Spontaneous processes
Chapter 19
We have a general idea of what we
consider spontaneous to mean:
Spontaneous Change: Entropy and Free Energy
A spontaneous process WILL OCCUR in a
system WITHOUT any outside action
being performed on the system.
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
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Spontaneous processes
Spontaneous processes
Ice will melt above zero Celcius.
We don’t have to DO anything!
Objects will fall to earth.
We don’t have to DO anything!
Ice melting above zero
Celcius is spontatneous
Object falling to
earth is
spontaneous
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Spontaneous processes
Non-spontaneous processes
We have a general idea of what we
consider non-spontaneous to mean:
Since we DON’T have to DO
anything for these spontaneous
processes to occur it APPEARS
that an overall energy change
from potential energy
to kinetic energy
IS SPONTANEOUS
A non-spontaneous process WILL NOT
OCCUR in a system UNTIL an outside
action is performed on the system.
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Non-spontaneous processes
Ice freezing above zero
Celcius is
non-spontatneous
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Non-spontaneous processes
We can make water freeze
above zero Celcius by
increasing the pressure.
We can make an object rise
from the earth by picking it
up.
Object rising from
earth is
non-spontaneous
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Non-spontaneous processes
Chemistry and spontaneity
We know there are chemical processes
that are spontaneous because we can put
the chemical system together and
reactants become products without us
having to do anything.
Since we DO have to ACT for
these non-spontaneous
processes to occur it APPEARS
that an overall energy change
from kinetic energy
to potential energy
IS NON-SPONTANEOUS
H3O+ (aq) + OH- (aq) → 2 H2O (l)
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Chemistry and non-spontaneity
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Spontaneous vs. non-spontaneous
It is obvious by the examples we’ve looked at
that the opposite of every spontaneous process
is a non-spontaneous process.
We know there are chemical processes
that are non-spontaneous because we
can put the chemical system together and
reactants DO NOT become products.
In chemical systems we’ve seen that if we put a
chemical system together a reaction occurs
until the system reaches equilibrium.
The system we put together stays like it is
UNTIL WE CHANGE SOMETHING!
Whether the forward reaction or the reverse
reaction dominates depends on which of the
two reactions is spontaneous at those
conditions!
2 H2O (l) → 2 H2 (g) + O2 (g)
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Spontaneity and energy
In our examples it
APPEARED that the
spontaneous process
ALWAYS takes a system to a
lower potential energy.
Equilibrium and
spontaneity are
related!
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Spontaneity and energy
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Recall the First Law
The First Law of thermodynamics stated
that the energy of an ISOLATED system is
constant.
What’s the largest ISOLATED system we
can think of?
If this were true all exothermic processes
would be spontaneous and all
endothermic processes would be nonspontaneous.
THIS ISN’T TRUE!
NH4NO3 (s) → NH4+ (aq) + NO3- (aq)
is spontaneous even though
It’s the UNIVERSE!
The energy of the universe is
constant!
H2O
∆H = +25.7 kJ
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4
Recall the First Law
Further proof lower energy isn’t enough
If an ideal gas expands into
a vacuum at a constant
temperature, then
On the universal scale, there is no
overall change in energy, and so lower
energy CANNOT be the only requirement
for spontaneity.
no work is done
There must be something
else as well!
and
no heat is
transferred
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Further proof lower energy isn’t enough
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Further proof lower energy isn’t enough
No work done and no
heat transferred means
This
spontaneous
process has
no overall
change in
energy!
NO OVERALL
CHANGE in
energy
of the system
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5
Entropy
Entropy
Entropy (from Greek, meaning
“in transformation”)
is a thermodynamic property
that relates
the distribution of the total energy
of the system
to the available energy levels of the
particles.
A general way to envision entropy is
“differing ways to move”
Consider mountain climbers on a mountain.
Two factors affect the distribution of
mountain climbers on a mountain:
Total energy of all the climbers
and how many places can you stop
on the mountain
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Consider hungry mountain climbers
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Consider well fed mountain climbers
Hungry mountain
climbers have
little total energy
amongst themselves to
climb a mountain, so
most of them are near the
bottom, while some are
distributed on the lower
parts of the mountain.
Well-fed mountain
climbers have
more total energy
amongst themselves to
climb a mountain, so
the climbers will be
more spread out on
the whole mountain.
