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Chapter 15
(not much on E)
Thermodynamics:
Enthalpy, Entropy
& Gibbs Free
Energy
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mo Thermodynamics: thermo = heat (energy)
dynamics = movement, motion
Some thermodynamic terms chemists use:
System: the portion of the universe that we areconsidering
open system: energy & matter can transfer
closed system: energy transfers only
isolated system: no transfers
Surroundings: everything else besides the system
Isothermal: a system that is kept at a constant
temperatureby adding or subtracting heat from the
surroundings.Heat Capacity: the amount of heat energy required to
raise the temperature of a certain amount of material by
1C (or 1 K).
Specif ic Heat Capacity: 1 g by 1C
Molar H eat Capacity: 1 mole by 1C
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mo Calorie: the amount of heat required to raise the
temperature of 1g of water by 1C.
specific heat of water = 1 cal/g C
1 calorie = 4.18 joules
Specific Heats and Molar Heat Capacities
Substance
Al
Specific Heat
(J/C
g Molar Heat (J/C
mol)
.0.380.450.84
.24.425.183.8
CuFe
CaCO3
Ethanol
WaterAir
2.43
4.181.00
112.0
75.3~ 29
important to: engineers chemists
EXAMPLE: How many joules of energy are needed raise
the temperature of an iron nail (7.0 g) from 25C to125C?The specific heat of iron is 0.45 J/CHeat energy = (specific heat)(mass)(T)
Heat energy = (0.45 J/C g)(7.0 g)(100C) = 315
J
g.
Note that T can be C or K, but NOT F. When jusbeing used
in a scientific formula it will usually be kelvin
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Problem: How much energy does it take to raise
the body temperature 2.5C (a fever of just over
103F) for someone who weighs 110 pounds (50 kg).
Assume an average body specific heat capacity of3 J/C.g.
Problem: What would be more effective at melting
a frozen pipehot water or a hair dryer (hot air
gun). Why?
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mo State Functions
System properties, such as pressure (P), volume (V),
and temperature (T) are called state functions.
The value of a state function depends only on the state
of the system and not on the way in which the system
came to be in that state.
A change in a state function describes a difference
between the two states. It is independent of the
process or pathway by which the change occurs.
For example, if we heat a block of iron from room
temperature to 100C, it is not important exactly how
we did it. Just on the initial state and the final state
conditions. For example, we could heat it to 150C,
then cool it to 100C. The path we take is unimportant,so long as the final temperature is 100C.
Miles per gallon for a car, is NOT a state function. It
depends highly on the path: acceleration, speed, wind,
tire inflation, hills, etc.
Most of the thermodynamic values we will discuss in
this chapter are state functions.
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mo Energy: " The capacity to do work
and/or transfer heat"
Forms of Energy:
Kinetic (Ekinetic = mv2)
Heat
Light (& Electromagnetic)
Electricity
Sound
Potential
Gravitational
Chemical
Nuclear - Matter (E = mc2)
WORK
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mo
First Law of Thermodynamics:
The total amount of energy (and
mass) in the universe is constant.
In any process energy can be
changed from one form to
another; but it can never be
created nor destroyed.
You can' t get something for
nothing
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mo
Enthalpy (Heats) of Reaction
The amount of heat released or absorbed by a
chemical reaction at constant pressure (as one woulddo in a laboratory) is called the enthalpy or heat or
reaction. We use the symbol H to indicate
enthalpy.
Sign notation (EXTREMELY IMPORTANT!!):
+H indicates that heat is being absorbed in the
reaction (it gets cold)
H indicates that heat is being given off in the
reaction (it gets hot) exothermic
endothermic
Standard Enthalpy = H ( is called a not)
Occurring under StandardConditions:
Pressure 1 atm (760 torr)
1.0 MConcentration
Temperature is notdefined or part of Standard
Conditions, but is often measured at 298 K (25C).
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mo Standard Enthalpy of Formation -- Hf
The amount of heat absorbed (endothermic) or
released (exothermic) in a reaction in which one
mole of a substance is formed from its elements in
their standard states, usually at 298 K (25C).
