# Thermodynamics - Bergen.org

Mechanics

Oct 27, 2013 (4 years and 7 months ago)

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Thermodynamics

7

Thermodynamics is a funny subject. The first time you go through it, you don’t understand it at all.

The second time you go through it, you think you understand it except for one or two small points.

The third time you go through it, you know you don’t understand it, but by that time you are so

accustomed to it that it doesn’t bother you anymore.

-
Arnold Sommerfeld

1. The First Law

The First Law of Thermodynamics, also known as the law of

conservation of energy, states that energy can
neither be created nor destroyed. In physics its use often involves the interconversion of kinetic energy and
various forms of potential energy (e.g. chemical, gravitational or electrostatic energy). However
in
chemistry the first law is used with quantities of chemical concern such as the total energy of a system, the
heat flow in or out of it, and the work done on or by the system. So the First Law is typically expressed as:

ΔE = q + w

The total energy of th
e system, E, is the sum of the kinetic and the potential energy. It is sometimes referred
to as the
internal energy
, U. Of course the actual energy of a system cannot be measured. All that can be
measured is the change in the energy, ΔE. The heat flow is r
eferred to as "q," and the work done on or by
the system is "w." While a different sign convention is sometimes used in physics or engineering, in
chemistry the sign convention is always the same. Energy which flows into a system

as heat flow or
work bei
ng done on the system

is positive; energy which flows out of a system is negative. In a chemical
system, where work is pressure times the change in volume, the sign change can be thought of as follows:
If work is done on the system, it will be squeezed i
nto a smaller volume. Hence +work is associated with
-
ΔV. Similarly a system whose volume increases is doing work on the surroundings. Thus
-
w (which is
what you have when the system does work on the surroundings) is associated with a +ΔV. This is
summariz
ed by:

w =
-

P ΔV

and therefore

ΔE = q
-

P ΔV

2. Calorimetry

The heat flow, q, cannot itself be measured. What we
can

measure is temperature change, which is a
consequence

of heat flow. The process of using temperature change to measure heat flow is known as
calorimetry, and the device in which this is done is known as a calorimeter.

The fundamental equation of calorimetry is:

q = c m ΔT

Here q is the heat flow (usually th
e heat given off in a chemical reaction), m is the mass of the system, and
ΔT is the change in the system's temperature. The specific heat, c, is the amount of heat required to change
the temperature of a gram of substance by one degree. Specific heat was
once measured in calories, the
amount of heat required to raise the temperature of a gram of water by one degree Celsius. Thus the
specific heat of water was one. Specific heats are now measured in joules, with the specific heat of water
being 4.184 Joules
/g

K.

-
2
-

In AP Chemistry, a calorimeter usually consists of water in a styrofoam
coffee cup. A coffee cup is used because it insulates well and, thus, prevents
heat from leaking into or out of the system. Styrofoam cups also have the
ery light and therefore absorbing so little heat that they
can
be ignored in our calculations.

Suppose, for example, a chemical reaction takes place in a coffee cup
ca
lorimeter containing 60.0 g of
water. The initial temperature of the water is
24.0

C an
d its final temperature is 27.4

C. How much heat is given off in
the reaction?

q = c m ΔT

q = 4.184 J/g

deg

60.0 g

3.4 deg

q = 853 joules = 850 joules

Sometimes a larger calorimeter, heavy enough to affect the experiment, must be used. Rather than

weighing the calorimeter and estimating its specific heat, the product of the weight and the specific heat is
determined. This product, known as the heat capacity and abbreviated as "C," can then be used in
calorimetric calculations. This can be stated as
:

q = (c
water

m
water

+ C) ΔT

The heat capacity of a calorimeter is usually determined once and then inscribed on the calorimeter.

The open calorimeter described above, in which the pressure remains constant, is only one way of
measuring heat flow. Calori
metric calculations at constant pressure are written as:

q = c
p

m ΔT

Since this is what chemists normally do, the c
p

is usually written as just "c."

