1
THERMODYNAMICS
- The study of energy in matter
- Thermodynamics allows us to predict whether a chemical reaction occurs or
not.
- Thermodynamics tells us nothing about how fast a reaction occurs.
- i. e., thermodynamics can’t explain kinetics and vice versa
STATE FUNCTIONS
A State Function is a thermodynamic quantity whose value depends only on
the state at the moment, i. e., the temperature, pressure, volume, etc…
The value of a state function is independent of the history of the system.
The fact that internal energy is a state function is extremely useful because it
we can measure the energy change in the system by knowing the initial energy
and the final energy.
TYPES OF ENERGY AND ENERGY CHANGES
Two types of energy changes
1. Heat – q
- chaotic change in molecular motion
- related to temperature
- heating increases (or decreases) molecular motion in all directions
- not a state function (must know history)
- sign convention
+q = heat gained by system
- q = heat lost by system
2. Work – w
- concerted change in molecular motion
- Work = Force  distance  w = F  d
- movement against force is work
- work increases (or decreases) molecular motion in a specific direction
- in gases, w = - pV
- not a state function (must know history)
- sign convention
+w = work done on system
(compression)
- w = work done by system
(expansion)
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Three types of energy
1. Internal Energy – E
- Internal Energy is the sum of kinetic and potential energy in a
thermodynamic system.
- state function
2. Enthalpy – H
- modified form of internal energy
- H = E + pV
- state function
- value is very close to value of internal energy for most chemical systems.
- change in enthalpy for constant pressure process is equivalent to heat
- H = Hf – Hi = qp
2. Gibbs Free Energy – G
- modified form of enthalpy
- G = H – TS (S – entropy)
- 3 very helpful uses
1. Predict whether reaction is spontaneous
2. Give amount of useful work
3. Relate how completely a reaction will proceed
- much more later in the chapter!
REVIEW OF FIRST LAW OF THERMODYNAMICS
- The change in the energy of a system is due to heat and/or work.
E = q + w
- The first law is a conservation of energy statement.
- We will see that although energy is conserved, not all of it will be useful.
- i. e., some energy in a process will always be wasted.
- Be sure to review first law, especially sign convention of heat and work from
Chapter 5.
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DEFINITIONS OF PROCESSES
Spontaneous Processes
- A process that occurs without outside help.
Spontaneous
Pants fall to floor
Raw egg becomes hard boiled in water
Dry ice sublimates at room temp.
CH4 + 2 O2  CO2 + 2 H2O
Nonspontaneous
Pants hang themselves in closet
Hard-boiled egg becomes raw egg
CO2 gas deposits as solid at room temp.
CO2 + 2 H2O  CH4 + 2 O2
Reversible Processes
- A reversible process is a process that is always in equilibrium (in balance).
Examples
1.) Ice melting at 0 C
H2O (s)  H2O (l)
-
Equilibrium is adjusted by adding or subtracting heat.
Water can go from solid to liquid to solid to liquid etc…
2.) Haber process in a closed container
At high temperature and pressure, nitrogen and hydrogen will combine to form
ammonia.
N2 (g) + 3 H2 (g)  2 NH3 (g)
However, ammonia also spontaneously decomposes into hydrogen and
nitrogen.
2 NH3 (g)  N2 (g) + 3 H2 (g)
The net result is a balance (equilibrium) between all three substances.
N2 (g) + 3 H2 (g)  2 NH3 (g)
- Production of ammonia can increase or decrease by adjusting the external
pressure. (also temperature)
The Haber process is extremely important for the support of human population. Natural
fertilizer is insufficient. Thank God for synthetic fertilizer.
Irreversible Processes
- A process not in equilibrium
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Examples:
1.) Ice melting at 25 C
- Process only happens in one direction
- i. e., melting occurs, never freezing
2.) Precipitation of AgCl
Ag+ (aq) + Cl- (aq)  AgCl (s)
- As a solution of Ag+ and a solution of Cl- are mixed, precipitation occurs.
- The solid AgCl does not redissociate into ions.
- However, once precipitate forms, system is in equilibrium. Now the
formation of solid can be considered reversible (e. g. addition or
subtraction ammonia causes amount of solid to change.
AgCl(s)  Ag+ (aq) + Cl- (aq)
AgCl(s) + 2 NH3 (aq)  [Ag(NH3)2]+ (aq) + Cl- (aq)
Recall that dissolving an unknown silver precipitate by adding ammonia is the identification
test for chloride.
ENTROPY
- Review notes from Chapter 13 about entropy.
- Entropy is measure of how energy can be dispersed or spread out.
- Entropy makes systems more disordered.
- Increasing the number of ways that a particle in a system can distribute
energy, increases the entropy of the system.
- Entropy is a state function.
Example: Which has more entropy: 1 mole of gas in 10 L or 1 mole of gas in 20 L?
- Larger volume means molecules can be in more places, the gas in the 20 L
has more entropy.
When heat exchange is not involved, spontaneous processes always increase
entropy.
Consider diffusion of gas in two connected gas bulbs.
