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Consider this Reaction
Thermodynamics
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
Concerning this reaction:
1. Does this reaction naturally occur as written?
2. Will the reaction mixture contain sufficient
amount of product at equilibrium?
1
4
Thermodynamics
We can answer these questions with
heat measurements only!!!
• Study of energy changes and flow of energy
• Answers several fundamental questions:
– Is it possible for reaction to occur?
– Will reaction occur spontaneously (without outside
interference) at given T?
– Will reaction release or absorb heat?
1. We can predict the natural direction.
2. We can determine the composition of the
mixture at equilibrium.
• Tells us nothing about time frame of reaction
– Kinetics
• Two major considerations—must be balanced
How?
– Enthalpy changes, ∆H (heats of reaction)
• Heat exchange between system and surroundings
– Nature's trend to randomness or disorder
2
5
Consider the Laws of Thermodynamics
1st Law: The change in internal energy of a
system ∆U, equals q + w
∆E = ∆U = q + w
You can’t win
Examples
of
Spontaneous Reactions
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
You cant break even.
3rd Law: A substance that is perfectly crystalline
at 0 K has an entropy of zero.
You can’t quit the game
3
6
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Review of First Law of Thermodynamics
First Law of Thermodynamics
• Internal energy, E or U
• Energy can neither be created nor destroyed
• It can only be converted from one form to another
– System's total energy
– Sum of KE and PE of all particles in system
– KE ↔ PE
– Chemical ↔ Electrical
– Electrical ↔ Mechanical
E system = (KE )system + (PE )system
∆E = E final − E initial
• E and ∆E are state functions
– or for chemical reaction
∆E = Eproducts − E reactants
– ∆E +
– ∆E −
– Path independent
– ∆E = -∆
∆U = q + w
energy into system
energy out of system
E system = (KE )system + (PE )system
7
Two Methods of Energy Exchange Between
System and Surroundings
• Heat
q =∆H
Work
Work in Chemical Systems
1. Electrical
2. Pressure-volume or P∆
∆V
– w = − P∆
∆V
• Where P = external pressure
– If P∆
∆V only work in chemical system, then
w = P∆
∆V
• ∆U = ∆E = q + w
• Conventions of heat and work
q + Heat absorbed by system
q − Heat released by system
w + Work done on system
w − Work done by system
10
Esystem ↑
Esystem ↓
Esystem ↑
Esystem ↓
∆ E = q + ( − P∆ V ) = q − P∆ V
8
11
∆U(Internal Energy) is a state function:
Most often we are interested in the change:
Internal Energy = ∆Ε = ∆U = Uf - Ui
Heat vs. Work
q = Energy that moves in and out of a system
w = Force x distance = F x d = P∆V
9
12
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Showing work is P∆V
P∆V = -92 J
w = -F xh
= - F x ∆V/A
q = 165 J
= -F/A x ∆V
= -P∆V
13
∆U = q + w = (+165) + (-92) = +73 J
16
Heat of Reaction and Internal Energy
Heat at Constant Volume
Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)
Reaction done at constant V
∆V = 0
P∆V = 0, so
∆E = qV
Entire E change due to heat absorbed or
lost
14
17
• Here the system expands and evolves
heat from A to B.
Heat at Constant Pressure
• More common
• Reactions open to atmosphere
– Constant P
• Enthalpy
– H = E + PV
• Enthalpy change
– ∆H = ∆E + P∆V
• Substituting in first law for ∆E gives
– ∆H = (q − P∆V) + P∆V = qP
– ∆H = qP
Zn2+(aq) + 2Cl-(aq) + H2(g)
• Heat of reaction at constant pressure
15
∆V is positive, so work is negative.
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q=-
w =-
Enthalpy and Enthalpy Change
Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)
Enthalpy is defined as qp
Enthalpy = H = U + PV
All are state functions.
∆H = Hf - Hi
∆H = (Uf + PVf) – (Ui + PVi) = (Uf –Ui) + P(Vf –Vi)
∆U = qp - P∆V
∆H = (qp - P∆V) + P∆V = qp
∆Hof = Σn∆Hof (products) - Σm∆Hof (reactants)
19
19
w
∆Hof = Standard enthalpy change (25oC)
22
∆Hof = Σn∆Hof (products) - Σm∆Hof (reactants)
=
-P∆V
=
-(1.01 x 105 pa)(24.5 l)
Calculate ∆Hof for the reaction in slide 15
105
=
-1.01 x
pa)(24.5 x
=
-2.47 x 103 J
=
-2.47 kJ
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
10-3m3)
q
=
-152 kJ
∆U
=
-152 kJ + (-2.47 kJ) = -154.9 kJ
∆Hof =
=
=
=
for NH3 (g)
for CO2 (g)
for NH2CONH2
for H2O (l)
=
=
=
=
-45.9 kJ
-393.5 kJ
-319.2 kJ
-285.8 kJ
∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7
Since ∆Ho has a negative sign, heat is evolved
20
Converting Between ∆E and ∆H For Chemical
Reactions
Diagram and explain the change in internal energy
for the following reaction.
• ∆H ≠ ∆E
• Differ by
∆ H − ∆ E = P∆ V
• Only differ significantly when gases formed or
consumed
• Assume gases are ideal
nRT
 nRT 
V =
∆V = ∆ 

P
 P 
• Since P and T are constant
 RT 
∆V = ∆n 

 P 
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
q = -890.2 kJ
w
= P∆V = +(1.01 x 105 pa)(24.5 l)(2)
= +(1.01 x 105 pa)(24.5 x 10-3m3)(2)
= +4.95 kJ
∆U
= -890.2 kJ + (+4.95 kJ) = -885.2 kJ
23
21
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Converting Between ∆E and ∆H For Chemical
Reactions
Ex. 1. finish
• When reaction occurs
Step 4: Calculate % difference
7.43 kJ
% difference =
∗ 100 % = 6.6%
113 kJ
Bigger than most, but still small
– ∆V caused by ∆n of gas
• Not all reactants and products are gases
– So redefine as ∆ngas
Where ∆ngas = (ngas)products – (ngas)reactants
• Substituting into ∆H = ∆E + P∆V gives
 RT 
∆H = ∆E + P ∗ ∆ngas 

