Chapter 19
Thermodynamics And Equilibrium
I. THERMODYNAMICS
A) Definition of Thermodynamics - It is the
study of energy changes that accompany
physical and chemical changes.
B) Most of the energy produced in modern
society comes from chemical reactions.
1) The burning of petroleum products gasoline, fuel oil
2) The burning of natural gas
3) The burning of coal
C) Thermodynamics is based on our
experience with how nature behaves. Our
study of nature has resulted in Three Laws of
Thermodynamics.
D) You studied the first law previously, I
hope.
1) Energy is conserved in physical and
chemical changes.
2) The total quantity of energy in the universe
is assumed to be a constant.
3) Energy is converted from one form to
another while the total remains the same.
4) You should have studied Hess's Law and
DH in conjunction with the first law. (The
change in enthalpy is the heat of reaction at
constant pressure.)
II. THE SECOND LAW OF
THERMODYNAMICS AND
SPONTANEITY
A) One of the basic goals of chemistry is to
predict whether or not a reaction will occur
when substances are brought together; that
is, whether the reaction is or is not
spontaneous.
A spontaneous process is a physical or
chemical change that occurs by itself. It can
start by itself, or once started it continues by
itself without further exertion.
B) THIS DOES NOT MEAN RAPIDLY.
Remember rate or speed of reaction is the
concern of kinetics. It could be a fraction of a
second or 100 years before a measurable
amount of products are formed and still be
called a spontaneous reaction.
C) Some examples of spontaneous processes:
1) Iron rusts in moist air.
2) Ice melts at 25oC
3) Natural gas continues to burn once it is lit.
4) Water runs downhill.
5) A wound clock (watch) runs down.
6) Heat flows from hotter to cooler objects.
7) Hydrogen burns in oxygen to form water.
8) Sodium reacts with water.
D) An early criterion for predicting whether
or not a process was spontaneous was that it
had to be exothermic.
1) If this were true, the only information we
would need to predict whether or not a
process was spontaneous is whether DH is
___________________________ .
2) But just looking at the set of spontaneous
processes above, we see an endothermic one.
Which one is that?
H2O(s) ----> H2O(l) at 25oC
DH = + 6.0 kJ/mol
Other spontaneous processes are:
H2O(l) ----> H2O(g) at 100oC
DH = + 40.7 kJ/mol
CaCO3(s) ----> CaO(s) + CO2(g) at 1100oC
DH = + 178 kJ/mol
3) The sign of DH is not enough information
to predict accurately whether or not a process
is spontaneous.
III. ENTROPY - THE THERMODYNAMIC
QUANTITY CENTRAL TO THE SECOND
LAW OF THERMODYNAMICS
A) Entropy, symbol S, is a thermodynamic
quantity that is a measure of the
randomness or disorder in a system.
John's Backpack
B) An example of a spontaneous change:
The distribution of the "gas" between the
two flasks has been a spontaneous change.
For mixing of two ideal gases, after a certain
amount of time, each gas will be evenly
distributed between the two flasks. The
mixing will be a spontaneous change. There
has been no temperature change, no changes
in energy, and no outside stirring was
required.
C) How do we explain this phenomenon?
1) There must be a natural tendency to
mixed-up-ness. A higher probability is
associated with the mixed state compared
with the unmixed state.
2) The final mixed state is more probable
than the separated state.
3) The driving force of the process is the
tendency toward mixed-up-ness. The random
motion of the molecules tends to produce
disorder.
IV. THE SECOND LAW OF
THERMODYNAMICS
A) The total entropy of a system and its
surroundings always increases for a
spontaneous process. The entropy of the
universe increases in a spontaneous process.
B) The driving force for a spontaneous
process is an increase in the entropy of the
universe. This is based on experience, but has
not been proven rigorously.
C) Two fundamental laws of nature are
particularly important in thermodynamics:
1) Systems tend to attain a state of minimum
potential energy.
2) Systems tend toward a state of maximum
disorder.
D) For a given substance, the ___________ is
the state of lowest entropy, the __________ is
intermediate, and the ______________ is the
highest entropy state.
E) Why should a substance spontaneously
freeze at a temperature below its melting
point? Doesn't this appear to violate the 2nd
law? It seems to be spontaneous yet there is a
decrease in the entropy of the system, the
water is going from a higher to a lower
entropy state.
F) All entropy effects must be considered.
1) When two ideal gases mix, there is no
exchange of matter or energy between the
system and the surroundings. The only
entropy effect is the increase in the isolated
system itself.
2) Usually the system is not isolated from the
surroundings.
3) When a liquid freezes, the heat of fusion
is involved. What is that?______________ .
This heat is absorbed by the surroundings
and the motion of the molecules of the
surroundings _______________ , therefore
the entropy of the surroundings __________ .
The decrease in the entropy of the system is
more than offset by the increase in the
entropy of the surroundings so there is a net
_________ in the randomness of the universe.
4) Mathematically we can express the Second
Law of Thermodynamics as follows:
For a spontaneous process:
DSuniverse = DSsystem + DSsurroundings must be
greater than zero.