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open stopcock
1 atm
0 atm
½ atm
½ atm
- Gas spontaneous expands, increasing entropy.
Thermodynamic definition of entropy
- Entropy is defined via a change in heat (i. e., a change in choatic motion)
S 
q rev
T
qrev – heat in a reversible process
- q is often difficult to measure for irreversible processes, but is usually
easy to measure for reversible processes.
SECOND LAW OF THERMODYNAMICS
In any spontaneous processes, the total entropy of the universe increases for
irreversible processes and is zero for reversible processes.
S univ  0
For a thermodynamic process, the universe can be broken into two parts.
1.) System – contains what we’re interested in
2.) Surroundings – everything else
Example: For a beaker of melting ice
System – beaker and ice
Surroundings – laboratory and everything else
Example: Measuring the enthalpy of neutralization in a coffee cup calorimeter
System – reaction e.g. NaOH + HCl
Surroundings – water and calorimeter
or
System – reaction, water and calorimeter
Surroundings – laboratory
example of an
isolated system
*Defining system and surroundings can be a matter of perspective.*
Since universe can be broken into two parts, 2nd Law can be rewritten as
Ssys  Ssurr  0
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Note that Ssys can be negative (system becomes more ordered), if Ssurr is
more positive.
Example: Water freezing at –10 C.
System – water
Surroundings – refrigerator
Water spontaneously decreases entropy, but 2nd law says Suniv > 0
Therefore, Ssurr must be much greater than zero
- i. e., Ssurr >> 0
Water transfers heat (choatic motion) to air in refrigerator.
Heating of air increases entropy of “surroundings”.
Molecular motion and entropy
- As the motion of molecule increases, its entropy increases.
- According to kinetic theory of matter, motion is proportional to temperature.
- Therefore, as the temperature of a system increases, its entropy increases.
Types of molecular motion
1. Translational
- Particle moves in a straight line, i. e., particle is translated.
- Accounts for most entropy in gases.
- Accounts for substantial amount of entropy in liquids.
- Has no contribution to the entropy of solids.
2. Vibrational
- Atoms in a bond vibrate as if on a spring.
- Accounts for substantial part of entropy of liquid.
- vibrations occur within molecules
- vibrations occur between molecules (librations)
- Accounts for all entropy in solid.
Vibrations of water
H
H
H
H
O
H
H
H
O
O
symmetric stretch
H
H
O
antisymmetric stretch
H
H
H
3. Rotational
O
- Molecule rotates on axis
O
bending
- Minor contribution to entropy except at low and very high temperatures
for gases and liquids.
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Molecular motion and degrees of freedom
Degrees of freedom – number of independent options for movement
One atom has 3 DOF
Three translational DOF
Two atoms have 6 DOF
Consider diatomic molecule
How many translational DOF?
3
How many rotational DOF?
2
- note molecule can’t rotate along bond axis
How many vibrational DOF?
1
Total DOF = 3 + 2 + 1 = 6
Three atoms have 9 DOF
Consider water
Translational DOF
Rotational DOF
Vibrational DOF
3
3
3
In general for N atoms in a molecule
Translational DOF
3
Rotational DOF
Linear molecule
2
Nonlinear molecule
3
Vibrational DOF
Linear molecule
3N – 5
Nonlinear molecule
3N – 6
Total DOF
3N
Miscellaneous notes on entropy and molecular motion
1.) Entropy increases during phase changes
s  l
l  g
s  g
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2.) Entropy increases when number of particles increases.
BF3 (g) + NH3 (g)  BF3NH3 (s)
[negative S]
N2O4 (g)  2 NO2 (g) [positive S]
3.) Entropy increases when temperature increases.
- Number of modes of motion increases
4.) Entropy increases when a gas is produced in a chemical reaction.
HNO3 (aq) + Rb2CO3 (s)  2 RbNO3 (aq) + CO2 (g) + H2O (l)
2 HBr (aq) + Cd (s)  CdBr (aq) + H2 (g)
5.)
Entropy increases when the volume of a gas increased.
Example: Which has more entropy, one mole of argon in a 1L bottle or 1 mole
of argon in a 2L bottle?
The argon atoms in the larger bottle have more places to move; therefore, they
have more energy states available to them. Thus a mole of gas in a 2L bottle
has more entropy than a mole of gas in a 1L bottle.
Entropy for a phase transition
Since a phase change at the transition temperature is a reversible process,
calculation of the entropy change of phase transition is a straight forward
application of the thermodynamic definition of entropy.
Example: R-134a is a refrigerant used automotive air conditioning and
has the formula, CH2FCF3. A typical air conditioning unit
holds 28 oz (800 g) of refrigerant. If the molar enthalpy of
vaporization of R-134a is 22.0 kJ/mol at -26.6 C,
a) Calculate the entropy of the refrigerant as it evaporates at -26.6 C.
Using the thermodynamic definition of entropy, we need to find qrev and T.
T = -26.6 C = 246.55 K
Note: Always use Kelvin for
thermodynamic calculations.
The evaporation of a liquid at its boiling point is a reversible process; thus
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q rev  n  H vap  800 g 
1mol 22.0 kJ 1000 J


