Chapter 19. Chemical Thermodynamics.

1
Chapter 19. Chemical Thermodynamics.
?G ?H T?S
2
The First Law of Thermodynamics
transfer
Surroundings
System
?E q w
?H suggested that highly exothermic reactions
proceeded very strongly, while the reverse
reaction was highly endothermic, did not proceed
spontaneously at all. Enthalpy is not the
only driving force in reactions. There is a
second contribution, the entropy. Entropy relates
to probability, and the increase in disorder of a
system.

• To what extent do physical chemical processes
  occur?
• In what direction do they occur?

3
Spontaneous Processes

• A spontaneous process is one that proceeds on its
  own without any outside assistance.
• Processes that are spontaneous in one direction
  are non-spontaneous in the opposite direction.
• A gas will expand from one flask into a second as
  a spontaneous process, but the reverse will never
  occur.

4
Examples of spontaneous processes
The Spontaneous Expansion of a Gas.
We connect two flasks as shown at the right by a
closed stopcock, one of which is evacuated.
When the stopcock is opened, the air will rush
into the evacuated flask to equalize the
pressure. What is the driving force for this?
Probability, or entropy.
5
The Spontaneous Expansion of a Gas

• Consider the simple case where two gas molecules
  are in the flask on the right hand side before
  the stopcock is opened.
• Once the stopcock is open, there is a higher
  probability that one molecule will be in each
  flask than both molecules being in the same flask.

For Avogadros number of molecules it is much
more probable that the molecules will distribute
between both flasks instead of remaining in a
single flask. i. e., there is greater disorder
or Entropy.
6
Spontaneous rusting of nails in air.
Examples of spontaneous processes
spontaneous
not spontaneous

rusted nails
4 Fe(s) 3 O2(g) ? 2Fe2O3(s)
rust
7
Examples of spontaneous processes
Ice at 25 oC will spontaneously melt
spontaneous
not spontaneous
water
ice

The reverse will never happen at this
temperature. Water at 25 oC will never freeze.
NB Direction of physical change!
8
Examples of spontaneous processes
At -10oC water spontaneously freezes
spontaneous
not spontaneous
water
ice

A spontaneous process is one that proceeds on its
own without any external interference.
9
Spontaneity and Entropy
Spontaneity depends on the path taken when
changing the system from one state to another.

• Recall
• T (temperature), E (internal energy) H
  (enthalpy) are state functions.
• q (heat) w (work) are not state functions.

A reversible process allows a system to change
between states along the same path such that the
original conditions of both system and
surroundings is restored.
A irreversible process cannot be reversed to
restore both system and surroundings to their
original conditions.
10
Spontaneity and Entropy

• A mixture of water and ice at 0 oC is at
  equilibrium.

H2O(l) ? H2O(s) T 0C

• The relative quantity of
• ice and water remains
• the same if we dont add
• or remove heat.
• There is no spontaneous
• process in either direction.

ice
water (at 0oC)
The two phases interconvert at the same rate and
there is no preferred direction of spontaneity.
11
Spontaneity and Entropy

• Adding an infinitely small amount of heat
• causes some of the ice to melt while removing
• an infinitely small amount of heat will cause
• some of the water to freeze at 0 oC.
• i.e., the process is isothermal since the
• temperature does not change but remains at
• 0C.

T 0C

• The melting and freezing of ice at 0 oC is
  therefore reversible.
• At all other temperatures an irreversible process
  occurs.
• Melting and freezing of a substance at its
  melting point are reversible.
• Boiling and condensation of a substance at its
  boiling point are also reversible.
• Reversible processes are not spontaneous and
  occur with infinitesmal changes in the system.

12
Entropy (S)
H2O(s) ? H2O(l) T 0C
Greater disorder. More random distribution of
molecules.
?S gt0
Randomness increases
Order increases
?S lt0
Entropy is a state function and is related to
the disorder or randomness of a system. The
change in entropy is given by
?S Sfinal - Sinitial
13
Relating Entropy (S), heat transfer (q) and
temperature(T)
I propose to name the quantity S the entropy of
the system, after the Greek word trope, the
transformation. I have deliberately chosen the
word entropy to be as similar as possible to the
word energy. (1865)

• Rudolf Clausius (1822-1888)

• To calculate the entropy change of a system we
  need to know the heat
• change, qrev, and divide it by the temperature in
  K. The term qrev is the
• reversible flow of heat in a reversible path such
  as enthalpy of melting
• of a solid at its melting point, or of boiling of
  a liquid at its boiling
• point.