However, measurements can also be made at constant volume in a sealed calorimeter. A constant
-
volume
calo
rimeter, usually referred to as a "bomb" calorimeter, often gives a different set of values. Calorimetry
under conditions of constant volume is governed by the equation:

q = c
v

m ΔT

A reaction in a bomb calorimeter proceeds under conditions of constant vol
ume. This means that no work
is done by (or to) the system. Thus:

ΔE = q + w = q

Finally it should be said that processes which, like calorimetry, do not exchange heat with their
surroundings are known as

processes.

3. Enthalpy

Chemical measurements are usually carried out in open containers, where the pressure is constant. If the
reaction pulls air into the flask, work is being done
on

the system by the surroundings. If gas goes from the
flask to the surroundings, work is being
done
by

the system. We can still measure the amount of heat given
off or taken in by the reaction, but it is no longer equal to the change in the internal energy of the system.
Some of the heat is converted to work.

We get around this problem by introduci
ng a term called enthalpy (H). Enthalpy is the sum of the system's
internal energy plus the product of pressure and volume. Thus:

H = E + PV

-
3
-

The change in the enthalpy of a system is equal to the change in its internal energy plus the change in the
produc
t of the pressure times the volume of the system.

ΔH = ΔE + P ΔV

Combining this equation with the first law of thermodynamics, ΔE = q
-

P ΔV, we end with:

ΔH = q

That is, the change in enthalpy will be equal to the heat flow. T
his will prove useful later o
n.

4. Hess's Law

Hess's Law states that if a reaction occurs in steps, even if only theoretically, the enthalpy change
associated with the reaction is equal to the sum of the enthalpy changes of the individual steps. The rules
for using Hess's Law are as
follows:

1) The equations corresponding to the steps must add up to the total reaction

2) Whatever is done to a step (e.g., doubling it) must be done to the ΔH for that step.

3) The ΔH values of the steps must add up to the ΔH of the reaction.

Let us, for

example, use the enthalpies of combustion for carbon, hydrogen and ethanol to calculate the
enthalpy of formation for ethanol. This corresponds to the reaction:

(1) 2 C + 3 H
2

+ ½ O
2

C
2
H
5
OH

The three enthalpies of combustion correspond to the following
reactions:

(2) C + O
2

CO
2

ΔH

=
-
394 kJ/mol

(3) H
2

+ ½ O
2

H
2
O

ΔH

=
-
286 kJ/mol

(4) C
2
H
5
OH + 3 O
2

2 CO
2

+ 3 H
2
O

ΔH

=
-
1366 kJ/mol

To get the desired reaction, we add two times the enthalpy change of reaction 2, three times the ΔH of
reaction 3 and subtract the ΔH of reaction 4. This gives us, for reaction 1, ΔH =
-
278 kJ/mol.

Hess's Law works because enthalpy is a
state function
. That is, a given substance under a given set of
conditions will always have the same enthalpy. Enthalpy d
epends on the state of the substance, not on how
the substance got to where it is. Other state functions include entropy and free energy, which we will
discuss shortly, as well as internal energy, volume, temperature and pressure. Work and heat are
not

sta
te
functions. You can talk about the entropy, the internal energy or the volume of a system. You cannot talk
about the work of a system or the heat of a system. It doesn't sound right.

5. Standard Enthalpies (Heats) of Formation

Since enthalpy is a state

function, tables of enthalpy values for various substances have been compiled. The
conditions at which these values are obtained, 25

C and unit concentration (1 atm or 1 M), are called
standard conditions and these values are referred to as standard enth
alpies. Actually the tabulated values
are not enthalpies but are enthalpy changes, since enthalpy values can only be calculated relative to an
arbitrary zero. The zero of enthalpy is defined as the enthalpy of an element in its standard state. So the
entha
lpy of water is the enthalpy change which occurs in this reaction:

H
2
(g) + ½ O
2
(g)

H
2
O(g)

The enthalpy change is called the "standard enthalpy of formation" (or the standard heat of formation) and
is abbreviated ΔH

f
. Notice in the above equation that

the standard heat of formation for the reactants is
zero, since both are in their standard states.

