Entropyx

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Oct 27, 2013 (3 years and 11 months ago)

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Entropy


Explanation and theory

What is Entropy?


Entropy is the measure of the disorder of
a system.



Ex, When a liquid turns into gas by heating
it, the intermolecular forces decrease and
the
speed of the particles increase
. The
particles become more
disordered
.

Numerical Values for Entropy


Entropy is given the symbol
S


∆S (Entropy change) is as important as
∆H (Enthalpy change)



∆S is defined in a system as:

∆S = q / T

q: Heat added to the
system from the
surrounding

T: Thermodynamic temperature
(in Kelvin) at which heat is
transferred (from the surrounding
to the system)


Ex, heating a system that is under going a
change of matter (
e.g
, ice melting to
liquid).



This liquid, it’s heat does not increase, it
only makes the matter (water molecules)
in the system more disordered.

Entropy changes in system and
surroundings


Types of entropy changes:


∆S (System)


∆S (Surrounding)



In terms of entropy changes ∆S, the
significance of the heat released by an
exothermic reaction into the surrounding
depends on how hot the surroundings
already are.


The size of entropy change is
proportional to the temperature.



∆S (surr) = q/T




=
-

∆H/T



∆S (tot) = ∆S (sys)+ ∆S (surr)




∆S (tot) = ∆S (sys)


(∆H/T)



∆S (tot) = ∆S


(∆H/T)

Becomes

By conversion

Laws of Thermodynamics


First law of Thermodynamics states:

Energy cannot be created or destroyed.



Second law of Thermodynamics states:

The total entropy of the universe [∆S (tot)]
tends to increase when any change takes
place, it never goes down.



As a result to the second law of
Thermodynamics, ∆S (tot) for any change
is always greater than (or equal to) zero.
A negative entropy change isn’t possible.


For a spontaneous change to occur:



∆S (tot) > 0


So substituting the equation above gives:


∆S


(∆H/T) > 0


If we multiply through by T, the criterion for
spontaneous change becomes:


T∆S
-

∆H > 0

For a spontaneous reaction

Predicting entropy changes

Change

∆S

Solid



Liquid

Increase (+)

Solid


Gas

Increase (+)

Liquid


Gas

Increase

(+)

Liquid


Solid

Decrease (
-
)

Gas


Solid

Decrease (
-
)

Gas



Liquid

Decrease

(
-
)

When predicting entropy changes, the change due to a
change in the number of particles in the gaseous state
is usually greater than any other possible factor.

What is a spontaneous process?




Processes
that proceed in a definite direction
when left to themselves and in the absence of
any attempt to drive them in reverse


are
known as
natural processes
or
spontaneous

changes.


I
s a process in which matter moves
from a less
stable state to a more stable
state.

Spontaneity: Examples


Drop

a

teabag

into

a

pot

of

hot

water,

and

you

will

see

the

tea

diffuse

into

the

water

until

it

is

uniformly

distributed

throughout

the

water
.

What

you

will

never

see

is

the

reverse

of

this

process,

in

which

the

tea

would

be

sucked

up

and

re
-
absorbed

by

the

teabag
.

The

making

of

tea,

like

all

changes

that

take

place

in

the

world,

possesses

a

“natural”

direction
.


A closer look at disorder: microstates
and
macrostates




How

can

we

express

disorder

quantitatively?

From

the

example

of

the

coins,

you

can

probably

see

that

simple

statistics

plays

a

role
:

the

probability

of

obtaining

three

heads

and

seven

tails

after

tossing

ten

coins

is

just

the

ratio

of

the

number

of

ways

that

ten

different

coins

can

be

arranged

in

this

way,

to

the

number

of

all

possible

arrangements

of

ten

coins
.











The
greater the number of microstates that correspond to a given
macrostate
, the
greater the probability of that
macrostate
.



All

of

the

changes

described

above

take

place

spontaneously
,

meaning

that
:


Once

they

are

allowed

to

commence,

they

will

proceed

to

the

finish

without

any

outside

intervention
.



It

would

be

inconceivable

that

any

of

these

changes

could

occur

in

the

reverse

direction

(that

is,

be

undone)

without

changing

the

conditions

or

actively

disturbing

the

system

in

some

way
.






Disorder
is more probable than
order because there are so many
more ways of achieving it.

What determines the direction in
which spontaneous change will occur?


It

is

clearly

not

a

fall

in

the

energy
,

since

in

many

cases

the

energy

of

the

system

does

not

change
.

I
f

there

is

no

net

loss

of

energy

when

these

processes

operate

in

the

forward

or

natural

direction,

it

would

not

require

any

expenditure

of

energy

for

them

to

operate

in

reverse
.

In

other

words,

contrary

to

the

common

sense,



The direction of a spontaneous
process is not governed by the
energy change...



...
and thus the First Law of Thermodynamics cannot
predict the direction of a natural process.


EXAMPLE


NA OVERVIEW


WORD DOCUMENT








𝑆
(
𝑢 𝑢 𝑑𝑖 𝑔
)



1
/T



𝑆
(
𝑢 𝑢 𝑑𝑖 𝑔
)



-

H(system)


𝑆
𝑢 𝑢 𝑑𝑖 𝑔
=
-

H(system)
/T


Defining

Standard
Gibbs

Energy
Change


THE EFFECT OF TEMPERATURE ON THE
SPONTANEOUS REACTIONS FOR DIFFERENT
REACTIONS IS SUMMARIZED IN THE TABLE BELLOW


http://www.chem.ufl.edu/~itl/2045/lecture
s/lec_u.html


http://www.chem1.com/acad/webtext/ther
meq/TE1.html#PageTop

Calculus



EX: Calculate ∆
G
reaction

for the reaction 2Al

(s) + fe
2
O
3

(s)


2fe(s)
+ Al

2
O
3
(s)


from the following data.


Compound ∆Gf/kJmol
-
1



fe2O3 (s)
-
742


Al 2O3(s)
-
1582


Comment on the significance of the value obtained


Solution



2Al

(s) + fe
2
O
3

(s)


2fe(s) + Al

2
O
3
(s)


2 (0)
-
742 2 (0)
-
1582



G
reaction

=∑∆
G
f

(products)
-

∑∆
G
f

(reactant)


=
-
1582


(
-

742) =
-
840 KJ mol
-
1


The reaction is spontaneous under stander conditions
.

Using ∆
S
reaction

and ∆H

reaction

values to
calculate ∆G
reaction

at all temperatures



Stander value = 298 K


The expression = ∆
G
reaction

= ∆
H
sys

-

T∆S
sys

Worked example


Calculate ∆
G
reaction

at

298 K for the thermal decomposition of
calcium carbonate from the following data


Compound ∆
H
f

/KJ mol
-
1

S/
Jk

-
1
mol

-
1


CaCo3

(s)
-
1207 92.9


Cao

(s)
-
635 39.7


Co2

(g)
-
394 214

Solution


First, calculate ∆
H
reaction

CaCo3 (s)
→ Cao (s) + Co2 (g)

-
1207
-
635
-
394


H
reaction

=
∑∆
H
f

(products)
-

∑∆
H
f

(reactants)

= (
-
635 +
-
394)


(
-
1207) = +178 KJ mol
-
1

Second, calculate ∆
S
reaction

=
∑S (products)
-

∑S

(reactant)

= (39.7 + 214)


92.9 = 160.8 KJ
-
1

mol
-
1


Now calculate the change in Gibbs’ free
energy of the reaction.



G
reaction

=

H
reaction

-

T ∆
S
reaction



=

-

178


(298) ( 160.8 * 10
-
3
)


= + 130 KJ mol
-
1