14
(?S ) for a phase change Practice exercise The
normal boiling point of ethanol is 78.3C, and
its molar enthalpy of vaporisation is 38.56
kJ/mol. What is the change in entropy in the
system when 68.3 g of C2H5OH(g) at 1
atm condenses to liquid at the normal boiling
point?
Vaporisation
C2H5OH(l) ? C2H5OH(g) ?Hvap T 78.3C
isothermal
qrev -?Hvap (since we are dealing with a
condensation)
C2H5OH(g) ? C2H5OH(l) -?Hvap T 78.3C
isothermal
15
qrev -?Hvap -38.56 kJ/mol
qrev 68.3g 1mol (-38.56) kJ
1000 J 46.07g
mol kJ - 57166 J
T 78.3 273.15 351.45 K
?Ssys qrev - 57166 J
T 351.45 K
- 162.6 J/K -163
J/K
Note that units of entropy are J/K.
NB
?S lt0
Order increases!!!!
C2H5OH(g) ? C2H5OH(l)
16
Entropy and the Second Law of Thermodynamics
Entropy is not conserved and in any spontaneous
process the entropy of the universe increases.
Example The enthalpy of fusion for H2O is
?Hfusion 6.01kJ/mol. What is the entropy
change when a mole of ice (273K) melts in the
palm of your hand at 310K(body temperature)?
NB Process is irreversible. System and
surroundings are not at the same temperature.
qrev ?Hfusion (melting of a substance)
?Ssystem q 6010J 22.0 J/K (ice
cube) T 273K
?S is a state function therefore can use qrev.
?Ssurrounding q - 6010J -19.4 J/K
(our hand) T 310K
?Stotal ?Ssystem ?Ssurrounding 22.0 19.4
2.6 J/K
?Stotal ?Suniv gt 0
17
The Second Law of Thermodynamics

• Reversible process
• ?Suniverse ?Ssystem ?Ssurroundings 0
• (non-spontaneous)
• Irreversible process
• ?Suniverse ?Ssystem ?Ssurroundings gt 0
  
• (spontaneous)

The total entropy of the universe increases in
any spontaneous process.
i.e., the sum of the entropy change of the system
and the surroundings is always positive.
18
Example What happens to the entropy of the
universe when you dissolve sugar in a hot cup of
coffee?
Answer
The entropy of the universe increases. ?Suniverse
gt 0 Sugar is soluble in hot coffee. Therefore
dissolving the sugar in the hot coffee is a
spontaneous process. Cooling the coffee is not
going to give you the sugar back in solid form.
Process is irreversible.
19
The Molecular Interpretation of Entropy
Statistical thermodynamics relates the behaviour
of atoms and molecules (microscopic realm) to
entropy as observed in the macroscopic realm.
Vibrational
Rotational
Translational

• Energy can be stored by the motion of molecules
  in different arrangements.
• Each arrangement is called a Microstate (W).

20
The Molecular Interpretation of Entropy
The number of ways a system can be arranged
indicates the degree of disorder the system has
and is indicated by the number of possible
microstates.
The entropy of a system is given by

• S k ln W
• W is the number of possible
• Microstates.
• K is Boltsmanns constant
• (1.38 x 10-23 J/K)

Ludwig E Boltzmann (1844-1906) Boltzman
n defined the nature of entropy.
NB Entropy (S) increases with an increased
number of microstates (W).
21
Qualitative Predictions About ?S.

• Entropy increases with
• Temperature molecules vibrate and move more
  rapidly resulting in greater disorder.
• Volume greater number of possible positions for
  molecules and hence greater disorder.
• Number of independently moving particles
  greater number of microstates and more ways of
  dispersing energy. i.e., greater randomness.
• Entropy increases from
• solid lt liquid lt gas.

