Transcript Entropy

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)
Becomes
By conversion

∆S (tot) = ∆S – (∆H/T)
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
For a spontaneous reaction
If we multiply through by T, the criterion for
spontaneous change becomes:
T∆S - ∆H > 0
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.
 Is 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. If 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(𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠) ∝ 1/T
 ∆𝑆(𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠) ∝ -∆H(system)
 𝑆 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 =-∆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 ∆Greaction for the reaction 2Al (s) + fe2O3 (s) → 2fe(s)
+ Al 2O3(s)

from the following data.

Compound

fe2O3 (s)
-742

Al 2O3(s)
-1582
∆Gf/kJmol-1
 Comment on the significance of the value obtained
Solution

2Al (s) + fe2O3 (s) → 2fe(s) + Al 2O3(s)

2 (0)

∆Greaction =∑∆Gf (products) - ∑∆Gf (reactant)

= -1582 – (- 742) = -840 KJ mol-1
-742
2 (0)
-1582
 The reaction is spontaneous under stander conditions.
Using ∆Sreaction and ∆H reaction values to
calculate ∆G reaction at all temperatures

Stander value = 298 K

The expression = ∆Greaction = ∆Hsys - T∆Ssys
Worked example

Calculate ∆Greaction at 298 K for the thermal decomposition of
calcium carbonate from the following data

Compound ∆Hf /KJ mol -1
 CaCo3
(s)
 Cao
(s)
 Co2
(g)
S/Jk -1 mol -1
-1207
92.9
-635
39.7
-394
214
Solution
First, calculate ∆Hreaction
CaCo3 (s) → Cao (s) + Co2 (g)
-1207
-635
-394
∆Hreaction = ∑∆Hf (products) - ∑∆Hf (reactants)
= (-635 + -394) – (-1207) = +178 KJ mol -1
Second, calculate ∆Sreaction = ∑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.

∆Greaction = ∆Hreaction - T ∆Sreaction
=

- 178 – (298) ( 160.8 * 10-3 )
= + 130 KJ mol-1