Few energy levels
can be reached!
More energy levels
can be reached!
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Temperature and total energy
Higher mountain means more places to stop
The total energy shared by molecules
is related to the temperature.
A given number of molecules at a low
temperature (less total energy) have
less “differing ways to move” than
the same number of molecules at a
high temperature (more total energy).
If a well-fed mountain climber tries to climb
Signal Hill, they will most likely reach the
top. They have only a few places to
stop (levels) because Signal Hill is a small
mountain.
The same well-fed mountain climber on
Mount Everest has a greater number of
places to stop (levels) because it is a
larger mountain.
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Volume and energy levels
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Entropy
The number of levels of energy
distribution of molecules is related to
the volume.
A given number of molecules in a
small volume have less “differing
ways to move” than the same
number of molecules in a larger
volume.
The greater the number of “differing
ways to move” molecules can take
amongst the
available energy levels of a system
of a given state (defined by temperature
and volume, and number of molecules),
the greater the entropy of
the system.
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Expansion into vacuum
Expansion into vacuum
A gas expands into a
vacuum because the
increased volume
allows for a greater
number of “differing
ways to move” for
the molecules, even
if the temperature is
the same.
That is, the entropy
increases when the gas
is allowed to expand into
a vacuum.
Entropy increase
plays a role in
spontaneity!
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Entropy is a state function
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Boltzmann equation and entropy
Entropy, S, is a state function like
enthalpy or internal energy.
The entropy of a system DEPENDS
ONLY on the current state
(n, T, V, etc.) of the system, and
NOT how the system GOT TO BE
in that state.
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More available energy levels
when the size of a box increases –
like expanding a gas into a vacuum
– ENTROPY INCREASES!
More energy levels are accessible when the
temperature increases – ENTROPY INCREASES!
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8
Hess’s Law
Change in entropy is a state function
Recall Hess’s Law – as long as we get
from the same initial state to the same
final state then ∆H will be the same
regardless of the steps we add together.
Because entropy is a state function,
then change in entropy ∆S is ALSO
a state function.
The difference in entropy between
two states ONLY depends on the
entropy of the initial and final states,
and NOT the path taken to get there.
Change in entropy ∆S will work exactly
the same way! As long as we get from the
same initial state to the same final state
then ∆S will be the same regardless of
the steps we add together.
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Boltzmann equation and entropy
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Playing cards
Say we have a deck of 52 playing cards.
Choosing one playing card is a state.
n, T, V help define the number of states
(number of available energy levels) the system
can have.
The many “different ways to move” of
molecules in a particular state are called
microstates.
Hopefully it makes sense that more total states
should automatically mean more total
microstates.
If we choose the first card out of the pack,
there are 52 microstates for this first state.
The second card (second state) we choose
has 51 microstates, and so on.
The number of microstates is often
symbolized by W.
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Playing cards
Playing cards and coin flips
If we flip 52 coins (a coin is one
state), with two possible microstates
(heads or tails) each, there are
Overall there are
W = 52! ≈ 8 x 1067
possible distributions
(total microstates) for
52 playing cards!
W = 252 = 4.5 x 1015
possible distributions.
(total microstates)
A deck of 52 playing cards has
greater entropy than 52 coins!
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Boltzmann equation and entropy
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Boltzmann equation and entropy
S = k ln W
where k = R / NA
Ludwig Boltzmann formulated the
relationship between the number of
microstates (W) and the entropy (S).
S = k ln W
The constant k is the Boltzmann
constant which has a value equal to the
gas constant R divided by Avagadro’s
number NA
k = (8.3145 J⋅K-1⋅mol-1) / (6.022 x 1023 mol-1)
k = 1.381 x 10-23 J⋅K-1
We can see the units for entropy
will be Joules per Kelvin (J⋅K-1)
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Measuring entropy change
Measuring entropy change
From the
units for entropy (J⋅K-1)
we get an idea of how we might
measure entropy change ∆S
∆S = qrev / T
The change in entropy is
the heat involved in a
reversible process at a
constant temperature.
It must involve some sort of
energy change relative to the
temperature change!
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Heat IS NOT a state function
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Reversible processes
Since heat IS NOT a
state function we need
a reversible process
to make it ACT LIKE a
state function.
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In a reversible process a change in one
direction is exactly equal and opposite
to the change we see if we do the
change in the reverse direction.