Slope increases
Translational
Figure Increase in entropy of a substance with
increasing temperature from solid ? liquid ?
gas.
22
Qualitative Predictions About ?S.
Ssolid
lt Sliquid
lt Sgas
gas
liquid
solid
In a gas, molecules are more randomly distributed
and molecular motion enjoys the highest degree of
freedom. i.e., greatest kinetic energy and hence
greater number of possible microstates.
Gases have the highest entropy of all phases.
23
Qualitative Predictions About ?S.
Larger Molecules generally have a larger entropy
Ssmall
lt Smedium
lt Slarge
Larger molecules have more internal motion
(vibrational and rotational) therefore have more
ways of storing energy.
24
Qualitative Predictions About ?S.
Dissolving a solid or liquid will increase entropy
solution
liquid
dissolves
lower entropy
Higher entropy
(more disorder, molecules move more freely)
Solid (region of more order)
25
Qualitative Predictions About ?S.
Dissolving a gas in a liquid will decrease entropy
gas
dissolves
overall more disordered arrangement higher
entropy
solution of gas in liquid, lower S
26
Example Predict whether ?S is positive or
negative for the following processes assuming
each occurs at constant temperature.
FeCl2(s) H2(g) ? Fe(s) 2 HCl(g)
DS ve
solid
solid
gas
gas
1 mol
2 mol
Ba(OH)2(s) ? BaO(s) H2O(g)
DS ve
solid
solid
gas
2 SO2(g) O2(g) ? 2 SO3(g)
DS -ve
gas
gas
gas
2 mol
1 mol
2 mol
Ag(aq) Cl- (aq) ? AgCl(s)
DS -ve
in aqueous solution
Insoluble precipitate
27
Standard Molar Entropy of a substance (Sº)

• Third Law of Thermodynamics
• The entropy of a pure crystalline substance at
  absolute zero (0 K) is zero.
• Reference point for tabulation of absolute
  entropies for substances.

No thermal motion. W 1 (one microstate)
The entropy of one mole of a substance in its
standard state (25oC 1atm) is known as the
standard molar entropy (Sº). Sº Units J/mol K.
28
Tabulated Examples
p. 820 in 10th Ed. See also Appendix C

size of molecules increases i.e., increase of
molar mass

Sgas gt Sliquid

dissolving a gas in a liquid is accompanied by a
lowering of the entropy

dissolving a liquid in another liquid is
accompanied by an increase in entropy.
O2(g)
205
Standard molar entropies of elements are not zero
29
Example Which of the following pairs of
substances has a higher molar entropy at 25C?
HCl(l) HCl(s)
Li(s) Cs(s)
C2H2(g) C2H6(g)
Pb2(aq) Pb(s)
O2(g) O2(aq)
HCl(l) HBr(l)
CH3OH(l) CH3OH(aq)
N2(l) N2(g)
30
Entropy Changes for Chemical Reactions
?So for a reaction Using the standard molar
entropies of reactants and products, the entropy
change for a reaction can be calculated .
?Sorxn S n So(products) S m So(reactants)
Stoichiometric coefficients of reactants
Stoichiometric coefficients of products
Sum
NB This equation is easily compared to the
equation for the standard enthalpy change for a
reaction.
31
Example What is ?So for the following reaction?
Do you expect ?So to be positive or negative?
C2H4(g) H2(g) ? C2H6(g)
gas
gas
gas
DS -ve
1 mol
1 mol
1 mol
DSorxn 229.5 219.4 130.6
-120.5 J/K
32
Example What is ?So for the following reaction?
Do you expect ?So to be positive or negative?
CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
gas
gas
gas
gas
1 mol
2 mol
1 mol
2 mol
substance So (J/K-mol) O2(g) 205.0 CH4(g)
186.3 CO2(g) 213.6 H2O(g) 188.3
?Sorxn S n So(products) S m So(reactants)
?Sorxn (213.6 2 ? 188.3) (186.3 2 ?
205.0)
? 6.1 J/K
33
Entropy Changes for the Surroundings.
We saw earlier that for a reversible process
?Ssystem qrev T
Irrespective of whether or not a spontaneous
process is taking place in a system (set of
chemicals reacting) we always assume the
surroundings of the system to be at equilibrium
while processes take place in the system.
We may therefore calculate ?Ssurr from
?Ssurr qsurr Tsurr
34
If we visualize the system and the surroundings
as follows
surroundings
heat
T
system
T
heat
heat
Any thermal energy leaving the system as heat
must be entering the surroundings as heat.
- qsys qsurr
Therefore
?Ssurr qsurr - qsys
Tsurr T
35
?Ssurr qsurr - qsys
Tsurr T
The heat associated with a chemical reaction is
given by the value of ?H for the reaction.
Therefore
?Ssurr - qsys - ?H
T T
Recall that the enthalpy change of a reaction can
be calculated from standard enthalpies of
formation of reactants and products from
tabulated data.
36
Example Calculate the entropy change at 25C (?S)
for the reaction involved in the corrosion of
iron
4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)
Discuss whether or not your answer is in accord
with the second law of thermodynamics.
?Sorxn S n So(products) S m So(reactants)
substance So (J/K-mol) O2(g) 205.0 Fe(s) 2
7.15 Fe2O3(s) 89.96
?Sorxn (2 ? 89.96) (4 ? 27.15 3 ? 205.0)
? 543.7 J/K
?S is ve, implying a large decrease in entropy
accompanies the rusting of iron. What is the
reason for the negative value obtained for
?S? One of the reactants is O2(g). The reaction
consumes a gas (high entropy substance) and
produces a solid (low entropy substance).
37
4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)
?Sorxn - 543.7 J/K

