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```Chapter 19
Chemical Thermodynamics
Grace Ngubeni
Office: C106
Email: [email protected]
2
What is Thermodynamics?
THERMODYNAMICS
Heat
Temperature
Energy
“Not static” from
the Greek Word
for “power”
Thermodynamics is the study of energy and its transformation.
3
Recap
FIRST LAW OF THERMODYNAMICS
• You will recall that energy cannot be created nor destroyed.
• Therefore, the total energy of the universe is a constant.
• Energy can, however, be converted from one form to another
or transferred from a system to the surroundings or vice versa.
4
Recap
FIRST LAW OF THERMODYNAMICS
Two ways for energy to be transferred
Heat lost by the system
Heat gained by the system
q
+
q
SYSTEM
w
−
w
Work done on the system
Work done by the system
or
ΔE = q + w
5
Recap
Are q and w State Functions? No,
they are Path Functions
Whether the battery is shorted out
or is discharged by running the fan,
its E is the same.
6
SPONTANEOUS PROCESSES
• Spontaneous processes are
those that can proceed without
any outside intervention, i.e.
without ongoing input of energy
from outside the system.
• The gas in vessel B will
spontaneously effuse into vessel
A, but once the gas is in both
vessels, it will not
B.
7
SPONTANEOUS PROCESSES
• Processes that are spontaneous
in one direction are
nonspontaneous in the reverse
direction.
• Spontaneity does not predict the
rate (speed) of the reaction.
8
SPONTANEOUS DOESN’T MEAN FAST
• Rusting of an iron nail is spontaneous and slow.
Fe + O2 → Fe2O3
9
SPONTANEOUS PROCESSES
• Processes that are spontaneous at one temperature may be
nonspontaneous at other temperatures.
– Above 0 oC, it is spontaneous for ice to melt.
– Below 0 oC, the reverse process is spontaneous.
10
QUESTION
Melting a cube of water ice at 10°C and atmospheric pressure
is a ________ process.
a) Spontaneous
b) Nonspontaneous
Attempt the question first.
The solution will be provided in a separate document.
11
REVERSIBLE PROCESSES
System at higher T
than Surroundings
In a reversible process the system
changes in such a way that the
system and surroundings can be
put back in their original states by
exactly reversing the process.
System at lower T
than Surroundings
The direction of heat flow can be
changed by an Infinitesimal
change in T or another property.
12
IRREVERSIBLE PROCESSES
• Irreversible processes cannot be undone by exactly reversing
the change to the system.
• Spontaneous processes are irreversible.
13
SPONTANEOUS PROCESSES
Reversible
Irreversible
• Chemical systems in equilibrium are
reversible.
• Both the system and its surroundings
can be restored to their original state.
• In an spontaneous process, the path
between reactants and products is
irreversible.
original state without there being a
permanent change in the
surroundings.
Thermodynamics gives us the direction of a process. It cannot predict the speed
at which the process will occur.
Direction of a process can be exothermic or endothermic.
14
ENTROPY
• Entropy (S) is a term coined by Rudolph Clausius in the 19th
century.
• Clausius was convinced of the significance of the ratio of heat
(q) delivered and the temperature (T) at which it is delivered.
𝑞
𝑆=
𝑇
ENTROPY
15
gas
increasing disorder
• Entropy – measure of
the
disorder/randomness
of a system.
• It is related to the
various modes of
motion in molecules.
liquid
solid
Entropy increases
with the freedom of
motion of molecules.
Therefore,
S(g) > S(l) > S(s)
16
ENTROPY
• Like internal energy, E, and enthalpy, H, entropy (S) is a state
function.
• Therefore,
S = Sfinal  Sinitial
17
ENTROPY
• For a process occurring at constant temperature (an
isothermal process), the change in entropy is equal to the heat
that would be transferred if the process were reversible
divided by the temperature:
q rev
S =
T
Heat transferred under
reversible conditions
The units of entropy are J K-1
Temp at which process
takes place
18
The sign of entropy change, ΔS, associated with the
boiling of water is _______.
a) Positive
b) Negative
c) Zero
Attempt the question first.
The solution will be provided in a separate document.
