Chapter 19: Thermodynamics Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Thermodynamics Study of energy changes and flow of energy Answers several fundamental questions:
Is it possible for a given reaction to occur? Will the reaction occur spontaneously (without
outside interference) at a given T ? Will reaction release or absorb heat?
Tells us nothing about time frame of reaction Kinetics
Two major considerations Enthalpy changes, H (heats of reaction)
Heat exchange between system and surroundings
Nature's trend to randomness or disorder Entropy
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Review of First Law of Thermodynamics
Internal energy, E System's total energy Sum of KE and PE of all particles in system
or for chemical reaction
E + energy into system E – energy out of system
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Two Methods of Energy Exchange Between System and
Surroundings Heat q Work w E = q + w Conventions of heat and work
q + Heat absorbed by system
Esystem increases
q – Heat released by system
Esystem decreases
w + Work done on system Esystem increases
w – Work done by system Esystem decreases
4
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
First Law of Thermodynamics Energy can neither be created nor destroyed It can only be converted from one form to
another Kinetic Potential Chemical Electrical Electrical Mechanical
E is a state function E is a change in a state function Path independent E = q + w
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Work in Chemical Systems
1. Electrical2. Pressure-volume or PV
w = –PV Where P = external pressure
If PV only work in chemical system, then
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Heat at Constant Volume Reaction done at constant V
V = 0 PV = 0, so E = qV
Entire energy change due to heat absorbed or lost
Rarely done, not too useful
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Heat at Constant Pressure More common Reactions open to atmosphere
Constant P Enthalpy
H = E + PV Enthalpy change
H = E + PV Substituting in first law for E gives
H = (q – PV) + PV = qP H = qP
Heat of reaction at constant pressure8
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Converting Between E and H For Chemical Reactions
H E Differ by H – E = PV Only differ significantly when gases
formed or consumed Assume gases are ideal
Since P and T are constant
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Converting Between E and H For Chemical Reactions
When reaction occurs 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 + PV gives
or
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 1 What is the difference between H and E for the following reaction at 25 °C?
2 N2O5(g) 4 NO2(g) + O2(g)
What is the % difference between H and E ?Step 1: Calculate H using data (Table 7.2) Recall
H ° = (4 mol)(33.8 kJ/mol) + (1 mol)(0.0 kJ/mol) – (2 mol)(11 kJ/mol)
H ° = 113 kJ 11
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 1. H and E (cont.)Step 2: Calculate ngas
ngas = (ngas)products – (ngas)reactants
ngas = (4 + 1 – 2) mol = 3 mol
Step 3: Calculate E using R = 8.31451 J/K mol T = 298 K E = 113 kJ –
(3 mol)(8.314 J/K mol)(298 K)(1 kJ/1000 J)E = 113 kJ – 7.43 kJ = 106 kJ
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 1. H and E (cont.)
Step 4: Calculate percent difference
Bigger than most, but still small
Note: Assumes that volumes of solids and liquids are negligible
Vsolid Vliquid << Vgas
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Is Assumption that Vsolid Vliquid << Vgas Justified?
Consider CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)
37.0 mL 2×18.0 mL 18 mL 18 mL 24.4 L
Volumes assuming each coefficient equal number of moles
So V = Vprod – Vreac = 24.363 L 24.4 L Yes, assumption is justifiedNote: If no gases are present reduces
to 14
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check Consider the following reaction for picric acid:8O2(g) + 2C6H2(NO2)3OH(l ) → 3 N2(g) + 12CO2(g) + 6H2O(l )
Calculate Η °, Ε °
8O2(g) + 2C6H2(NO2)3OH(l ) → 3 N2(g) + 12CO2(g) + 6H2O(l )
Η °f
(kJ/mol) 0.00 3862.94 0.00 –393.5 –241.83
Ε ° = H ° – ngasRT = H ° – (15 – 8) mol × 298 K × 8.314 × 10–3 kJ/(mol K)
H ° = 12 mol(–393.5 kJ/mol) + 6 mol(–241.83 kJ/mol) + 6 mol(0.00 kJ/mol) – 8 mol(0.00 kJ/mol) – 2 mol(3862.94 kJ/mol)
15
ΔH ° = –13,898.9 kJ
Ε ° = –13,898.9 kJ – 29.0 kJ = –13,927.9 kJ
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Given the following:3H2(g) + N2(g) → 2NH3(g) Η °= –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
Η = E + nRT E = Η – nRT E = –46.19 kJ mol –
(–2 mol)(8.314 J K–1mol–1)(298 K)(1 kJ/1000 J) E = –51.14 kJ mol–1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Enthalpy Changes and Spontaneity What are relationships among factors
that influence spontaneity? Spontaneous Change
Occurs by itself Without outside
assistance until finished e.g.