– or
 P 
∆H = ∆ E + ∆ngas RT
Note: Assumes that volumes of solids and liquids
are negligible
Vsolid ≈ Vliquid << Vgas
25
Ex. 1. What is the difference between ∆H and ∆E for
the following reaction at 25 °C?
28
Is Assumption that Vsolid ≈ Vliquid << Vgas Justified?
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
• Consider
What is the % difference between ∆H and ∆E?
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
37.0 mL
2*18.0 mL 18 mL
18 mL 24.4 L
Step 1: Calculate ∆H using data (Table 7.2)
Recall
∆ H o = ( ∆ H fo ) products
o
∆H =
Volumes assuming each coefficient equal
number of moles
− ( ∆ H fo ) reactants
• So ∆V = ∆Vprod – ∆Vreac = 24.363 L ≈ 24.4 L
4 ∆Hfo (NO2 ) + ∆Hfo (O2 ) − 2 ∆Hfo (N2 O5 )
∆H° = (4 mol)(33.8 kJ/mol) + (1 mol)(0.0 kJ/mol)
– (2 mol)(11 kJ/mol)
∆H° = 113 kJ
• Yes, assumption is justified
Note: If No gases are present reduces to
∆E ≈ ∆H
26
Ex. 1. (cont.)
Learning Check
• Consider the following reaction for picric acid:
Step 2: Calculate
8O2(g) + 2C6H2(NO2)3OH(ℓ) → 3 N2(g) + 12CO2(g) + 6H2O(ℓ)
∆ngas = (ngas)products – (ngas)reactants
• What type of reaction is it?
• Calculate ΔΗ°, ΔΕ°
∆ngas = (4 + 1 – 2) mol = 3 mol
8O2(g) + 2C6H2(NO2)3OH(ℓ) → 3N2(g) + 12CO2(g) + 6H2O(ℓ)
Step 3: Calculate ∆E using
R = 8.31451 J/K·mol
29
ΔΗ°f
(kJ/mol)
T = 298 K
∆E = 113 kJ –
(3 mol)(8.31451 J/K·mol)(298 K)(1 kJ/1000 J)
∆E = 113 kJ – 7.43 kJ = 106 kJ
27
0.00
3862.94
0.00
-393.5
-241.83
ΔH0 = 12mol(–393.5 kJ/mol) + 6mol(–241.83kJ/mol) +
6mol(0.00kJ/mol) – 8mol(0.00kJ/mol) – 2mol(3862.94kJ/mol)
ΔH0 = – 13,898.9 kJ
ΔΕ° = ΔH° – ΔngasRT = ΔH° – (15 – 8)mol*298K*
8.314×10–3kJ/(mol·K)
∆Ε° = –13,898.9 kJ – 29.0 kJ = – 13,927.9 kJ
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Enthalpy Changes and Spontaneity
Your Turn!
Given the following:
3H2(g) + N2(g) → 2NH3(g) ∆Ho = -46.19 kJ mol-1
Determine ∆E for the reaction.
A. -51.14 kJ mol-1
B. -41.23 kJ mol-1
C. -46.19 kJ mol-1
D. -46.60 kJ mol-1
•
•
•
•
• What are relationships among factors that influence
spontaneity?
• Spontaneous Change
– Occurs by itself
– Without outside
assistance until
finished
• Ex.
– Water flowing over waterfall
– Melting of ice cubes in glass
on warm day
∆H = ∆E + ∆nRT
∆E = ∆H - ∆nRT
∆E =-46.19 kJ mol – (∆NRT)
-(-2 mol)(8.314 J K-1mol-1)(298K)(1 kJ/1000 J)
∆E = -41.23 kJ mol-1
31
Nonspontaneous Change
∆Hof = Σn∆Hof (products) - Σm∆Hof (reactants)
Calculate
∆Hof
for the reaction in slide 15
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
∆Hof =
=
=
=
for NH3 (g)
for CO2 (g)
for NH2CONH2
for H2O (l)
=
=
=
=
Since
• Occurs only with outside assistance
• Never occurs by itself:
– Room gets straightened up
– Pile of bricks turns into a brick wall
– Decomposition of H2O by electrolysis
-45.9 kJ
-393.5 kJ
-319.2 kJ
-285.8 kJ
∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7
∆Ho
34
has a negative sign, heat is evolved
• Continues only as long as outside
assistance occurs:
– Person does work to clean up room
– Bricklayer layers mortar and bricks
– Electric current passed through H2O
32
Spontaneity and Entropy
Definition of Spontaneous Process:
Physical or chemical process that occurs by itself.
Why??
Give several spontaneous processes:
Still cannot predict spontaneity…..
33
35
Entropy and the 2nd Law of Thermodynamics
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
Entropy = S = A thermodynamic quantity that is a
measure of how dispersed the energy
is among the different possible ways
that a system can contain energy.
Consider:
1.
A hot cup of coffee on the table
2.
Rock rolling down the side of a hill.
3.
Gas expanding.
4.
Stretching a rubber band.
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Reaction Rate and Spontaneity
Direction of Spontaneous Change
• ∆H indicates if reaction has tendency to occur
• Rate of reaction also plays role
– Some very rapid:
• Some endothermic reactions occur
spontaneously:
– Ice melting
– Evaporation of water from lake
– Expansion of CO2 gas into vacuum
• Neurons firing in nerves in response to pain
• Detonation of stick of dynamite
– Some gradual:
• ∆H and ∆E are positive
• Erosion of stone
• Ice melting
• Iron rusting
– Heat absorbed
– Energy entering system
– Some so slow, appear to be nonspontaneous:
• Clearly other factors influence spontaneity
• Gasoline and O2 at RT
• Many biochemical processes
40
Flask connected to an evacuated flask by a
valve or stopcock
Concept Check:
You have a sample of solid iodine
at room temperature. Later you notice
that the iodine has sublimed. What
can you say about the entropy change
of the iodine?
∆S = Sf - Si
H2O (s) → H2O (l)
∆S = (63 – 41) J/K = 22 J/K
38
41
Direction of Spontaneous Change
• Many reactions which occur spontaneously are
exothermic:
– Iron rusting
– Fuel burning
There are how many Laws of thermodynamics?
Early thoughts were that exo
were spontaneous)
a.
b.
c.
d.
e.
• ∆H and ∆E are negative
– Heat given off
– Energy leaving system
1
2
3
4
5
• Thus, ∆H is one factor that influences spontaneity
39
42
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Heat Transfer Between Hot and Cold Objects
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
• Consider system of two objects
– Initially one hot and one cold
– “Hot” = higher KEave of molecules = faster
– “Cold” = lower KEave of molecules = slower
Process occurs naturally as a result of energy dispersal
In the system.
∆S = entropy created + q/T
∆S > q/T
For a spontaneous process at a given temperature, the
