Chapter 18
Thermodynamics
Chapter 18
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Spontaneous Processes
Thermodynamic Entropy
Third Law of Thermodynamics
Calculating Entropy Changes
Free Energy
Temperature and Spontaneity
Free Energy and Equilibrium
Temperature and Equilibrium Constants
Coupled Reactions
Microstates: Quantized view of Entropy
Spontaneous Processes
We have all observed how some processes tend to
spontaneously occur while others don’t.
• Objects roll downhill but not up.
• Our rooms effortlessly become disorganized
• We age rather than grow younger.
• Milk eventually turns sour.
• Air rushes out of a punctured tire.
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill.
• Our rooms effortlessly become organized.
• We do not grow younger.
• Sour milk does not become fresh again.
• Air does not rush in to refill a punctured tire.
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized.
• We do not grow younger.
• Sour milk does not become fresh again.
• Air does not rush in to refill a punctured tire.
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized. Physical
• We do not grow younger.
• Sour milk does not become fresh again.
• Air does not rush in to refill a punctured tire.
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized. Physical
• We do not grow younger. Chemical
• Sour milk does not become fresh again.
• Air does not rush in to refill a punctured tire.
Nonspontaneous Processes
Which of the above process are chemical and which are
physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized. Physical
• We do not grow younger. Chemical
• Sour milk does not become fresh again. Chemical
• Air does not rush in to refill a punctured tire.
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized. Physical
• We do not grow younger. Chemical
• Sour milk does not become fresh again. Chemical
• Air does not rush in to refill a punctured tire.
Physical
Nonspontaneous Processes
Which of the following process are chemical and which
are physical?
For example:
• Objects do not roll up hill. Physical
• Our rooms effortlessly become organized.
Physical
• We do not grow younger. Chemical
• Sour milk does not become fresh again. Chemical
• Air does not rush in to refill a punctured tire.
Physical
Spontaneous Processes
It would be very useful to predict what physical and
chemical changes will occur spontaneously and
what ones won’t?
Spontaneous Processes
It would be very useful to predict what physical and
chemical changes will occur spontaneously and
what ones won’t?
thermodynamics will be used to predict spontaneity.
Spontaneity “A Closer Look”
Some spontaneous process require activation energy
to get them started. For example, to start a fire you
need a match, but once the fire is started it will burn until
one of the reactants are exhausted.
Spontaneity “A Closer Look”
Some spontaneous process require activation energy
to get them started. For example, to start a fire you
need a match, but once the fire is started it will burn until
one of the reactants are exhausted.
Does this mean that exothermic reactions are
spontaneous and endothermic reactions are
nonspontaneous?
Spontaneity “A Closer Look”
Some spontaneous process require activation energy
to get them started. For example, to start a fire you
need a match, but once the fire is started it will burn until
one of the reactants is exhausted.
Does this mean that exothermic reactions are
spontaneous and endothermic reactions are
nonspontaneous?
Consider ice melting is this spontaneous? Exothermic?
Spontaneity “A Closer Look”
Some spontaneous process require activation energy
to get them started. For example, to start a fire you
need a match, but once the fire is started it will burn until
one of the reactants is exhausted.
Does this mean that exothermic reactions are
spontaneous and endothermic reactions are
nonspontaneous?
Consider ice melting is this spontaneous? Exothermic?
Ice melting is spontaneous, but endothermic, right?
Spontaneity “A Closer Look”
Some spontaneous process require activation energy
to get them started. For example, to start a fire you
need a match, but once the fire is started it will burn until
one of the reactants is exhausted.
Does this mean that exothermic reactions are
spontaneous and endothermic reactions are
nonspontaneous?
Consider ice melting is this spontaneous? Exothermic?
Ice melting is spontaneous, but endothermic, right?
Only at room temperature and pressure!
Spontaneity “A Closer Look”
How about ice stored in a freezer? Is ice melting
spontaneous or nonspontaneous?
Spontaneity “A Closer Look”
How about ice formation in the freezer? Is this
spontaneous or nonspontaneous? Yes, spontaneous
good choice.