For an equilibrium process:
DSuniverse = DSsystem + DSsurroundings must be
equal to zero.
5) It is unfortunate, but it is not always
easy to assess the change in randomness
of the surroundings.
V. THE THIRD LAW OF
THERMODYNAMICS
A) THE ENTROPY OF A PERFECT
CRYSTALLINE SOLID AT 0 K IS ZERO
(0).
B) Why should that seem reasonable?
C) Since we have the above definition we can
obtain entropies, S's, of pure substances.
D) This can be contrasted to enthalpies for
which we could obtain only what?
E) Another word of caution is necessary. So at
standard state conditions has a numerical
value for a pure element. Under these
conditions, DH = 0.
F) The standard entropy of a substance or
ion, also called its absolute entropy, So, is the
entropy value for the standard state of the
species. Table 18.1 gives the standard
entropies of various substances at 25oC and 1
atm. See also appendix C.
G) Example: For the reaction below, calculate
DS.
Hg(l) + ½ O2(g) ---> HgO(s)
The first thing I want you to do is guess the
sign for DS.
It appears that the entropy is decreasing. The
system is going from a random to a more
ordered system. You should guess that the
sign of DS for the reaction is ___________.
The Standard Entropy Change is calculated
from the following:
DSoreaction = DSoproducts - DSoreactants
DSor = SoHgO(s) - [SoHg(l) + 1/2 SoO 2(g)]
DSor = 1 mol(70.27 J/molK) - [1 mol(76.027
J/molK) + 1/2 mol(205.0 J/molK)]
DSor = -108.3 J/K
The negative sign indicates that the entropy
associated with the reaction is decreasing. We
predicted this earlier.
H) The entropy usually increases in the
following situations:
1) A reaction in which a molecule is broken
into two or more smaller molecules or atoms.
2) A reaction in which there is an increase in
the number of moles of gas.
3) A process in which a solid changes to a
liquid or gas or a liquid changes to a gas.
VI. GIBBS FREE ENERGY
A) Since it is difficult to assess all of the
entropy changes in the system and
surroundings, the entropy increasing
criterion for spontaneity is often difficult to
obtain. Our goal is to have a criterion for
spontaneity prediction that focuses only on
the system because it is easy __________ .
B) Our criterion for spontaneity which relies
only on our understanding of what is
happening to the system is related to the store
of useful work in a system. What do we mean
by useful work?
C) Theoretically, spontaneous reactions can
be used to obtain useful work.
D) The maximum useful work which can be
obtained in driving a turbine or work
produced by a battery is the change in the
Free Energy, DG.
There is a relationship among the enthalpy
change, DH, the entropy change, DS, and the
change in free energy, DG.
DG = DH - TDS
The T must be in Kelvins.
E) A process or reaction will occur
spontaneously at constant T and P only if it
is capable of doing useful work, i.e. DG is
negative. The free energy of the system
must decrease
F) If DG = zero, the system is at equilibrium.
G) If DG is positive the reaction as written is
not spontaneous, but the reverse reaction is
spontaneous.
H) Look at DG = DH - TDS
1) DH reflects the heat energy transferred
between the system and surroundings.
2) TDS is a measure of the change in the
randomness of the system. The energy
consumed or evolved with respect to the
change in order of a system.
DG = DH – TDS
3) The more negative DH is, the more likely
the reaction will be spontaneous. This reflects
the drive toward minimum energy. Also, DH
is always in kJ/mol and DS is in J/molK. BE
CAREFUL This has caused more lost points
than I care to remember.
4) If the reaction is exothermic (DH is -) and
proceeds with an increase in entropy (DS is +)
the reaction is spontaneous regardless of the
temperature. Why?___________________
5) If DH is + ( what kind of reaction?) and DS
is - (the reaction proceeds with a decrease in
entropy), then DG must be ________________
no matter what the temperature. The reaction
must be __________________.
6) If DH is -, the reaction is _____________
and DS is -, then DG is - at low
temperatures, reaction is _______________ .
DG is ____________ at high temperatures
and _______________ .
7) If DH is + (_____________________), and
DS is +, then DG is ______ at low T and
_______________________. DG is ______ at
high T and ______ ___________________ as
written.
REMEMBER SPONTANEOUS DOES NOT
MEAN _____________.
8) Free Energy (DG ) is the link between the
drive toward minimum energy and maximum
entropy.
I) Calculation of DG from DGo= DHo – TDSo
Calculate DGo for the following reaction:
Hg(l) + ½ O2(g) -----> HgO(s)
DHfo for HgO(s) = -90.79 kJ/mol
DSo for the reaction is - 108.3 J/molK
DGo = -90.79 kJ - (-32.3 kJ) = -58.5 kJ
The reaction is spontaneous even though the
entropy is decreasing. DH is large enough to
overcome this decrease in entropy.
J) DGof is the standard free energy of
formation. This is the free energy change
that occurs when one mole of substance is
formed from its elements in their most
stable states at 1 atm and at a specified
temperature, usually 25oC, 298 K.
1) The DGof of an element in its standard state
is zero.
