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Thermodynamics
1. Review of the First Law of Thermodynamics
What was the first law of thermodynamics?
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What is the system in the following scenarios? What are the surroundings?
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A candle burns in a room
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Butter melts in a hot pan
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A plant absorbs CO2 from the air
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Sodium chloride dissolves in water
How do we apply the first law?
2. Spontaneous Processes and Entropy
A process is said to be spontaneous if it occurs without outside intervention
Spontaneous does not mean fast
Thermodynamics can tell us the direction in which a process can occur but can say nothing
about the speed of the process
Often a reaction that is potentially spontaneous does not occur without some sort of stimulus
to provide the required activation energy. Once started a spontaneous reaction continues by
itself without further input of energy from the outside.
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What are the requirements for spontaneity?
The direction of spontaneous change is not always determined by the tendency for a system
to go to a state of lower energy.
A characteristic common to all spontaneous processes is an increase in a property called
entropy, S. The driving force for a spontaneous process is an increase in the entropy of the
universe.
Entropy is a thermodynamic function that describes the number of arrangements (position
and/or energy levels) that are available to a system existing in a given state.
Several factors influence the amount of entropy that a system has in a particular state.
1. A liquid has a higher entropy that the solid from which it is formed. Why?
2. A gas has a higher entropy than the liquid from which it is formed. Why?
3. Increasing the temperature of a substance increases its entropy. Why?
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Graph: Molar entropy of NH3 as a function of temperature
The Third Law of Thermodynamics tells us that a completely ordered pure crystalline solid
has an entropy of 0 at 0K
S = k lnΏ
Example 1: For each of the following pairs, choose the substance with the higher entropy per
mole
a. Solid carbon dioxide and gaseous carbon dioxide
b. Nitrogen gas at 1 atm and nitrogen gas at 0.001 atm
c. 1 mole H2O(s) at 0oC or 1 mole H2O(l) at 20oC
Example 2: Which of the following involve an increase in the entropy of the system?
a.
b.
c.
d.
e.
Melting of a solid
Sublimation
Freezing
Mixing
Separation
Example 3: Predict the sign of the entropy change for the following processes
a.
b.
c.
d.
Solid sugar is added to water to form a solution
Iodine vapor condenses on a cold surface to form crystals
The thermal decomposition of solid calcium carbonate
The oxidation of sulfur dioxide in air
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Standard Molar Entropy J/molK
Note: Always positive quantities. Elements as well as compounds have nonzero entropies.
Calculating So for a reaction
The standard entropy values S of many substances are found in the back of your text
To calculate So = ΣnpSo - ΣnrSo
Example 3: Calculate So at 25.0oC for the reaction
2NIS(s) + 3O2(g) → 2SO2(g) + 2NIO(s)
Given the following standard entropy values
Substances
S (J/molK)
SO2(g)
248
NiO(s)
38
O2(g)
205
NiS(s)
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The Second Law of Thermodynamics
1. In a spontaneous process there is a net increase in entropy taking into account both
system and surroundings
2. In any spontaneous process there is always an increase in the entropy of the universe
Notice that the Second Law refers to the total entropy change involving both systems and
surroundings.
Sign of ΔSsurroundings depend on the direction of the heat flow.
Expression for the entropy change of the surroundings
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Magnitude of ΔS dependent upon two factors.
ΔSsurroundings =
−ΔH
T
(constant T, P)
Example 4: In the metallurgy of antimony, the pure metal is recovered via different reactions,
depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide
ores.
Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s)
ΔH = -125kJ
Carbon is used as the reducing agent for oxide ores.
Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(g)
ΔH = 778 kJ
Calculate ΔSsurroundings for each of these reactions at 25oC and 1 atm.
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Free Energy
ΔHo - TΔSo < 0
ΔGo = ΔHo - TΔSo
1. If ΔGo is negative, the reaction is spontaneous
2. If ΔGo is positive, the reaction will not occur spontaneously.
3. If ΔGo is 0, the reaction system is at equilibrium.
Putting it another way, we can think of ΔGo as a measure of the driving force of a reaction.
Calculation of ΔGo
2H2(g) + O2(g) → 2H2O(l)
1. Using standard free energies of formation
2. Using standard enthalpies of formations and standard entropies
Example 4: Calculate ΔGo at 25.0oC for the following reactions
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
1. Using standard free energies of formation
2. Using standard enthalpies of formation and standard entropies
Example 5: At what temperatures is the following process spontaneous at 1 atm?
Br2(l) → Br2(g)
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Example 6: Calculate the ΔGo for the reaction Cu(s) + H2O(g) → CuO(s) + H2(g) at 500K
Example 7: At what temperature does ΔGo become zero for the reaction of the decomposition of
calcium carbonate?
Example 8: A chemical engineer wants to determine the feasibility of making ethanol by reacting
water with ethylene. Is this reaction spontaneous under standard conditions?