Spontaneity “A Closer Look”
Enthalpy and Entropy are the thermodynamic terms
responsible for predicting spontaneity.
Spontaneity “A Closer Look”
Enthalpy and Entropy are the thermodynamic terms
responsible for predicting spontaneity.
In Chapter #5 enthalpy was defined, as well as the
system, surroundings, pv work, and energy transfer
in general.
Spontaneity “A Closer Look”
Enthalpy and Entropy are the thermodynamic terms
responsible for predicting spontaneity.
In Chapter #5 enthalpy was defined, as well as the
system, surroundings, pv work, and energy transfer
in general.
Also, discussed was the first law of thermodynamics;
energy cannot be, created nor destroyed.
Laws of Thermodynamics
• The First Law of Thermodynamics states that
energy cannot be created nor destroyed.
∆E = q +w = ∆H – P∆V
• The Second Law of Thermodynamics states that the
energy to do useful work is constantly decreasing. Or
that the total entropy of the universe increases in any
spontaneous process.
• Entropy (S) is a measure of the distribution
(spreading) of energy in a system at a specific
temperature. Or the randomness of a system.
Entropy, the spreading of Energy
Consider the spontaneous change of air escaping a car
tire. In order to do so we make the following
assumptions:
• The air is an ideal gas
• The temperature inside and outside the tire is the
same
• Temperature of the gas does not change as it
escapes the tire
Since average kinetic energy is directly proportional to
the absolute temperature, the kinetic energy of the air
inside and outside of the tire is the same.
Entropy, the spreading of Energy
Since kinetic energy is the energy of motion, then we
must examine the different types of motion contained by
air particles.
• Translational movement; moving from one point to
another
• Rotational movement; spinning about an axis
• Vibratonal movement; atoms vibrating about their
covalent bonds.
Degrees of Motion
Three types of motion.
1. Translational
2. Rotational
3. Vibrational
As the temperature of a sample
increases, the amount of
motion increases.
Entropy and Microstates
• Quantum mechanics teaches that energy is not
continuous at the atomic scale; only certain levels of
quantized energy are possible. (Remember electrons,
no in between levels are allowed?)
• Example, potential energies of books in the library.
• The motion of molecules is quantized, which means
different states are separated by specific energies.
• An energy state, also called an energy level, is an
allowed value of energy.
• A microstate is a unique distribution of particles
among energy levels.
Energy States
Translational energy levels are the closest together,
compared to vibrational and rotational levels for
molecules. So close that we will assume them to be
continuous and that is why they are excluded on the
next slide. This also means not much energy is required
to move a molecule.
First rotational energy level is in the microwave region
First vibrational energy level is in the infrared region
First electronic energy level is found in the ultraviolet
region.
Energy States
Statistical Entropy
• Entropy is related to the number of microstates
by the following equation:
S = k ln(W)
 S is entropy
 W is the number of microstates
 k is the Boltzmann constant (k = 1.38 x 10-23 J/K)
• This equation indicates that entropy increases
as the number microstates increases.
Statistical Entropy
Consider a new deck of playing cards and comment on
the order.
Statistical Entropy
Consider a new deck of playing cards and comment on
the order.
After shuffling are they still ordered?
Statistical Entropy
Consider a new deck of playing cards and comment on
the order.
After shuffling are they still ordered?
How many shuffles to return them to the same order?
Statistical Entropy
Consider a new deck of playing cards and comment on
the order.
After shuffling are they still ordered?
How many shuffles to return them to the same order? 1064
Thermodynamic Entropy
• Consider an isothermal process (a process that
takes place at a constant temperature).
 S = qrev/T
 qrev reversible process is one that can be made
to reverse direction by just an infinitesimal
change in a system property, in this case heat
flowing into or out of a system.
 An example of qrev would be the heat of fusion,
which is the amount of heat added to a
substance to melt it or released when a
substance freezes. Here the direction change is
melting to freezing or freezing to melting.
Thermodynamic Entropy
Remember the total entropy of the universe
increases in any spontaneous process.
∆Suniv = ∆Ssys + ∆Ssurr
• If the sign of ∆Suniv is positive then the reaction is
spontaneous.