Also called heat of formation.
Hf = 0 for any element in its standard state (the
natural elemental form at 1 atm or 1 M) at 298 K.
EXAMPLES:
grap e, s +O2(g) (g)H = 0 kJ/mol 0kJ/mol
.kJ/mol
rxn
product(one mole)elements in
theirstandard states
nega ve s gnheat released --exotherm ic rxn
H (CO2) = 393.5kJ/mol
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mo10g +(g) (g)
H = 0 kJ/mol 0kJ/mol
.mol
rxn
elements intheirstandardstates
product(twomoles
v e yto puton per molebasis!!
nega ve s gnheat released --
exotherm ic rxn H (H O) = 241.8kJ/mol
ote that I usually will nothave you calculate Hf onhomeworks or testsso you generally dont have to worry about normalizinanswer to a
per mole basis.Hess's Law -- Adding Reactions
The overall heat of reaction (Hrxn) is equal to the
sum of the Hf (products) minus the sum of the Hf
(reactants):
(# eqiv)H(reactants)
f
= eq v(products)rxn
Therefore, by knowing Hf of the reactants and
products, we can determine the Hrxn for any
reaction that involves these reactants and products.
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mo11EXAMPLE: CO2 is used in certain kinds of fireextinguishers
to put out simple fires. It works by smothering the fi" heavier" CO2 that replaces oxygen needed to maintfire.CO2 is not good, however, for more exotic electrical
chemical fires.g s
COg s
C s =kJ/mol
-kJ/mol
- mokJ/mol
TS TS
(# eqiv)H(reactants)
f
= eq v (products)rxn
(2 eqiv)(-602 kJ/mol) + (1
eqiv)(0 kJ/mol)
= rxn
(2 eqiv)(0 kJ/mol) + (1 eqiv)
393 kJ/mol)= -
kJ/mol)-
kJ/mol)rxn
= - mo +
393 kJ/mol
rxn
} highly exothermic
rxn !!
=kJ/mol
x
Therefore, Mg will " burn"CO2 !
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mo12You can also add two reactions together to get the Hfor
another new reaction:Calculate for the
following reaction:
HrxnH O l OH l =??4 5 nven ese wo reac ons an
thermod namic data:a+ 3O2(g)b) C H (g) +
3O (g)
g3H2O(l)2CO2(g) +
2H2O(l)
= -kJ/moln = -kJ/mol4 n
ow osolve:
s on e pro uc s e o e rsreaction -- so we want to5sw c equa on a aroun o ge a son the roduct side:5
g3H2O l
g +1367 kJ/mol
rxn
*
no e awhen we
chan es si n! ! !
reverse ereaction, H n
ow we can a e wo reac ons oge erto ive us the desirednereaction:
C2H5OH(l) + 3O2(g) H= +1367 kJ/mol
rxn
g3H2O l
+H = -1411kJ/molrxn
g3O
g2H2O l4
H = -44
kJ/molrxn
gH O l OH l4 5
If we have to multiply one (or more) of the reactions someconstant to get them to add correctly, then we also wohaveto multiply Hrxn for that reaction by the same amo
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mo14Problem: Calculate H for the followingrxnreactions given the following Hf values:
Hf (SO2, g) = 297kJ/mol
Hf (H2SO4, l) = 814
kJ/mol
Hf (H2O, l) = 286
kJ/mol
Hf (SO3, g) = 396kJ/mol
Hf (H2SO4, aq) = 90
kJ/mol
Hf (H2S, g) = 20
kJ/mol
a) S(s) + O2(g) SO2(g)
b) 2SO2(g) + O2(g)
c) SO3(g) + H2O(l)
d) 2H2S(g) + 3O2(g)
2SO3(g)
H2SO4(l)
2SO2(g) + 2H2O(l)
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mo15Internal Energy -- E
The internal energy, E, represents all the energy
contained within a material. It includes kinetic
energy (heat), intra- and intermolecular forces
(bond energies, electrostatic forces, van der Waals),
and any other forms of energy present. As with
enthalpy, H, the absolute value cant (or is extremely
difficult) to define.What we can track is the change in E:
E = EfinalEinital = Eproducts
Ereactants
A key relationship is:E = q + w
Where q = heat and w = work performed on or by
the system. Sign notations:
q= positive = heat added to system (adds energy)
q= negative = heat removed (removes energy)
w= positive = work done on system (adds energy
w= negative = work done by system (removes ener
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mo16The most common type of work involvespressure/volume changes: e.g., explosion of gasoline
vapors in an internal combustion engine. The
explosion creates a dramatic pressure and volumeincrease that pushes the piston and creates work.
