Spontaneity, Entropy, &
Free Energy
Chapter 16
1st Law of Thermodynamics
The first law of thermodynamics is a
statement of the law of conservation of
energy: energy can neither be created nor
destroyed. The energy of the universe is
constant, but the various forms of energy
can be interchanged in physical and
chemical processes.
Spontaneous Processes
and Entropy
Thermodynamics lets us predict whether a
process will occur but gives no information
about the amount of time required for the
process.
A spontaneous process is one that occurs
without outside intervention.
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Kinetics & Thermodynamics
Chemical kinetics focuses on the pathway
between reactants and products--the
kinetics of a reaction depends on
activation energy, temperature,
concentration, and catalysts.
Thermodynamics only considers the initial
and final states.
To describe a reaction fully, both
kinetics and thermodynamics are
necessary.
Domain of
thermodynamics
(the initial and
final states)
16_343
Energy
Domain of kinetics
(the reaction pathway)
Reactants
Products
Reaction progress
The rate of a reaction depends on the pathway from
reactants to products. Thermodynamics tells whether
the reaction is spontaneous and depends on initial
& final states only.
What are some examples of
spontaneous processes?
A process that does occur under a given set of
conditions is considered spontaneous.
– For example, a waterfall flows
downhill, but never up,
spontaneously.
– Heat flows from a warmer object to
a cooler one, but the reverse never
happens spontaneously.
– Iron exposed to water and oxygen
forms rust, but rust does not
spontaneously change back into iron.
Entropy
The driving force for a spontaneous
process is an increase in the entropy of the
universe.
Entropy, S, can be viewed as a measure of
randomness, or disorder.
Nature spontaneously proceeds toward the
states that have the most “spread out
energy”, or the highest probabilities of
existing. In other words, towards an
arrangement where energy can be
contained in the greatest number of ways.
16_349
The expansion of an ideal gas into an evacuated bulb.
As the number of molecules increases, the probability of
finding the molecules only in one bulb becomes VERY small!
Positional Entropy
A gas expands into a vacuum because the
expanded state has the highest positional
probability of states available to the system.
Therefore,
Ssolid < Sliquid << Sgas
Solutions also have high entropy-accounting
for the solubility of many solids into water to
form ions in solution.
Entropy
Which of the following has higher entropy?
a) Solid CO2 or gaseous CO2?
b) N2 gas at 1 atm or N2 gas at 1.0 x 10-2 atm?
Entropy
What is the sign of the entropy change for
the following?
a) Solid sugar is added to water to form a
solution?
S is positive
b) Iodine vapor condenses on a cold
surface to form crystals?
S is negative
Predict the sign of the entropy
change for each:
Solid sugar is added to water to form a solution + ΔS
Iodine vapor condenses on a cold surface to
form crystals
Water forms H2O vapor
+ ΔS
Water freezes
- ΔS
A gas expands
+ ΔS
Student breaks a pyrex beaker
+ ΔS
- ΔS
Temperature and Entropy
Entropy is directly affected by temperature
changes.
Recall that kinetic molecular theory tells us that
matter is made up of particles in motion.
Temperature is a measurement of the kinetic
energy of particles in a system.
Increasing temperature increases particle
movement, increasing the disorder in a
system, increasing its entropy.
Entropy Values
We can make generalizations about a reaction’s entropy;
2KClO3(s)  2KCl(s) + 3O2(g)
two mol solids  two mol solids + 3 mol gases
Entropy appears to increase in this reaction.
CaO(s) + CO2(g)  CaCO3(s)
one mol solid + one mol gas  one mol solid
Entropy appears to decrease in this reaction.
More
factors
that
influence
entropy
Although many
factors influence
the entropy of a
system, there is
usually a
DOMINANT
source.
Entropy for a reaction
We can assign values to the entropy of formation of a
substance, called standard entropy, and calculate a
reaction’s entropy quantitatively. Standard entropy
is determined at 25°C and 1 atm (gas partial Check out
pressure) or 1 M (sol’n concentration).
values in
2KClO3(s)  2KCl(s) + 3O2(g)
143.7 J/mol*K  82.6 J/mol*K + 205.1
Appendix…
what are
values for
elements?