• This means that either the system or the
surroundings must be the larger positive, to get the
sum to be positive
• If the sign of ∆Suniv is negative then the reaction is
nonspontaneous, but spontaneous in the opposite
direction.
• If ∆Suniv is zero then the reaction is at equilibrium
Thermodynamic Entropy
The change in entropy of the surroundings
depends on the amount of heat transferred to the
surroundings.
• An exothermic system releases heat to the
surroundings, thus increasing the kinetic energies of
the particles, and increasing their randomness.
∆Ssurr = +∆Hsurr /T
• If a system is endothermic, then surroundings must
transfer heat to the system, thus decreasing the
kinetic energy of the surroundings; surroundings less
disordered (more order, right?)
∆Ssurr = -∆Hsurr/T
• Important to note: If ∆Hsys<0, then ∆Hsurr>0, required
from the first law right?
Thermodynamic Entropy
Now for the tricky part
• When heat flows from the system to the surroundings,
the heat is exothermic relative to the system and
endothermic relative to the surroundings.
• When heat flows to the surroundings, then
∆Ssurr= -∆H /T for increasing entropy of the
surroundings, this then is our definition for positive
∆Ssurr. Note since . -∆H/T does not have a subscript, it is
relating to the system. .
Free Energy
The two thermodynamic factors controlling spontaneity
are enthalpy and entropy. Enthalpy and entropy can
be thought of as thermodynamic forces. If they are
pulling together in their preferred direction, then a
reaction is spontaneous, if they are pulling opposite to
their preferred direction, then nonspontaneous. If they
are opposing each other, the larger one wins. This is
like a tug of war. Another useful thermodynamic term for
predicting.
spontaneity is Free Energy (G). The (G) for
Gibbs an instructor at Yale University
Who Wins?
S
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a
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u
s
Entropy
Enthalpy
n
o
n
s
p
o
n
t
a
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s
Gibbs Free Energy
Since enthalpy and entropy are finite, then
numerical values can be assigned to them. In order
to find the winner subtracting their values would
identify the larger one, except they do not have the
same units. Enthalpy has units of J while entropy
has units of J/K. Multiplying the entropy by the
absolute temperature, resolves the subtraction issue
to give a difference in Joules. This process was first
done by Josiah Willard Gibbs of Yale University
(1839-1903)
Gibbs Free Energy
The mathematical statement for
the difference between enthalpy
and entropy is stated below and
is equal to G in honor of Josiah
Gibbs.
G = H – TS
Since chemistry involves changes,
then this statement most often is
stated relative to changes in
Josiah Willard Gibbs enthalpy and entropy at constant
temperature
ΔG = Δ H – TΔS
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
We have previously defined -∆H/T = ∆Ssurr which
substitutes into equation (2) to give
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
We have previously defined -∆H/T = ∆Ssurr which
substitutes into equation (2) to give
-∆G/T = ∆Ssurr + ∆S
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
We have previously defined -∆H/T = ∆Ssurr which
substitutes into equation (2) to give
-∆G/T = ∆Ssurr + ∆S
remember no subscript means system
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
We have previously defined -∆H/T = ∆Ssurr which
substitutes into equation (2) to give
-∆G/T = ∆Ssurr + ∆S
remember no subscript means system
-∆G/T = ∆Ssurr + ∆Ssyst.
Gibbs Free Energy
The change in Free Energy from the previous slide
contains no subscripts and therefore is for the
system.
∆G = ∆H - T∆S (1)
Dividing both sides of this equation by –T gives the following
-∆G/T = -∆H/T + ∆S (2)
We have previously defined -∆H/T = ∆Ssurr which
substitutes into equation (2) to give
-∆G/T = ∆Ssurr + ∆S
remember no subscript means system
-∆G/T = ∆Ssurr + ∆Ssyst.
From a previous definition ∆Suniv = ∆Ssys + ∆Ssurr
Gibbs Free Energy
In the previous slide we have shown
∆Suniv = -∆G/T
Since the value of T does not affect the sign of -∆G,
then whenever ∆G is less than zero a chemical or
physical change will be spontaneous.