In reality it is impossible to make a
reversible process without making an
infinite number of infinitesimally small
changes.
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Reversible processes
Endothermic increases in entropy
We can however imagine
the process is done
reversibly and calculate
the heat involved in it, so
we can calculate the
reversible entropy
change that could be
involved in a process.
In these three
processes the
molecules gain greater
“differing ability to
move.”
The molecules occupy
more available
microstates at the given
temperature, and so the
entropy increases in
all three processes!
∆Srev = qrev / T
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Problem
Generally entropy increases when…
…we go from solid to liquid.
…we go from solid or liquid to gas.
…we increase the amount of gas in a
reaction.
…we increase the temperature.
…we allow gas to expand against a
vacuum.
…we mix gases, liquids, or otherwise
make solutions of most types.
Predict whether entropy increases,
decreases, or we’re uncertain for the
following processes or reactions:
a) 2 H 2 S (g) + SO 2 (g) ⎯
⎯→ 3 S (s) + 2 H 2 O (g)
b) 2 HgO (s) ⎯
⎯→ 2 Hg (l) + O 2 (g)
c) Zn (s) + 2 Ag 2 O (s) ⎯
⎯→ ZnO (s) + 2 Ag (s)
is
d) 2 Cl - (aq) + 2 H 2 O (l) ⎯electrolys
⎯ ⎯⎯
→ 2 OH - (aq) + H 2 (g) + Cl 2 (g)
Answers: a) decreases b) increases
c) uncertain d) increases
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Evaluating entropy and entropy changes
Evaluating entropy and entropy changes
Phase transitions – In phase transitions the
heat change does occur reversibly, so we can
use the formula
Phase transitions – For water going from ice
(solid) to liquid, ∆H°fus = 6.02 kJ⋅mol-1 at the
melting point (transition temp.) of 273.15 K (0
°C)
∆Srev = qrev / T
∆S°fus = ∆H°fus / Tmp
∆S°fus = 6.02 kJ⋅mol-1 / 273.15 K
∆S°fus = 22.0 J⋅K-1⋅mol-1
to calculate the entropy change. In this case the
heat is the enthalpy of the phase transition and
the temperature is the transition temperature
∆S = ∆Htr / Ttr
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Problem
Evaluating entropy and entropy changes
Phase transitions – For water going from liquid
to gas, ∆H°vap = 40.7 kJ⋅mol-1 at the boiling point
(transition temp.) of 373.15 K (100 °C)
What is the standard molar entropy
of vapourisation ∆S°vap for CCl2F2 if
its boiling point is -29.79 °C and
∆H°vap = 20.2 kJ⋅mol-1?
∆S°vap = ∆H°vap / Tbp
∆S°vap = 40.7 kJ⋅mol-1 / 373.15 K
∆S°vap = 109 J⋅K-1⋅mol-1
Answer: ∆S°vap = 83.0 J⋅K-1⋅mol-1
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Problem
Absolute entropies
The entropy change for the transition from
solid rhombic sulphur to solid monoclinic
sulphur at 95.5 °C is ∆S°tr = 1.09 J⋅K-1⋅mol-1.
What is the standard molar enthalpy change
∆H°tr for this transition?
Answer: ∆H°tr = 402 J⋅mol-1
Say we imagine a system of molecules
that has no total energy. At this zeropoint energy there can ONLY be ONE
possible distribution of microstates, as
no molecule has the energy to occupy a
higher energy level.
The entropy CAN NEVER get smaller than
its value in this situation, so we define
the entropy S of this situation as ZERO.
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Absolute entropies
This kind of imagining is the Third Law of
Thermodynamics which states that
The entropy of a pure perfect
crystal at 0K is zero.
Methyl chloride
entropy as a function
of temperature
At conditions other than at absolute zero, our
entropy is that of the perfect system (zero)
PLUS any entropy changes that come
changing temperature and/or volume.
These are absolute entropies!
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Standard molar entropies
Standard molar entropies
One mole of a substance in its standard
state will have an absolute entropy that we
often call the standard molar entropy S°.
∆S° = [ΣνpS°(products) - ΣνrS°(reactants)]
Hopefully this looks somewhat familiar!