-
4
-

To predict the enthalpy change for a reaction you add the ΔH

f

values of the products and subtract those of
the reactants. Remember:
products minus
reactants
. Take for example the reaction:

C
2
H
5
OH(
l
) + 3 O
2
(g)

2 CO
2
(g) + 3 H
2
O(
l
)

The enthalpy change for this reaction can be calculated as:

ΔH

= ΔH

f

(products)
-

ΔH

f

(reactants)

ΔH

= 2

ΔH

f

(
CO
2
) + 3

ΔH

f

(H
2
O)
-

1

ΔH

f

(
C
2
H
5
OH)
-

3

H

f

(O
2
)

ΔH

= 2

(
-
393.5) + 3

(
-
285.8)
-

1

(
-
277.1)
-

3

(0)

ΔH

=
-
1367.3 kJ/mol

6. Bond Energy

One way of calculating the enthalpy change in a reaction is from the bond energies. One adds the energy of
the bonds in the products and subtracts
the energy of the bonds in the reactants. So the enthalpy change is
the energy of the broken bonds minus the energy of the bonds which are formed. Please remember "
broken
bonds minus formed bonds
."

Let us consider, as an example, the reaction:

N
2
(g) + 3 H
2
(g)

2 NH
3
(g)

The relevant bond energies are:

N

N

941 kJ/mol

H
-

H

432 kJ/mol

N
-

H

391 kJ/mol

The enthalpy change in the reaction is calculated as below. Notice the sign change.

ΔH = broken bonds
-

formed bonds

ΔH = (3

E
H
-
H
) + (1

E
N
-
N
)
-

(6

E
N
-
H
)

ΔH = (3

432) + (1

941)
-

(6

391)

ΔH =
-
109 kJ/mol

7. Entropy

If the energy in the Universe remains constant, if energy cannot be consumed, then what drives chemical
reactions? It is the
dispersal

of energy. Reactions proceed when they allow energy to spread out. This
tendency of energy to disperse is measured by something called "entropy."

When a hot object cools, the heat energy which was concentrated in the object spreads to the
molecules

ai
r and table molecules, for example

which touch the hot object.

When you drop an object its potential energy is converted to kinetic energy which then
changes

very quickly

to thermal energy when it hits the ground.

When a gas expands, the energy of
the gas molecules is spread out more widely.

When ice melts, thermal energy from a (warm) external source spreads into the ice and
breaks the bonds holding the water molecules in their crystal lattice.

Even a spontaneous, exothermic reaction, a case wher
e entropy doesn't seem necessary, is
still a case where energy is spread out. Heat is transferred from the bonds of reactant to
thermal energy in the system and its surroundings.

-
5
-

The Second Law of Thermodynamics states that “in any spontaneous process, the

entropy of the
Universe increases.” All of the above processes are spontaneous (or irreversible) since energy is being
dispersed. An example of a non
-
spontaneous (reversible) process would be an ice cube at 0° C in a glass of
water at 0.0001° C. Energy
is being (slowly) transferred, but the temperature difference is so small that the
system is virtually at equilibrium.

The change in entropy caused by one of these processes is calculated by taking the heat which is transferred
and dividing it by the temp
erature at
which the transfer occurs. Thus
:

Δ
S
=
q
T

This equation is Clausius' original, thermodynamic definition of entropy, proposed around 1850. You will
often see it written with a negative sign! This is because in physics and engineering a positive heat flow is
from
the system to the surroundings.

What is the entropy change associated with melting 1 gram of ice? Melting ice requires that 6.01
kjoules/mol flow into the system. (This is known as the enthalpy or the heat of fusion.) Thus:

ΔS = ___________
_________

273 K

ΔS
=
1.22
J
K

Notice the entropy is measured in joules per degree, not in kilojoules. This is not an obvious point.