If one has a constant volume situation, then no
pressure/volume work will be done and w = 0.
So under constant volume conditions:
E = qThe change in internal energy is, therefore, equal to
the amount of heat added or removed from the
material (system).
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mo17Relationship Between E & H
E = q (at constant volume and temperature)
H = q (at constant pressure and temperature)
The difference is that volume changes occur for H athat typically involves work of some type. Remembethat significant volume changes only occur when gassare involved. So we only need to be concerned about
volume work when there is a change in the amount oproduced in a chemical reaction.
The relationship between H and E is defined,
therefore, as:
H = E + (n)RT
Where R = gas constant; T = temperature in kelvin,and:
n = equivalents (moles) of product gas
equivalents (moles) of reactant gas
If n = 0, then H = E.
But even when n 0, the PV work component is ussmall. See example in textbook (pages 574-575).
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mo18Entropy
The final state of a system is more energetically
favorable if:
1. Energy can be dispersed over a greater number
and variety of molecules.
2. The particles of the system can be more
dispersed (more disordered).
The dispersal of energy and matter is described by
the thermodynamic state function entropy, S.
The greater the dispersal of energy or matter in a
system, the higher is its entropy. The greater the
disorder (dispersal of energy and matter, both in
space and in variety) the higher the entropy.Adding heat to a material increases the disorder.
-structure
-disordered
-disordered
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mo19Unlike H, entropy can be defined exactly becauseof the Third Law of Thermodynamics:
Third Law of Thermodynamics: Any
pure crystalline substance at a
temperature of absolute zero (0.0 K) has
an entropy of zero (S = 0.0 J/Kmol).
Sign notation (EXTREMELY IMPORTANT!!):
+S indicates that entropy is increasing in the
reaction or transformation (it's getting more
disordered -- mother nature likes)
S indicates that entropy is decreasing in thereaction or transformation (it's getting less
disordered {more ordered} -- mother nature doesn't
like, but it does happen)
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mo20Qualitative " Rules" About Entropy:
1) Entropy increases as one goes from a solid to a
liquid, or more dramatically, a liquid to a gas.
0d s
0
15
0
10
0
50
uid
p asetransi t ions
re K2) Entropy increases if a solid or liquid is dissolved
in a solvent.
3) Entropy increases as the number of particles
(molecules) in a system increases:
N2O4(g) 2NO2(g)
S= 304 J/K(1
mole)
S= 480 J/K (2
moles)These first 3 above are most important forevaluating Srxn.
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mo21The rules below are mainly for comparing theentropy of individual molecules or materials.
4) The Entropy of any material increases with
increasing temperature
5) Entropy increases as the mass of a molecule
increases
S(Cl2(g)) > S(F2(g))
S= 165 J/K mol S= 158 J/Kmol
6) Entropy is higher for weakly bonded compounds
than for compounds with very strong covalent
bonds
S(graphite) > S(diamond)
S= 5.7 J/K mol S= 2.4 J/Kmol
Note that for individual molecules(materials) the
higher the entropy, the more likely the molecule will
want to fall apart to produce a number of smallermolecules.
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7) Entropy increases as the complexity (# of atoms,
# of heavier atoms, etc.) of a molecule increases
ntropy o a er es o Gaseous
H drocarbons
S = 186 J/KmolHe a
ne
S = 201 J/Kmol
S = 220 J/Kmol
ce y en e
Hy e
ne
S = 230 J/Kmol
S = 270 J/KmolH
H H ane
C
H
C
HH ropane
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mo23What are the biggest factors for evaluating Srxn for
a chemical rxn?