Spontaneity and Entropy
∆Suniv = ∆Ssys + ∆Ssurr
1. If Ssys > 0 and Ssurr > 0, then
Suniv > 0
2. If Ssys < 0 then Ssys + Ssurr > 0
for Suniv > 0
3. If Ssys > 0 and Ssurr < 0 as long
as Ssys + Ssurr> 0
Spontaneity and Free Energy
∆G = ∆H - T∆S
Free energy possibilities (∆G<0, spontaneous)
1. ∆S>0, ∆H<0; Spontaneous at all temperatures
2. ∆S>0, ∆H>0; Spontaneous at high temperatures
3. ∆S<0, ∆H<0; Spontaneous at low temperatures
4. ∆S<0, ∆H>0; Not spontaneous at any temperature
Free Energy/Equilibrium
We learned in Chapter #15 that when ΔG=0, then the
system is at equilibrium. Also, discussed was the
reaction quotient Q. If Q>K, then the products are too
high, meaning the equilibrium will shift left. This implies
that the reverse reaction is spontaneous and the forward
reaction is non spontaneous. From this the following
holds.
When Q=K, ΔGrxn=0, i.e. the reaction is at equilibrium
When Q<K, ΔGrxn<0, i.e. the reaction is spontaneous
When Q>K, ΔGrxn>0, i.e. the reaction is nonspontaneous
Free Energy/Equilibrium
Also it is important to note that since free energy is
related to the equilibrium constant and that the
equilibrium constant is related to concentration, then free
energy should be related to concentration. ΔG° refers to
free energy at standard conditions, i.e. at 25 °C and 1
atm. This implies that free energy at any temperature
should be higher or lower than ΔG°. In order to make
free energy smaller or larger then ΔG° should be
combined with K, but K and ΔG° do not have the same
units. Multiplying the reaction quotient by RT makes the
units compatible.
ΔG = ΔG° + RT ln Q
note: RT is added to make the
units compatible
Free Energy vs.Reactants/Products
97.8 kj
N2O4(g)
51.3 kj
2 NO2(g)
ΔG = products – reactants = 2(51.3) – 97.8 = 4.8 kj
This means the reaction will not go to completion since ΔG>0, but an
equilibrium does exist with minimum free energy
Point #1 System contains only reactants; Point #2 System contains only
products; Point #3 Minimum free energy and defines the equilibrium mixture
Thermodynamics and Equilibrium
• G = Go + RTlnQ
• At equilibrium: Q = K
G = Go + RTlnK = 0
Go = -RTlnK
K = e-G /RT
o
Relationship
Between K
and G
K<1
K>1
For exergonic reactions K>1
For endogonic reactions K<1
Neither exergonic, nor
endogonic, then K=1
K=1
Practice
Calculate the value of Go and the
equilibrium constant at 298 K for
2NO(g) + Cl2(g)
2NOCl(g)
Practice
Calculate the value of Go and the
equilibrium constant at 298 K for
86.6
105.3
2NO(g) + Cl2(g)
66.1
2NOCl(g)
Practice
Calculate the value of Go and the
equilibrium constant at 298 K for
86.6
105.3
2NO(g) + Cl2(g)
66.1
2NOCl(g)
Go = [2(66.1)]-[2(86.6) + 105.3] = -146.3 kj
Practice
Calculate the value of Go and the
equilibrium constant at 298 K for
86.6
105.3
2NO(g) + Cl2(g)
66.1
2NOCl(g)
Go = [2(66.1)]-[2(86.6) + 105.3] = -146.3 kj -Go
Go
= - RTlnK
Go
lnK= RT
K=e
RT
Practice
Calculate the value of Go and the
equilibrium constant at 298 K for
86.6
105.3
2NO(g) + Cl2(g)
66.1
2NOCl(g)
Go = [2(66.1)]-[2(86.6) + 105.3] = -146.3 kj -Go
Go
= - RTlnK
Go
lnK= RT
K=e
3j -146.3 kj
-Go
mole-K
10
= 8.314 j 298K kj 2 mole = 29.52
RT
K = e 29.52 = 7 X 1012
RT
Spontaneity and Free Energy
Example: Talk about melting/freezing of Ice
Ice melting is an endothermic process at temperatures
above zero degrees with ∆H>0 and ∆S>0, this fits
category #2 from the previous slide. If T>273K, then --T∆S is larger than ∆H, making ∆G<0, thus
spontaneous.
