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Ch. 16: Thermodynamics
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Geysers are a dramatic display of thermodynamic principles in nature. As water
inside the earth heats up, it rises to the surface through small channels. Pressure
builds up until the water turns to steam, and steam is expelled forcefully
through a hole at the surface. (credit: modification of work by Yellowstone
National Park)
Chemistry: OpenStax
!  Principles of General Chemistry (CC BY-NC-SA 3.0):
http://2012books.lardbucket.org/pdfs/principles-of-generalchemistry-v1.0.pdf
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1
Thermodynamics
16.1 Spontaneity
• 
!  Recall thermochemistry (study of heat changes) from
CHM 150/151: Chapter 5 sections:
!  1 (Conservation of energy),
!  2 (Energy change, system, surroundings)
!  3 (Enthalpies of Physical and Chemical Changes),
!  3 (Calculating ΔHorxn using ΔHof – Standard Enthalpy
• 
• 
Thermodynamics allows us to determine IF a reaction will occur
(is spontaneous) as well as telling us the direction in which (and
extent to which) a reaction will occur (like Q vs K).
A spontaneous process is one that proceeds on its own without
any external influence.
A spontaneous process always moves a reaction mixture toward
equilibrium.
of Formation).
!  We’ll introduce the concept of Entropy this semester.
Figure 16.4: An isolated system consists of an ideal gas in one flask that is connected
by a closed valve to a second flask containing a vacuum. Once the valve is opened,
the gas spontaneously becomes evenly distributed between the flasks.
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4
16.1 Spontaneity
• 
• 
• 
• 
• 
• 
Spontaneity
Other spontaneous processes include:
Ice melting at room temperature
Solid sodium reacting violently with water
A ball rolling downhill
An egg boiling in hot water
Heat flowing from a hot object to a cold one
!  Spontaneity is determined by enthalpy and entropy of
a reaction.
!  A system tends to move toward a lower state of energy.
!  The enthalpy (ΔH in kJ/mol) of many spontaneous processes
are exothermic (e.g., combustion reactions), but some
spontaneous processes can be endothermic under the right
conditions (e.g., melting of ice above 0oC).
!  A system tends to become more disordered.
!  The entropy (ΔS in J/mol·K) of a spontaneous system tends to
increase. There is an increase in the disorder or randomness of
a system (i.e., your room or office naturally becoming messier
over time).
Figure 16.5: When two objects at different temperatures come in contact, heat
spontaneously flows from the hotter to the colder object.
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Predicting Sign of ΔHorxn
150/151 Thermochemistry Review
!  Predict signs of enthalpy for the following processes:
!  1st Law of Thermodynamics: Energy is conserved, it cannot
be created or destroyed
!  If a system / reaction gives off heat, the surroundings /
universe must absorb it (and vice versa).
!  Heat flow (enthalpy, ΔH) is always defined with reference
to the system:
!  System absorbs heat, ΔH > 0
!  Associated with breaking bonds or Intermolecular forces.
!  Heat energy goes into a system to break the bonds or pull
Decomposition
Combination
Acid-Base Neut.
Combustion
• 
atoms/molecules apart.
• 
!  System gives off heat, ΔH < 0
!  Associated with making bonds or Intermolecular forces
Melting/fusion
+
Boiling/vaporization +
Freezing
Condensation
+
Sublimation
Deposition
In Chemistry reference manuals, only ΔHofusion and
ΔHovaporization are listed. The opposite processes have the
same values but are opposite in sign.
ΔHosublimation = ΔHofusion + ΔHovaporization
+
-
7
8
Defining ΔHof; Calculating ΔHorxn
Defining ΔHof, Calculating ΔHorxn
!  Heat of formation, ΔHof , is the enthalpy change when 1
!  Write the equation for the heat of formation of water.
mole of a substance is made from its elements in their
standard state (defined in section 5.3)
H2 (g) + ½ O2 (g) ! H2O (l)
!  The symbol refers to the standard state, 1.00 atm pressure,
o
25.0oC, 1.00 M for solutions (see section 16.3)
0
0
-285.83 kJ/mol
!  This is how the values in Appendix G are obtained.
ΔHorxn = Σ n·ΔHof (products) - Σ n·ΔHof (reactants)
!  We then use these values to calculate the enthalpy of a
reaction.