8. The Statistical Definition of Entropy

A more recent (around 1900) def
inition of entropy is the statistical definition. This definition, which was
put forth by Ludwig Boltzmann, gives the same results but in a different way. The statistical definition says
that entropy is determined by the number of microstates

the number
of different arrangements

in
which a system can exist. The relevant equation is:

S = k ln W

Here k is a constant, known as the Boltzmann constant, which is equal to R divided by Avogadro's
Number. W is the number of microstates

the ways in which a system can be arranged.

The impor
tance of the number of arrange
ments
can be seen in a sy
stem containing four

molecules (A, B, C, and D) in two connected
bulbs (1 and 2). In the first column, all four
molecules are in bulb #1 and there is only one
possible arrangement (one possible microstate).
In the second column, where one molecule is in
b
ulb #2 and the remainder are in bulb #1, there
are four possible states. The column with the

Possible Arrangements of 4 molecules in 2 bulbs.

most possible arrangements, and hence the most

likely configuration, is column 3, where the number of molecules
in each bulb is equal. This column has
the greatest entropy.

A
B
C
D

A
B
C
D

A
B
C
D

A
B
C
D

A
B
C
D
1
1
1
1
2
1
1
1
2
2
1
1
2
2
2
1
2
2
2
2

1
2
1
1
2
1
2
1
2
2
1
2

1
1
2
1
1
2
2
1
2
1
2
2

1
1
1
2
2
1
1
2
1
2
2
2

1
2
1
2

1
1
2
2

6010

J

x
1 mol

x 1 g

mol 18.0 g

A B C D

A B C D

A B C D

A B C D

A B C D

1 1 1 1 2 1 1 1

2 2 1 1 2 2 2 1 2 2 2 2

1 2 1 1 2 1 2 1 2 2 1 2

1 1 2 1 1 2 2 1 2 1 2 2

1 1 1 2 2 1 1 2 1 2 2 2

1 2 1 2

1 1 2 2

-
6
-

Consider a crystal of CO at 0 K. Because carbon monoxide is symmetrical, each molecule of CO can be
oriented either as C

O or as O

C. Assuming that each configuration has the same energy, we can
calculate
the entropy of one mole of CO at 0 K as follows:

S = k ln W

Since k = R/N, where N is Avogradro's Number

S = (R/N) ln W

Since the number of ways in which one mole of CO molecules can be arranged is 2
N

S = (R/N) ln 2
N

= (R/N) (N

ln 2)

S = R ln 2

S = 8.31

0.693 = 5.76 joules

This is fairly close to the experimental value of 4.0 joules/mole, and even this small difference can be
explained. Since carbon monxide has a slight dipole, adjacent molecules which are oriented CO CO are
engergetically fav
ored over molecules which are CO OC. Therefore states with an excess of CO CO
orientations are favored and the crystal is not truly random, as our calculation assumes.

9. Entropy Summarized

Often it is possible to predict the sign of the entropy change in

a reaction. Consider the reaction:

2 H
2
S(
g
) + SO
2
(
g
)

3 S(
s
) + 2 H
2
O(
g
)

Here the number of gas molecules decreases. Therefore the entropy of the system decreases. This is in fact
quite general. The translational entropy of a gas is so large that a reacti
on in which the number of gas
molecules increases will necessarily be associated with an entropy increase. Similarly a decrease in the
number of gas molecules will be associated with decreasing entropy.

You can usually predict whether entropy will increas
e or decrease by knowing the following:

Entropy of a gas > entropy of a liquid > entropy of a solid

Dissolving a solute increases its entropy greatly.

Heating something, a solid, liquid or gas, increases its entropy.

Allowing a gas to expand increases its
entropy.

Mixing two substances, whether they are solids, liquids or gases, increases their entropy.