We have seen a special treatment of
Hess’s Law in Chem 1050 where
These are usually tabulated at 298.15 K
In a chemical process we can then use
these standard molar entropies to
calculate the entropy change in the
process.
∆H° = [Σνp∆Hf°(products) - Σνr∆Hf°(reactants)]
We can do something similar with ANY
thermodynamic property that IS A STATE
FUNCTION!
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Standard molar entropies
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Problem
∆S° = [ΣνpS°(products) –
ΣνrS°(reactants)]
∆H° = [Σνp∆Hf°(products) –
Σνr∆Hf°(reactants)]
Enthalpies of formation ARE
NOT absolute!
Use the data given to calculate the
standard molar entropy change for the
synthesis of ammonia from its elements.
N2 (g) + 3 H2 (g) → 2 NH3 (g)
S°298 for N2 = 191.6 J⋅K-1⋅mol-1
S°298 for H2 = 130.7 J⋅K-1⋅mol-1
S°298 for NH3 = 192.5 J⋅K-1⋅mol-1
Answer: -198.7 J⋅K-1 ⋅ mol-1 (per mole of rxn)
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Problem
The second law of thermodynamics
We’ve seen that entropy MUST
play a role in spontaneity,
because the total energy of the
universe doesn’t change.
We could say that an entropy
increase leads to spontaneity,
but we have to be careful.
N2O3 is an unstable oxide that readily
decomposes. The decomposition of 1.00 mol
N2O3 to nitrogen monoxide and nitrogen dioxide
at 25 °C is accompanied by the entropy change
∆S° = 138.5 J⋅K-1⋅mol-1. What is the standard
molar entropy of N2O3 (g) at 25 °C?
S°298 for NO (g) = 210.8 J⋅K-1⋅mol-1
S°298 for NO2 (g) = 240.1 J⋅K-1⋅mol-1
Answer: 312.4 J⋅K-1 ⋅mol-1
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The second law of thermodynamics
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The second law of thermodynamics
Ice freezing below 0 Celcius is
spontaneous, but the entropy of the
water decreases in the process!
The water is ONLY the system.
The rest of the universe (the
surroundings) must experience an
opposite heat change as it takes the heat
the freezing water gave off (a +ve ∆H for
the surroundings), which means
the entropy of the REST OF THE
UNIVERSE INCREASES in the process
of water freezing!
∆S°freeze = -∆H°fus / Tmp
Since ∆H°fus and Tmp are +ve,
then ∆S°freeze is –ve!
-ve since “freezing” is the reverse of
“fusion” (like we do in Hess’s Law)
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The second law of thermodynamics
The second law of thermodynamics
The total entropy change in any process
is the entropy change for the system
PLUS
the entropy change for the surroundings
In any spontaneous process
the entropy of the universe
INCREASES.
∆Suniverse = ∆Ssys + ∆Ssurr >
∆Suniverse = ∆Stotal = ∆Ssys + ∆Ssurr
Now we can connect entropy
and spontaneity!
0
This is the Second Law of
Thermodynamics!
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Water freezing
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Free energy
So while water freezing below zero
Celcius decreases the entropy of the
system, the heat given off to the
surroundings increases the entropy of
the surroundings to a greater extent.
We’ve seen entropy
increases when molecules
have more ways to
distribute themselves
amongst the
energy levels.
The total entropy change of the
universe is positive and the process of
water freezing below 0 Celcius is
spontaneous!
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Free energy
Free energy
∆Suniverse = ∆Ssys + ∆Ssurr
However, some of the energy a molecule
uses to put itself at a higher energy level
CAN NO LONGER be used to to do work
because doing work would put the
molecule back at a lower energy level,
which would automatically decrease
entropy.
T∆Suniverse = T∆Ssys + T∆Ssurr
T∆Suniverse = T∆Ssys + ∆Hsurr
The energy is NOT free
(or available) to be used!
∆S = ∆Hrev / T
then
Reversible since the rest
of the universe is SO BIG
T∆S = ∆H
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Free energy
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Free energy
T∆Suniverse = T∆Ssys + ∆Hsurr
T∆Suniverse = T∆Ssys - ∆Hsys
T∆Suniverse = T∆Ssys – ∆Hsys
-T∆Suniverse = ∆Hsys - T∆Ssys
∆G = ∆Hsys – T∆Ssys
∆G is the free energy
(Gibbs free energy)
Since ∆Hsys = -∆Hsurr
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Free energy
Free energy
∆G < 0 is spontaneous
∆G = -T∆Suniv
For a spontaneous
process ∆Suniv > 0, which
means for a spontaneous
process ∆G < 0!