Water flowing over waterfall Melting of ice cubes in glass
on warm day
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Nonspontaneous Change 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
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Nonspontaneous Change Occur only when
accompanied by some spontaneous change You consume food,
spontaneous biochemical reactions occur to supply muscle power to tidy up room or to build wall
Spontaneous mechanical or chemical change to generate electricity
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Direction of Spontaneous Change Many reactions which occur
spontaneously are exothermic: Iron rusting Fuel burning
H and E are negative Heat given off Energy leaving system
Thus, H is one factor that influences spontaneity
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Direction of Spontaneous Change Some endothermic reactions occur
spontaneously: Ice melting Evaporation of water Expansion of CO2 gas into vacuum
H and E are positive Heat absorbed Energy entering system
Clearly other factors influence spontaneity
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!We can expect the combustion of
propane to be:A. spontaneousB. non-spontaneousC. neither
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy (Symbol S ) 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 possible ways to
distribute energy and that it is a more probable result
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Fig. 19.6 - Entropy Distribution
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Low Entropy – (a) A absorbs E in units of
10 Few ways to distribute
E ● represent E ’s of
molecules of A
High Entropy – (b) More ways to distribute E B absorbs E in units of 5 ●represent E ’s of
molecules of B
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy
If Energy = money Entropy (S ) describes number of different
ways of counting it25
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Examples of Spontaneity Spontaneous reactions
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy, S State function Independent of path S = Change in entropy
For chemical reactions or physical processes
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Effect of Volume on Entropy
For gases, entropy increases as volume increasesa) Gas separated from vacuum by partition b) Partition removed, more ways to distribute energy c) Gas expands to achieve more probable particle
distribution More random, higher probability, more positive S
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Effect of Temperature on Entropy As T increases, entropy increases
(a) T = 0 K, particles in equilibrium lattice positions and S relatively low (b) T > 0 K, molecules vibrate, S increases(c) T increases further, more violent vibrations occur and S higher than in (b)
29
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Effect of Physical State on Entropy Crystalline solid very low entropy
Liquid higher entropy, molecules can move freely More ways to distribute KE among them
Gas highest entropy, particles randomly distributed throughout container Many, many ways to distribute KE
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy Affected by Number of Particles Adding particles to system
Increase number of ways energy can be distributed in system
So all other things being equal Reaction that produces more particles will
have positive S
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which represents an increase in entropy?A. Water vapor condensing to liquidB. Carbon dioxide sublimingC. Liquefying helium gasD. Proteins forming from amino acids
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy Changes in Chemical Reactions
Reactions without gases Calculate number of mole molecules
n = nproducts – nreactants
If n is positive, entropy increases More molecules, means more disorder Usually the side with more molecules, has less
complex molecules
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Reactions involving gases Calculate change in number of moles of
gas, ngas
If ngas is positive , S is positivengas is more important than nmolecules
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Entropy Changes in Chemical ReactionsEx. N2(g) + 3H2(g) 2NH3(g)
nreactant = 4 nproduct = 2
n = 2 – 4 = –2Predict Srxn < 0
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Higher positional probability
Lower positional probability
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 2 Predict Sign of S for Following Reactions
CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)
ngas = 1 mol – 0 mol = 1 mol
since ngas is positive, S is positive