J/mol*K
Cmpds?
Using ΔS = Sproducts – Sreactants, the reaction has a total
entropy change of +493.1 J/mol*K
NOTE that the units are joules/mol K rather than kJ/mol
as for enthalpies. Typically smaller energies.
The Third Law of
Thermodynamics
. . . the entropy of a perfect crystal at 0 K is zero.
Because S is known (= 0) at 0 K,
S values at other temps can be calculated.
See Appendix 4 for values of S0, which is the entropy
value of substances at 1 atm when heated to 298K.
Review: What is the “zero” for enthalpy values?
Practice problems
Calculate the standard entropy changes
for the following reactions at 25۫C:
(a) CaCO3(s) →CaO(s) + CO2(g)
 ΔSorxn
= [So(CaO) + So(CO2)] – So(CaCO3)
 ΔSorxn
= [39.8 J/K.mol + 213.6 J/K.mol] – (92.9 J/K.mol)
 ΔSorxn
= 160.5 J/K.mol
+ΔS= favorable
(b) N2(g) + 3H2(g) → 2NH3(g)
 ΔSorxn
= 2So(NH3) – [So(N2) + 3So(H2)]
 ΔSorxn
= (2)(193) J/K.mol – [192 J/K.mol] + (3)(131 J/K.mol)]
 ΔSorxn
= -199 J/K.mol
-ΔS= unfavorable
More Practice Soreaction
Calculate S at 25 oC for the reaction
2NiS(s) + 3O2(g) ---> 2SO2(g) +2NiO(s)
S = npS(products)  nrS(reactants)
S = [(2 mol SO2)(248 J/Kmol) + (2 mol NiO)(38
J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol
O2)(205 J/Kmol)]
S = 496 J/K + 76 J/K - 106 J/K - 615 J/K
S = -149 J/K # gaseous molecules decreases!
Comparing So values
Substances with a greater freedom of motion (or
number of possible ways to move) have a
greater absolute entropy.
example: I2(g) (So = 261 J/K.mol) and
I2(s) (So = 117 J/K.mol)
example: CH4(g) (So = 186 J/K.mol) and
C2H6(g) (So = 230 J/K.mol)
The Second Law of
Thermodynamics
. . . in any spontaneous process there is
always an increase in the entropy of the
universe.
Suniv > 0
for a spontaneous process.
SUniverse
Suniverse is positive -- reaction is spontaneous.
Suniverse is negative -- reaction is spontaneous
in the reverse direction.
Suniverse = 0 -- reaction is at equilibrium.
How can we predict whether a
chemical reaction will be
spontaneous?
Thermodynamics can help us determine the
direction of a spontaneous reaction, but cannot
tell us about the speed of the process (for that
we need kinetics…).
Is this chemist observing
an exothermic or endoreaction?
What is the sign of ΔH?
an an
thermic
What factors are important in
predicting the spontaneity of a
process?
We know that exothermic reactions
(negative ΔH) are favored since the
system moves to a lower energy state.
However, not ALL exothermic reactions
will occur spontaneously, and
endothermic reactions may also be
spontaneous.
Example: ice will spontaneously melt
(endothermic) at room temperature.
ENTROPY
The other driving force for a
spontaneous process is an increase
in entropy (S) in the universe.
A reaction tends to be spontaneous if
the change in entropy (ΔS) is
positive, and change in enthalpy
(ΔH) is negative.
Note: -ΔH is really increasing entropy in
disguise! Where is entropy increasing?
How can we determine
spontaneity using ΔS and ΔH?
Two tendencies exist in nature:
•
tendency toward higher entropy -- S
•
tendency toward lower energy -- H
If the two processes oppose each other (e.g.
melting ice cube), then the direction is
decided by the Free Energy, G, and
depends upon the temperature.
G -- Free Energy!!!
Gibbs Free Energy
G = H  TS
(allows us to focus
from the standpoint
of the system)
An American physicist and
founder of thermodynamics at Yale! (18391903. 2005 stamp)
A process (at constant T, P) is spontaneous in
to
the direction in which free energy decreases:went
Hopkins and
Gsys means +Suniverse
But, how do exothermic processes increase the
entropy of the universe??
graduated
from Yale at
the age of
19. He was
praised by
Albert
Einstein as
"the greatest
mind in
American
history".