!  n is moles (coefficient in balanced equation)
!  ΔHof values, Appendix G, reported in kJ/mol
!  ΔHof for elements is 0 (no energy needed!)
ΔHorxn = Σ n·ΔHof (products) - Σ n·ΔHof (reactants)
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Calculating ΔHorxn
Group Work, Calculating ΔHorxn
!  Calculate ΔHorxn for the following reaction:
!  Calculate ΔHorxn for the following reaction:
4 NH3(g) + 5 O2(g) ! 4NO(g) + 6 H2O(g). What
1C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2 H2O(g)
information do you need?
!  Find ΔHof values in Appendix G.
!  NH3(g) = - 45.9 kJ/mol
!  O2(g) = 0 kJ/mol
!  NO(g) = 90.25 kJ/mol
!  H2O(g) = - 241.82 kJ/mol
Predict sign of ΔHorxn first.
!  Find ΔHof values in Appendix G.
!  C2H4(g) = 52.4 kJ/mol
!  O2(g) = 0 kJ/mol
!  CO2 (g) = - 393.51 kJ/mol
!  H2O(g) = - 241.82 kJ/mol
! Answer: -906.3 kJ (sig figs by decimal places!)
! Watch physical states when looking up values
! (l) versus (g) has a different value!
10
! Answer: -1323.1 kJ
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16.2 Entropy
Entropy of Molecules
To predict spontaneity, both the enthalpy and the entropy of a
process must be known.
Entropy (S) of a system is a measure of how spread out or how
dispersed the system’s energy is (measured in J/mol·K).
ΔS (entropy change for a system) = Sfinal - Sinitial
If randomness increases, ΔS is positive.
If randomness decreases, ΔS is negative.
The atoms, molecules, or ions that compose a chemical system can
undergo several types of molecular motion, including translation, rotation,
and vibration. The greater the molecular motion of a system, the greater the
number of possible microstates and the higher the entropy.
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Entropy of a System
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-v1.0/s19-chemical-equilibrium.html, CC-BY-NC-SA 3.0 license
Which phase has greatest entropy?
!  Solid, liquid or gas? (ice, water, or steam)
Figure 16.8: The sixteen microstates associated with placing four
particles in two boxes are shown. The microstates are collected into
five distributions—(a), (b), (c), (d), and (e)—based on the numbers of
particles in each box. Scenario c is most probable because it is most
disperse.
Which side has greater entropy?
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Entropy and Temperature
Figure 16.11: As temperature increases, entropy increases
because of increased kinetic energy and molecular motion.
Entropy increases as a substance melts (s ! l); even
more increase is seen as it boils (l ! g).
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Entropy of Dissolution
Entropy Trends
Dissolving a substance can lead to an increase or a decrease in
entropy, depending on the nature of the solute.
Molecular solutes (i.e. sugar): entropy increases
Entropy increases as the number of gas particles increases:
N2O4(g) ! 2 NO2(g)
Comparing entropy of molecules – Entropy increases as:
1) temperature increases.
H2O at 10oC vs H2O at 100oC
2) mass of an atom or molecule increases (more
electrons moving around)
F2 (158 J/K"mol) < Cl2 (165 J/K"mol)
3) molecular complexity increases (i.e., number of
atoms, number of heavier atoms, etc.)
CH4 (186 J/K"mol) < C2H6 (230 J/K"mol)
Ionic compounds: entropy can decrease or increase
NaCl (s): ΔSsoln = +43.4 J/K
AlCl3 (s): ΔSsoln = -253.2 J/K
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Group work: Which has greater entropy?
Greater entropy?
!  Determine which example in each scenario will have more
!  Determine which example in each scenario will have more
entropy.
entropy.
!  Temperature:
C(s) at
!  Volume:
1 mol O2(g) in 1 L
!  Phase:
O2(g)
O2(l)
!  Molar mass:
He(g)
Ne(g)
25oC
!  Molecular Complexity:
C(s) at
35oC
1 mol O2(g) in 2 L
C4H10(g)
!  Temperature:
C(s) at 25oC
!  Volume:
1 mol O2(g) in 1 L
!  Phase:
O2(g)
O2(l)
!  Molar mass:
He(g)
Ne(g)
!  Molecular Complexity:
C3H6(g)
C4H10(g)
C(s) at 35oC
1 mol O2(g) in 2 L
C3H6(g)
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Processes that Increase Entropy
22
Predict the sign of ΔS
!  Does entropy increase (ΔS > 0) or decrease (ΔS < 0) for
Entropy usually increases for the following:
the following? Why?