2 N2O5(g) 4 NO2(g) + O2(g)
ngas = 4 mol + 1 mol – 2 mol = 3 mol
since ngas is positive, S is positive
OH–(aq) + H+(aq) H2O
ngas = 0 mol n = 1 mol – 2 mol = –1 mol since ngas is negative, S is negative
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Predict Sign of S in the Following:Dry ice → carbon dioxide gas
Moisture condenses on a cool window
AB → A + B
A drop of food coloring added to a glass of water disperses
2Al(s) + 3Br2(l ) → 2AlBr3(s)
positive
positive
positive
negative
CO2(s) → CO2(g)
H2O(g) → H2O(l )
negative36
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!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)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Both Entropy and Enthalpy Affect Reaction Spontaneity
Sometimes they work together Building collapses PE decreases H is negative Stones disordered S is positive
Sometimes work against each other Ice melting (ice/water mix) Endothermic
H is positive nonspontaneous Increase in disorder of molecules
S is positive spontaneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 Next few slides will develop the
relationship between H, S, and T that defines a spontaneous process
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Second Law of Thermodynamics When a spontaneous event occurs, total
entropy of universe increases (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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Law of Conservation of Energy
Says q lost by system must be gained by surroundings qsurroundings = –qsystem
If system at constant P, then qsystem = H
So qsurroundings = –Hsystem
and
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Spontaneous Reactions (cont.)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Thus Entropy for Entire Universe is
Multiplying both sides by T we getTStotal = TSsystem – Hsystem
orTStotal = – (Hsystem – TSsystem)
For reaction to be spontaneous TStotal > 0 (entropy must increase)
So, (Hsystem – TSsystem) < 0 must be negative for reaction to be
spontaneous 43
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Gibbs Free Energy Would like one quantity that includes all
three factors that affect spontaneity of a reaction
Define new state function, G 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 – TS
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
G = H – TS
G state function Made up of T, H and S = state
functions Has units of energy Extensive property
G = Gfinal – Ginitial
Gibbs Free Energy
G < 0 Spontaneous processG = 0 At equilibriumG > 0 Nonspontaneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Criteria for Spontaneity? At constant P and T, process spontaneous
only if it is accompanied by decrease in free energy of system
H
S Spontaneous?
– + G = (–) – [T (+)] = –
Always, regardless of T
+ – G = (+) – [T (–)] = +
Never, regardless of T
+ + G = (+) – [T (+)] = ?
Depends; spontaneous
at high T, –G
– – G = (–) – [T (–)] = ? Depends; spontaneous
at low T, –G
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Summary When H and S have
same sign, T determines whether spontaneous or nonspontaneous
Temperature-controlled reactions are spontaneous at one temperature and not at another
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!At what temperature (K) will a reaction become nonspontaneous when H = –50.2 kJ mol–1 and S = +20.5 J K–1 mol–1?A. 298 KB. 1200 KC. 2448 KD. The reaction cannot become non-spontaneous at any temperature
48
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Third Law of Thermodynamics 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)49
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Consequences of Third Law1. All substances have positive entropies as
they are more disordered than at 0 K Heating increases randomness S ° is biggest for gases—most disordered
2. For elements in their standard states S ° 0 (but Hf° = 0)
Units of S ° J/(mol K)Standard Entropy Change
To calculate S ° for reaction, do Hess's Law type calculation
Use S ° rather than entropies of formation
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
Calculate S ° for the following: CO2(s) → CO2(g)
S °: 187.6 213.7 J/mol KS ° = (213.7 – 187.6) J/mol KS ° = 26.1 J/mol K
CaCO3(s) → CO2(g) + CaO(s)
S °: 92.9 213.7 40 J/mol KS ° = (213.7 +40 – 92.9) J/mol KS ° = 161 J/mol K
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 3. Calculate S° for reduction of aluminum oxide by hydrogen gasAl2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
Substance S ° (J/ K mol)
Al(s) 28.3
Al2O3(s) 51.00
H2(g) 130.6
H2O(g) 188.7
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 3 Calculate ΔS° (cont.)
S ° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 )
S ° = 179.9 J/K
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!What is the entropy change for the
following reaction?
Ag+(aq) + Cl–(aq) → AgCl(s)So 72.68 56.5 96.2 J K–1 mol–1
A. +32.88 J K–1 mol–1
B. –32.88 J K–1 mol–1
C. –32.88 J mol–1
D. +112.38 J K–1 mol–1 S ° = [96.2 – (72.68 + 56.5)] J K–1 mol–1 S ° = –32.88 J K–1 mol–1
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Standard Free Energy Changes 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 Method 1: G ° = H ° – TS ° Method 2:
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 1
Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
Step 1: Calculate H° for reaction using heats of formation below
56
Substance (kJ/mol)
Al(s) 0.0
Al2O3(s) –1669.8
H2(g) 0.0
H2O(g) –241.8
Calculate G ° for reduction of aluminum oxide by hydrogen gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 1
H ° = 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ) H ° = 944.4 kJ
57
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 1
Step 2: Calculate S ° see Example 3S ° = 179.9 J/K
Step 3: Calculate G ° = H ° – (298.15 K)S °
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 ° is positive Indicates that the reaction is not
spontaneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 2 Use Standard Free Energies of
Formation
Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 2Calculate G ° for reduction of aluminum
oxide by hydrogen gas. Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
60
Substance (kJ/mol)
Al(s) 0.0
Al2O3(s) –1576.4
H2(g) 0.0
H2O(g) –228.6
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 4. Calculate G ° Method 2
61
G° = 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ) G° = 890.6 kJ
Both methods same within experimental error
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Spontaneous Reactions Produce Useful Work
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
heat62
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Thermodynamically Reversible 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
63
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
G ° = Maximum Possible Work G ° is maximum amount of energy
produced during a reaction that can theoretically be harnessed as work 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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 5 Calculate G ° Calculate G ° for reaction below at 1 atm and 25 °C, given H ° = –246.1 kJ/mol, S
° = 377.1 J/(mol K). H2C2O4(s) + ½O2(g) → 2CO2(g) + H2O(l )
G °25 = H – TS
G ° =(–246.1 – 112.4) kJ/mol G ° = –358.5 kJ/mol
65
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Calculate G ° for the following reaction,
H2O2(l ) → H2O(l ) + O2(g)
given H °= –196.8 kJ mol–1 and S °= +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 × 105 kJ mol–1
G ° = –196.8 kJ mol–1 – 298 K (0.12572 kJ K–1 mol-1)
G ° = –234.3 kJ mol 66
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
G° and Position of Equilibrium 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
67
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
G ° and Position of Equilibrium 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 completion68
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
At EquilibriumG ° = –RT lnK and K = e–G
°/RT
Provides connection between G ° and K Can estimate K at various temperatures if G ° is
known Can get G ° if K is known
Relationship between K and G ° Keq G ° Reaction
> 1 – SpontaneousFavored
Energy released
< 1 +non-
spontaneousUnfavorable
Energy needed
= 1 0 At Equilibrium69
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check
Ex. 6 Given that H ° = –97.6 kJ/mol, S ° = –122 J/(mol K), at 1 atm and 298 K, will the following reaction occur spontaneously?
