Entropy | AP Chemistry
Entropy
• Entropy is nature’s tendency to
favor disorder
o i.e. an ice cube melts in a
warm room because the
haphazard molecular
arrangement of a liquid is
more random than the
structured arrangement of
a solid
• Entropy is merely a
manifestation of probability
o Molecules in a beaker can
assume a practically
infinite number of
possible arrangements
o It’s most probable that
they’ll disperse randomly
o Not quite. To freeze the
water, the freezer had to
perform work, and in so
doing, it threw off heat
o The heat induces an
entropy increase that’s
greater in magnitude than
the entropy loss inside the
freeze
Net Entropy during Freezing
When water is frozen to ice in a freezer, the
entropy inside the freezer decreases. However,
the work performed by the freezer causes an
entropy increase that supersedes the entropy
decrease inside the freezer. The overall process
yields a net increase in entropy.
(a)
(b)
Entropy as a Manifestation of Probability
It’s much more probable that molecules in a
beaker will assume (a) a haphazard arrangement
than (b) a structured lattice. So we say that
nature “favors” randomness.
•
•
All processes increase the
entropy of the universe
o That is, all processes
increase the universe’s
randomness
But what about a freezer, which
freezes water into ice?
o When the water freezes,
the arrangement of water
molecules becomes less
random
Does the entropy
decrease?
System and Surroundings
• We say that the universe consists
of a system and its surroundings:
universe = system + surroundings
• For any process, the change in
entropy, ∆S, of the universe will
equal ∆S of the system and that
of the surroundings:
∆Suniverse = ∆Ssystem + ∆Ssurroundings
• ∆Suniverse will always be positive,
since net entropy always
increases
o ∆Ssystem may decrease, but
only if ∆Ssurroundings
increases with greater
magnitude
As in our freezer
example
1
© 2017 J Co Review, Inc., Accessed by Guest on 06-14-2017
Entropy | AP Chemistry
Thermodynamic Reversibility
• For a thermodynamically
reversible reaction, ∆Suniverse = 0
• When ∆Suniverse = 0, notice that
∆Ssystem = -∆Ssurroundings
•
Forward Reaction
For this truly thermodynamically reversible
reaction, ∆Ssystem and ∆Ssurroundings are equal in
magnitude, but opposite, such that ∆Suniverse = 0.
•
When that reaction is reversed,
∆Suniverse is still equal to 0
o After all, the reverse
reaction would have ∆S <
0, which we know is
impossible
To demonstrate, consider a
reaction in which blue molecules
are converted to red molecules,
and ∆S > 0
Arbitrary Reaction where ∆S > 0
•
By default, the reverse reaction
will have a negative ∆S
Reverse Reaction Where ∆S < 0
Reverse Reaction
When the reaction from the previous illustration
is reversed, , ∆Ssystem and ∆Ssurroundings reverse
their signs, and ∆Suniverse is still equal to 0.
•
•
•
In both the forward and reverse
reactions, notice that ∆Suniverse = 0
o That is, neither reaction is
entropically favored, so
the forward and reverse
reactions are equally
likely to occur
This reaction is entirely
theoretical; no known reactions
actually have ∆Suniverse = 0
o Practically speaking, a
reaction’s ∆S is always
positive, making its
reverse reaction’s ∆S
negative
But if ∆Suniverse > 0 for all known
reactions, how is it that reverse
reactions are possible?
•
•
•
From a thermodynamic
perspective, the reaction will be
irreversible
o A reaction cannot have a
negative ∆S
But in a laboratory, we can
supply heat to the system
Adding heat may increase the
system’s ∆S, making it positive,
and allowing the reaction to
proceed
Adding Heat to the System
Heating the system increases its change in
entropy, thereby giving the reaction a positive
∆S. Now, this reverse reaction will be possible.
2
© 2017 J Co Review, Inc., Accessed by Guest on 06-14-2017
Entropy | AP Chemistry
•
•
So, the reaction is chemically
reversible
o We can force the reaction
to run in either direction
But the reaction isn’t
thermodynamically reversible
o The forward and reverse
reactions aren’t equally
entropically favored
Entropy and Equilibrium
• Entropy is a major driving force
behind reactions, pushing them
towards equilibrium
o At equilibrium, a
system’s entropy is
maximized
• At equilibrium, ∆S will be 0, as
the system’s entropy has been
maximized and won’t continue to
change
•
“random” than the
reactants
Entropy therefore increases
during the forwards reaction,
making ∆S > 0
Entropy and Temperature
• Entropy varies with temperature
o After all, more heat =
more randomness
• So, a reaction’s entropic energy
(∆S) is measured in joules per
Kelvin
Sample Entropy Question
• Suppose you’re given the
following reaction:
Sample Reaction
•
•
•
•
You are asked whether ∆S of the
forwards reaction is positive,
negative, or 0
We know that when randomness
increases, entropy increases
o So, we compare the
reactants and products
The reactants and products both
consist of 2 moles
However, the products contain 2
moles of gas, whereas the
reactants contain 1 mole of gas
and 1 mole of solid
o Gases are much more
entropic than solids, and
so the products are more
3
© 2017 J Co Review, Inc., Accessed by Guest on 06-14-2017