!  2SO2 (g) + O2 (g) → 2SO3 (g)
•  Melting, vaporization, or sublimation
!  2O3 (g) → 3O2 (g)
•  Temperature increase
!  CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
!  I2 (s) → I2 (g)
•  Reaction resulting in a greater number of gas
molecules (or greater molar mass of solid)
!  C2H4(g) + 3O2 (g) → 2CO2 (g) + 2H2O(g)
!  4 Al(s) + 3 O2(g) → 2Al2O3(s)
•  Decomposition reactions
Example 16.3
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16.3 2nd and 3rd Laws of Thermo
Standard Entropy Values (So)
Which phase has the lowest entropy? Highest?
!  3rd Law of Thermodynamics: The entropy of a
perfectly ordered crystalline substance at 0 K is
zero.
!  This allows us to calculate entropy (So: (J/mol•K))
and changes in entropy (ΔSo); unlike enthalpy where we can only measure change!
!  Use Standard Molar Entropies, So, (Table 16.2 and
Appendix G) - entropy of 1 mole of a pure
substance at 1 atm and 25oC
ΔSorxn
=Σ
n·So
(products) -
Σ
n·So
(reactants)
25
Substance
So (J mol-1 K-1)
Substance
So (J mol-1 K-1)
C (s, diamond)
2.38
H (g)
114.6
C (s, graphite)
5.740
H2 (g)
130.57
Al (s)
28.3
CH4 (g)
186.3
Al2O3 (s)
50.92
HCl (g)
186.8
BaSO4 (s)
132.2
H2O (g)
188.8
CO (g)
197.7
H2O (l)
70.0
CO2 (g)
213.8
Hg (l)
75.9
C2H4 (g)
219.3
H2O2 (l)
109.6
C2H6 (g)
229.5
Br2 (l)
152.23
CH3OH (g)
239.9
CCl4 (l)
214.4
CCl4 (g)
309.7
Note: The standard entropy for a pure element in its most stable
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form is NOT zero (like for ΔHof).
Entropy and the 2nd Law
Calculate Entropy of Reaction
The second law of thermodynamics states that in any
spontaneous process, the total entropy of a system and its
surrounding always increases.
ΔSuniverse (or ΔStotal) = ΔSsys + ΔSsurr > 0
ΔSsys (or ΔSrxn) (that we just calculated)
ΔSsurr (or ΔSH2O) is entropy change for surroundings
!  Calculate ΔSorxn for:
!  1 C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2 H2O(g)
!  C2H4(g) = 219.3 J/mol•K,
!  O2(g) = 205.2 J/mol•K,
!  CO2(g) = 213.8 J/mol•K
!  H2O(g) = 188.8 J/mol•K
ΔStotal > 0; reaction is spontaneous
ΔStotal < 0; reaction is nonspontaneous
ΔStotal = 0; system is at equilibrium
!  ΔSorxn = - 29.7 J/K (close to zero because same
number of moles of gas on both sides)
Examples 16.5, 16.6
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28
Entropy Changes in Surroundings
Entropy Changes in Surroundings
!  ΔSsurr is related to the enthalpy of the reaction; if
!  Temperature is inversely related to ΔSsurr:
heat enters surroundings, molecules have more
energy and entropy increases.
!  If surroundings have a low temperature, adding heat results
in a substantial change in randomness. (Like throwing a
huge rock into a calm lake.)
!  Conversely, if surroundings have a high temperature, adding
Entropy
Entropy
Heat
heat makes a marginal difference in change in randomness.
(Like throwing a pebble into the ocean during a hurricane.)
!  Putting it all together we get:
Heat
ΔSsurr =
-ΔH sys
T
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Total Entropy Changes
16.4 Free Energy & Spontaneity
•  Gibbs free energy (G) or free energy can be used to express
spontaneity from the perspective of the system.