∆G > 0 is non-spontaneous
∆G = 0 is at equilibrium
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Problem
Predict the spontaneity at low and high
temperatures for:
N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆H° = -92.22 kJ
2 C (s) + 2 H2 (g) → C2H4 (g) ∆H° = 52.26 kJ
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Standard free energy change ∆G°
Problem
Just like we can have a standard
enthalpy change ∆H° for chemicals,
we can also define the standard free
energy change…
∆G° = ∆H° - T∆S°
N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆H° = -92.22 kJ
Spontaneous @ low T and nonspontaneous @ high T
2 C (s) + 2 H2 (g) → C2H4 (g) ∆H° = 52.26 kJ
Nonspontaneous @ all T
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Problem
Standard free energy of formation ∆Gf°
Standard enthalpies of formation ∆Hf° of
elements in their standard states are zero:
∆H°rxn = [Σνp ∆Hf°(products) –
Σνr ∆Hf°(reactants)]
Standard free energies of formation ∆Gf° of
elements in their standard states are zero:
∆G° = [Σνp ∆Gf°(products) –
Σνr ∆Gf°(reactants)]
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Calculate ∆G° at 298.15 K for the reaction
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
by using the two following sets of data.
Compare your answers.
a) ∆H° = -1648 kJ ⋅mol-1
and ∆S° = -549.3 J⋅K-1 ⋅mol-1
b) ∆Gf° (Fe) = 0 kJ⋅mol-1
∆Gf° (O2) = 0 kJ⋅mol-1
∆Gf° (Fe2O3) = -742.2 kJ⋅mol-1
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Problem answer
Free energy and equilibrium
a) ∆G° = -1484 kJ⋅mol-1
b) ∆G° = -1484.4 kJ⋅mol-1
The answers are the same
because free energy is a state
function.
We’ve already seen that
∆G < 0 is spontaneous
∆G > 0 is non-spontaneous
One process in a written reaction is
spontaneous while the reverse is not, as
long as ∆G ≠ 0. Therefore
∆G = 0 is at equilibrium
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Water and steam
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Water and steam
H2O (l, 1 atm) ' H2O (g, 1 atm)
∆G°373.15 K = 0 kJ⋅mol-1
The system is at equilibrium
at 1 atm (standard
conditions) and at the
boiling point temperature!
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H2O (l, 1 atm) ' H2O (g, 1 atm)
∆G°298.15 K = 8.590 kJ⋅mol-1
The system is not at
equilibrium at 1 atm
(standard conditions) and
at the room temperature!
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Water and steam
Water and steam
H2O (l, 0.03126 atm) '
H2O (g, 0.03126 atm)
H2O (l, 1 atm) ' H2O (g, 1 atm)
∆G°298.15 K = 8.590 kJ⋅mol-1
∆G298.15 K = 0 kJ
The system is at equilibrium at
0.03126 atm (non-standard
conditions) and at the room
temperature!
The forward process is nonspontaneous (∆G° > 0) so the
reverse process is spontaneous
and condensation occurs.
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Water and steam
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Non-standard conditions
H2O (l, 0.03126 atm) '
H2O (g, 0.03126 atm)
∆G298.15 K = 0 kJ⋅mol-1
Water CAN evaporate at room
temperature,
just not to give an equilibrium
pressure of 1 atm!
As we’ve seen with the previous water example,
our interest in an equilibrium system is
often at non-standard conditions,
so knowing ∆G°
is usually not as useful as
knowing ∆G.
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Non-standard conditions
Non-standard free energy
For an ideal gas ∆H does not change
if pressure changes, so at all nonstandard conditions ∆H = ∆H°.
Because of these facts, the nonstandard free energy change is
For an ideal gas ∆S does change if
pressure changes (expansion into
vacuum shows us this!), so at all
non-standard conditions ∆S ≠ ∆S°.
