measure the heat capacity, a calorimeter is used.
A calorimeter measures heat transfers, heats of
reaction or heats of dissolution.
32
Chemistry 120
Thermochemistry
Specific Heats, Molar Heats and Calorimetry
In principle, they consist of an insulated chamber
and an accurate way of measuring temperature (a
thermocouple or thermometer).
Insulation ensures that the only heat involved in the
temperature rise is that inside the calorimeter.
Thermochemistry
Heat capacity measurements
Chemistry 120
A sample with a known temperature is placed into a
fluid of known heat capacity and known temperature
and allowed to come to thermal equilibrium.
Thermochemistry
Heat capacity measurements
Chemistry 120
A sample with a known temperature is placed into a
fluid of known heat capacity and known temperature
and allowed to come to thermal equilibrium.
At thermal equilibrium, Tsample = Tfluid and so we
know ∆T for the sample and for the fluid.
33
Thermochemistry
Heat capacity measurements
Chemistry 120
A sample with a known temperature is placed into a
fluid of known heat capacity and known temperature
and allowed to come to thermal equilibrium.
At thermal equilibrium, Tsample = Tfluid and so we
know ∆T for the sample and for the fluid.
We also know Cfluid and therefore we know qfluid, the
heat transferred into the fluid - q = Cfluid∆Tfluid
Thermochemistry
Heat capacity measurements
Chemistry 120
A sample with a known temperature is placed into a
fluid of known heat capacity and known temperature
and allowed to come to thermal equilibrium.
At thermal equilibrium, Tsample = Tfluid and so we
know ∆T for the sample and for the fluid.
We also know Cfluid and therefore we know qfluid, the
heat transferred into the fluid - q = Cfluid∆Tfluid
As this is the only source of heat in the calorimeter,
we know qfluid and ∆Tsample, so we can calculate
Csample
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A?
34
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5 oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A? Cwater = 4.18 Jg-1K-1
1.
Calculate qwater
2.
qwater = - qA from conservation of energy
3.
Calculate CA from qA
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5 oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A? Cwater = 4.18 Jg-1K-1
1.
Calculate qwater:
∆Twater = Tfinal – Tinitial = (25.7 – 22.5) oC = 3.2 oC
qwater= 25 x 4.18 x 3.2 = 334 J
Note: qwater is positive as heat is entering the water
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5 oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A? Cwater = 4.18 Jg-1K-1
1.
qwater = 334 J
2.
qwater = - qA thus qA = - 334 J
35
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5 oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A? Cwater = 4.18 Jg-1K-1
1.
qwater = 334 J
2.
qwater = - qA thus qA = - 334 J
3.
qA = mCA∆TA
∆TA = Tfinal – Tinitial = (25.7 – 98.9) oC = -73.2 oC
Thermochemistry
Example
Chemistry 120
15.5g of alloy A has a temperature of 98.9 oC. It is
placed into a calorimeter containing 25 g of water at
22.5 oC. Thermal equilibrium is achieved at 25.7 oC.
What is the heat capacity of A? Cwater = 4.18 Jg-1K-1
1.
qwater = 334 J
2.
qwater = - qA thus qA = - 334 J
3.
qA = mCA∆TA; ∆TA = -73.2 oC
CA = qA/m∆TA = -334/(15.5 x –73.2) = 0.29 Jg-1K-1
Thermochemistry
Chemistry 120
Bomb Calorimetry
For reactions which generate gas, the P∆V work
makes a significant contribution and the quanitiy we
will measure in an open calorimeter is the enthalpy.
We cannot easily measure the P∆V work in this case.
We can measure ∆U in a bomb calorimeter – one
where the volume change is zero and therefore ∆V =
0.
The calorimeter is calibrated using a known sample.
36
Thermochemistry
Chemistry 120
Hess’ Law of Summation
If we wish to determine the heat of reaction or
formation of a compound which is not stable, cannot
be isolated or cannot be measured for some reason,
we use Hess’ Law to determine this quantity.
Hess’ law states that the
the heat of reaction is constant and is not
determined by the path of the reaction.
We know this as ∆U (and ∆H) is a state function
Thermochemistry
Chemistry 120
Hess’ Law of Summation
Practically, if we can find a cycle of reactions that is
measureable, then we can derive the unmeasurable
quantity as we know the total sum of all the energy
changes in the cycle.