change in entropy of the system is greater than the heat
divided by the absolute temperature.
43
• When they collide, what is most likely to occur?
– Faster objects bump into colder objects and
transfer energy so…
– “Hot” objects cool down and slow down
– “Cold” objects warm up and speed up
– The reverse doesn’t occur
Entropy and Molecular Disorder
Heat Transfer Between Hot and Cold Objects
Result:
• Heat flows spontaneously from hot to colder object
• Heat flows because of probable outcome of
intermolecular collisions
• Spontaneous processes tend to proceed from states
of low probability to states of higher probability
• Spontaneous processes tend to disperse energy
44
Thermodynamic vs. Kinetics
•
• Thermodynamics tells us:
Domain of
Thermodynamics
(initial and final states)
Domain of Kinetics
(reaction pathway)
Entropy (symbol S)
Energy
– Direction of reaction
– Is it possible for reaction to
occur?
– Will reaction occur
spontaneously at
given T?
– Will reaction release
Reactants
or absorb heat?
– Kinetics tells us:
– Speed of reaction
– Pathway between reactants
and products
Products
Reaction Progress
45
• Thermodynamic quantity
• Describes number of equivalent ways that energy
can be distributed
• Quantity that describes randomness of system
• Greater statistical probability of particular state
means greater the entropy!
– Larger S, means more random and ∴ more probable
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Entropy
Entropy
• Sproducts > Sreactants means ∆S +
– Entropy ↑'s
– Probability of state ↑'s
– Randomness ↑'s
– Favors spontaneity
• Sproducts < Sreactants means ∆S −
– Entropy ↓'s
– Probability of state ↓'s
– Randomness ↓'s
– Does not favor spontaneity
• Any reaction that occurs with ↑ in entropy tends
to occur spontaneously
• If Energy = money
• Entropy (S) describes number of different ways of
counting it
Criterion for Spontaneity
52
Effect of Volume on Entropy
• Clear order to things
– Things get rusty spontaneously
• Don't get shiny again
– Sugar dissolves in coffee
• Stir more—it doesn't undissolve
– Ice → liquid water at RT
• Opposite does NOT occur
– Fire burns wood, smoke goes up chimney
• Can't regenerate wood
• Common factor in all of these:
– Increase in randomness and disorder of system
– Something that brings about randomness more
likely to occur than something that brings order
• For gases, Entropy ↑ as Volume ↑
A. Gas separated from vacuum by partition
B. Partition removed
C. Gas expands to achieve more probable particle
distribution
• More random, higher probability, more positive S
53
Entropy, S
Effect of Temperature on Entropy
•
•
•
•
•
•
•
Measure of randomness and disorder
Measure of chaos
State function
Independent of path
∆S = Change in Entropy
Also state function
∆S = Sfinal − Sinitial
As T ↑, entropy ↑
A. T = 0 K, particles (●) in equilibrium lattice positions and S
relatively low
B. T > 0 K, molecules vibrate, S ↑
C. T ↑ further, more violent vibrations occur and S higher than in B
• For chemical reaction
∆S = S products − Sreactants
51
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Effect of Physical State on Entropy
Entropy Change for a Reaction
• Crystalline solid very low S
• Liquid higher S, molecules can move freely
– More ways to distribute KE among them
• Gas highest S, particles randomly distributed throughout container
– Many, many ways to distribute KE
∆So may be positive for reactions if/with the following:
1. The reaction is one in which a molecule is
broken into two or more smaller molecules.
2. The reaction is one in which there is an increase
in moles of gas.
3. The process is one in which a solid changes to
a liquid or a liquid changes to a gas.
55
Entropy Affected by Number of Particles
•
•
•
•
Did You Get It!
Which represents an increase in entropy?
Adding particles to system
↑ number of ways energy can be distributed in system
So all other things being equal
Reaction that produces more particles will have positive ∆S
A. water vapor condensing to liquid
B. carbon dioxide subliming
C. liquefying helium gas
D. proteins forming from amino acids
56
Summary
59
Entropy Changes in Chemical Reactions
• Larger V, greater ∆S
– Expansion of gas ∆S +
• Higher T, greater ∆S
– Higher T, means more KE in particles, move more, so random
distributions favored
• Ssolid < Sliquid << Sgas
Ex. N2 (g) + 3 H2(g) → 2 NH3(g)
nreactant = 4
nproduct = 2
∆n = 2 – 4 = –2
Predict ∆Srxn < 0
– Solids more ordered than liquids, which are much more ordered
than gases
Reactions involving gases
• Simply calculate change in number of mole gas,
∆ngas
– If ∆ngas + , ∆S +
– If ∆ngas − , ∆S −
Higher positional
probability
Lower positional
probability
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Entropy Changes in Chemical Reactions
Reactions without gases
• Simply calculate number of mole molecules
∆n = nproducts – nreactants
–
–
–
–
If ∆n + , ∆S +
If ∆n − , ∆S −
More molecules, means more disorder
Usually the side with more molecules, has less complex
molecules
• Smaller, fewer atoms per molecule
Did You Get It?
Which of the following has the most entropy at standard conditions?
A. H2O(ℓ)
B. NaCl(aq)
C. AlCl3(s)
D. Can’t tell from the information
Which reaction would have a negative entropy?
A. Ag+(aq) + Cl-(aq)→AgCl(s)
B. N2O4(g) → 2NO2 (g)
C. C8H18(l ) + 25/2 O2(g) → 8CO2(g) + 9H2O(g)
D. CaCO3(s) → CaO(s) + CO2(g)
64
Ex. 2 Predict Sign of ∆S for Following Reactions
CaCO3(s) + 2H+ (aq) → Ca2+(aq) + H2O(l) +
– ∆ngas = 1 mol – 0 mol = 1 mol
– ∴ ∆ngas +, ∆S +