•  ΔG (kJ/mol) is the maximum amount of energy available to
do work on the surroundings and takes into account both
enthalpy and entropy.
•  In a spontaneous process at constant temperature and
pressure, the free energy of the system always decreases.
!  ΔStotal = ΔSrxn + ΔSsurr
!  ΔStotal = ΔSrxn – ΔHrxn/T
!  Multiply both sides by –T
!  -TΔStotal = ΔGrxn (note the opposite signs!)
•  The actual amount of work (wmax) obtained is always less than the
maximum available because of energy lost in carrying out a
process (given off as heat, light, sound, etc. energy).
!  ΔGrxn = ΔHrxn – TΔSrxn
!  ΔGrxn is Gibbs Free Energy; ΔG < 0 indicates
spontaneous reaction
ΔGsys = ΔHsys - TΔSsys
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Predicting Spontaneity
32
Gibbs Free Energy and Temperature
ΔGsys = ΔHsys - TΔSsys
ΔG = ΔH – TΔS: Notice that ΔG is temperature dependent!
Temperature plays a role in determining spontaneity. Also
note the absence of standard conditions notation (o).
ΔG < 0 The reaction is spontaneous (still moving forward).
ΔG > 0 The reaction is nonspontaneous (moves in reverse
direction).
ΔG = 0 The reaction is at equilibrium
Endothermic reactions will be spontaneous at higher
temperatures (where TΔS > ΔH)
Figure 16.12
Exothermic reactions will be spontaneous at lower
temperatures (where ΔH > TΔS)
33
“Higher” and “lower” temps are reaction-specific. Every
reaction has a cut-off temperature depending on values of
ΔH and ΔS.
Standard Free Energy Changes
Standard Free Energy Changes
The standard free energy of reaction (ΔGorxn) is freeenergy change for a reaction when it occurs under
standard-state conditions.
We can calculate ΔGorxn = ΔHorxn – TΔSorxn for the reaction:
1 C2H4(g) + 3O2 (g) ! 2 CO2 (g) + 2 H2O(g)
ΔHorxn = -1323.1 kJ
ΔSorxn = - 29.7 J/K
Standard-state conditions are:
•  Solids, liquids, and gases in their pure form
•  1 M solutions, 1 atm gases, 25oC (298 K)
ΔGorxn = ΔHorxn – TΔSorxn
ΔGorxn = -1314.2 kJ
ΔG (non-standard) predicts spontaneity
ΔGo (standard – at 25oC) predicts reactant- or productfavored at equilibrium; relates to K
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Example 16.7
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Gibbs Free Energy
Temperature Dependence
ΔG = Gibbs Free Energy (or free energy), kJ/mol: Uses
both enthalpy and entropy to determine if a reaction will be
spontaneous.
!  ΔG = 0: 1) turning point between spontaneous and non-
Calculate ΔG when ΔH = -227 kJ, ΔS = - 309 J/K, T = 1450 K.
Are you calculating Standard or Non-Standard ΔG?
Is this process spontaneous?
!  0 = ΔH - TΔS
ΔG = ΔH – TΔS = -227 kJ – (1450 K * -0.309 kJ/K) = 221 kJ;
Non-spontaneous at this temperature.
spontaneous; 2) phase changes (occur at equilibrium)
!  ΔG = 0
!  Solve for T ! T = ΔH / ΔS
!  This temperature tells us the temperature cut-off for “high”
and “low” temps.
!  We can also use this to determine the temperature for a
phase change (system at equilibrium). Note: this ΔG is the
non-standard value!
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Temperature of Spontaneity
Temperature of Phase Change
!  For the previous problem, we calculated ΔG = +221 kJ when
!  1) Calculate the melting point of ice? H2O(s) ! H2O(l)
ΔH = -227 kJ, ΔS = - 309 J/K, and T = 1450 K. At what
temperature (in oC) will the system become spontaneous?
Will it be spontaneous above or below this temp?