• Is this reaction a spontaneous or
  non-spontaneous reaction?
• Spontaneous!!!

• What does the 2nd law of thermodynamics say
  about spontaneous
• reactions?
• ?Suniv gt 0 !!! Do we have a contradiction???

• Notice the 2nd law of thermodynamics refers to
  ?Suniv.
• We have a calculated ?Srxn which is ?Ssys.

• The two quantities are related by
• ?Suniv
  ?Ssys ?Ssurr
• 
  ?Suniv -543.7 J/K ?Ssurr

38
?Ssurr - qsys - ?H T
T
Enthalpy change for the reaction (?Hsys)
We have shown that
Calculate the enthalpy change of a reaction from
tabulated standard enthalpies of formation of
reactants and products.
substance ?Hof (kJ/mol) O2(g) 0 Fe(s)
0 Fe2O3(s) -822.16
4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)
DHosys S n DHof (products) S m DHof
(reactants) (2 ? -822.16) (0
0) -1644.3 kJ
?Ssurr - (-1644.3 kJ) 5.515 kJ/K
298.15 K
5515 J/K
NB The entropy of the surroundings increases
substantially as a result of the heat liberated
during the rusting of the iron.
39

• Now we know both ?Ssys and ?Ssurr can
  calculate the entropy
• change of the universe ?Suniv.

?Suniv ?Ssys ?Ssurr ?Suniv -543.7
J/K 5515 J/K 4971.3 J/K

• NB
• Although the entropy of the system ?Ssys
  (reacting substances)
• decreases during the corrosion of iron, the
  heat liberated by the
• reaction leads to a substantial increase in
  the entropy of the
• surroundings ?Ssurr that more than
  compensates for the loss of
• entropy in the system. The overall entropy
  of the universe ?Suniv
• increases the 2nd Law of Thermodynamics.

?Suniv ?Ssys ?Ssurr gt 0 (spontaneous)
40
Focusing on the system
We have seen how it is possible through
calculation of ?Suniv, to predict whether or not
a chemical reaction can proceed spontaneously.
The calculation requires that we consider both
the chemical system ?Ssys and its surroundings
?Ssurr and is therefore a cumbersome
calculation. Prediction of reaction spontaneity
can be done in a simpler way that is based on the
2nd Law of Thermodynamics but focuses only on the
properties of the system. This method has the
advantage of being simpler to apply.
41
Focusing on the system

• Calculating the entropy change of the universe
  ?Suniv

?Suniv ?Ssys ?Ssurr 0

• Calculating the entropy change of the
  surroundings ?Ssurr

?Ssurr - ?H T

• Substituting

?Suniv ?Ssys - ?H
T

• Multiplying by T

-T ?Suniv ?H - T ?Ssys
42
Focusing on the system
-T ?Suniv ?H - T ?Ssys

• It is sensible to define a new quantity, with
  symbol G, such that

-T ?Suniv ?G
?G ?H - T ?Ssys

• NB
• ?G can be calculated from the properties of
  the
• system only, i.e., from ?H, T, and ?S.