QUESTION
H2O (l) → H2O (g)
19
S FOR PHASE CHANGES
Melting of a substance at its melting point and vaporization at its
boiling point are isothermal processes.
Consider melting of ice at 0 oC,
q = Hfusion
for a small change in heat by raising the temp of the surrounding
infinitely slowly,
qrev = Hfusion
20
S FOR PHASE CHANGES
• For the system,
qrev = Hfusion
q rev
Sfusion =
(constant T)
T
Hfusion
=
(units:
T
J/K)
21
EXAMPLE
• Consider ΔHfusion for H2O is 6.01 kJ/mol (take note: the values is
positive because melting is an endothermic process). Therefore,
calculate the ΔSfusion for melting 1 mol of ice at 0°C.
𝑞𝑟𝑒𝑣 ∆𝐻𝑓𝑢𝑠𝑖𝑜𝑛
∆𝑆𝑓𝑢𝑠𝑖𝑜𝑛 =
=
𝑇
𝑇
1 𝑚𝑜𝑙 (6.01 × 103 𝐽. 𝑚𝑜𝑙−1 )
=
= 22.0 𝐽. 𝐾 −1
273 𝐾
22
SECOND LAW OF THERMODYNAMICS
States:
• the entropy of the universe increases for spontaneous
processes, and
• the entropy of the universe does not change for reversible
processes.
23
SECOND LAW OF THERMODYNAMICS
In other words:
For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
These last truths mean that as a result of all spontaneous
processes, the entropy of the universe increases.
ENTROPY IS NOT
CONSERVED
24
EXAMPLE 1
The normal melting point of ice is 0 oC and its
molar enthalpy of fusion is ∆Hfus = 6.01 kJ/mol.
What is the entropy of the system when 1 mol
of ice melts at the normal freezing point?
25
•
∆S is a state function so we can calculate the heat
transferred if the process is reversible
•
∆Ssys = qrev/T = ∆Hfus /T
={(1 mol)(6.01x103 J/mol)} / 273K
= 22.0 J/K
•
∆Suniv= ∆Ssys + ∆Ssurr
= 22.0 J/K + (-22.0 J/K)
=0 J/K
26
EXAMPLE 2
Calculate the change in entropy of the system
and the change in entropy of the surrounding,
when our system is 1 mol of ice melting in the
palm of our hand, which is part of the
surrounding.
27
•
Note: The process is not reversible because the system and
surroundings are at different temperatures.
∆Ssys = qrev/T= ∆Hfus /T
={(1 mol)(6.01x103 J/mol)} / 273 K
= 22.0 J/K
The surroundings immediately in contact with the ice is
your hand. (Assume 370C = 310K)
∆Ssurr = qrev/T= ∆Hfus /T
={(1 mol)(-6.01x103 J/mol)} / 310 K
= -19.4 J/K
28
∆Suniv= ∆Ssys + ∆Ssurr
= 22.0J/K + (-19.4J/K)
=2.6 J/K
• The entropy of the universe increases for any
spontaneous process.
• This can be used to predict whether a process is
spontaneous or not.
 Reversible: Suniv = Ssystem + Ssurroundings = 0Example 1
 Irreversible: Suniv = Ssystem + Ssurroundings > 0Example 2
29
Which of these properties are state functions?
a) i) and ii) only
b) all of the above
c) ii) iii) iv)
d) iii) and iv) only
QUESTION
Consider the following thermodynamic properties.
i) Work done on a system
ii) Heat absorbed
iii) Entropy
iv) Enthalpy
Attempt the question first.
The solution will be provided in a separate document.
30
ENTROPY ON A MOLECULAR SCALE
•
•
Ludwig Boltzmann described the concept of entropy
on the molecular level.
Temperature is a measure of the average kinetic
energy of the molecules in a sample.
31
ENTROPY ON A MOLECULAR SCALE
• Molecules exhibit several types of motion:
– Translational: Movement of the entire molecule from one
place to another.
– Vibrational: Periodic motion of atoms within a molecule.
– Rotational: Rotation of the molecule about an axis or
32
ENTROPY ON A MOLECULAR SCALE
• Boltzmann envisioned the motions of a sample of
molecules, at a particular instant in time.
• This would be akin to taking a snapshot of all the
molecules.