Ice freezing is an exothermic process at temperatures
below zero degrees with ∆H<0 and ∆S<0, which fits
category #3. If T<273K, then ∆G<0 and spontaneous.
How about when T=273K?
Spontaneity and Free Energy
Example: Talk about melting/freezing of Ice
Ice melting is an endothermic process at temperatures
above zero degrees with ∆H>0 and ∆S>0, this fits
category #2 from the previous slide. If T>273K, then --T∆S is larger than ∆H, making ∆G<0, thus
spontaneous.
Ice freezing is an exothermic process at temperatures
below zero degrees with ∆H<0 and ∆S<0, which fits
category #3. If T<273K, then ∆G<0 and spontaneous.
How about when T=273K? ∆G=0, Equilibrium
Effects of Temperature on K
Go = -RTlnK
Go = Ho - TSo
H 1
S
lnK = +
R T
R
o








o
K2
H  1
1 
ln  = K1
R T 2
T1
o

K2
ΔH
ln
=
K1
R
T1 – T2
T1T2
Clausius-Clapeyron Equation: K1 and K2 are
the equilibrium constant at T1 and T2
respectively.
Clausius-Clapeyron Equation
Ho  1  So
lnK = +
 
R T 
R
Third Law of Thermodynamics
• Third Law of Thermodynamics - the entropy
of a perfect crystal is zero at absolute zero.
• S is explicitly known (=0) at 0 K, S values at
other temps can be calculated.
• Absolute entropy is the entropy change of a
substance taken from S = 0 (at T = 0 K) to
some other temperature.
• Standard molar entropy (So) is the absolute
entropy of 1 mole of a substance in its
standard state (25oC and 1 atm).
Select Standard Molar Entropy Values (1 atm, 298 K)
Formula
So
(J/(mol•K)
Formula
Name
So
(J/(mol•K)
Br2(g)
245.5
CH4(g)
methane
186.2
Br2(l)
152.2
CH3CH3(g)
Ethane
229.5
Cdiamond(s)
2.4
CH3OH(g)
Methanol
239.7
Cgraphite(s)
5.7
CH3OH(l)
Methanol
126.8
CO(g)
197.7
CH3CH2OH(g)
Ethanol
282.6
CO2(g)
213.8
CH3CH2OH(l)
Ethanol
160.7
H2(g)
130.6
CH3CH2CH3(g)
Propane
269.9
N2(g)
191.5
CH3(CH2)2CH3(g)
n-Butane
310.0
O2(g)
205.0
CH3(CH2)2CH3(l)
n-Butane
231.0
H2O(g)
188.8
C6H6(g)
Benzene
269.2
H2O(l)
69.9
C6H6(l)
Benzene
172.8
NH3(g)
192.3
C12H22O11(s)
Sucrose
360.2
Trends
• Ssolid < Sliquid < Sgas
 H2O(g) is 188.8 J/(mol•K) and H2O(l) is 69.9 J/(mol•K)
Other Trends
• Entropy increases as the
complexity of molecular structure
increases.