Although these can all be explained in terms of the dispersal of energy, explanations are often made by
equating entropy with disorder. Strictly speaking e
ntropy
is

associated with disorder, but the analogy is
often overstated. Consider the use of entropy to describe a macroscopic system such as a messy desk.
Messing up a desk does not occur spontaneously and does not involve a heat flow. It is not truly an entropy
i
ncrease.

10. The Third Law and the Zero of Entropy

Enthalpies are defined in terms of an arbitrary zero

an element in its standard state at 25

C. Thus there is
no such thing as an absolute enthalpy, only an enthalpy change. Entropy, however, is differe
nt. Since
entropy is the dispersal of energy, the less energy there is to disperse, the lower the entropy. In other words
as temperature approaches zero, so does entropy. This is so fundamental that it makes up the third law of
thermodynamics. "The zero of

entropy is a perfect crystal of a pure substance at zero degrees Kelvin."
(This makes sense since a perfect crystal at absolute zero has only one possible configuration.) So enthalpy
is given as a standard enthalpy of formation, ΔH

f
.

However entropy is n
ot given as a change but as a
standard entropy, S

.

-
7
-

11.Free Energy

A quantity called the Gibbs Free Energy (G) incorporates both entropy and enthalpy. The sign of ΔG
determines the spontaneity of a reaction and is therefore of great interest to chemists
. The Gibbs Free
Energy is calculated from the equation:

ΔG = ΔH
-

TΔS

A reaction with a negative ΔG is spontaneous; a reaction with a positive ΔG is not spontaneous; and a
reaction with zero ΔG is at equilibrium.

Looking at the above equation and the aff
ect of the sign of ΔG on spontaneity, we can make the following
table. Make sure that you agree.

ΔH

ΔS

Is it spontaneous?

1)

+

+

At high temperature

2)

+

Never

3)

At low temperature

4)

+

Always

Case 1, where high temperature is needed to make the

reaction spontaneous, is of greatest concern to the
chemist. An example would be the boiling of water. Steam has a both higher entropy and a higher enthalpy
than water, and the boiling process requires a high temperature. Let us calculate the temperature
at which it
happens.

From a table of thermodynamic data w
e find that for boiling water

ΔH =
-
242 kJ
-
(
-
286 kJ) = 44 kJ/mol.

Similarly

ΔS = 189 J
-

70 J = 119 J/mol.

The boiling point of water is the temp
erature at which the liquid and

gaseous forms of water are in equilib
rium. At higher temperature the

gas phase is favored and at lower temperatu
re the liquid phase is

favored.
But what we are looking for is t
he
equilibrium temperature,

the
temperature at which ΔG = 0.

We can calculate the transition temperature, the temperature at which equilibrium occurs and at which

ΔG = 0. But in order to do so, we need values for ΔH and ΔS. We can take them from a
thermodynamic
table, but these data are correct only at 273

K. So we must assume that both ΔH and ΔS are independent
of temperature. This is a good assumption for the enthalpy but is less good for entropy. Nevertheless, we
will assume it to be true for b
oth. We can calculate:

ΔG = ΔH
-

TΔS

0 = 44 kJ
-

T (.119 kJ)

T = 369 K = 96

C

This is clearly wrong, since the normal boiling point of water is 100

C. The error comes from using S

values at temperatures other than 25

C. But the result is close and thi
s is how the calculation is done. You
should know the method.

ΔH

f

S

H
2
O(
l
)

-
286 kJ

70 J

H
2
O(
g
)

-
242 kJ

189 J

Thermodynamic properties
of
water

-
8
-

The thermodynamic tables found in chemistry textbooks not only contain enthalpy values (ΔH

f
)

and
entropy values (S

), they also contain free energy values (ΔG

f
).

Like energy and enthalpy,
free energy is a
state function. These tables are used to calculate the ΔG associated with a reaction, but only at 25

C. If you
need to calculate free energy change at any other temperature, calculate ΔH and ΔS from the tables and
then use them to calcula
te ΔG.