MgO(s) + 2HCl(g) → H2O(l ) + MgCl2(s)
G ° = H ° – TS ° = –97.6 kJ/mol – 298 K(–0.122 kJ/mol K)G ° = –97.6 kJ/mol + 36.4 kJ/mol = –61.2 kJ/mol
70
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
This important expression allows for any concentration or pressure
Recall:
71
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 7 Calculating G at Nonstandard Conditions
Calculate G at 298 K for the Haber process N2(g) + 3H2(g) 2NH3(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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 7 Calculating G at Nonstandard conditions
73
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Free Energy Diagrams G is different than
G° G interpreted as the
slope of the free energy curve and tells which direction reaction proceeds to equilibrium
Minimum, at G = 0, indicates composition of equilibrium mixture
Because G ° is positive equilibrium lies closer to reactants
74
Equilibrium occurs here at Ptotal = 1 atm with
16.6% N2O4 decomposed
Non-spont. G ° > 0
N2O4(g) 2NO2(g)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Free Energy Diagrams
75
Spontaneous Rxn G ° < 0 Reaction proceeds toward equilibrium from either A or B where slope, G , is not zero
The curve minimum, where G = 0, lies closer to the product side for spontaneous reactions. This is determined by G °
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
K = e–G ° /RT
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 8 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
G ° = –RT lnK = –(8.3145 J/K mol)(298 K) × ln(5.6 109)(1
kJ/1000 J) G ° = –56 kJ/mol Negative G ° indicates precipitation will occur
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910
106.5108.1
11
spKK
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
System at Equilibrium Neither spontaneous nor nonspontaneous In state of dynamic equilibrium Gproducts= Greactants
G = 0 Consider freezing of water at 0˚C H2O(l ) H2O(s)
System remains at equilibrium as long as no heat added or removed
Both phases can exist together indefinitely Below 0 °C, G < 0 freezing spontaneous Above 0 °C, G > 0 freezing nonspontaneous
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
No Work Done at Equilibrium G = 0 No “free” energy available to do work Consider fully charged battery
Initially All reactants, no products G large and negative Lots of energy available to do work
As battery discharges Reactants converted to products ∆G less negative Less energy available to do work
At equilibrium G = Gproducts – Greactants = 0 No further work can be done Dead battery
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Phase Change = Equilibrium H2O(l ) H2O(g)
G = 0 = H – TS Only one temperature possible for phase
change at equilibrium Solid-liquid equilibrium
Melting/freezing temperature (point) Liquid-vapor equilibrium
Boiling temperature (point) Thus H = TS and or
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TH
S
SH
T
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 9 Calculate Tbp
Calculate Tbp for reaction below at 1 atm and 25 °C, given H ° = 31.0 kJ/mol, S ° = 92.9 J/ mol K
Br2(l ) → Br2(g)
For T > 334 K, G < 0 and reaction is spontaneous (S ° dominates) For T < 334 K, G > 0 and reaction is nonspontaneous (H ° dominates) For T = 334 K, G = 0 and T = normal boiling point
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckWhat is the expected melting point for Cu?
Hf° (kJ/mol) Gf° (kJ/mol) S (J/mol K)
Cu(l ) 341.1 301.4 166.29
Cu(s) 0 0 33.1
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H = 1mol(341.1 kJ/mol – 1mol(0 kJ/mol)
Cu(s) Cu(l )
H = +341.1kJ
S = 133.19 J/KS = 1mol(166.29 J/mol K – 1mol(33.1 J/mol K)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Effect of Temperature on G ° Reactions often run at temperatures
other that 298 K Position of equilibrium can change as
G ° depends on temperatureG ° = H ° – TS °
For temperatures near 298 K, expect only very small changes in H ° and S °
For reaction at temperature, we can write:
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 10 Determining Effect of Temperature on Spontaneity
Calculate G ° at 25 °C and 500 °C for the Haber process
N2(g) + 3H2(g) 2NH3(g)
Assume that H ° and S ° do not change with temperature
Solving strategyStep 1. Using data in Tables 6.2 and 18.2
calculate H ° and S ° for the reaction at 25 °C H ° = – 92.38 kJ S ° = – 198.4 J/K 84
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 10 Determining Effect of Temperature on Spontaneity
Step 2. Calculate G ° for the reaction at 25 °C using H ° and S ° N2(g) + 3H2(g) 2NH3(g)
H ° = –92.38 kJ S ° = –198.4 J/K