Example 9: Using the following data at 25.0oC
Cdiamond(s) + O2(g) → CO2(g)
ΔGo = -397kJ
Cgraphite(s) + O2(g) → CO2(g)
ΔGo = -394kJ
Calculate the ΔGo for the reaction Cdiamond → Cgraphite
Example 10: Water gas is a fuel made by exposing steam to red hot coke. Coke is the product of
coal distillation – it contains carbon.
H2O(g) + C(s) → CO(g) + H2(g)
Calculate the temperature at which the reaction becomes favorable. Assume ΔHo and ΔSo are
independent of temperature.
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Example 11: The heat of fusion of gold is 12.36 kJ/mol and its entropy of fusion is 9.25 kJ/mol
K. What is the melting point of gold?
Example 12: The heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol
respectively. Calculate the entropy change for solid to liquid and liquid to vapor transitions for
benzene. At 1.00 atm benzene melts at 5.5oC and boils at 80.1oC.
Example 13: Calculate the standard free energy for formation at 25.0oC for methane using the
following data.
C(s) + 2H2(g) → CH4(g)
ΔHf CH4 = -74.8kJ
SoCH4 = 186.2J/K
SoC = 5.70 J/K
SoH2 = 130.6J/K
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Summary of relationship between ΔG, ΔH and ΔS
The Dependence of Free Energy on Pressure
A system at constant temperature and pressure will proceed spontaneously in the direction that
lowers its free energy.
The Equilibrium position represents the lowest free energy value available to a particular
reaction system. The free energy of a reaction system changes as the reaction proceeds, because
free energy is dependent on the pressure of a gas or on the concentration of a species in solution.
To understand the pressure dependence of free energy, we need to know how pressure affects the
thermodynamic functions that comprise free energy.
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G = Go + RTln(P)
Example: N2(g) + 3H2(g) → 2NH3(g)
ΔG = ΣnpGproducts - ΣnrGreactants
ΔG = 2G NH3 – G N2 – 3G H2
Where: GNH3 = GoNH3+ RTln(PNH3)
GN2 = GoN2+ RTln(PN2)
GH2 = GoH2+ RTln(PH2)
ΔG = 2[GoNH3+ RTln(PNH3)] – [GoN2+ RTln(PN2)] – 3[GoH2+ RTln(PH2)]
= [2GoNH3 - GoN2 - 3GoH2] + RT[2ln(PNH3) - ln(PN2) - 3 ln(PH2)]
This is ΔGo of reaction
= ΔGo reaction + RT [2 ln(PNH3) - ln(PN2) - 3 ln(PH2)]
Since 2ln (PNH3) = ln (PNH3)2 and -ln(PN2) = ln(1/PN2) and – 3ln(PH2) = ln (1/PH2)3
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Then: ΔG = ΔG + RTln
o
PNH3
(PN2)( PH2)
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which is the reaction quotient
Therefore: ΔG = ΔGo + RTln(Q)
Example 1: Solid calcium oxide reacts with sulfur trioxide gas
CaO(s) + SO3(g) → CaSO4(s)
The ΔGo at 25oC for the reaction is -345kJ. Calculate the ΔG for the reaction at 25oC when the
pressure of sulfur trioxide is 0.15 atm.
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Example 2: One method for synthesizing methanol involves reacting carbon monoxide and
hydrogen gases.
CO(g) + H2(g) → CH3OH(l)
Calculate ΔG at 25oC for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas
at 3.0 atm are converted to liquid methanol.
The meaning of ΔG for a chemical reaction
The system can achieve the lowest possible free energy by going to equilibrium, not
completion.
The relationship between free energy and equilibrium constants
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Calculate the equilibrium constant for the following reaction at 25oC
2H2O (l) ↔ 2H2(g) + O2(g)
Example 2: Consider the ammonia synthesis reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
Where ΔGo = -33.3kJ per mole N2 consumed at 25oC. For each of the following mixtures of
reactant and products at 25oC, predict the direction in which the system will shift to reach
equilibrium.
1. P NH3 = 1.00atm, P N2 = 1.47 atm, P H2 = 0.0100 atm
2. P NH3 = 1.00atm, P N2 = 1.00 atm, P H2 = 1.00 atm
Example 3: The overall reaction for the rusting of iron by oxygen is
4Fe(s) + 3O2(g) ↔ 2Fe2O3(s)
Calculate the equilibrium constant for this reaction at 25oC
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The temperature dependence of K
In our equilibrium chapter we used Le Chatelier’s Principle to predict qualitatively how the value
of K for a given reaction would change in temperature. Now we can specify the quantitative
dependence of the equilibrium constant on temperature.
ΔGo = -RTlnK = ΔHo – TΔSo
lnK =
−ΔHo
RT
+
ΔSo
R
=
(−ΔHo )(l)
(R)(T)
ΔSo
+
R
Free energy and work
The maximum possible useful work obtainable from a process at constant temperature and
pressure is equal to the change in free energy
Achieving the maximum work available from a spontaneous process can occur only via a
hypothetical pathway. Any real pathways waste energy.
Reversible process
Irreversible process
All real processes are irreversible
First Law: You can’t win, you can only break even
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Second Law: You can’t break even.