Entropy changes in the surroundings are primarily determined
by the heat flow. An exothermic process in the system
increases the entropy of the surroundings.
G, H, & S
Spontaneous reactions (shift to RIGHT) are
indicated by the following signs:
G = negative
H = negative
S = positive
How are you feeling at this point?
The entropy of your brain must be increasing…
Temperature Dependence
Ho is not temperature dependent. So is
entropy measured at constant 298K.
Go is temperature dependent, and can be
calculated for any constant temperature
using the equation:
G ◦ = H ◦  TS ◦
G ◦ = H ◦  TS ◦
Calculations showing that the melting of ice is temperature
dependent. As temperature increases, the Free Energy
becomes lower (more negative) -the process is spontaneous
above 0oC.
Predicting spontaneity…
ΔH
ΔS
ΔG
-
+
- ALWAYS SPONTANEOUS
+
-
+ NEVER SPONTANEOUS
+
+
- only spontaneous when TΔS is
greater than ΔH (at high temp.)
-
-
- only spontaneous when TΔS is less
than ΔH (at low temp.)
Free Energy G
G ◦ = H ◦  TS ◦
G = negative – spontaneous (RIGHT shift)
G = positive -- spontaneous in opposite
direction (SHIFT left)
G = 0, system is at equilibrium (with
products and reactants in std states)
Summary of 3 ways to
calculate ΔG at std state
•
1. ΔGrxn = ΔHo - TΔSo
•
2. G = npGf(products)  nrGf(reactants)
•
3. Hess’s Law
1. Free Energy Change and
Chemical Reactions
(a) Calculate H, S, & G for the
following reaction at 25⁰C
2 SO2(g) + O2(g) ----> 2 SO3(g)
H = npHf(products)  nrHf(reactants)
H = [(2 mol SO3)(-396 kJ/mol)]-[(2 mol
SO2)(-297 kJ/mol) + (0 kJ/mol)]
H = - 792 kJ + 594 kJ
H = -198 kJ
G Calculations
Continued
S = npS(products)  nrS(reactants)
S = [(2 mol SO3)(257 J/Kmol)]-[(2 mol
SO2)(248 J/Kmol) + (1 mol O2)(205
J/Kmol)]
S = 514 J/K - 496 J/K - 205 J/K
S = -187 J/K
G Calculations
Continued
Go = Ho  TSo
Go = - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J)
Go = - 198 kJ + 55.7 kJ
Go = - 142 kJ
The reaction is spontaneous in forward
direction at 25 oC and 1 atm.
Now try this using the second method…
(b) Find ΔG for the following reaction at constant
pressure (1 atm) and temperature (900˚C):
CaCO3(s) → CaO(s) + CO2(g)
ΔGrxn = ΔHo - TΔSo
 ΔSorxn
= [So(CaO) + So(CO2)] – So(CaCO3)
 ΔSorxn = [39.8
J/K.mol)
 ΔSorxn
J/K.mol + 213.6 J/K.mol] – (92.9
= 160.5 J/K.mol
 ΔHorxn = [ΔHfo(CaO) +
ΔHfo(CaCO3)
 ΔHorxn
ΔHfo(CO2)] –
= [(-635.6 kJ/mol) + (-393.5 kJ/mol)] –
(-1206.9 kJ/mol)
 ΔHorxn
= 177.8 kJ/mol
Continued…
 ΔGrxn
= ΔHo – TΔS
 ΔGrxn = 177.8
kJ/K.mol)
 ΔGrxn
 Try
kJ/mol – 1173K(0.1605
= -10.5 kJ/mol
this: at what Temp (in C) will this
become nonspontaneous?
2. More G Calculations
G = standard free energy change that occurs if
reactants in their standard state are converted
to one mole of product in its standard state.
Values given in textbook appendix. Note that
Gf for elements in their standard states is
zero. (formation)
G = npGf(products)  nrGf(reactants)
The more negative the value of G, the further a
reaction will go to the right to reach equilibrium.