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
– ∆ngas = 4 mol + 1 mol – 2 mol = 3 mol
– ∴ ∆ngas +, ∆S +
OH− (aq) + H+ (aq) → H2O (l)
– ∆ngas = 0 mol
– ∆n = 1 mol – 2 mol = –1 mol
– ∴ ∆n − , ∆S −
CO2(g)
Predict Sign of ∆S in Following:
• Dry ice → carbon dioxide gas
positive
CO2(s) → CO2(g)
• Moisture condenses on a cool window
negative
H2O(g) → H2O(ℓ)
• AB → A + B
positive
Both Entropy and Enthalpy Can Affect Reaction
Spontaneity
• Sometimes they work together
– Building collapses
– PE ↓ ∆H −
– Stones disordered
∆S +
• Sometimes work against each other
– Ice melting (ice/water mix)
– Endothermic
• ∆H +
nonspontaneous
– ↑ Disorder of molecules
• ∆S +
spontaneous
Which Prevails?
• Hard to tell—depends on temperature!
– At 25 °C, ice melts
– At −25 °C, water freezes
• So three factors affect spontaneity:
– ∆H
– ∆S
– T
• A drop of food coloring added to a glass of water
positive
disperses
• 2Al(s) + 3Br2(ℓ) → 2AlBr3(s) negative
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What do you think?
Second Law of Thermodynamics
• When a spontaneous event occurs, total entropy of
universe ↑’s
– (∆Stotal > 0)
• In a spontaneous process, ∆Ssystem can decrease as
long as total entropy of universe increases
– ∆Stotal = ∆Ssystem + ∆Ssurroundings
• It can be shown that
∆S surroundings =
When ice melts in your hand (assume your hand is 30o C),
A. the entropy change of the system is less then the entropy
change of the surroundings.
B.the entropy change of the surroundings is less than the
entropy change of the system.
C.the entropy change of the system equals the entropy
change of the surroundings.
• ∆Ssys = ∆H/273K
• ∆Ssys > ∆Ssurr
q surroundings
T
∆Ssurr = ∆H/303K
Law of Conservation of Energy
2nd Law: The total entropy of a system and its
surroundings increases for a spontaneous
process.
• Says q lost by system must be gained by surroundings
– qsurroundings = − qsystem
• If system at constant P, then
– qsystem = ∆H
• So
– qsurroundings = − ∆Hsystem
• and
Process occurs naturally as a result of energy dispersal
In the system.
∆S = entropy created + q/T
∆S > q/T
∆S surroundings =
For a spontaneous process at a given temperature, the
change in entropy of the system is greater than the heat
divided by the absolute temperature.
q surroundings − ∆H system
=
T
T
68
Does This Make Sense?
Thus Entropy for Entire Universe is
∆S total = ∆S system −
• Yes?
• As heat added
– Disorder or entropy ↑, ∴ S ∝ q
• But how much ↑ T, ↑ S, depends on T at
which it occurs
– Low T, larger ↑ S
– High T, smaller ↑ S
• ∴ S ∝ 1/T
∆H system
T
Multiplying both sides by T we get
T∆
∆Stotal = T∆
∆Ssystem – ∆Hsystem
or
T∆
∆Stotal = – (∆
∆Hsystem – T∆
∆Ssystem)
• For reaction to be spontaneous
– T∆
∆Stotal > 0 (+)
So,
(∆
∆Hsystem – T∆
∆Ssystem) < 0
– (–) for reaction to be spontaneous
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Third Law of Thermodynamics
Standard Entropy (Absolute Entropy): Entropy
value for the standard state of a species.
See Table B-13 p. B-17 and Table B-16 p. B-28
• At absolute zero (0 K),
– Entropy of perfectly ordered, pure crystalline
substance is zero
• S = 0 at T = 0 K
• Since S = 0 at T = 0 K
– Define absolute entropy of substance at higher
temperatures
• Standard entropy, S°
– Entropy of 1 mole of substance at 298 K (25°C) and
1 atm pressure
– S°= ∆S for warming substance from 0 K to 298 K
(25°C)
Entropy values of substances must be positive.
So must be >0 but Ho can be plus or minus
(Why?)
How about ionic species?
So for H3O+ is set at zero.
73
76
Consequences of Third Law
Predict The Entropy sign for the following reactions:
a. C6H12O11 (s) → 2CO2 (g)+ C2H5OH (l)
1. All substances have positive entropies as they are more
disordered than at 0 K
–
–
Heating ↑ randomness
S°is biggest for gases—most disordered
b. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
2. For elements in their standard states
c. CO (g) + H2O (g) → CO2 (g) + H2 (g)
• Units of S°⇒ J/(mol·K)
Standard Entropy Change
d. Stretching a rubber band
–
–
–
Exercise 13.2 p 578
S°≠ 0 (but ∆Hf°= 0)
To calculate ∆S°for reaction, do Hess's Law type calculation
Use S°rather than entropies of formation
∆S o = n ∑ S o (products ) − m ∑ S o (reactants )
74
77
Calculating ∆So for a reaction
∆So
=
Σn∆So
(products) -
Σm∆So
Learning Check
(reactants)
Calculate the entropy change for the following reaction
At 25 oC.
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
So 2 x 193
214
174
70
Calculate ∆S0 for the following:
• CO2(s) →
CO2(g)
187.6
213.7
S0 (J/mol·K)
• ∆S0 = (213.7 – 187.6) J/mol·K
• ∆S0 = 26.1 J/mol·K
•CaCO3(s) → CO2(g) + CaO(s)
∆So = Σn∆So (products) - Σm∆So (reactants)
92.9
∆So= [(174 + 70) - (2x 193 + 214)] J/K= -356 J/K
213.7
40 S0 (J/mol·K)
•∆S0 = (213.7 +40 – 92.9) J/mol·K
See exercise 13.3 p 584 and problems 8-12 and 23-26
75
•∆S0 = 161 J/mol·K
78
13
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Ex. 3. Calculate ∆S°for reduction of aluminum oxide by
hydrogen gas
Al2O3 (s) + 3 H2 (g) → 2 Al (s) + 3 H2O (g)
Substance
Al (s)
Al2O3 (s)
H2 (g)
H2O (g)
∆S o =
∆S
o
Gibbs Free Energy
∆G = ∆H – T∆
∆S
∆G
∆G
∆G
S° (J/ K·mol)
28.3
51.00
130.6
188.7
Spontaneous process
At equilibrium
Nonspontaneous
• G ⇒ state function
– Made up of T, H and S = state functions
– Has units of energy
– Extensive property
• ∆G = Gfinal – Ginitial
o
o
− ∑ n r S reactants
∑ n p S products
o
= [2S Al
+ 3S Ho 2O ( g ) ]
(s )
o
− [S Al
+ 3S Ho 2 ( g ) ]
2O 3 ( s )
<0
=0
>0
79
82
Criteria for Spontaneity?
Ex. 3