 173000 J
102.03 g mol
kJ
Therefore the increase of the entropy of the R-134a is
S 
q rev 173000 J

 700 J K
T
246.55 K
b) Calculate the molar entropy of the phase change.
To calculate the molar entropy, we need the molar heat, which in this case is
the enthalpy of vaporization.
Svap 
H vap
T
22.0 kJ 1000 J

mol
kJ  89.2 J

246.55 K
mol  K
THE THIRD LAW OF THERMODYNAMICS
The entropy of a perfect crystal is zero at absolute zero (0 K).
In other words, if we have a perfect crystal, we must be at absolute zero.
Consequences of the Third Law
- Absolute zero is unattainable.
- Entropy of all substances at absolute zero is zero.
- At temperature above zero, crystal will not be perfect.
- Vibrational motion introduces imperfections
- To remove imperfection takes some sort of motion
- But introducing motion keeps crystal imperfect
- It’s a no-win situation!
CALCULATION OF ENTROPY CHANGES
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The entropy change of chemical reaction can be calculated from a table of
standard entropies.
- Standard (absolute) entropies are measured at 1 atm or 1 M and 298 K
- Note: S is not zero for elements in standard state. (Different than
H and G)
S0  T   S  0 K   Ssolid 0m  Smelt  Tm   Sliquid
T
Tb
Tm
 Svap  Sgas
T
Tb
For the general reaction a A + b B  c C + d D
Srxn = c S(C) + d S(D) - a S(A) - b S(B)
Example: Calculate the entropy change for the reaction
4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)
S(Fe) = 27.15 J/molK
S(O2) = 205.0 J/molK
S(Fe2O3) = 89.96 J/molK
Srxn = 2 S(Fe2O3) – 4 S(Fe) – 3 S(O2)
= 2  89.96 J/molK – 4  27.15 J/molK – 3  205.0 J/molK = – 543.7 J/molK
GIBBS FREE ENERGY
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Reconsider 2nd Law of Thermo.
Ssys  Ssurr  0
Using the thermodynamic definition of entropy, the entropy change of the
surroundings can be related to the heat of the system.
Ssurr 
q surr  q sys

