Second Law of Thermodynamics

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27 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

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Thermodynamics

First Law of Thermodynamics
-

Conservation of Energy

Energy cannot be created nor destroyed.

The energy of the universe is constant

Energy is only converted between forms.

An exothermic reaction is converting chemical potential energy into
energy

An endothermic reaction converts thermal energy into chemical potential energy.

Spontaneous Processes

A spontaneous process occurs without outside intervention. They may be either fast or
slow.

Under the right conditions, combust
ion occurs spontaneously and quite rapidly. However,
the rusting of iron is also spontaneous but is quite slow.

Whether a reaction is spontaneous is determined by the thermodynamics of the reaction,
which is determined by the energy of the initial and fina
l states.

The rate of the reaction is determined by the reaction pathway including the properties of
activation energy, temperature, concentration, etc.

Examples of Spontaneous processes (The reverse is highly unlikely or impossible)

An apple falls from
a tree

A hot rock cools down in cold water

Heat always flows from a hot object to a colder object

[This is the heat formulation of the second law of thermodynamics]

A candle burns to form carbon dioxide and water

Steel in contact with air and moisture slo
wly rusts

A gas fills a container completely and uniformly

Water put into a
-
5

C environment freezes

A chemical reaction moving toward equilibrium

Spontaneous processes have two components, flow of heat, and flow of entropy.

Entropy (S)

Many events occur
in one direction only. This is a common property is that is shared
with that of increasing entropy.

Entropy is a measure of the number of arrangements (states) available to a system. The
more states there are, the less predictable the system is.

Entropy i
s commonly viewed as a measure of randomness or disorder

(the more disorder, the less predictable it is).

The more ways a particular state can be achieved, the higher it’s probability of occurring,
and the higher it’s entropy.

Nature spontaneously moves
toward the states that have the highest probabilities of
existing.

Entropy is the movement toward the highest probability state.

A driving force for a spontaneous process in an increase in entropy of the universe.

Positional Probability/Entropy

Depends on the configuration in space that yield a particular state.

Gases have more positional entropy than liquids

Liquids have more positional

entropy than solids

Solutions have more positional entropy than pure compounds

Materials have more positional entropy as temperature increases, and as volume
increases.

Second Law of Thermodynamics

In any nonreversible process

there is always an increase in the entropy of the
universe

The entropy of the universe is increasing (in time)

A reversible process results in no overall change in the entropy of the universe (an
idealization)

System/Surroundings

When the entropy of a sy
stem decreases (ie making crystal from rock) the entropy of the
universe still increases. So the entropy of the universe is the sum of the entropy of the
system and surroundings.

The system is that portion of the universe that we are examining, the surroun
dings is the
rest of the universe.

∆S
univ

= ∆S
sys

+ ∆S
surr

If ∆S
univ

is positive, then the process is spontaneous

If ∆S
univ

is zero, then the process has no tendency to occur, the system is stable (at
equilibrium)

If ∆S
univ

is negative then the process is spontaneous in the opposite directi
on (or in
negative time)

∆S
sys

can be negative (such as the organizing effect of cells, or the organizing effect of
modern technology) as long as ∆S
surr

is a larger positive resulting in a positive ∆S
univ
.

The Effect of Temperature on Spontaneity

The entro
py changes in the surroundings are determined primarily by the flow of energy
into or out of the system as heat.

In an
exothermic
process, heat is transferred to the surroundings where it becomes
thermal energy. It increases the random motions of the atom
s in the surrounding and
thereby increases entropy.

∆S
surr

is positive.

In an
endothermic

process, heat flows from the surroundings to the system. This slows
down the random motions of atoms in the surroundings and thereby decreases entropy.

∆S
surr

is n
egative.

The significance of exothermicity as a driving force depends on the temperature at which
the process occurs.

The sign of ∆S
surr

depends on the direction of the heat flow

The driving force of entropy is often described in terms of energy.
Nature te
nds to seek
the lowest possible energy.

The magnitude of ∆S
surr

depends on the temperature

The transfer of a given amount of energy as heat produces a much greater percent change
in the randomness of the surroundings at a low temperature than it does at a
high
temperature.

Thus ∆S
surr

depends directly on the amount of heat transferred and inversely on
temperature.

Exothermic Process:

∆S
surr

= +heat

(J)/Temperature (K)

Endothermic Process:

∆S
surr

=
-
heat (J)/Temperature (K)

Enthalpy (

H) is heat flow at constant pressure. We can express ∆S
surr

in terms of
enthalpy.

∆S
surr

=
-

䠯H

This definition works at constant pressure and temperature. The minus s
ign is to change
the direction of heat flow since,

H is defined with respect to the system, and now we are
looking at the surroundings.

∆S
sys

∆S
surr

∆S
univ

Spontaneous?

+

+

+

Yes

-

-

-

No

+

-

Varies

If ∆S
sys

is larger than ∆S
surr

-

+

Varies

If ∆S
surr

is

larger than ∆S
sys

Determining ∆S
surr

Example

Carbon is used to reduce oxide ores of antimony according to the reaction:

Sb
4
O
6

(s) + 6C(s)

4Sb(s) + 6CO(g)

H = 778 kJ

Calculate ∆S
surr

at 25C and 1 atm.