• He referred to this sampling as a microstate of the
thermodynamic system.
• Microstate is a single possible arrangement of the
positions and kinetic energies of the molecules when
the molecules are in a specific thermodynamic state.
33
ENTROPY ON A MOLECULAR SCALE
• Each thermodynamic state has a specific number of
microstates, W, associated with it.
• Entropy is
S = k ln W
where k is the Boltzmann constant, 1.38  1023 J/K.
Entropy is a measure of how many microstates are associated
with a specific macroscopic state.
34
S = k ln W
R
8.3154 J K -1 mol1
-23
-1
k


1.38

10
J
K
N A 6.022  1023 mol1
The Boltzmann constant (k) is in effect the
equivalent of the universal gas constant, R,
but per molecule rather than per mole.
35
ENTROPY ON A MOLECULAR SCALE
Expansion of a Gas at the Molecular Level
The two molecules are colored red
and blue to keep track of them
Four possible arrangements (microstates)
once the stopcock is opened
Possible arrangements of two gas molecules in two flasks. (a) Before the stopcock is
opened, both molecules are in the left of the flask. (b) After the stopcock is opened,
there are four possible arrangements of the two molecules.
36
ENTROPY ON A MOLECULAR SCALE
Expansion of a Gas at the Molecular Level
• Probability of red molecule being in the left flask is
two in four (2/4) or ½.
• Probability of both molecules being in the left flask is
(½)2 = 1/4.
• For three gas molecules, the probability of all three
molecules being in the left flask is (1/2)3 = 1/8.
37
ENTROPY ON A MOLECULAR SCALE
Expansion of a Gas at the Molecular Level
• For one mole of gas, the probability of all molecules
being in the left flask at the same time is (1/2)N where
N = 6.022 x 1023.
• The gas spontaneously expands to fill both the left and
right flasks, and it does not spontaneously all go back
• With the stopcock opened, the arrangement of gas
molecules is more random of dispersed than when
the molecules are confined in the left flask.
38
ENTROPY ON A MOLECULAR SCALE
• The change in entropy for a process, then,
is
S = k ln Wfinal  k ln Winitial
S = k
ln
•
Wfinal
Winitial
Entropy increases with the number of
microstates in the system.
39
ENTROPY ON A MOLECULAR SCALE
• The number of microstates and, therefore,
the entropy tends to increase with
increases in
– temperature
– volume
– the number of molecules
40
QUESTION
Predict the sign of ΔS for each the following
processes:
a) Dissolving a solute in a solvent to produce a
solution.
b) Freezing a liquid.
c) Condensing vapour.
d) Evaporating a liquid.
Attempt the question first.
The solution will be provided in a separate document.
41
ENTROPY AND PHYSICAL STATES
Entropy increases with the freedom of motion of
molecules. Therefore, S(s) < S(l) < S(g)
42
INCREASE vs. DECREASE IN ENTROPY
2NO(g) +O2(g)2NO2(g)
Generally, when a solid is
dissolved in a solvent,
entropy increases.
Entropy for the reaction?
Entropy decreases when
NO(g) is oxidized by O2(g)
to NO2(g)
43
CHANGES IN ENTROPY
• In general, entropy increases when
– Gases are formed from liquids and
solids.
– Liquids or solutions are formed from
solids.
– The number of gas molecules
increases.
– The number of moles increases.
44
QUESTION
If a process is exothermic (loss of energy), does the entropy of
the surroundings:
A) always increase
B) always decrease
C) sometimes increase and sometimes decrease, depending on
the process?
Attempt the question first.
The solution will be provided in a separate document.
45
Predict whether change in entropy (ΔS) is positive
or negative for each of the processes assuming
each occurs at constant temperature.
CO2(g)
(ii) HCl (g) + NH3(g)
a)
b)
c)
d)
NH4Cl(s)
(i) Negative (ii) Positive
(i) Negative (ii) Negative
(i) Positive (ii) Negative
(i) Positive (ii) Positive
QUESTION
(i) CO2(s)
Attempt the exercise first.
The solution will be provided in a separate document.
46
EXERCISE
Choose the sample of matter that has greater
entropy in each pair and explain your choice:
(i) 1 mole of NaCl(s) or 1 mole of HCl (g) at 25oC.