Practice Problem
Calculate the standard free energy changes for the
following reactions at 1 atm and 25˚C.
(a) CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔGorxn = ΔGproducts – ΔG reactants
ΔGorxn = [Gof(CO2) + 2Gof (H2O)] – [Gof(CH4) + 2Gof(O2)]
ΔGorxn = [(-394 kJ/mol) + (2)(-237 kJ/mol)] –
[(-51 kJ/mol) + (2)(0 kJ/mol)]
ΔGorxn = -818 kJ/mol
ΔG is -, SPONTANEOUS
(b) 2MgO(s) → 2Mg(s) + O2(g)
ΔGorxn = ΔGproducts – ΔG reactants
ΔGorxn = [2Gof(Mg) + Gof (O2)] – 2Gof(MgO)
ΔGorxn = [(2)(0 kJ/mol) + (0 kJ/mol)] – (2)(-570 kJ/mol)
ΔGorxn = 1139 kJ/mol
ΔG is +, NOT SPONTANEOUS
3. Hess’s Law & Go
Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ
Cgraphite(s) + O2(g) ---> CO2(g) Go = -394 kJ
Calculate Go for the reaction
Cdiamond(s) ---> Cgraphite(s)
Cdiamond(s) + O2(g) ---> CO2(g) Go = -397 kJ
CO2(g) ---> Cgraphite(s) + O2(g) Go = +394 kJ
Cdiamond(s) ---> Cgraphite(s) Go = -3 kJ
Diamond is kinetically stable, but thermodynamically unstable.
ΔG◦ =0, Boiling Point Calculations
(a) What is the normal boiling point for liquid Br2?
Br2(l) ---> Br2(g)
Ho = 31.0 kJ/mol & So = 93.0 J/Kmol
At equilibrium (a phase change), Go = 0
Go = Ho  TS0 = 0
Ho = TS0
T = Ho/S0
T = 3.10 x 104 J/mol/(93.0J/Kmol)
T = 333K
(b) Find the increase in entropy of the phase
transition: H2O(s) → H2O(l)
During a phase transition, ΔG=0 since the system is at
equilibrium.
This transition occurs when T=273K
 ΔGrxn
0
= ΔH – TΔS
= ΔH – TΔS
 ΔSrxn
= ΔHfusion
T
 ΔSrxn
= 6010 J/mol
273 K
 ΔSrxn
= 22 J/K.mol
Calculating Ssurroundings
Ssurr is positive -- heat flows into the
surroundings out of the system.
Ssurr is negative -- heat flows out of the
surroundings and into the system.
Ssurr = - Hsystem
T
Ssurroundings Calculations
Sb2S3(s) + 3Fe(s) ---> 2Sb(s) + 3FeS(s) H = -125 kJ
Sb4O6(s) + 6C(s) ---> 4Sb(s) + 6CO(g) H = 778 kJ
What is Ssurr for these reactions at 250C & 1 atm.
Ssurr = - Hsystem
T
Ssurr = -(-125kJ/298K)
Ssurr = 419 J/K
Ssurr = - Hsystem
T
Ssurr = -(778kJ/298K)
Ssurr = -2.61 x 103 J/K
Free Energy and partial pressure
For reactions that occur under partial
pressures other than standard (1
atm), the G at those pressures is
calculated as follows:
G = G + RT ln(Q)
Q = reaction quotient from the law of
mass action, using non-standard
pressures.
R= 8.314 J/mol K
ΔG and Equilibrium
Although a reaction with a negative ΔG will
move forward spontaneously, it does NOT
mean that a reaction will go 100% to
completion.
Remember, equilibrium occurs when the forward
and reverse reaction rates are equal (kinetics).
From a thermodynamic point of view, this occurs
when the reaction system is also at the lowest
value of ΔG.
The relationship between K and ΔGo:
ΔGo = -RTln(K)
Making connections:
At any point in the reaction, the following
equation is used:
Q=[prod]a
G = G + RT ln(Q)
[react]b
It follows that when all concentrations are
standard (1M),
Q=1
ln(Q)=0
and G = G
At equilibrium,
G = 0
and
G = -RT ln(Q)
352
A
a)
A
B
B
C
(b)
A system can achieve the lowest possible free energy
by going to equilibrium, not by going to completion.