1) phase change 2) change in # of molecules
Problem: For the following reactions, is the entropy
of the reaction increasing or decreasing?
a) Ag+(aq) + Cl-(aq)
b) H2CO3(aq)
AgCl(s)
H2O + CO
2(g)
c) Ni(s) + 4CO(g)
d) H2O(s)
Ni(CO)4(l)
H2O(l)
e) graphite diamond
f) 2Na(s) + 2H2O
g) H2S(g) + O2(g)
h) 2H2O(l)
2Na+(aq) + 2OH (aq) +H2(g)
-
H2O(l) + SO(g)
2H2(g) + O2(g)
i) CO2(g) + CaO(s)
) CaCl2(s) + 6H2O(l)
CaCO3(s)
CaCl26H2O(s)
k) 2NO2(g) N2O4(g)
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Just as with enthalpies, one can calculate entropies
of reaction.
(# eqiv)S(reactants)
S = (# eqiv)S
(products)rxn
EXAMPLE:
g s +CO
g s +C s
S = 32 J/Kmol 215
J/Kmol J/Kmol J/Kmol
TS TS
(# eqiv)S(reactants)
S = (# eqiv)S
(products)rxn
= eq v mo + eq v
(2 eqiv)(32 J/Kmol) + (1 eqiv
J/Kmol)
rxn
=
J/K.mol) J/K.mol)rxn
}S = 218J/K.mol
entropy s ecreas ng(reaction is becomingmore ordered
rxn
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mo25Spontaneous Processes
A process that takes place without the net input of
energy from an external source is said to be
spontaneous (notinstantaneous).
1) Rxn of sodium metal with water:
-2Na(s) + 2H2O 2Na+(aq) + 2OH (aq) +
H2(g)
2) Combustion rxns:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
2H2O(l)2H2(g) + O2(g)
3) Expansion of a gas into a vacuum
xCO2(g) yCO2(s) + zCO2(g) (x = y + z)
4) A salt dissolving into solution:
NH4NO3(s) + H2O(l) NH4+(aq) + NO3-(aq)
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mo26Second Law of Thermodynamics: In
any spontaneous process the entropy
of the universeincreases
Suniverse = Ssystem + Ssurroundings
Second Law (variant): in trying to do
work, you alwayslose energy to the
surroundings.
You can' t even break even!
Neither entropy (S) or enthalpy (H)
alone can tell us whether a chemical
reaction will be spontaneous or not.
An obvious (?) conclusion is that one
needs to use some combination of the two.
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mo27Gibbs Free Energy
The combination of entropy, temperature
and enthalpy explains whether a reaction is
going to be spontaneous or not. The symbol
Gis used to define the Free Energyof a
system. Since this was discovered by J.
Willard Gibbs it is also called the Gibbs
Free Energy. "Free" energy refers to the
amount of energy available to do work once
you have paid your price to entropy. Note
that this is not given simply by H, the heat
energy released in a reaction.
G = H TS
When G is negative, it indicates that a
reaction or process is spontaneous. A
positive G indicates a non-spontaneous
reaction.
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mo28
G = H TS
S
+
G = negative G = ??
spontaneous
spontaneousat all
temperaturesat hightemperatures
- +0 H
G = ?? G = positive
spontaneousat lowtemperatures
non-spontaneousat alltemperatures
-
Spontaneous = exoergic (energy releasing)
Non-spontaneous = endoergic (energy releasing)
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mo29Remember that entropies are
given in units of J/Kmol while
enthalpies and free energies are in
kJ/mol.
DON'T forget to convert all units to
kJ or J when using both S and H
in the same equation! !
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mo30G vs. G: Standard vs. Non-Standard Condition
Remember that the (not) on G indicates that th
numerical value of G is based on the reaction atstandard conditions (1 M solution concentration, 1 atgas pressure). Temperature is NOT part of standardconditions!
As soon as one has a concentration different than 1 M1 atm pressure, the not goes away and one has G
Consider the reaction:
Initial: 1 atm 1 atm 1 atm
2SO2(g) + O2(g) 2SO3(g)
Grxn = 142 kJ/mol
The Grxnof 142 kJ/mol is for when each gas is pr
with a concentration of 1 atm. This indicates that thereaction under these conditions will proceed to makeproducts (spontaneous).