DGoreaction = SDGoproducts - SDGoreactants
3 O2(g) + 6 C(gr) + 6 H2(g) ----> C6H12O6(s)
DGof = - 910.56 kJ/mol
For the reaction below, calculate the
DGo reaction:
2 NO(g) + O2(g) ---> 2 NO2(g)
DGoreaction = SDGoproducts - SDGoreactants
DGoreaction = 2 mol(DGofNO2) - [2 mol(DGofNO)
+ 1mol(DGofO2)]
DGoreaction = 2mol(51 kJ/mol) - [2 mol(86.6
kJ/mol) + 1mol(0)]
DGoreaction = ___________
The reaction is __________________________
under standard state conditions.
VII. FREE ENERGY AND EQUILIBRIUM
A) A system at constant temperature and
pressure will proceed spontaneously in the
direction that lowers the free energy. This is
why reactions proceed to equilibrium. The
equilibrium position represents the lowest
free energy value available to a particular
reaction system.
B) For a reaction at conditions which are not
standard state, the following equation has
been derived:
DG = DGo + RT ln Q
What is Q?
For the general reaction
A(g) + B(g) ---> 2 C (g)
R = 8.31 J/molK
DGo is the free energy change for reactants
and products at 298K and 1 atm.
DG is the free energy change for the reaction
at specified pressures of reactants and
products.
For the reaction:
CO(g) + 2 H2(g) ---> CH3OH(l)
Calculate DG at 25oC when the pressure of
CO is 5.0 atm and that of H2 is 3.0 atm.
DG = DGo + RT ln Q
We first have to compute DGo from the
thermodynamic tables.
DGoreaction = DGoproducts - DGoreactants
DGoreaction = [(1 mol)(-166.2 kJ/mol)] - [(1 mol)
(-137.2 kJ/mol) + (2 mol)(0 kJ/mol)]
DGoreaction = ______________
DG = DGo + RT ln Q
DG = - 2.90 X 104 J + (-9.5 X 103 J) = ______ J
DG = ______ kJ
Therefore the reaction is ________________ .
We can use the equation to find DrG, the
instantaneous rate of change of the Gibb's
function at specific concentrations (more
correctly activities) and you will use these
values to predict whether or not a precipitate
will occur in your experiment.
For the reaction:
Ag+(aq) + Cl-(aq) ---> AgCl(s)
You will want to know if a precipitate forms
upon the addition of 5 mL each of 2M AgNO3
and 2 M HCl.
You will use the equation
DrG = DGo + RT ln Q
The first thing you will have to do is calculate
DGo from values in the table. For the above
reaction that is - 55.6 kJ.
To compute DrG, we need to take into account
the concentrations of the reactants, to
calculate the reaction quotient, Q.
Since we are mixing equal volumes of 2 M
solutions, the concentrations to use in Q are
not 2 M. We have doubled the volume by
mixing so we have halved the concentrations
of the ions. Thus the [Ag+] and [Cl-] are now
1 M. Make certain you understand this
point as you did not quite get it in the
solubility chapter.
Thus, DrG = - 55.6 kJ Why?
C) DG's relationship with Kp
DG = DGo + RT ln Q
at equilibrium G = 0; Q = Kp
Then 0 = DGo + RT ln Kp
DGo = - RT ln Kp
for DGo to be -, the ln Kp must be +, Kp > 1
for DGo to be +, the ln Kp must be -, Kp < 1.
D) If DGo is a large negative number (more
negative than about -10 kJ), reactants are
transformed almost entirely to products
when equilibrium is reached.
E) If DGo is a large positive number (more
positive than about +10 kJ), the reactants do
not give significant amounts of products when
equilibrium is reached.
F) If DGo is between - 10 and + 10 kJ the
reaction gives an equilibrium mixture with
significant amounts of both reactants and
products.
G) Calculate Kp for the following reaction at
298 K:
2 C2H2(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g)
DGorxn = [4 mol(DGof CO2(g) + 2 mol(DGof
H2O(g)] - [2 mol(DGof C2H2(g) + 5 mol(DGof
O2(g)]
DGorxn = [4 mol(-394.4kJ/mol) + 2 mol
(-228.6kJ/mol)] - [2mol(209kJ/mol) + 5 mol
(0)]
DGorxn = [-1578 -457.2]kJ - [+418]kJ =
- 2453 kJ
DGorxn = - RT ln Kp
- 2453 kJ = -2.49 kJ ln Kp
taking the antiln of such a big number will
result in an error message on some
calculators.
If that is the case, you will have to go to
log base 10.
2.303 log Kp = ln Kp = 985
What does this mean?
This method also works for net ionic
equations – problem 18.52 in the text.
VIII. APPLICATION OF DG TO PHASE
CHANGES - MELTING AND BOILING
A) At the melting point and boiling point we
have an equilibrium situation for change of
phase, therefore DG = 0.
DG = DH - TDS
0 = DH - TDS
DH = TDS
B) The boiling point of CHCl3 is 61.7oC.
The DHvap is 31.4 kJ/mol. Calculate the DS
of vaporization.
C) Estimate the melting point of benzene in
oC, if DH
fus is 10.9 kJ/mol and the DSfus is
39.1 J/molK.