 28.3 J 
 188.7 J 
∆S o = 2 mol * 
 + 3 mol * 

mol
⋅
K


 mol ⋅ K 


 51.00 J 
 130.6J 
− 1 mol * 
 + 3 mol * 

 mol ⋅ K 
 mol ⋅ K 

∆S°= 56.5 J/K + 566.1 J/K – 51.00 J/K – 391.8 J/K
• At constant P and T, process spontaneous only if it is
accompanied by ↓ in free energy of system
∆H
∆S
–
+
∆G = (–) – [T(+)] = –
Always, regardless of T
+
–
∆G = (+) – [T(–)] = +
Never, regardless of T
+
+
∆G = (+) – [T(+)] = ?
–
–
∆G = (–) – [T(–)] = ?
Depends; spontaneous at
high T, –∆
∆G
Depends; spontaneous at
low T, –∆
∆G
Spontaneous?
∆S°= 179.9 J/K
80
Gibbs Free Energy
83
Summary
• Would like one quantity that includes all
three factors that affect spontaneity of
a reaction
• Define new state function
• Gibbs Free Energy
– Maximum energy in reaction that is "free" or
available to do useful work
G ≡ H – TS
• At constant P and T, changes in free energy
∆G = ∆H – T∆
∆S
• When ∆H and ∆S have
same sign, T determines
whether spontaneous or
non-spontaneous
• Temperature-controlled
reactions are spontaneous
at one temperature and
not at another
81
84
14
2/7/2012
Standard Free Energy Changes
Free Energy Concept
G=H–TS
∆Ho
–
T∆So
• Standard Free Energy Change, ∆G°
– ∆G measured at 25 °C (298 K) and 1 atm
• Two ways to calculate, depending on what data is
available
Can Serve as a criteria for Spontaneity
2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)
∆Ho = -119.7 kJ
∆So = -365 J/K = -0.365 kJ/K
•Method 1. ∆G°= ∆H°– (298.15 K)∆
∆S°
∆Ho – T∆So = (-119.7 kJ) – (298 K) x (-0.365kJ/K
= -13.6 kJ
Ex. 4. Calculate ∆G°for reduction of aluminum oxide by
hydrogen gas
∆Ho – T∆So is a negativity quantity, from which we can
conclude that the reaction is spontaneous
under standard conditions.
Al2O3(s) + 3 H2(g) → 2 Al(s) + 3H2O(g)
85
88
Ex. 4. Method 1
Free Energy and Spontaneity
Al2O3 (s) + 3 H2 (g) → 2 Al (s) + 3 H2O (g)
Step 1: Calculate ∆H°for reaction using
∆Hfo
Substance
∆Hfo(kJ/mol)
Free Energy: Thermodynamic quantity defined by
the equation G = H - TS
∆G = ∆H - T∆S
If you can show that ∆G for a reaction at a given
temperature and pressure is negative, you can
predict that the reaction will be spontaneous…
Al (s)
0.0
Al2O3 (s)
H2 (g)
H2O (g)
–1669.8
0.0
–241.8
∆H o = ∑ np ∆Hfo products − ∑ nr ∆Hforeactants
∆H o = [2∆H fo Al(s ) + 3∆H foH O( g ) ]
2
86
− [∆H fo Al O (s )
2 3
+ 3∆H foH
2 (g )
]
89
Ex. 4. Method 1 Step 1 (∆H°)
Standard Free Energy Changes
0.00 kJ 

 − 241.8 kJ  
∆H o = 2 mol* 
 + 3 mol* 

mol
 mol 



−
1669
.
8
kJ
0
.
00
kJ




− 1 mol* 
 + 3 mol* 

mol


 mol  

Standard Conditions:
1 atm pressure
1 atm partial pressure
1 M concentration
Temperature of 25 oC or 298 K
Standard free energy is free-energy change that takes
place when reactants in their standard states are
converted to products in their standard states.
∆H°= 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ)
∆H°= 944.4 kJ
∆Go = ∆Ho - T∆So
87
90
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Ex. 4. Method 2
Ex. 4 Method 1
0.00 kJ 

 − 228.6 kJ  

 + 3 mol* 
mol


 mol 

.
kJ
.
kJ
1576
4
0
00
−





− 1 mol* 
 + 3 mol* 

mol


 mol  

∆G o = 2 mol* 
Step 2: Calculate ∆S° see Ex. 4
∆S°= 179.9 J/K
Step 3: Calculate
∆G°= ∆H°– (298.15 K)∆
∆S°
∆G°= 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ)
∆G°= 944.4 kJ – (298 K)(179.9 J/K)(1 kJ/1000 J)
∆G°= 944.4 kJ – 53.6 kJ = 890.8 kJ
∆G°= 890.6 kJ
– ∆G°= +
– ∴ not spontaneous
Both methods same
within experimental error
91
∆Go as a Criterion for Spontaneity
Method 2
• Use Standard Free Energies of Formation
∆Gfo
• Energy to form 1 mole of substance from its
elements in their standard states at 1 atm and 25
°C
∆G o = ∑ np ∆Gfo products − ∑ nr ∆Gforeactants
92
Ex. 4. Method 2
∆G
Substance
∆Gfo(kJ/mol)
Al (s)
Al2O3 (s)
H2 (g)
H2O (g)
0.0
–1576.4
0.0
–228.6
=
[2 ∆ G fo Al( s )
− [ ∆G
+
o
f Al2 O 3 ( s )
∆Go = ∆Ho - T∆So
∆Go
1. When
is a large negative number (more
negative than about – 10 kJ), the reaction is spontaneous
as written, and reactants transform almost entirely into
products when equilibrium is reached.
2. When ∆Go is a large positive number (more
positive than about + 10 kJ), the reaction is not
spontaneous as written, and reactants transform almost
entirely into products when equilibrium is reached.
3. When ∆Go has a small positive or negative value(less
than about 10 kJ), the reaction mixture gives an
equilibrium mixture with significant amounts of both
reactants and products..
95
Interpreting the Sign of ∆Go
Calculate ∆G°for reduction of aluminum oxide
by hydrogen gas.
Al2O3 (s) + 3 H2 (g) → 2 Al (s) + 3 H2O (g)
o
94
Calculate ∆Ho and ∆Go for the following reaction
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
Interpret the signs of ∆Ho and ∆Go
∆Hfo are as follows: KClO3 (s) = -397.7 kJ/mol
KCl (s) = -436.7 kJ/mol
O2 (g) = 0
∆Gfo are as follows: KClO3 (s) = -296.3 kJ/mol
KCl (s) = -408.8 kJ/mol
O2 (g) = 0
3∆ G fo H O( g ) ]
2
+ 3 ∆ G fo H
2 (g
)]
93
96
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2/7/2012
2 KClO3 (s)
∆Hfo
2 x (-397.70)
∆Gfo 2 x (-296.3)
→ 2 KCl (s)
+
2 x (-436.7)
2 x (-408.8)
3 O2 (g)
∆G = Maximum Possible Work
0 kJ
• ∆G is maximum amount of energy produced
during a reaction that can theoretically be
harnessed as work
0 kJ
Then:
∆Ho = [2 x (-436.7) – 2 x (-397.7)] kJ = -78 kJ
∆Go = [2 x (-408.8) – 2 x (-296.3)] kJ = -225 kJ
The reaction is exothermic, liberating 78 kJ of heat. The
large negative value of ∆Go indicates that the equilibrium
is mostly KCl and O2.
– Amount of work if reaction done under reversible
conditions
– Energy that need not be lost to surroundings as heat
– Energy that is “free” or available to do work
97
100
Spontaneous Reactions Produce Useful Work
Ex. 5
• Fuels burned in engines to power cars or heavy machinery
• Chemical reactions in batteries
– Start cars
– Run cellular phones, laptop computers, mp3 players
• Energy not harnessed if reaction run in an open dish
– All energy lost as heat to surroundings
• Engineers seek to capture energy to do work
– Maximize efficiency with which chemical energy is
converted to work
– Minimize amount of energy transformed to unproductive
heat
Calculate ΔG° for reaction below at 1 atm and
25°C, given ΔH° = –246.1kJ/mol, ΔS° = 377.1
/(mol·K).
H2C2O4(s) + ½ O2(g) → 2CO2(g) + H2O(ℓ)
∆G25 = ∆H – T∆
∆S
∆G o = −246.1
kJ
J