T
T
At constant pressure: q sys  H sys
Therefore the second law can be rewritten as
Ssys 
H sys
T
 0  TS  H  0
Define Gibbs Free Energy as G = H – TS
At constant temperature:
G = H – T S
Return to 2nd Law
TS  H  0   H  TS  0  H  TS  0  G  0
Second Law in terms of Gibbs Free Energy: Gsys < 0
G is useful to decide if a reaction occurs
G < 0 rxn is spontaneous
G = 0 system is at equilibrium Recall Suniv = 0 for reversible process.
G > 0 rxn is nonspontaneous; i.e., rxn does not happen
Note: When -10 kJ/mol < G < 10 kJ/mol, reactions do not go to completion but
may favor left side (-G) or right side (+G) of the chemical equation.
Negative sign of G is consistent with idea that nature always chooses lowest energy.
- System going from high energy to low energy must release energy.
Magnitude of G is also important.
G = wmax
- As G gets more negative, more work can be done by the system.
Consider combustion of three fuels
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2 H2 (g) + O2 (g)  2 H2O (g)
G = - 241.8 kJ/mol = - 119.7 kJ/g
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
G = - 801.14 kJ/mol = - 49.92 kJ/g
C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (g)
G = - 5214.1 kJ/mol = - 45.65 kJ/g
Octane does the most work per mole (the chemist’s perspective), but hydrogen
does the most work per gram (the engineer’s perspective).
Fuel cells (with hydrogen) have the potential to be nonpolluting and more
efficient than the gasoline engine.
CALCULATING STANDARD GIBBS FREE ENERGY CHANGES
For the general reaction a A + b B  c C + d D
Grxn = c Gf(C) + d Gf(D) - a Gf(A) - b Gf(B)
Gf = 0; for elements in their standard state by definition.
Example: Calculate the Gibbs free energy for the following reaction at 25 C.
C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (g)
Grxn = 8 Gf(CO2 (g)) + 9 Gf(H2O (g))
– Gf(C8H18 (l)) – 25/2 Gf(O2 (g))
= 8(-394.4 kJ/mol) + 9(-228.57 kJ/mol) – (1.77 kJ/mol) – 25/2 (0 kJ/mol)
= – 5214.1 kJ/mol
Note: Negative sign tells reaction is spontaneous.
2,2,4-trimethylpentane burns better than octane.
Example: Calculate the Gibbs free energy for the following reaction at 25 C.
Cu (s) + H2O (g)  CuO (s) + H2 (g)
Grxn = Gf(CuO (s)) – Gf(H2O (g)) = (–129.7 kJ/mol) – (–228.6 kJ/mol)
= 98.9 kJ/mol
Note: Positive sign tells reaction is nonspontaneous.
i. e., reverse reaction is spontaneous.
(Of course! Copper is not an active metal.)
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Concentration dependence of Gibbs Free Energy
Standard concentrations are
p = 1 atm
c = 1 M
The Gibbs free energy for the partial pressure of a gaseous substance is
G i  G 0i  RT ln pi p0 
The Gibbs free energy for the molar concentration of an aqueous substance is
G i  G 0i  RT lnci c0 
Note: The higher the concentration, the higher the free energy
Note the distinction between G and G.
G is calculated for any conditions
G is calculated only for standard conditions.
For a general reaction: a A + b B  c C + d D
G rxn  cG C  dG D  aG A  bG B
 c  G 0C  RT ln  C c 0    d  G 0D  RT ln  D  c 0  
 a  G 0A  RT ln  A  c0    b  G 0B  RT ln  B c 0  
 cG 0C  dG 0D  aG 0A  bG 0B
 cRT ln  C c0   dRT ln  D  c 0 
 aRT ln  A  c0   bRT ln  B c 0 
ln  C c0 c  ln  D  c 0 d 
 G  RT 

0 a
0 b
 ln  A  c   ln  B c  
  C  c 0 c  D  c 0 d 
0
 G rxn  RT ln 

   A  c 0  a   B c 0  b 


0
 G rxn  RT ln Q
0
rxn
G  G 0  RT ln Q
- Where G and G0 are stoichiometric sums and Q is reaction quotient
- As reaction goes to right, Q increases; thus G increases
- As reaction goes to left, Q decreases; thus G decreases
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Example: For the reaction N2O4 (g)  2 NO2 (g) at 298 K the standard free
energy of reaction is 5.40 kJ/mol.
a) calculate the free energy of the reaction when p(N2O4) = 1 atm and
p(NO2) = 1 atm
G  G 0  RT ln Q
Q
p2NO2
p N 2 O4
1 atm

2
1
1 atm
G  5.40 kJ / mol  0.008314 kJ / mol  K  298 K ln1
 5.40 kJ / mol  0 kJ / mol
 5.40 kJ / mol
b) calculate the free energy of the reaction when p(N2O4) = 0.905 atm
and p(NO2) = 0.115 atm
G  G 0  RT ln Q
Q
p 2NO2
p N 2O4
.
atm
0115

2
 0.0146
0.905 atm
G  5.40 kJ / mol  0.008314 kJ / mol  K  298 K ln 0.0146
 5.40 kJ / mol  10.47 kJ / mol
 5.06 kJ / mol
c) calculate the free energy of the reaction when p(N2O4) = 0.444 atm
and p(NO2) = 0.224 atm
G  G 0  RT ln Q
Q
p 2NO2
p N 2O4
0.224 atm