But the standard free energy change
is
∆G = ∆H° - T∆S
∆G° = ∆H° - T∆S°
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Non-standard free energy
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Boltzmann distribution
The difference between standard and
non-standard free energy is totally
due to the difference in entropy
change between the standard and
non-standard conditions
By the Boltzmann distribution
S = R ln W for one mole of particles
∆G = ∆G° + T(R ln W - R ln W°)
∆G = ∆G° + T(R ln W / W°)
We are comparing a real
system with a standard one.
We are dealing with activities!
∆G - ∆G° = - T(∆S-∆S°)
∆G = ∆G° + T(∆S°-∆S)
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Boltzmann distribution
Boltzmann distribution
∆G = ∆G° + RT ln Qeq
In the comparison, we are looking at a
reaction quotient!
In other words, we are looking at
how our non-standard system is
different from the system we have at
standard conditions!
∆G = ∆G° + RT ln Qeq
If our system
is at equilibrium
then Qeq = Keq
and ∆G = 0
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Boltzmann distribution
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Equilibrium constant
∆G = ∆G° + RT ln Keq = 0
which means
The thermodynamic equilibrium
constant for a reaction is directly
related to the standard free
energy change!
∆G° = -RT ln Keq
∆G° = -RT ln Keq
Keq = e-∆G°/RT
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Equilibrium constant
∆G° ≈ 0
Keq ≈ 1
∆G° >> 0
then Keq is
very small
Equilibrium constant at 298 K
∆G°
∆G° << 0
then Keq is
very large
Keq
Meaning
Keq
= e-∆G°/RT
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Predicting reaction direction
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Predicting reaction direction
∆G < 0 means the forward
reaction is spontaneous at the
given non-standard conditions
∆G = 0 means the system is at
equilibrium at the given nonstandard conditions
∆G° < 0 means the forward
reaction is spontaneous at
standard conditions and so K > 1
∆G° = 0 means the system is at
equilibrium at standard
conditions and so Keq = 1 which
only occurs at one specific T!
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Predicting reaction direction
Predicting reaction direction
∆G = ∆G°
∆G > 0 means the forward
reaction is non-spontaneous at
the given non-standard
conditions
∆G° > 0 means the forward
reaction is non-spontaneous at
standard conditions and so K < 1
ONLY at standard
conditions!
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Thermodynamic equilibrium constant Keq
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Thermodynamic equilibrium constant Keq
The thermodynamic equilibrium constant
Keq is expressed in terms of activities, which
are unitless quantities.
Activities relate properties like concentration or
pressure compared to a standard property
value, like 1 M for concentration or 1 bar for
pressure.
aA+bB'cC+dD
K eq
(a C )c (a D )d
=
(a A )a (a B )b
where ax = [X] / c0 (c0 is a standard concentration of
1 M)
or ax = Px / P0 (P0 is a standard pressure of 1 atm)
Note that ax = 1 for pure solids and liquids
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Problem
Free energy and equilibrium constants
∆G° = -RT ln Keq
IS ALWAYS TRUE
∆G° = -RT ln Kc
and
∆G° = -RT ln Kp
DO NOT have to be true!
Write thermodynamic equilibrium constant
expressions for each of the following
reactions and relate them to Kc and Kp
where appropriate:
a) Si (s) + 2 Cl2 (g) ' SiCl4 (g)
b) Cl2 (g) + H2O (l) ' HOCl (aq) + H+ (aq) + Cl- (aq)
105
Problem answer
Problem
a) Si (s) + 2 Cl2 (g) ' SiCl4 (g)
K eq =
(a
SiCl 4
(a Si )(a Cl
)
⎛ PSiCl 4 ⎞
⎜
⎟
PSiCl 4
P° ⎠
= ⎝
=
= Kp
2
2
PCl 2
⎛ PCl 2 ⎞
(1)⎜ P ° ⎟
⎝
⎠
( )
( )
)
2
2
b) Cl2 (g) + H2O (l) ' HOCl (aq) + H+ (aq) + Cl- (aq)
K eq =
(a HOCl )(a H )(a Cl
+
(a )(a )
Cl 2
H 2O
⎛ [HOCl]
-
) = ⎜⎝
106
+
⎞⎟⎛ [H ] ⎞⎛ [Cl ] ⎞
⎜
⎟⎜
⎟
c ° ⎠⎝
c ° ⎠⎝
c ° ⎠ [HOCl][H + ][Cl - ]
=
PCl 2
⎛ PCl 2 ⎞
⎜
° ⎟(1)
P ⎠
⎝
( )
107
Use the given data to determine if the
following reaction is spontaneous at
standard conditions at 298.15 K:
N2O4 (g) ' 2 NO2 (g)
∆G°f (N2O4) = 97.89 kJ⋅mol-1
∆G°f (NO2) = 51.31 kJ⋅mol-1
Answer: ∆G° = 4.73 kJ⋅mol-1, not
spontaneous at standard conditions.