 1 J

− (298K ) 377.1


mol
K ⋅ mol  1000kJ 

ΔG° =(–246.1 – 112.4)kJ/mol
ΔG° = –358.5kJ/mol
98
Thermodynamically Reversible
101
You Try!
• Process that can be reversed and is always very close to
equilibrium
– Change in quantities is infinitesimally small
• Example - expansion of gas
– Done reversibly, it does most work on surroundings
Calculate ∆Go for the following reaction,
H2O2(l ) → H2O(l ) + O2(g)
given ∆Ho = -196.8 kJ mol-1 and ∆So = +125.72 J K-1
mol-1.
A.
-234.3 kJ mol-1
B.
+234.3 kJ mol-1
C.
199.9 kJ mol-1
D.
3.7 x 105 kJ mol-1
•∆Go = -196.8 kJ mol-1 – 298 K (0.12572 kJ K-1 mol-1)
•∆Go = -234.3 kJ mol
99
102
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System at Equilibrium
•
•
•
•
•
•
Free energy change during reaction
Neither spontaneous nor nonspontaneous
In state of dynamic equilibrium
Gproducts= Greactants
∆G = 0
Consider freezing of water at 0o C
H2O(ℓ) H2O(s)
– System remains at equilibrium as long as no heat added
or removed
– Both phases can exist together indefinitely
– Below 0oC, ∆G < 0 freezing spontaneous
– Above 0oC, ∆G > 0 freezing nonspontaneous
•Define equilibrium
103
Ex. 6
Free Energy and Equilibrium Constant
Calculate Tbp for reaction below at 1 atm and 25°C, given
ΔH° = 31.0kJ/mol, ΔS° = 92.9 J/(mol·K)
Br2(l) → Br2(g)
T bp ≈
∆H o
∆S o
≈
106
31.0kJ / mol
= 334K
0.0929kJ /(mol ⋅ K )
Very important relation is the relation between
free energy and the equilibrium constant.
Thermodynamic Equilibrium Constant- the equilibrium
constant in which the concentration of gases are
expressed in partial pressures in atmospheres,
whereas the concentration of solutes in liquid
are expressed in molarities.
• For T > 334 K, ΔG < 0 and reaction is
spontaneous (ΔS° dominates)
• For T < 334 K, ΔG > 0 and reaction is
nonspontaneous (ΔH° dominates)
• For T = 334 K, ΔG = 0 and T = normal
boiling point
K = Kc for reactions involving only liquid solutions
K = Kp for reactions involving only gases
104
107
Free energy change during reaction
Equilibrium Expression
aA + bB ⇔ cC + dD
Kc =
[C]c[D]d
[A]a[B]b
Solids and liquids considered unity (1)
105
108
18
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No Work Done at Equilibrium
Write Kp and Kc
• ∆G = 0
• No “free” energy available to do work
• Consider fully charged battery
– Initially
• Consider again
• All reactants, no products
• ∆G large and negative
• Lots of energy available to do work
N2(g) + 3H2(g) ⇄ 2 NH3(g)
– As battery discharges
Kc =
[NH3]2
• Reactants converted to products
• ∆G less negative
• Less energy available to do work
[N2][H2]3
– At Equilibrium
109
• ∆G = Gproducts – Greactants = 0
• No further work can be done
• Dead battery
112
Phase Change = Equilibrium
• In terms of partial pressures,
• H2O(ℓ) H2O(g)
• ∆G = 0 = ∆H – T∆S
• Only one temperature possible for phase
change at equilibrium
N2(g) + 3H2(g) ⇄ 2 NH3(g)
Kp =
– Solid-liquid equilibrium
• Melting/freezing temperature (point)
– Liquid-vapor equilibrium
• Boiling temperature (point)
[PNH3]2
[PN2][PH2]3
110
•
•
•
•
• Thus ∆H = T∆S and
• or T = ∆H
∆S
∆S =
∆H
T
113
Kc and Kp are related.
PV = nRT
P = [n/V]RT
Kp=Kc × (RT)∆n
• ∆n = (number of moles of product gas) –
(number of moles of reactant gas)
• For this reaction, ∆n = -2.
N2(g) + 3H2(g) ⇔ 2 NH3(g)
111
114
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Effect of Temperature on ∆G°
∆G°and Position of Equilibrium
• Reactions often run at T’s other that 298 K
• Position of equilibrium can change as ∆G°depends on T
– ∆G°= ∆H°−
− T∆S°
• For T’s near 298 K, expect only very small changes in ∆H
and ∆S°
• For reaction at T, we can write:
• When ∆G°> 0 (positive)
– Position of equilibrium lies close to reactants
– Little reaction occurs by the time equilibrium is reached
– Reaction appears nonspontaneous
• When ∆G°< 0 (negative)
– Position of equilibrium lies close to products
– Mainly products exist by the time equilibrium is reached
– Reaction appears spontaneous
∆GTo = ∆HTo − T∆S To
o
o
∆G To ≈ ∆H 298
− T∆S 298
115
∆G°and Position of Equilibrium
118
Ex. 8 Determining Effect of T on Spontaneity
• When ∆G°= 0
– Position of equilibrium lies ~ halfway between products
and reactants
– Significant amount of both reactants and products
present at time equilibrium is reached
– Reaction appears spontaneous, whether start with
reactants or products
• Can Use ∆G°to Determine Reaction Outcome
– ∆G°large and positive
• No observable reaction occurs
– ∆G°large and negative
• Reaction goes to completion
• Calculate ∆G° at 25°C and 500°C for the Haber
process
N2 (g) + 3 H2 (g) 2 NH3 (g)
• Assume that ∆H° and ∆S° do not change with T
• Solving Strategy
Step 1. Using data from Tables, calculate ∆H° and ∆S°
for the reaction at 25°C
– ∆H° = – 92.38 kJ
– ∆S° = – 198.4 J/K
116
Learning Check
119
Ex. 8 Determining Effect of T on Spontaneity
Ex. 7 Given that ΔH° = –97.6 kJ/mol,
ΔS° = –122 J/(mol·K), at 1atm and 298K,
will the following reaction occur spontaneously?
• MgO(s) + 2HCl(g) → H2O(ℓ) + MgCl2(s)
∆G° = ∆H° – T∆S°
= –97.6kJ/mol – 298K(–0.122 kJ/mol⋅K)
∆G° = –97.6kJ/mol +36.4kJ/mol
= –61.2 kJ/mol
117
Step 2. Calculate ∆G°for the reaction at 25°C using
∆H° and ∆S°
N2 (g) + 3 H2 (g) 2 NH3 (g)
 1kJ 
– ∆H° = – 92.38 kJ


 1000 J 
°
– ∆S = – 198.4 J/K
• ∆G° = ∆H° – T∆
∆S°
°
• ∆G = –92.38 kJ – (298 K)(–198.4 J/K)
• ∆G° = –92.38 kJ+ 59.1 kJ = –33.3 kJ
• So the reaction is spontaneous at 25°C
120
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Ex. 9 Calculating ∆G at Nonstandard conditions
Ex. 8 Determining Effect of T on Spontaneity
Step 3. Calculate ∆G°for the reaction at 500°C using
∆H° and ∆S°.
– T = 500°C + 273 = 773 K
 1kJ 