∆S
surr

=
-
(778 kJ)/298 K =
-
2.61 kJ/K =
-
2.61 x 10
3

J/K

Practice

Iron is used to reduce sulfide ores of antimony according to the reaction:

Sb
2
S
3
(s) + 3Fe(s)

2Sb(s) + 3FeS(s)

H =
-
125 kJ

Calculate ∆S
surr

at 25C and 1 atm.

Microstates

Molecular motion con
sists of translation motion (motion through space); vibrational
motion (atoms move independently creating flexing, stretching, bending motions); and
rotational motion (molecule rotates in space).

If we could take a snapshot of a system consisting of the po
sitions and motions of all the
molecules, that set would be called a microstate of the system. The connection between
entropy and the number of microstates, W, of a system is expressed by the Boltzmann's
Equation:

S = k lnW

Where k is the Boltzmann's const
ant, 1.38 x 10
-
23

J/K

The entropy change during a process is:

∆S = k lnW
final

k lnW
initial

= k ln(W
final
/W
initial
)

Entropy increases with the number of microstates of the system.

Free Energy

Another property besides entropy can also be used to predict sp
ontaneity and is
particularly useful in dealing with the temperature dependence of spontaneity. This is
called free energy (G).

Free energy is defined by the relationship

G = H
-

TS

Where H is enthalpy, T is temperature in Kelvin, and S is entropy.

For a p
rocess that occurs at constant temperature, the change in free

energy (∆G) is given
by:

∆G = ∆H
-

T∆S

We can get the relationship to entropy by dividing by T

∆S
univ

= ∆S
surr

+ ∆S
sys

=
-
∆H/T + ∆S =
-
∆G/T (at constant T and P)

∆Suniv =
-
∆G/T

This means that a process is
spontaneous only if ∆G is negative
. (at constan
t T and P)

A process is spontaneous in the direction in which the free energy decreases.

Free Energy and Spontaneity

∆S

∆H

∆G

Spontaneous?

+

-

-

Yes at all temperatures

-

+

+

No, reverse process spontaneous at all T's

+

+

Varies

Spontaneous at high temper
ature

(entropy dominant)

-

-

Varies

Spontaneous at low temperature

(exothermicity dominant)

∆G = negative

spontaneous

∆G = 0

at equilibrium

∆G = positive

Not spontaneous (spontaneous in reverse direction)

We can use free energy to determine regi
ons of spontaneity or positions of equilibrium.

Example

What is the normal boiling point of bromine?

Br
2
(l)

Br
2
(g)

∆H

= 31.0 kJ/mol and ∆S

= 93.0 J/K mol

The normal boiling point is when the gas and liquid states are at equilibrium at 1 atm.

So, ∆G

= 0 = ∆H

-

T∆S

Which gives ∆H

= T∆S

Solve for T = ∆H

/∆S

= 3.10 x 10
4

J/mol
/

93.0 J/K mol = 333 K = 60

C

Notice in, ∆G

= ∆H

-

T∆S

that a larger T will produce a negative free energy and the
vaporization process is spontaneous. With a smaller T, t
he free energy will be positive
and the process will not be spontaneous. (Condensation will be spontaneous)

Note:
The superscript degree symbol means that the substances are in their standard
states.

Standard states
are:

For an element:

The standard state
is the form the element exists in under 1 atm pressure at 25

C.

For a compound:

The standard state of a gas is 1 atm.

The standard state of a solid or liquid is the pure solid or liquid

The standard state of a substance in solution is a concentration of 1
M

Practice

For Hg, ∆H

vap

= 58.51 kJ/mol and ∆S

vap

= 92.92 J/K mol;

What is the normal boiling point of mercury?

Standard free en
ergy change
-

∆G
o

Standard free energy change is the free energy change that occurs for reactants to
products when both are in their standard states (298K, 1 atm).

G
o
is not related at all to rate.

G
o

is related to the equilibrium position

When a reacti
on is reversed, the sign of ∆
G
o

is reversed.

If ∆
G

is negative, the reaction is spontaneous in the forward direction.

If

G

is zero, the reaction is at equilibrium

If

G

is positive, the reaction is spontaneous in the reverse direction

In any spontaneou
s process at constant temperature and pressure, the free energy always
decreases.

Free Energy and Chemical Reactions

Entropy Changes in Chemical Reactions

The entropy changes in the system (reactants and products) is determined by positional
probability.

The change in positional entropy in a reaction is dominated by the relative # of molecules
of gaseous reactants and products.

An increase in molecules/moles increases entropy

A decrease in molecules/moles decreases entropy

Predicting the Sign of ∆S
o

Example:

Predict the sign of ∆S

for the thermal decomposition of solid calcium carbonate
(limestone)

CaCO
3
(s)

CaO(s) + CO
2
(g)

∆S

is positive since a gas is produced from a solid

Practice

Predict the sign of ∆S

for the oxidat
ion of SO
2

in air:

2SO
2
(g) + O
2
(g)

2SO
3
(g)

Third Law of Thermodynamics
-

The entropy of perfect crystal at 0K is zero

This statement allows absolute entropy values to be assigned to compounds.