!  ΔHfusion = 6.01 kJ/mol; ΔSfusion = 22.0 J/K•mol
!  For ice:
!  263K (-10oC): ΔG = +0.22 kJ/mol (non-spontaneous)
!  283K (10oC): ΔG = -0.22 kJ/mol (spontaneous)
!  T = ΔH / ΔS = 734 K (461oC); spontaneous below this
temperature
!  2) Calculate the boiling point (in oC) of ethanol if
the entropy of vaporization is 111.9 J/K and the
enthalpy of vaporization is 39.3 kJ
!  T = ΔH / ΔS = 39.3 kJ / 0.1119 kJ/K = 351.206 K =
78.2oC
Examples 16.9, 16.10
Predicting Spontaneity
40
Predicting Spontaneity
! 
!  Write the equation for the formation of rust.
!  Calculate ΔHorxn and ΔSorxn using the table below.
!  Is the reaction spontaneous or nonspontaneous at 500oC?
ΔHo
rxn
= [2 mol * -824.2 kJ/mol] – [0] = -1648.4 kJ
!  ΔSorxn = [2 mol * 87.40 J/K·mol] – [4 mol * 27.3 J/K·mol + 3
mol * 205.2 J/K·mol] = -550.0 J/K = -0.5500 kJ/K
!  ΔG = ΔHo – TΔSo = -1648.4 kJ – (773.15 K * -0.5500 kJ/K)
= -1223.2 kJ, spontaneous
ΔHo
Fe(s)
O2(g)
Fe2O3(s)
Example 16.7
f (kJ/mol)
0
0
-824.2
So
!  At what temperature does the reaction become
(J/K·mol)
27.3
205.2
87.40
spontaneous? Above or below this temperature?
!  T = ΔH / ΔS = 2997 K (FYI, surface of the sun is 5,778 K!)
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Standard Free Energy of Formation
!  Calculate ΔGorxn for the following reaction:
Alternatively, we can use the equation
1C2H4(g) + 3 O2 (g) ! 2CO2 (g) + 2H2O(g)
ΔGorxn = Σ n·ΔGof (products) - Σ n·ΔGof (reactants)
to calculate
ΔGo
Practice Calculation of ΔGorxn
rxn.
ΔGof is the free energy change for the formation of 1 mole
of a substance from elements in their standard states.
ΔGof for any element in its most stable form at 1 atm is defined
as zero (just like ΔHof values).
Predict the sign of ΔGorxn first.
!  Find ΔGof values in Appendix G.
!  C2H4(g) = 68.4 kJ/mol
!  O2(g) = 0 kJ/mol
!  CO2 (g) = -394.36 kJ/mol
!  H2O(g) = - 228.59 kJ/mol
! Answer: -1314.3 kJ (we calculated -1314.2 kJ using ΔG =
ΔHo – TΔSo) Should they be the same or different?
Example 16.8
Standard vs Non-Standard ΔG
• 
• 
Free Energy and Equilibrium
The sign of ΔG determines spontaneity (whether a
reaction is still moving forward to reach equilibrium).
ΔGo determines whether a reaction is reactant- or
product-favored.
The relationship between ΔG
and ΔGo
44
ΔG is determined by the slope of the tangent at any point
on the graph.
Figure 16.14
is:
ΔGo < 0 =
Products – Reactants
o
ΔG = ΔG + RT ln Q
What is K?
R is the gas constant (8.314 J/K·mol).
T is the Kelvin temperature.
Q is the reaction quotient (from Ch. 13).
45
Free Energy and Equilibrium
46
Standard vs Non-Standard ΔG
Left of equilibrium, ΔG < 0, spontaneous;
Right of equilibrium, ΔG > 0, non-spontaneous;
At equilibrium, ΔG = 0
ΔG is the actual free energy change under non-standard
conditions. It changes as a reaction proceeds (as
concentrations, pressures, etc. change).
ΔG = ΔG o + RT ln Q
ΔGo > 0 =
Products – Reactants
ΔG changes as a reaction proceeds (initially decreases).
This value determines the spontaneity of a reaction at a
given point in time.
What is K?
ΔGo is standard and does NOT change. It is a constant
for a reaction (like ΔHo and ΔSo). ΔGo < 0 is spontaneous
and product-favored.
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Free Energy and Equilibrium
Free Energy and Equilibrium
What would the graph look like for a reaction where
ΔGo = 0?
At equilibrium, ΔG = 0 and Q = K:
ΔG = ΔGo + RT ln K
ΔGo = - RT ln K
What is K?