G ° = H ° – TS° 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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 10 Determining Effect of Temperature on Spontaneity
Step 3. Calculate G ° for the reaction at 500 °C using H ° and S °. T = 500 °C + 273 = 773 K H ° = – 92.38 kJ S ° = – 198.4 J/K
G ° = H ° – TS ° 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 86
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 10 Does this answer make sense? G ° = H ° – TS °
H ° = –92.38 kJ S ° = –198.4 J/K
Since both H ° and S ° are negative At low temperature
G ° will be negative and spontaneous At high temperature
TS ° will become a bigger positive number and
G ° will become more positive and thus eventually, at high enough temperature, will become nonspontaneous 87
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 11 Calculating K from G° Calculate K at 25 °C for the Haber process
N2(g) + 3H2(g) 2NH3(g) G ° = –33.3 kJ/mol = –33,300 J/mol
Step 1 Solve for exponent
Step 2 Take e x to obtain K
Large K indicates NH3 favored at room temp. 88
54.13/ 107 eeK RTG
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Calculate the equilibrium constant for the decomposition of hydrogen peroxide at 298 K given G ° = –234.3 kJ mol.A. 8.5 × 10-42
B. 1.0 × 10499
C. 3.4 × 10489
D. 1.17 × 1041
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
H3CC
C
O
OH
O
H3CC
O
H+ CO2
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 12 First Calculate ∆G° from ∆Gf°
Compound ∆Gf°, kJ/mol
CH3COH –133.30
CH3COCOOH –463.38
CO2 –394.36
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 12 Next Calculate Equilibrium Constant
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K = 1.85 1011
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Temperature Dependence of K G ° = –RT lnK = H ° – TS °
Rearranging gives
Equation for line Slope = –H °/RT Intercept = S °/R
Also way to determine K if you know H ° and S °
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ex. 12 Calculate K given H ° and S ° Calculate K at 500 °C for Haber process
N2(g) + 3H2(g) 2NH3(g)
Given H ° = –92.38 kJ and S ° = –198.4 J/K
Assume that H ° and S ° do not change with T
ln K = + 14.37 – 23.86 = –9.49 K = e–9.49 = 7.56 × 10–5
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Bond Energy Amount of energy needed to break chemical
bond into electrically neutral fragments Useful to know Within reaction
Bonds of reactants broken New bonds formed as products appear
Bond breaking First step in most reactions One of the factors that determines reaction rate e.g. N2 very unreactive due to strong NN bond
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Bond Energies Can be determined spectroscopically for
simple diatomic molecules H2, O2, Cl2
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
Energy required to rupture chemical bonds of one mole of gaseous molecules to give gaseous atoms
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Determining Bond Energies Ex. CH4(g) C(g) + 4H(g)
H atom = energy needed to break all bonds in molecule
H atom /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? Use H °f for forming gaseous atoms from elements
in their standard states Hess’s Law
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Determining Bond Energies Path 1: Bottom
Formation of CH4 from its elements = H°f
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 = 4H °f (H,g)
2.C(s) C(g) H °2 = H °f (C,g)
3.4H(g) + C(g) CH4(g) H °3 = –H °atom
2H2(g) + C(s) CH4(g) H ° = H °f(CH4,g)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Calculating H atom and Bond Energy H °f(CH4,g) = 4H °f(H,g) + H °f(C,g) – H atom
Rearranging givesH atom= 4H °f(H,g) + H °f(C,g) – H °f(CH4,g) Look these up in Table 18.3, 6.2 or
appendix CH atom= 4(217.9kJ/mol) + 716.7kJ/mol –
(–74.8kJ/mol)H atom= 1663.1 kJ/mol of CH4
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= 415.8 kJ/mol of C—H bonds
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Table 19.4 Some Bond Energies
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Using Bond Energies to Estimate H
˚f Calculate H ˚f for CH3OH(g) (bottom reaction)
Use four step path Step 1: break C—C bonds Step 2: break H—H bonds Step 3: break O—O bond Step 4: form 4 C—H bonds
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Using Bond EnergiesH °f(CH3OH,g) =
H ˚f (C,g) + 4H ˚f (H,g) + H ˚f (O,g) – H atom (CH3OH,g)
H °f(C,g) + 4H ˚f (H,g) + H ˚f (O,g) = [716.7 +(4 × 217.9) + 249.2] kJ = +1837.5 kJ
H atom (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
Experimentally find H ˚f (CH3OH,g) = –201 kJ/mol So bond energies give estimate within 10% of actual
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