Thermochemistry
Chemistry 120
Hess’ Law of Summation
Example
The combustion of C results in the formation of
CO2 in a bomb calorimeter. The heat of formation
of CO is therefore hard to measure.
We can measure the heat of combustion of CO and
that of C both to give CO2.
37
Thermochemistry
Chemistry 120
Hess’ Law of Summation
∆Hf(CO)
CO + 1/2O2
Cgraphite + O2
∆Hcombustion(CO)
∆Hf(CO2)
CO2
Thermochemistry
Chemistry 120
Hess’ Law of Summation
∆Hf(CO)
CO + 1/2O2
Cgraphite + O2
∆Hcombustion(CO)
∆Hf(CO2)
CO2
Of the reactions in this cycle, the heats of
combustion of CO and C are known, but the heat of
formation of CO from C is not.
Thermochemistry
Chemistry 120
Hess’ Law of Summation
∆Hf(CO)
Cgraphite + O2
∆Hf(CO2)
CO + 1/2O2
∆Hcombustion(CO)
CO2
∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO)
38
Thermochemistry
Chemistry 120
Hess’ Law of Summation
∆Hf(CO)
Cgraphite + O2
∆Hf(CO2)
CO + 1/2O2
∆Hcombustion(CO)
CO2
∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO)
∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO)
Thermochemistry
Hess’ Law of Summation C
Using the lower equation
and the values for the
heats of combustion of
CO and C, we can
calculate the unknown
heat in the cycle
Chemistry 120
∆Hf(CO)
graphite
CO + 1/2O2
+ O2
∆Hcombustion(CO)
∆Hf(CO2)
CO2
∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO)
∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO)
Thermochemistry
Hess’ Law of Summation C
Using the lower equation
and the values for the
heats of combustion of
CO and C, we can
calculate the unknown
heat in the cycle
Chemistry 120
∆Hf(CO)
graphite
+ O2
∆Hf(CO2)
CO + 1/2O2
∆Hcombustion(CO)
CO2
∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO)
∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO)
∆Hf(CO2) = - 393.5 kJ ∆Hcombustion(CO) = - 283.0 kJ
∆Hf(CO2) = ∆Hf(CO2) - ∆Hcombustion(CO)
39
Thermochemistry
Hess’ Law of Summation C
Using the lower equation
and the values for the
heats of combustion of
CO and C, we can
calculate the unknown
heat in the cycle
Chemistry 120
∆Hf(CO)
graphite
+ O2
CO + 1/2O2
∆Hcombustion(CO)
∆Hf(CO2)
CO2
∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO)
∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO)
∆Hf(CO2) = - 393.5 kJ ∆Hcombustion(CO) = - 283.0 kJ
∆Hf(CO2) = (- 393.5) – (- 283.0) = -110.5 kJ
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
Just as we cannot determine the absolute value for
the internal energy of a system and so concentrate
on the change in internal energy, so we cannot fix
an absolute zero-point for reaction and formation
enthalpies.
We chose the Standard state of a material as that at
1 bar pressure (1 bar = 1 x 105 Pa) and the
temperature of interest.
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
The standard enthalpy of formation of an element in
the standard state is defined as zero.
Using these two facts, we can calculate the heats of
formation and, through Hess’ cycles, the heats of
reaction for all substances.
40
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
When we combine different reactions, we must take
account of the stoichiometry of the reaction.
Remember that ∆H can be thought of as a product
of reaction and must be combine with the correct
stoichiometry.
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
For the reaction
∆Hdimerization(NO2)
2NO2
N2O4
We can construct a Hess’ cycle:
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
For the reaction
∆Hdimerization(NO2)
2NO2
N2O4
∆Hdimerization(NO2)
We can construct a
Hess’ cycle:
2NO2
∆Hf(NO2)
N2O4
/2∆Hf(NO2)
1
1
/2N2+ O2
41
Thermochemistry
Chemistry 120
Standard enthalpies of formation and reaction
For the reaction
∆Hdimerization(NO2)
2NO2
N2O4
∆Hdimerization(NO2)
We can construct a
Hess’ cycle.
Note that we must
include
the
stoichiometry in the
calculation.
2NO2
∆Hf(NO2)
N2O4
/2∆Hf(NO2)
1
1
/2N2+ O2
42