(ii) 1 mole of HCl (g) or 1 mole of Ar (g) at 25oC.
Attempt the exercise first.
The solution will be provided in a separate document.
47
THIRD LAW OF THERMODYNAMICS
The entropy of a pure crystalline substance
at absolute zero is 0.
S (0K) = 0
At T = 0 K there is only one microstate; W = 1
So S = k ln W = 0
48
THIRD LAW OF THERMODYNAMICS
Since, the entropy of a pure crystalline substance
at absolute zero is 0.
Remember that S = k ln W
k = 1.38  1023 J/K
When W = 1,
S = 1.38  1023 J/K ln 1
=0
49
ENTROPY CHANGES IN CHEMICAL REACTIONS
Entropy vs Temperature
graphs can be obtained
by carefully measuring
how the heat capacity
(C) of a substance varies
with temperature and
we can use the data to
obtain absolute
entropies at different
temperatures.
50
STANDARD MOLAR ENTROPIES
• These are molar entropy
values of substances in their
standard states.
• Standard entropies tend to
increase with increasing
molar mass.
• Unlike enthalpies of
formation, standard molar
entropies of elements at the
reference temperature of
298K are not zero.
51
STANDARD MOLAR ENTROPIES
• Standard molar entropies
of gases are generally
greater than those of
liquids and solids.
• Standard entropies tend
to increase with increasing
number of atoms in the
formula of a substance.
52
STANDARD ENTROPIES
Larger and more complex molecules have
greater entropies.
53
Predict whether ΔS is positive or negative, assuming
each process occurs at constant temperature.
1.
2.
3.
4.
5.
6.
7.
8.
H2O(l) H2O(g)
Ag+(aq) + Cl-(aq) AgCl(s)
4Fe(s)+ 3O2(g) 2Fe2O3(s)
CO2(l) CO2(g)
2SO2(g) + O2(g)  2SO3(g)
N2(g) + O2(g)  2NO(g)
NaCl(s) Na+(aq) + Cl-(aq)
CaSO4(s) Ca2+(aq) + SO42-(aq)
QUESTION
•
Attempt the question first.
The solution will be provided in a separate document.
54
EXAMPLE
Calculate the standard entropy change of reaction at
25oC for the synthesis of ammonia.
N2(g)+ 3H2(g) → 2NH3(g)
ΔSo= ∑ nSo (Products) – ∑ mSo (Reactants)
ΔSo= 2So(NH3) – { So(N2) + 3So(H2)}
= {2 mol x 192.5 J/K mol} – {(1mol x 191.5 J/K
mol) + (3 mol x (130.6 J/K mol)}
= -198.3 J/K
55
QUESTION
Calculate the standard entropy
change for the following reactions:
a. H2O(l) H2O(g)
b. H2(g) + 1/2 O2(g) H2O(l)
c. CH3OH(l) CH3OH(g)
Attempt the question first.
The solution will be provided in a
separate document.
56
ENTROPY CHANGES IN SURROUNDINGS
• Heat that flows into or out of the system changes
the entropy of the surroundings.
• For an isothermal process:
Ssurr =
qsys
T
At constant pressure, qsys is simply H for the system.
∴
Ssurr =
 Hsys
T
57
ENTROPY CHANGE IN THE UNIVERSE
• The universe is composed of the system and
the surroundings.
• Therefore,
Suniverse = Ssystem + Ssurroundings
• For spontaneous processes
Suniverse > 0
58
ENTROPY CHANGE IN THE UNIVERSE
• It is preferable to predict the entropy on the basis of
the property of the system
• This becomes:
Suniverse = Ssystem + Hsystem
T
Multiplying both sides by (T),
TSuniverse =  TSsystem + Hsystem
TSuniverse is defined as the Gibbs energy
(also known as the “free energy” or Gibbs free energy), G
59
GIBBS FREE ENERGY
• Endothermic process that is spontaneous e.g.
dissolution of ammonium nitrate in water.
– must always be accompanied by an increase in
entropy of the system.
• Also processes that are spontaneous yet proceed
with a decrease in entropy of the system e.g.
formation of NaCl from its constituent elements.