As the reactants start reacting, however, theirconcentrations decrease (SO2twice as fast as O2) an
G turns into G and becomes less negative.
When G = 0 the reaction has reached equilibrium.Although for this rxn, SO2is probably the limiting
reagent (not enough present to complete the rxn).
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mo32Just as with enthalpies and entropies, one cancalculate free energies of reaction.
(# eqiv)
G(reactants)f
= eq v
(products)
rxn
EXAMPLE:
g s +CO (g)
g s +C(s)
G = 0kJ/mol
-kJ/mol
- mokJ/mol
TS S
(# mol) G(reactants)
f
= mo (products)rxn
= mo - mo + mo m (2 mol)(0 kJ/mol) + (1 mol)(
kJ/mol)
rxn
G = (-1140kJ)
-rxn = - +
394 kJrxn
} rxn!highly exothermicrxn !!
=kJrxn
ompare o w c was - orthe same rxn.
rxn
e mssng o energy wen oENTROPY!
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mo33Example: To make iron, a steel mill takes Fe2O3 (rust
or iron ore) and reacts it with coke (a complex, impure
form of carbon) to make iron and CO2. Based on the
data below, this is a non-spontaneous reaction at roomtemperature, but it becomes spontaneous at higher
temperatures. Assuming that H and S do not
change much with temperature, calculate the temp-
erature above which the reaction becomes spontaneous
(i.e., Grxn = 0).
Hrxn= +465 kJ/mol
Srxn= +552 J/molK (or0.552 kJ/molK
Grxn= +301 kJ/mol (at 298 K)
Grxn = Hrxn TSrxn
as we raise the temperature, G will eventuallyreach 0 andthen go negative & spontaneous, so let G = 0solve forT, the temperature at which this will happen:0 = Hrxn TSrxn
rearrang ng o so ve orgives:
/T = (Hrxn) (Srxn)
T = (465 kJ/mol) / (0.552 kJ/molK)
T = 842 K
(above this temperature Grxn will be negat ivewill have
a spontaneous reaction)
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mo34Problem: Calculate Grxn for the following.
Gf (SO2, g) = 300kJ/mol
Gf (SO3, g) = 371kJ/mol
Gf (H2SO4, l) = 690 kJ/mol Gf (H2SO4, aq) 742 kJ/molGf (H2O, l) = 237kJ/mol
Gf (H2S, g) = 34kJ/mol
a) S(s) + O2(g) SO2(g)
b) 2SO2(g) + O2(g)
c) SO3(g) + H2O(l)
d) 2H2S(g) + 3O2(g)
2SO3(g)
H2SO4(l)
2SO2(g) + 2H2O(l)
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mo35Comparisons of Hrxn and Grxn
S(s) + O2(g) SO2(g)
Hrxn = 297 kJ/molSrxn = 11 J/mol K
Grxn = 300 kJ/mol
2SO2(g) + O2(g) 2SO3(g)
Hrxn = 198 kJ/mol
Grxn = 142 kJ/mol
SO3(g) + H2O(l) H2SO4(l)
Hrxn = 132 kJ/mol
Grxn = 82 kJ/mol
2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(l)
Hrxn = 1126 kJ/mol
Grxn = 1006 kJ/mol
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mo36Entropy of Fusion and Vaporization
While the entropy of a substance increases steadily
with increasing temperature, there is a considerable
ump in the entropy at a phase transition:
250
o as
150
100
50
d
phasetransi t ions
Temperature(K)This jump at the melting point is called the entropy
of fusion, Sfusion, and as you might expect, it is
related to the enthalpy (or heat) of fusion, Hfusion:
HfusionSfusion
Tm
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Problem: Calculate the boiling point for ethanol
(CH3CH2OH) from the data in the thermodynamic
tables.
(literature value = 78.5C)
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mo39
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mo40
, , ,Davis, Peck & Stanley. Thomson Brooks/ColePublisher.