– ∆H° = – 92.38 kJ
 1000 J 
°
– ∆S = – 198.4 J/K
• ∆G° = ∆H° – T∆
∆S°
°
• ∆G = –92.38 kJ – (773 K)(–198.4 J/K)
• ∆G° = –92.38 kJ+ 153 kJ = 61 kJ
• So the reaction is NOT spontaneous at 500°C
• Calculate ∆G at 298 K for the Haber process
– N2 (g) + 3 H2 (g) 2 NH3 (g) ∆G°= –33.3 kJ
• For a reaction mixture that consists of 1.0 atm N2, 3.0
atm H2 and 0.5 atm NH3
Step 1 Calculate Q
Q =
2
PNH
3
PN 2 PH32
=
(0.50)2
= 9.3 × 10 − 3
3
(1.0)(3.0)
121
124
Ex. 9 Calculating ∆G at Nonstandard conditions
Ex. 8 Does this answer make sense?
• ∆G° = ∆H° – T∆
∆S°
– ∆H° = – 92.38 kJ
– ∆S° = – 198.4 J/K
• Since both ∆H° and ∆S° are negative
• At low T
– ∆G° will be negative and spontaneous
• At high T
– T∆
∆S° will become a bigger positive number and
– ∆G° will become more positive and thus eventually, at
high enough T, will become nonspontaneous
Step 2 Calculate ∆G = ∆G°+ RT lnQ
∆G = –33.3 kJ/mol + (8.314J/K·mol)*(1kJ/1000J)*
(298K)*ln(9.3 × 10–3)
= –33.3 kJ/mol + (2.479 kJ/mol)*ln(9.3 × 10–3)
= –33.3 kJ + (–11.6 kJ/mol)
= – 44.9 kJ/mol
• At standard conditions all gases (N2, H2 and NH3)
are at 1 atm of pressure
• ∆G becomes more negative when we go to 1.0 atm N2,
3.0 atm H2 and 0.5 atm NH3
• Indicates larger driving force to form NH3
– Preactants > Pproducts
122
Effect of Change in Pressure or Concentration on ∆G
• ∆G at nonstandard conditions is related to ∆G°at
standard conditions by an expression that includes
reaction quotient Q
∆G = ∆G o + RT ln Q
• This important expression allows for any concentration
or pressure
• Recall:
y
Q=
[products ]
125
How K is related to ∆G°
• Use relation ∆G = ∆G°+ RT lnQ to
derive relationship between K and ∆G°
• At Equilibrium
∆G = 0 and Q = K
• So 0 = ∆G°+ RT lnK
∆G°= −RT lnK
Taking antilog (ex) of both sides gives
[reactants ]x
K = e−∆G°/RT
123
126
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At Equilibrium
You Try!
• ∆G°= −RT lnK
and
K = e−∆G°/RT
• Provides connection between ∆G°and K
• Can estimate K’s at various T’s if know ∆G°
• Can get ∆G°in know K’s
Relationship between K and ∆G
Keq
>1
∆G°°
−
Reaction
Spontaneous
<1
+
non-spontaneous
=1
0
Calculate the equilibrium constant for the
decomposition of hydrogen peroxide at 298 K given
∆Go = -234.3 kJ mol.
A. 8.5 x 10-42
B. 1.0 x 10499
C. 3.4 x 10489
D. 1.17 x 1041
Favored
Energy released
Unfavorable
Energy needed
−∆G o
RT
=
−( −234, 300 J / mol )
= 94.56
(8.3145J / K ⋅ mol )(298K )
K = e 94.56 = 1.17 x 10 41
At Equilibrium
127
130
Ex. 10 Calculating ∆G° from K
• Ksp for AgCl(s) at 25°C is 1.8 × 10−10 Determine ∆G° for
the process
• Ag+ (aq) + Cl− (aq) → AgCl (s)
• Reverse of Ksp equation, so
K =
1
=
K sp
1
= 5.6 × 10 9
1.8 × 10 −10
Ex. 12 Calculating K from ∆G°, First
Calculate ∆G°
• Calculate the equilibrium constant at 25°C for the
decarboxylation of liquid pyruvic acid to form
gaseous acetaldehyde and CO2
O
• ∆G°= –RT lnK = –(8.3145J/K·mol)(298K)*
ln(5.6 × 109)*(1kJ/1000J)
• ∆G°= –56 kJ/mol
• Negative ∆G°indicates precipitation will occur
O
C
H3 C
OH
C
C
H3C
H
+
CO2
O
128
131
Ex. 11 Calculating K from ∆G°
First Calculate ∆G°from ∆Gf°
• Calculate K at 25°C for the Haber process
N2 (g) + 3 H2 (g) 2 NH3 (g)
∆G°= –33.3 kJ/mol = –33,300 J/mol
K = e − ∆G
o
/ RT
Step 1 Solve for exponent
− ∆G o
− (−33,300 J / mol )
=
= 13.4
RT
(8.3145 J / K ⋅ mol )(298K )
x
Step 2 Take e to obtain K
K = e − ∆G
o
/ RT
∆Gf°, kJ/mol
CH3COH
CH3COCOOH
−133.30
−463.38
CO2
−394.36
∆G o = ∆G fo (CH 3COH ) + ∆G fo (CO 2 ) − ∆G fo (CH 3COCOOH )
∆G o = −133.30 + (−394.36) − (−463.38 )
= e 13.4 = 7 × 10 5
Large K indicates NH3 favored at RT
Compound
∆G o = −64.28kJ
129
132
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Next Calculate Equilibrium Constant
Calculation of ∆Go at Various Temperatures
Consider the following reaction:
o
K = e − ∆G RT
∆G o
− 64.28kJ
1000J
=
×
= −25.94
RT (8.314 J / K )(298K )
kJ
At 25 oC ∆Go = +130.9 and Kp = 1.1 x 10-23 atm
What do these values tell you about CaCO3?
K = e −( −25.945 ) = e 25.945
K = 1.85 ×
CaCO3 (s) → CaO (s) + CO2 (g)
What happens when the reaction is carried out at a
higher temperature?
1011
133
Temperature Dependence of K
136
Calculating ∆Go and K at Various Temperatures
• ∆G°= –RT lnK = ∆H°– T∆S°
• Rearranging gives
a. What is ∆Go at 1000oC for the calcium carbonate reaction?
 ∆Ho  1 ∆S o
 +
ln K = −
 R T
R