Relationship between ΔG and K
ΔGo = -RTln(K)
Given this equation, it follows that when products
and reactants are in their std states and…

ΔGo = 0,
K=1

ΔGo < 0,
K>1

ΔGo > 0,
K<1
We can use ΔGo to calculate K, since
K= e
-ΔGo/RT
The relationship between K and ΔG at equilibrium:
ΔGo = -RTln(K) = 0 = ΔHo -TΔSo
Two cases approaching equilibrium
G = G + RT ln(Q)
Q= [products]
[reactants]
Case 1: A large (-) value for G will make G negative
also. Reaction will proceed to the right, creating MORE
products (so lnQ goes from (-) with Q<1 to (+) with Q>1 as
Q increases). Eventually, the RT ln(Q) term becomes (+)
enough that G is canceled out and G is zero (equilib.)
Case 2: A large (+) value for G will make G positive
also. Reaction will proceed right to left, creating MORE
reactants (so lnQ goes from (+) with Q>1 to (-) with Q<1 as
Q decreases). Eventually, the RT ln(Q) term becomes (-)
enough that G is canceled out and G is zero (equilib.)
Determine the sign of ΔG for each reaction.
Which reaction favors reactants? Products? Recall
G = Gf(prod)  Gf(rcts)
But….
If it looks like a free energy diagram
then you are in AP Chemistry!
Practice Problem
Calculate ΔG at 25˚C for the following reaction where
PCO = 5.0 atm and PH2 = 3.0 atm:
CO(g) + 2H2(g) →CH3OH(l)
ΔG = ΔGo + RTln(Q) Q =
1
(PCO)(PH2)2
=
1
= 2.2x10-2
(5.00)(3.00)2
ΔGo= ΔGo (CH3OH) – [ΔGo (CO) + ΔGo (H2) ]
ΔGo = -166 kJ/mol – [-137 kJ/mol + 2(0 kJ/mol)] = -29 kJ/mol
ΔG = -2.9x104 + [8.31 J/K.mol)(298K)ln(2.2x10-2)
= -2.9x104 J/mol - 9.4 x103 J/mol
= -3.8 x104 J/mol = -38 kJ/mol
Compare this ΔG to ΔGo: a more negative ΔG means this reaction is
more spontaneous (shift to right) than at one atm.
Equilibrium Calculations
4Fe(s) + 3O2(g) <---> 2Fe2O3(s)
Calculate K for this reaction at 25 oC.
Go = - 1.490 x 106 J *or find this using table, is
2(-740 KJ/mol)
G = RT ln(K)
K = e - G/RT
K = e 601 or 10261 *too big to handle in calculator
try (e200)3
K is very large because G is very negative.
AP 1971
Practice Problem #1
Given the following data for graphite and diamond at 298K:
S°(diamond) = 2 J/mol K
S°(graphite) = 6 J/mol K
ΔHf° CO2(from graphite) =
-395.3 kJ/mol
ΔHf° CO2(from diamond) = -393.4 kJ/mol
Consider the change: C (graphite) → C(diamond) at 298K and 1 atm.
(a) What are the values of ΔS° and ΔH° for the conversion of graphite
to diamond?
ΔS° = S°(dia.) - S°(graph.) = (2 - 6) J/mol K
= - 4 J/mol K
CO2  C(dia.) + O2
ΔH = + 393.4 kJ/mol
C(graph.) + O2 → CO2
ΔH = - 395.3 kJ/mol
C(graph.) → C(dia.)
ΔH = -1.9 kJ/mol
1971 AP (Continued…)
(b) Perform a calculation to show whether it is
thermodynamically feasible to produce diamond from
graphite at 298K and 1 atmosphere.
G° = ΔH° - TΔS°
= -1.9x103 J/mol - (298K)(-4 J/mol K)
= -708 J/mol;
a ΔG° is negative, indicates feasible conditions
(c) For the reaction, calculate the equilibrium constant
Keq at 298K. ***check this one
Keq = e-ΔG/RT
= e-(-708/(8.314)(298))
e0.285 = 1.3
AP 1999
Practice Problem #2
Answer the following question in terms of thermodynamic
principles and concepts of kinetic molecular theory.