108
27
Problem
Problem
Determine the equilibrium constant at
298.15 K for the following reaction using
the given data:
AgI (s) ' Ag+ (aq) + I- (aq)
Based on the problem of the previous
slide, determine which direction the
reaction will go in if 0.5 bar of each gas is
placed in an evacuated container:
N2O4 (g) ' 2 NO2 (g)
∆G°f (AgI) = -66.19 kJ⋅mol-1
∆G°f (Ag+) = 77.11 kJ⋅mol-1
∆G°f (I-) = -51.57 kJ⋅mol-1
Answer: Since ∆G° = 4.73 kJ⋅mol-1, then
Keq = 0.15 = Kp. Since Qp = 0.5 the
reaction should go from right to left.
109
110
∆G° and Keq are functions of T
Problem answer
Since ∆G° = 91.73 kJ⋅mol-1,
then Keq = 8.5 x 10-17
= Kc = Ksp.
If we compare to the Ksp value
in Table 18.1 (8.5 x 10-17) we
see we are definitely in the
right ballpark!
111
We’ve already seen that equilibrium
constants change with temperature.
Why?
Keq = e-∆G°/RT
and
∆G° = ∆H° - T∆S°
112
28
∆G° and Keq are functions of T
Problem
∆G° = -RT ln Keq = ∆H° - T∆S°
ln Keq= -(∆H°/RT) + (∆S°/R)
At what temperature will the following
reaction have Kp = 1.50 x 102?
2 NO (g) + O2 (g) ' 2 NO2 (g)
∆H° = -114.1 kJ⋅mol-1
and ∆S° = -146.5 J ⋅K-1⋅mol-1
Answer: T = 606 K
113
∆G° and Keq are functions of T
114
∆G° and Keq are functions of T
∆G° = -RT ln Keq = ∆H° - T∆S°
ln Keq= -(∆H°/RT) + (∆S°/R)
If ∆H and ∆S are constant
over a temperature range
then
115
ln K1= -(∆H°/RT1) + (∆S°/R)
minus
ln K2= -(∆H°/RT2) + (∆S°/R)
ln K1 - ln K2 =
-(∆H°/RT1) - (-∆H°/RT2)
116
29
∆G° and Keq are functions of T
van’t Hoff equation
ln K1 - ln K2 =
[-∆H°/R] [(1/T1) - (1/T2)]
OR
ln [K1/K2] =
[-∆H°/R] [(1/T1) - (1/T2)]
The van’t Hoff equation relates
equilibrium constants to
temperatures
ln [K1/K2] = [-∆H°/R] [(1/T1) - (1/T2)]
It looks very similar to the Arrhenius
equation and so a plot of ln K
versus 1/T should give a straight line
with a slope of [-∆H°/R]
117
118
Problem
Plot of ln Kp
versus 1/T for the
reaction.
2 SO2 (g) + O2 (g) '2 SO3 (g)
slope = -∆H° / R,
so for this
reaction
The following equilibrium constant data
have been determined for the reaction
H2 (g) + I2 (g) ' 2 HI (g)
Kp = 50.0 @ 448 °C
Kp = 66.9 @ 350 °C
Estimate ∆H° for the reaction.
Answer: ∆H° = -11.1 kJ⋅mol-1.
∆H° = -180
kJ⋅mol-1
119
120
30
Coupled reactions
Coupled reactions
This should make sense because if we
have two equilibria in one container, one
with a small K (like Ksp) and one with a
large K (like Kf for complex formation),
then the coupled reaction that is the sum
of the two reactions has a K value that is
larger than Ksp!
We have made the non-spontaneous
reaction occur!
If we have a non-spontaneous (∆G > 0)
reaction with a product that appears as
a reactant in a different reaction that is
spontaneous (∆G < 0) then we can
couple the two reactions (do them in
the same container at the same time) to
drive the non-spontaneous reaction!
121
122
31