0.444 atm
2
 0113
.
G  5.40 kJ / mol  0.008314 kJ / mol  K  298 K ln 0.113
 5.40 kJ / mol  5.40 kJ / mol
 0 kJ / mol
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Temperature dependence of Gibbs free energy
Recall for spontaneous processes, usually H is negative and S is positive.
However, the only reliable measure of spontaneity is G.
Putting negative H and positive S into equation for G yields negative G.
Gibbs-Helmholtz Equation
G = H – T S
Consider possibility when S is negative such as
4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)
S = - 0.5437 kJ/molK
H = - 822.16 kJ/mol
G = - 822.16 kJ/mol – (298.0 K)(- 0.5437 kJ/molK) = - 660.1 kJ/mol
In this reaction, heat evolved (increasing disorder of universe) compensates for
system becoming more ordered.
In increasing the temperature, the entropy has a greater effect on the system.
We can increase temperature until the entropy effects become more important
than enthalpy effects.
From the preceding we see that the spontaneity of a reaction is affected by
temperature.
Example: At what temperature is the rusting of iron nonspontaneous.
We need to find T where entropy effects are balanced with enthalpy effects
G = H – T S = 0
T
H
 822.16 kJ / mol

 1512 K
S  0.5437 kJ / mol  K
As long as we keep iron above 1512 K, it won’t rust.
In other words, the smelting of iron ore in a blast furnace must occur at
temperature above 1512 K.
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Example: The human body receives energy from glucose (sugar) by using
oxygen to decompose it into carbon dioxide and water. The first step
in a lengthy decomposition process is the decomposition of glucose
(C6H12O6) into pyruvate (C3H5O3-) and H+.
C6H12O6 (aq)  2 C3H5O3- (aq) + 2 H+ (aq)
Calculate the Gibbs free energy at 37 C using the Gibbs – Helmholtz
equation.
To use the Gibbs-Helmholtz equation, G = H – T S, we need to calculate Hrxn
and Srxn
Hrxn = 2 Hf(C3H5O3- (aq)) – Hf(C6H12O6 (aq))
= 2 (–686.6 kJ/mol) – (–1274.4 kJ/mol) = – 98.8 kJ/mol
Srxn = 2 S(C3H5O3- (aq)) – S(C6H12O6 (aq))
= 2  712.1 J/molK – 212.1 J/molK = 1212.1 J/molK
G = H – T S = – 98.8 kJ/mol – 310 K(0.2121 kJ/molK) = – 474.6 kJ/mol
All values are approximate or made-up.
Example: For the reaction CaCO3 (s)  CaO (s) + CO2 (g), The enthalpy of
reaction is 178.3 kJ/mol and the entropy of reaction is 160.5 J/molK.
a) Calculate the Gibbs free energy at 450 K using the Gibbs –
Helmholtz equation.
G = H – T S = 178.3 kJ/mol – 450 K(0.1605 kJ/molK) = 106.1 kJ/mol
b) Calculate the Gibbs free energy at 2110 K using the Gibbs –
Helmholtz equation.
G = H – T S = 178.3 kJ/mol – 2110 K(0.1605 kJ/molK) = – 160.4 kJ/mol
c) Calculate the temperature where the reaction becomes
spontaneous.
G = 0

T
H
178.3 kJ / mol

 1110 K
S 01605
.
kJ / mol  K
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Temperature dependence of free energy can be summarized in table.
G  H  TS
H
+
+
S
+
+
-
G
- or +
- or +
+
spontaneous
spontaneous below specific temp.
spontaneous above specific temp.
never spontaneous
FREE ENERGY AND EQUILIBRIUM
When G is zero; there is no change in the system; the reaction does not
proceed to the right; nor does the reaction proceed to the left.
In this state, a balance exists among all the chemical components and we say
that the system is in equilibrium.
When G is zero, the value of the reaction quotient is the equilibrium constant, K.
To repeat:
**At equilibrium, G = 0**
 0 = G + RT ln K

G = -RT ln K
Thus the free energy of a reaction is related to its equilibrium constant.
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Example: Calculate the eq. constant for the reaction:
Fe3O4 (s) + 2 C (s)  3 Fe (s) + 2 CO2 (g), given a table of
Gibbs free energies
Gf(Fe3O4(s)) = -1162.0 kJ/mol
Gf(CO2(g)) = -394.4 kJ/mol
a) at 25 C
G 0
G 0

0
ln K  
 K  e RT
G   RT ln K
RT
G = 2(–394.4 kJ/mol) – (-1162.0 kJ/mol) = 373.2 kJ/mol
Ke

372.2kJ/mol  1000J/kJ
8.314J/molK  298K
 e150  5.7 1066
 pCO2  K p  5.7 1066  2.4 1033 atm
2
K p  pCO
2
b) at 2500 C
Ke
2
K p  pCO
2

373.2kJ/mol  1000J/kJ
8.314J/molK  2773K
 e16.2  9.3 108
 pCO2  K p  9.3 108  3.1104 atm
Even though pCO2 is still small, a continuous “blast” of fresh air helps the reaction proceed.