(a) Consider the reaction represented below, which is
spontaneous at 298 K.
CO2(g) + 2 NH3(g)  CO(NH2)2(s) + H2O(l)
Hº = –134 kJ
(i) For the reaction, indicate whether the standard entropy
change, Sº, is positive, negative, or zero. Justify your
answer.
(i) Sº is negative because (1) two different gases make a solid and a
liquid (both with smaller entropies) and (2) three molecules of
reactant make two molecules of product (a decrease in entropy).
AP 1999 (Continued…)
(a) Consider the reaction represented below, which is spontaneous at 298
K.
CO2(g) + 2 NH3(g)  CO(NH2)2(s) + H2O(l)
Hº = –134 kJ
(ii) Which factor, the change in enthalpy, Hº, or the change in
entropy, Sº, provides the principle driving force for the reaction at
298 K? Explain.
(ii) natural tendency is to maximize entropy and since this reaction
decreases entropy and is spontaneous (-Gº), then Hº must be
negative to overcome the entropy change and drive this reaction.
(iii) For the reaction, how is the value of the standard free energy change,
Gº, affected by an increase in temperature? Explain.
(iii) Gº = Hº –TSº ; as T increases, the value of –TSº increases
(it becomes more positive since ΔS is negative) and the value of –
Gº becomes a smaller negative number (i.e., moves toward zero).
AP 1999 (Continued…)
b) Some reactions that are predicted by their sign of Gº to
be spontaneous at room temperature do not proceed at a
measurable rate at room temperature.
(i) Account for this apparent contradiction.
(i) the sign of Gº (thermodynamics) does not account for
activation energy (kinetics); a large activation energy
would effectively prevent a reaction even though there is
a favorable free energy change.
(ii) A suitable catalyst increases the rate of such a reaction.
What effect does the catalyst have on Gº for the
reaction? Explain.
(ii) a catalyst changes neither the Hº nor the Sº for a
reaction, therefore, it will have no effect on the Gº.
EXTRA: Temperature
Dependence of K
o
H
S
ln( K )  
(1 / T ) 
R
R
y = mx + b
(H and S  independent of temperature
over a small temperature range)
If the temperature increases, K decreases
for exothermic reactions, but increases for
endothermic reactions.
Summary
Entropy is usually described as a measure of the
disorder of a system. Any spontaneous process
must lead to an increase in entropy of the universe.
The standard entropy of a chemical reaction can be
calculated from the absolute entropies of reactants
and products.
Under conditions of constant temperature and pressure,
the free-energy change ΔG is less than zero for a
spontaneous process and greater than zero for a
non-spontaneous process. For an equilibrium
process, ΔG = 0.
Summary
4. For a chemical and physical change at constant
temperature and pressure, ΔG = ΔH – TΔS. This
equation can be used to predict the spontaneity of a
process.
5. The standard free energy change for a reaction, ΔGo,
can be calculated from the standard free energies of
formation of reactants and products.
6. The relationship between ΔG and equilibrium
position is given by ΔGo = -RTln(K).
7. The value of ΔG for conditions other than standard is
calculated from ΔG =ΔGo + RTln(Q).
Tutorials
http://www.wwnorton.com/college/chemist
ry/gilbert2/contents/ch13/studyplan.asp
THE END
Reversible vs. Irreversible
Processes
Reversible: The universe is exactly the same
as it was before the cyclic process.
Irreversible: The universe is different after
the cyclic process.
All real processes are irreversible -- (some
work is changed to heat).  w < G
Work is changed to heat in the surroundings
and the entropy of the universe increases.
Laws of Thermodynamics
First Law: You can’t win, you can only
break even.
Second Law: You can’t break even.
Review…. So
So increases with:
•
solid ---> liquid ---> gas
•
greater complexity of molecules (have a
greater number of rotations and
vibrations)
•
greater temperature (if volume
increases)
•
lower pressure (if volume increases)