ΔGo
K
ΔGo < 0
K > 1 Equilibrium mixture contains mostly products.
ΔGo > 0
K < 1 Equilibrium mixture contains mostly reactants.
ΔGo = 0
K = 1 Equilibrium mixture contains appreciable
amount of reactants and products at equilibrium
Comment
49
Free Energy and Equilibrium
50
Free Energy and Equilibrium
ΔGorxn for the formation of ethanol is -24.7 kJ/mol.
Consider the following equilibrium: H2(g) + I2(g) ⇌
2HI(g). ΔGo at 25oC = 2.6 kJ/mol. Calculate ΔG when
PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 3.0 atm.
C(s) + O2(g) + H2(g) ! CH3OH(g)
Calculate the equilibrium constant, K, at 25oC.
ln K = ΔGo/-RT = -24.7 kJ / -(0.008314 kJ/mol·K * 298K)
Qp =
ln K = 9.969; K = 2.14 x 104
ΔG =
(PHI )2
(3.0)2
=
= 2.25
(PH2 )(PI2 ) (2.0)(2.0)
2.6 kJ ⎛ 8.314 ×10−3 kJ ⎞
+⎜
⎟ 298 K ln2.25 = 4.6 kJ/mol
mol ⎝
K mol
⎠
(
)(
)
Example 16.11
Example 16.12
52
Practice Calculations
Practice Calculations
1) Calculate ΔGorxn for the following reaction at 25oC:
P4 (g) + 6 Cl2 (g) ! 4 PCl3 (g)
2)
!  1) Calculate ΔGorxn for the following reaction at 25oC:
for the melting of gold is 12.55 kJ/mol.
fusion for
the melting of gold is 9.38 J/mol·K. What is the melting
point (in oC) of gold? What is ΔGo for at 25oC? Is
!  P4 (g) + 6 Cl2 (g) ! 4 PCl3 (g)
this reactant- or product-favored?
!  ΔGof (Cl2 (g)) = 0 kJ/mol
ΔHofusion
ΔSo
3) Calculate ΔG for S(s) + O2(g) ! SO2(g) when [O2] = 0.140
M and [SO2] = 1.24 M at 25oC. In which direction will
the reaction proceed?
!  ΔGof (P4 (g)) = 24.4 kJ/mol
!  ΔGof (PCl3 (g)) = -269.6 kJ/mol
!  ΔGorxn = [4 mol(-269.6 kJ/mol] – [1 mol(24.4 kJ/mol)
+ 0] = - 1102.8 kJ
4) If K for the reaction CuO(s) ! Cu(s) + O2(g) at 25oC is
2.66 x 10-46, what is ΔGo?
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Practice Calculations
Practice Calculations
!  2) ΔHofusion for the melting of gold is 12.55 kJ/mol.
!  3) Calculate ΔG for S(s) + O2(g) ! SO2(g) when [O2]
!  T = ΔH / ΔS = 12.55 kJ/mol / 0.00938 kJ/molK
!  Need to calculate ΔGo and Q.
!  T = 1337.95 K – 273 = 1060 oC
!  ΔGorxn = -300.2 kJ
!  What is ΔGo for at 25oC? Is this reactant- or product-
!  Q = 1.24 M / 0.140 M = 8.8571
ΔSofusion for the melting of gold is 9.38 J/mol·K. What is
the melting point (in oC) of gold?
= 0.140 M and [SO2] = 1.24 M at 25oC. In which
direction will the reaction proceed?
!  ΔGof (SO2(g)) = -300.2 kJ/mol
favored?
!  ΔGo = ΔHo – TΔSo = 12.55 kJ/mol – (298K)(0.00938 kJ/
molK) = 9.75 kJ/mol; reactant-favored
55
!  ΔG = -294.8 kJ; spontaneous in the forward
direction.
56
Practice Calculations
!  4) If K for the reaction CuO(s) ! Cu(s) + O2(g) at
25oC is 2.66 x 10-46, what is ΔGo?
ΔGo = -RT ln K = -(8.314 J/K·mol)(298)(ln 2.66x10-46)
ΔGo = 2.60 x 105 J/mol = 260. kJ/mol
57
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