– always exothermic.
• Spontaneity therefore involves two
thermodynamic concepts – enthalpy and entropy.
60
GIBBS FREE ENERGY
G = H – TS
(G = Gibb’s free energy or just free energy)
For an isothermal process, the change in free
energy of the system, G, is
G = H - TS
Under standard conditions,
G° = H  TS
61
GIBBS FREE ENERGY
• TSuniverse is defined as the Gibbs free energy,
G.
G = TSuniverse
• When Suniverse is positive, G is negative.
• Therefore, when G is negative, a process is
spontaneous.
62
GIBBS FREE ENERGY
• If both T and P are constant, the relationship between
the sign of G and the spontaneity of a reaction is:
1. If G < 0 (negative), the forward reaction is spontaneous.
2. If G = 0, the system is at equilibrium.
3. If G > 0 (positive), the reaction is spontaneous in the
reverse direction.
It is more convenient to use G as a criterion for spontaneity than to
use Suniv because G relates to the system alone and avoids the
complication of having to examine the surroundings.
63
GIBBS FREE ENERGY
Potential energy
Free energy
Analogy between the gravitational potential energy change of a boulder rolling
down a hill and the free energy change in a spontaneous reaction.
In any spontaneous process carried out at constant temperature and pressure,
the free energy always decreases.
64
GIBBS FREE ENERGY
N2 (g) + 3H2 (g) ⇄ 2NH3 (g)
1. If G < 0, Q < K and
the forward reaction
is spontaneous.
2. If G = 0, Q = K and
the system is at
equilibrium.
3. If G > 0, Q > K and
the reverse reaction
is spontaneous
65
FREE ENERGY CHANGES
At temperature of 25°C,
G° = H  TS
How does G change with temperature?
66
FREE ENERGY AND TEMPERATURE
G° = H  TS
• There are two parts to the free energy equation:
 H— the enthalpy term
 TS — the entropy term
• The temperature dependence of free energy,
then comes from the entropy term.
67
FREE ENERGY AND TEMPERATURE
G° = H  TS
68
STANDARD FREE ENERGY CHANGES
Analogous to standard enthalpies of formation
are standard free energies of formation, G.
G = SnGf (products)  SmGf (reactants)
where n and m are the stoichiometric
coefficients.
69
Thermodynamic Quantities for Selected Substances at 298.15 K (25°C)
Substance
Carbon
C (s, diamond)
C (s, graphite)
C2H2 (g)
C2H4 (g)
C2H6 (g)
CO (g)
CO2 (g)
Hydrogen
H2( g)
Oxygen
O2 (g)
H2O (l)
ΔH°f (kJ/mol)
1.88
0
226.7
52.30
-84.68
-110.5
-393.5
0
0
-285.83
ΔG°f (kJ/mol)
S° (J/K-mol)
2.84
0
209.2
68.11
- 32.89
-137.2
-394.4
2.43
5.69
200.8
219.4
229.5
197.9
213.6
0
130.58
0
-237.13
205.0
69.91
70
EXAMPLE
Calculate ΔG° (in kJ) for the following reaction at
1 atm and 25°C:
2C2H6 (g) +7O2 (g) → 4CO2 (g)+6H2O (l)
ΔGf° C2H6 (g) = -32.89 kJ/mol
ΔGf° CO2 (g) = -394.4 kJ/mol
ΔGf° H2O (l) = -237.2 kJ/mol
What is the value of O2?
71
G = SnGf (products)  SmGf (reactants)
ΔGo={4ΔGf°(CO2(g))+6ΔGf°(H2O(l))}
– {2ΔGf°(C2H6 (g))+ 7ΔGf°(O2(g))}
={4×(-394.4) + 6× (-237.2)}
– {(-2×32.89)+(7×0)}
= -2935.0kJ
72
EXAMPLE
Calculate the standard free-energy change for
ammonia synthesis at 298K.