Gibbs Free Energy is not conserved and, despite its units, is not really energy. It is sometimes called the
"Gibbs Function." However, the Gibbs Free Energy does have a relationship to work. Work is not a state
function and the amount of work obtai
ned from a chemical process depends on how it is carried out. What
we can calculate is the
maximum

amount of work obtainable from a chemical process. THIS is equal to the
free energy change.

12. Equilibrium Constants

The free energy change determines whe
ther a reaction is spontaneous. The equilibrium constant, however,
is more generally useful and can be calculated from the free energy change. Thus the following equation,
where Q is the equilibrium quotient, gives a reaction's free energy change.

ΔG = ΔG

+ RT ln Q

For a system at equilibrium, K = Q, ΔG = 0, and the following equation is used:

ΔG

=
-
RT ln K

You use 8.31 J/mol

K as the gas constant and get a free energy in units of joules. Do NOT use 0.08206 as
the gas constant or expect the energy to be in kJ! Notice, also, that K depends on ΔG

and not ΔG. This is
because the free energy change at equilibrium is always zero,
which would lead to the equilibrium constant
being one.

Let's calculate the equilibrium constant for:

N
2
(g) + 3 H
2
(g)

2 NH
3
(g)

From the table on the right we know that the free energy

change is
-
34 kJ/mol at 25

C. So

-
34,000 =
-
8.31

298

(ln K)

ln K = 13.729 = 13.

Since this number has two significant digits, neither of which

come after the decimal we can calculate

K = 1.

10
6

Of course if we had mistakenly used
-
-
34,000 J, we would calculate ln K = 0.013 and

K = 1.01. Mistakes in this type of calculation often give you a K value of approximately one. So if you get
an answer of K = 1, CHECK I
T!

ΔH°
f

ΔG°
f

S

kJ/mol

kJ/mol

J/K mol

N
2
( g)

0

0

192

H
2
(g)

0

0

131

NH
3
(g)

-
46

-
17

193

Beware of equilibrium constants close to 1!

-
9
-

Now let us do the same calculation at 100

C. Since this is not at standard temperature we can no longer use
"G

" but will instead use "G'" to represent

the free energy at standard concentrations but at a non
-
standard
temperature. Also because of the non
-
standard temperature, we must calculate the free energy change from
the changes in enthalpy and entropy. Thus:

ΔG' = ΔH

-

T ΔS

ΔG' = (2

-
46,000)
-

373 (2

193
-

192
-

3

131)

ΔG' =
-
92,000 + 74,227

ΔG' =
-
17,773 J/mol

Convince yourself that
t
his is known to two significant digits. Now we calculate K

ΔG' =
-

RT lnK

-
17,773 =
-
8.31

373 ln K

ln K = 5.73

The ln K is known to two signific
ant digits, one of which is after the decimal. So K is only known to one
significant digit.

K = 309.2 = 300

13. Other Statements of the Three Laws

In this house we obey the laws of thermodynamics!

-

Homer Simpson

The Second Law can be stated in

a number of ways. These include:

A) Heat will not flow spontaneously from a cold object to a hot object.

B) A system which is free of external influences becomes more disordered with time. This disorder
can be expressed in terms of a quantity called entro
py.

C) In any irreversible process (which includes almost everything) the entropy of the Universe
increases.

D) You cannot create a heat engine which extracts heat and converts it to useful work. If you could,
an air conditioner would produce energy instead of consuming it.

A device which violates (D) is referred to as a "perpetual motion machine of the second
kind." In order to
convert heat to work, you must take heat from a hot object and move it to a colder object.

The Third Law can also be stated as:

It is impossible to reach absolute zero in a finite number of steps.

Now let us conclude with another versi
on of the three laws, attributed to author C. P. Snow:

1) You can't win.

2) You can't break even except at absolute zero.

3) You can't reach absolute zero.