Enth
alpy and Free Energy usually cannot be assigned an absolute value but are known as
values of change.

As temperature increases, vibration increases, so entropy increases with temperature

The more complex a molecule, the greater the standard entropy value. M
ore complex
molecules have more modes of motion (stretching, vibrating, rotating) which corresponds
to more entropy.

Calculating ∆S
o

Entropy is a state function
-

i.e. it is dependent on the state only and is not dependent on
pathway between states.

The entropy of a reaction is calculated by summing the entropy of the products and
subtracting the sum of the entropy of th
e reactants:

S
o
reaction

= ∑n
p
S
o
p

-

n
r
S
o
r

Where ∑ = sum over species and n = number of moles

Example

Calculate ∆S

at 25

C for the reaction:

2NiS(s) + 3O
2
(g)

2SO
2
(g) + 2NiO(s)

Given the standard entropy values:

Substance

S

(J/K mol)

SO
2
(g)

248

NiO(s)

38

O
2
(g)

205

NiS(s)

53

Solution

S

= ∑n
p
S
o
p

-

n
r
S
o
r

S

= (2(248) + 2(38))

(2(53) + 3(205)) =
-
149 J/K

Practice

Here

Calculating ∆H
o
, ∆S
o
, ∆G
o

G
o

is not measured directly but is calculated from other values.

G
o

= ∆
H
o

-

T ∆
S
o

G
o

= ∑n
p
G
o
f(p)

-

n
r
G
o
f(r)

That's sum of products minus sum of reactants (the n is from the balanced equation)

or ∆
G
o

is summed from several reactions that add to the reaction of interest.

Standard Free Energy of formation

G
o
f

= standard free energy of formation
-

the change
in free energy in the formation of 1
mole of material from the elements with all the species in their standard states

Defined zero's of enthalpy and free energy: Elements in their standard state (the physical
state they normally are in) at 25

C have a stan
dard free energy of formation, and a
standard enthalpy of formation, equal to zero.

Free energy and Spontaneity

Dependence of Free Energy on Pressure

Free energy depends on both pressure of a gas and/or concentration of solution

Entropy depends on volume o
f a gas and therefore the pressure of a gas

S(large V) > S(small V)

S(low P) > S(high P)

G = G
o

+ RTln(P)

Calculating ∆G
o

The meaning of ∆G
o

for a chemical Reaction

∆G = ∆G
o
reaction

+ RT(S
i
n
i
ln(P
i
)
-

S
j
n
j
ln(P
j
)) where i is product species and j is reactant

species

∆G = ∆G
o

+ RTln(Q) where Q is the reaction quotient

and R = 8.3145 J/K mol

Free Energy and Equilibrium

There is a relationship between the reaction quotient, Q, and free energy, ∆
G

.

When Q < K; ∆
G

is negative (reaction will go forward).

When Q

= K; ∆
G

is zero (reaction is at equilibrium).

When Q > K; ∆G

is positive (reaction will go in reverse).

Most reactions occur under nonstandard conditions. The relationship between standard
free energy and free energy under other conditions is:

∆G = ∆G

+ RTlnQ

At equilibrium:

∆G = 0

And;

∆G

=
-
RTln(Q)

But since Q = K at equilibrium

G

=
-
RTln(K)

or

K = e
-
∆G

/RT

Relationship between ∆G

and the Equilibrium Constant

∆G
o

= 0 then K = 1

∆G
o

< 0 then K > 1

∆G
o

> 0 then K < 1

Equilibrium Point

The equilib
rium point occurs at the lowest value of free energy available to the reaction system
(represented by ∆G = 0)

Temperature Dependence of K

∆G
o

=
-
RTln(K) = ∆
H
o

-

T ∆
S
o

ln(K) =
-

H
o
/R(1/T) + ∆
S
o
/R

which has the form of a straight line (ln(K) vs (1/T)) with
slope
-

H
o
/R

and intercept of ∆
S
o
/R

Free Energy and Work

Free energy of a process equals the maximum useful work that can be done by the system
on its surroundings.

G =
-

max

For a spontaneous process free energy represents the maximum energ
y free to do work

For a nonspontaneous process, free enrgy represents the minimum work required to do
the process.

The work obtained is always less than the maximum amount possible.

The rest is lost as heat.

Gas Expansion

If a gas undergoes a reversible is
othermal expansion, the work done on the surroundings
by the changing container can be calculated by

w
rev

-

-
nRT ln(V
2
/V
1
)

Since an the internal energy of an ideal gas only depends on temperature

∆E = 0 = q
rev

+ w
rev

So q
rev

=
-
w
rev

∆S
sys

= q
rev
/T =
-
w
rev
/T

Or

∆S
sys

= nR ln(V
2
/V
1
)

Reversible/Irreversible

Reversible processes can be returned to the original starting position of the universe
(system and surrounding).

Irreversible pro
cesses end up changing the universe when the system is returned to the
original starting position.

All real processes are irreversible

After any real cyclic process, the heat and entropy of the universe has increased. (2nd law
of thermodynamics)