N2 + 3H2  2NH3
Given ΔHo = -92.2kJ and ΔSo = -198.7J/K
G° = H  TS
= (-92.2 × 103 J) - (298 K) (-198.7 J/K)
= (-92.2 × 103 J) - (-59.2 × 103 J)
= -33 000 J
= -33.0 kJ
73
EXAMPLE
Calculate the value of ΔS° for the catalytic
hydrogenation of acetylene to ethene,
C2H2 (g) + H2 (g) → C2H4 (g)
S° = SnS°(products) - SmS°(reactants)
S° = {1mol × 219.4 J/mol-K} –
{(1mol × 200.8 J/mol-K) + (1mol × 130.58 J/mol-K)}
= 219.4 J/K - 331.38 J/K
= -112.0 J/K
74
EXAMPLE
Calculate ΔG° (in kJ) for the following reaction
at 1 atm and 25°C:
2C2H6 (g) + 7O2 (g) →4CO2 (g) +6 H2O (l)
• ΔHf° C2H6 (g) = -84.7 kJ/mol
S° C2H6 (g) = 229.5 J/K·mol
• ΔHf° CO2 (g) = -393.5 kJ/mol;
S° CO2 (g) = 213.6 J/K·mol
• ΔHf° H2O (l) = -285.8 kJ/mol;
S° H2O (l) = 69.9 J/K · mol
• S° O2 (g) = 205.0 J/K · mol
LONG METHOD
ΔH°=Sum Products – Sum of Reactants
ΔS°=Sum Products – Sum of Reactants
G° = H  TS
Different manifestation of earlier
example
75
EXAMPLE
Iron metal can be produced by reducing iron(III)oxide
with hydrogen.
Fe2O3(s) +3 H2(g) 2Fe(s) +3 H2O(g)
ΔHo = 98.8kJ; ΔSo = 141.5J/K
• Is the reaction spontaneous at 25oC?
• At what temperature will the reaction become
spontaneous?
76
ΔGo= ΔHo – TΔSo
= 98.8 kJ- (298 K)(0.1415 kJ/K)
= 56.6 kJ
Therefore, reaction is not spontaneous because
ΔG>0.
We can estimate the temperature at which ΔG
changes from a positive to a negative value by
setting
ΔGo= ΔHo – TΔSo=0
Therefore, this will give us reaction at equilibrium.
77
T=ΔHo/ΔSo
=(98.8 kJ)/(0.1415 kJ/K)
= 698 K
Note: that this calculation assumes that the value
of ΔH and ΔS remains unchanged over the
temperature range.
Therefore, the reaction will be spontaneous at a
temperature just above 698 K.
ΔGo= ΔHo – TΔSo
78
Find the temperature above which a reaction with a ΔHo
of 123.0 kJ and a ΔSo of 90.00 J/K becomes spontaneous.
QUESTION
Attempt the question first.
The solution will be provided in a separate document.
79
FREE ENERGY AND EQUILIBRIUM
Under any conditions, standard or nonstandard,
the free energy change can be found this way:
G = G + RT ln Q
(R = universal gas constant = 8.314Jmol-1K-1)
(Under standard conditions, all concentrations are 1 M,
so Q = 1 and ln Q = 0; the last term drops out.)
80
FREE ENERGY AND EQUILIBRIUM
Gf° = 0
81
THE REACTION QUOTIENT, Q
•
The reaction quotient, Q, is obtained by substituting the
reactant and product concentrations or partial
pressures at any point during a reaction into an
equilibrium constant expression.
aA + bB
•
dD + eE
The reaction quotient expression for this
reaction would be
Qc =
[D]d[E]e
[A]a[B]b
OR
PDdPEe
Qp =
PAaPBb
82
FREE ENERGY AND EQUILIBRIUM
At equilibrium, Q = K, and G = 0.
The equation {G = G + RT ln Q} becomes
0 = G + RT ln K
Rearranging, this becomes
G = RT ln K
ln K = G/-RT
or,
K = eG/RT
83
If Q = K,
the system is at equilibrium.
84
If Q > K,
there is too much product and the equilibrium shifts
to the left.
85
If Q < K,
there is too much reactant, and the equilibrium shifts
to the right.