CaCO3 (s) → CaO (s) + CO2 (g)
• Equation for line
Is this reaction spontaneous at 1000oC and 1 atm?
– Slope = – ∆H°/RT
– Intercept = ∆S°/R
b. What is the value of Kp at 1000oC for this reaction?
• Also way to determine
K if you know ∆H°and ∆S°
What is the partial pressure of CO2?
134
137
Ex. 12 Calculate K given ∆H° and ∆S°
Strategy for solution…
• Calculate K at 500 °C for Haber process
N2 (g) + 3 H2 (g) 2 NH3 (g)
– Given ∆H° = – 92.38 kJ and ∆S° = – 198.4 J/K
• Assume that ∆H° and ∆S° do not change with T
∆H o ∆S o
ln K = −
+
RT
R
− 92 ,380 J
− 198.4 J / K
ln K = −
+
(8.314 J / K )(773K ) (8.314 J / K )
lnK = + 14.37 – 23.86 = – 9.49
K = e–9.49 = 7.56 × 10–5
135
a. Calculate ∆Ho and ∆So at 25 oC using standard
enthalpies of formation and standard entropies.
Then substitute into the equation for ∆Gfo.
b. Use the ∆Gfo value to find K (=Kp ) as in
Exercise 13.16.
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Bond Energy
a. From Table B-16 you have the following:
→
CaCO3 (s)
∆Hfo: -1206.9
So:
∆Ho
CaO (s) +
CO2 (g)
-635.1
-393.5 kJ
92.9
38.2
• Amount of energy needed to break chemical
bond into electrically neutral fragments
• Useful to know
• Within reaction
213.7 J/K
– Bonds of reactants broken
– New bonds formed as products appear
= [(-635.1 – 393.5) – (-1206.9)] = 178.3 kJ
∆So = [(38.2 + 213.7) – (92.9)] = 159.0 J/K
• Bond breaking
∆GTo = ∆Ho –T∆So
= 178.3kJ – (1273 K)(0.1590kJ/K) = -24.1 kJ
∆Go is negative – reaction is spontaneous
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142
Bond Energies
b. Substitute the values of ∆Go at 1273 K, which
equals -24.1 x 103 J, into the equation relating
ln K and ∆Go.
• Can be determined spectroscopically for simple
diatomic molecules
– H2, O2, Cl2
-24.1 x 103 = 2.278
∆Go
=
-RT
-8.31 x 1273
ln K =
– 1st step in most reactions
– One of the factors that determines reaction rate
– Ex. N2 very unreactive due to strong N ≡ N bond
• More complex molecules, calculate using
thermochemical data and Hess’s Law
– Use ∆H°formation enthalpy of formation
• Need to define new term
• Enthalpy of atomization or atomization energy,
∆Hatom
K = Kp = e2.278 = 9.76
Kp = PCO2 = 9.76 atm
– Energy required to rupture chemical bonds of 1
mole of gaseous molecules to give gaseous atoms
140
Where does the reaction change from spontaneous
to non-spontaneous?
∆Go = 0 = ∆Ho - T∆So
Solve for T;
T =
∆Ho
∆So
=
178.3 kJ
0.1590 kJ/K
143
Determining Bond Energies
• Ex. CH4(g) → C(g) + 4 H(g)
• ∆Hatom = energy needed to break all bonds in
molecule
• ∆Hatom /4 = average bond C—H dissociation
energy in methane
– D = bond dissociation energy
• Average bond energy to required to break all bonds in
molecule
– How do we calculate this?
T = 1121 K = 848 oC
141
• Use ∆H°f for forming gaseous atoms from elements in
their standard states
• Hess’s Law
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Using Bond Energies to Estimate ∆H°f
Determining Bond Energies
• Calculate ∆H°f for CH3OH(g) (bottom reaction)
• Use 4 step path
• Path 1: bottom
– Formation of CH4 from its elements = ∆H°f
– Step 1 break 1C—C bonds
– Step 2 break 2H—H bonds
– Step 3: break 1O—O bond
• Path 2: top 3 step path
– Step 1 break H—H bonds
– Step 2 break C—C bonds
– Step 3: form 4 C—H bonds
1. 2H2(g) → 4H(g)
∆H°1 = 4∆H°f (H,g)
2. C(s) → C(g)
∆H°2 = ∆H°f (C,g)
3. 4H(g) + C(g) → CH4(g) ∆H°3 = –∆Hatom
2H2(g) + C(s) → CH4(g) ∆H°= ∆H°f(CH4,g)
– Step 4: form 3 C—H, 1 O—H, & 1O—C bonds
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148
Using Bond Energies
Calculating ∆Hatom and Bond Energy
∆H°f(CH3OH,g) = ∆H°f(C,g) + 4∆H°f(H,g) + ∆H°f(O,g)
∆H°f(CH4,g) = 4∆H°f(H,g) + ∆H°f(C,g) – ∆Hatom
– ∆Hatom (CH3OH,g)
• Rearranging gives
∆Hatom= 4∆H°f(H,g) + ∆H°f(C,g) –∆H°f(CH4,g)
• ∆H°f(C,g) + 4∆H°f(H,g) + ∆H°f(O,g)
= {716.7 +(4*217.9) + 249.2}kJ = +1837.5 kJ
• Look these up in Table 18.3, 6.2 or appendix C
∆Hatom= 4(217.9kJ/mol) + 716.7kJ/mol –
(–74.8kJ/mol)
∆Hatom= 1663.1 kJ/mol of CH4
• ∆Hatom (CH3OH,g) = 3DC—H + DC—O + DO—H
= (3*412) + 360 + 463 = 2059 kJ
• ∆H°f(CH3OH,g) = +1837.5 kJ – 2059 kJ = – 222 kJ
1663.1 kJ/mol
4
= 415.8 kJ/mol of C—H bonds
bond energy =
• Experimentally find ∆H°f(CH3OH,g) = – 201 kJ/mol
• So bond energies give estimate within 10% of actual
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149
Chapter 13/19 Chemical Thermodynamics
Table 19.4 Some Bond Energies
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
147
Spontaneous Chemical and Physical Processes
Entropy and Disorder
Entropy and the Second Law of Thermodynamics
Standard-State Entropies of Reaction
The Third Law of Thermodynamics
Calculating Entropy Changes for Chemical Reactions
Gibbs Free Energy
The Effect of Temperature on the Free Energy of a Reaction
Beware of Oversimplification
Stand-State Free Energies of Reaction
Equilibria Expressed in Partial Pressures
Interpreting Stand-State Free Energy of Reaction Data
Relationship Between Free Energy and Equilibrium Constants
Temperature Dependence of Equilibrium Constants
Gibbs Free Energies of Formation and Absolute Entropies
Calculate ∆H with bond energies
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Entropy Change for a Phase Transition
∆S > q/T
(at equilibrium)
What processes can occur under phase
change at equilibrium?
Solid to liquid
Liquid to gas
Solid to gas
151
Solution:
∆S = ∆Hvap/T = (39.4 x 103 J/mol)/ 298 K
= 132 j/(mol·K)
Entropy of Vapor = (216 + 132) J/(mol·K)
= 348 J/(mol·K)
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153
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