• The a new quantity G is formally defined as

G H - TS
state function
and is called the Gibbs free energy, or the free
energy, and sometimes the Gibbs function, of a
system.
J. Willard Gibbs (1839 1903)
43
Gibbs Free Energy and Reaction Spontaneity

• The change in the Gibbs Free Energy is related
  to the change of
• entropy of the universe

-T ?Suniv ?G

• Accordingly
• For a spontaneous reaction ?Suniv gt
  0 so ?G lt 0
• For a reaction at equilibrium ?Suniv
  0 so ?G 0
• For a non-spontaneous reaction ?Suniv lt 0
  so ?G gt 0

NB The Gibbs free energy of a system must
decrease during a spontaneous process at constant
temperature and pressure, ie., ?G lt 0 (products
favoured). If ?G gt 0 the reaction is
non-spontaneous but the reverse reaction is
spontaneous.
44
Standard Free-Energy Changes
Standard free energies of formation are tabulated
for substances as were the enthalpies of
formation. Standard free energy of formation
(?Gfo) Free energy change for the formation of
1 mol of compound from elements with all
substances in their standard states.
Standard state conditions Pure compound at 25oC
and 1 atm for solids, liquids and gases. For
solutions standard state is 1M solution.
Standard free energy change (?Go)
?Go S n ?Gfo (products) S m ?Gfo (reactants)
?Gfo for elements in their standard states are
set to zero.
45
Example a) What is ?Go for the following reaction
at 25oC using tabulated values of ?Gfo
provided?
CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
substance ?Gfo(kJ/mol) CH4(g)
-50.8 CO2(g) -394.4 H2O(g)
-228.57
DGo S n DGfo (products) S m DGfo (reactants)
-394.4 2 ? (-228.57) -50.8 0
- 801 kJ
b) Recalculate ?Go for the reaction at 25oC using
tabulated values of ?Ho and So.
46
CH4(g) 2O2(g) ? CO2(g) 2H2O(g) T
25oC
substance ?Hfo(kJ/mol) So
(J/K-mol) O2(g) 0
205.0 CH4(g) -74.8
186.3 CO2(g) -393.5
213.6 H2O(g) -241.8
188.3
?Sorxn S n So(products) S m So(reactants)
?Sorxn (213.6 2 ? 188.3) (186.3 2 ?
205.0)
- 6.1 J/K
DHorxn S n DHof (products) S m DHof
(reactants)
?Ho (-393.5 2? (-241.8)) (-74.80 0)
- 802.3 kJ
?G ?H - T ?Ssys
?Go -802.3 kJ/mol 298K ? (-6.1 J/mol K) /1000
- 801 kJ/mol
47
Free Energy and Temperature
?G ?H - T ?S
H2O(s) ? H2O(l) spontaneous at 298 K
(25oC)
?H gt 0, ?S gt 0
At low values of T the magnitude of ?H
dominates and ?G gt 0.
At high values of T the magnitude of T?S
becomes dominant and ?G lt 0.
48
Example Determine the temperature at which the
following reaction will become spontaneous in the
opposite direction.
CH4 (g) 2O2 (g) ? CO2 (g) 2H2O (g)
We already calculated ?Ho and ?So for this
reaction previously
?Ho - 802.3 kJ and ?S - 6.1 J/K
Need to calculate the temperature at which ?G
0. The temperature at which the spontaneity is
reversed!!!
?G ?H - T?S 0
-802.3 kJ - T(-6.1J/K)/1000 0
T 131525 K
T 13 104 K
49
Exercise 19.63 For a particular reaction ?H -
32 kJ and ?S - 98 J/K. Assume that ?H and ?S
do vary with temperature. a) At what temperature
will the reaction have ?G 0?
?G ?H - T?S 0
T ?H - 32 000 J ?S - 98
J/K
326.5 K 3.3 102 K 330K
b) If T is increased from that in part (a) will
the reaction be spontaneous or nonspontaneous?
The sign of ?S is ve. As T increases T?S
becomes more ve. i.e., ?G becomes more ve.
Therefore the reaction will become nonspontaneous.
50
Example Use the enthalpy of fusion (?Hofus) and
enthalpy of vaporization (?Hovap) to calculate
the values of ?Sofus and ?Sovap for 1 mol of
water.
(?Hofus 6.02 kJ/mol, ?Hovap 40.7 kJ.mol)
At a phase transition, ?G 0 ?H - T?S
?H ?S
T
For fusion, T 273 K (0 oC)
For vaporization, T 373 K
?Hfus 6.02kJ/mol ?Sfus
T
273K 22.1 J/mol K
?Hvap 40.7kJ/mol ?Svap
T
373K 109 J/mol K
51
End of Chapter 19Thermochemistry