86
EXAMPLE
Calculate the free-energy change for ammonia synthesis at 298K
given the following sets of partial pressures.
a. 1.00atm N2, 3.0 atm H2, 0.020 atm NH3
N2+ 3H2 2 NH3 Go=-33.0 kJ
Qp =(0.020)2/{(1.0)(3.0)3}
=1.5x10-5
aA + bB
PDdPEe
Qp =
PAaPBb
dD + eE
87
EXAMPLE CONT’D
G can be obtained from data tables. Value obtained from
earlier calculation & it is given in the question.
G = G + RT ln Q
=(-33.0 X103Jmol-1) + (8.314J.K-1mol-1)(298K)(ln 1.5x0-5)
= -60 519.5 J/mol
= -60.5 kJ/mol
88
QUESTION
Methanol an important alcohol used in the manufacturing of
adhesives, fibers and plastics is synthesized industrially by the reaction.
CO(g) + 2 H2(g)  CH3OH(g)
Calculate the equilibrium constant for this reaction using the data for
standard conditions:
ΔGo = ΔGof(CH3OH) - {ΔGof (CO)+2ΔGof(H2)}
89
RELATING Kp and Kc
The equilibrium constant obtained by this procedure is Kp
because the reaction and products are gases and their standard
states are defined in terms of pressure. If we want the value of Kc
we must calculate it from the relation
Kp=Kc(RT)Δn
where R must be expressed in the proper units
0.08206(L.atm)/(K.mol)
90
EXAMPLE
Given: Kp = 3.5; Δn = -2; Temperature = 474.15 K
Calculate Kc
Kp=Kc(RT)Δn
3.5 = Kc (0.08206 x 474.15 K)-2
Kc = 5299
91
EXAMPLE
The equilibrium constant for a reaction is 0.48 at
25 °C. What is the value of ΔG° (kJ/mol) at this
temperature?
ΔGo = -RT ln K
Ans: 1.8 kJ/mol
92
EXAMPLE
The value of ΔG° at 25 °C for a reaction is 1.8 kJ/mol.
Calculate the equilibrium constant for this reaction at
25 °C
ΔGo = -RT ln K
ln K = -ΔGo/RT
K = eG/RT
K = e-1.8x10^3/(8.314)(298)
(Ans: K=0.48)
93
DRIVING NON SPONTANEOUS REACTIONS
• Consider extraction of the copper metal from the
mineral chalcocite, which contains Cu2S. The
decomposition is nonspontaneous
Cu2S(s)2 Cu(s) + S(s) ΔGo=+86.2kJ
Cu cannot be obtained directly since ΔGo is positive.
The reaction can be coupled so that the overall
reaction is spontaneous
S(s) +O2(g)SO2(g) ΔGo=-300.4kJ
94
By coupling these reactions Cu can be extracted via a spontaneous
reaction.
CuS(s)+O2(g)2Cu(s)+SO2(g)ΔGo=-214.2kJ.
Biological reactions also use spontaneous reactions to drive non
spontaneous reactions.
The metabolism of food is the usual source of free energy
needed to do work of maintaining biological system.
C6H12O6(s)+6O2(g)6CO2(g)+6H2O(l) ΔGo=-2880 kJ.
Free energy of oxidation used to convert ADP to ATP.
The conversion of ATP to ADP releases free energy that can be used
to drive other reactions (Converts simple molecules to more
complex ones).
95
COUPLED REACTIONS
96
QUESTION
Calculate ΔG (the free-energy change under non standard
condition) at 298K for a reaction mixture that consists of 1.0atm
N2, 3.0 atmH2 and 0.50 atm NH3.
97
QUESTION
Consider the decomposition of N2O4.
N2O4 (g) 2NO2 (g)
ΔHo=57.1kJ ΔSo=175.8 J/K.
• Is the reaction spontaneous at 25oC?
• Estimate the temperature at which the reaction becomes
spontaneous?
Solution
• Reaction is not spontaneous. ΔGo= 4.71 kJ.
• Estimated T = 325 K
98
EXAMPLE
• Calculate ∆G at 298K for a reaction mixture
that consists of 0.50 atm N2, 0.075 atm H2 and
2.0atm NH3
• Remember :
• ∆Go from earlier calculation = -33.3 kJ/mol.
99
QUESTION
Calculate the standard free energy change,ΔGo, and the
equilibrium constant K, at 298K for the reaction:
H2(g)+Br2(l)  2HBr(g)
100
101
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