• More bonds, more branching,
more entropy
• CH4 < CH3CH3 < CH3CH2CH3 <
CH3(CH2)2CH3
Entropy Differences in Allotropes
Limited Modes of Motion
Additional Modes of Motion
Predicting Entropy Changes
Predict the sign of ΔSo for each of the following
reactions
a. The thermal decomposition of solid calcium carbonate:
CaCO3 (s)
CaO (s) + CO2 (g)
Predicting Entropy Changes
Predict the sign of ΔSo for each of the following
reactions
a. The thermal decomposition of solid calcium carbonate:
CaCO3 (s)
CaO (s) + CO2 (g)
Since in the reaction a gas is produced from a
solid reactant, the postiionsl entropy increases,
and ΔSo > 0
Predicting Entropy Changes
Predict the sign of ΔSo for each of the following
reactions
a. The thermal decomposition of solid calcium carbonate:
CaCO3 (s)
CaO (s) + CO2 (g)
Since in the reaction a gas is produced from a
solid reactant, the postiionsl entropy increases,
and ΔSo > 0
b. The oxidation of SO2 in air:
2SO2 (g) + O2 (g)
2 SO3 (g)
Predicting Entropy Changes
Predict the sign of ΔSo for each of the following
reactions
a. The thermal decomposition of solid calcium carbonate:
CaCO3 (s)
CaO (s) + CO2 (g)
Since in the reaction a gas is produced from a
solid reactant, the postiionsl entropy increases,
and ΔSo > 0
b. The oxidation of SO2 in air:
2SO2 (g) + O2 (g)
2 SO3 (g)
Here three molecules of gaseous reactants become two
molecules of gaseous products. Since the number of
gas molecules decreases, positional entropy decreases,
and ΔSo <0
Calculating Entropy Changes
• Entropy change for a reaction (Srxn) is
Srxn = nSoproducts - mSoreactants
where n and m are the coefficients of the
products and reactants in the balanced equation.
Practice
Calculate the standard entropy change for the
reaction using the Thermodynamic properties
from Apendix 4 (units in j/mole-K)
229.5
31.998
69.9
213.6
2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(l)
Practice
Calculate the standard entropy change for the reaction
using the Thermodynamic properties from Apendix 4
(units in j/mole-K)
229.5
31.998
69.9
213.6
2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(l)
Remember from 161 that change (Δ) is products minus
reactants.
Δ So = Δ So products - Δ So reactants
∆ So = [4(69.9)+6(213.6)] – [2(229.5) + 7(31.998)]
Practice
Calculate the standard entropy (∆ So change for the
reaction using the Thermodynamic properties from
Apendix 4 (units in j/mole-K)
∆ So = [4(69.9)+6(213.6)] – [2(229.5) + 7(31.998)]
∆ So = [279.6 + 1281.6] – [459.00 + 223.986]
∆S So = 878 j/K
Practice
Given the data below, calculate the temperature at
which the reactions becomes spontaneous. (H0 = -92
kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
∆Go = ∆Ho - T∆So
2NH3(g)
Practice
Given the data below, calculate the temperature at
which the reactions becomes spontaneous. (H0 = -92
kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
∆Go = ∆Ho - T∆So
2NH3(g)
Remembering:
∆G>0 nonspontaneous
∆G<0 spontaneous
∆G = 0 Equilibrium
Practice
Given the data below, calculate the temperature at
which the reactions becomes spontaneous. (H0 = -92
kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆Go = ∆Ho - T∆So
0 = ∆Ho - T∆So
Remembering:
∆G>0 nonspontaneous
∆G = 0 Equilibrium
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆Go = ∆Ho - T∆So
0 = ∆Ho - T∆So
∆Ho = T∆So
Remembering:
∆G>0 nonspontaneous
∆G<0 spontaneous
What number is inbetween?
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆Go = ∆Ho - T∆So
0=
∆Ho
-
T∆So
∆Ho = T∆So
T=
∆H
∆S
Remembering:
∆G>0 nonspontaneous
∆G<0 spontaneous
What number is inbetween?
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆Go = ∆Ho - T∆So
0 = ∆Ho - T∆So
∆Ho = T∆So
∆Ho
T=
=
o
∆S
-92 kj
mol
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆Go = ∆Ho - T∆So
0 = ∆Ho - T∆So
∆H = T∆S
-92 kj
∆Ho
=
T=
mol
∆So
mol-K
-199 j
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆G = ∆H - T∆S
0 = ∆H - T∆S
∆H = T∆S
∆H
T=
=
∆S
-92 kj mol-K
mol -199 j
103 j
kj
=
Practice
Given the data below, calculate the temperature
at which the reactions becomes spontaneous.
(H0 = -92 kJ/mol and S0 = -199J/mol-K)
N2(g) + 3H2(g)
2NH3(g)
∆G = ∆H - T∆S
0 = ∆H - T∆S
∆H = T∆S
∆H
T=
=
∆S
-92 kj mol-K
mol -199 j
103 j
kj
= 460 K = 190 C
The temperatures above are for equilibrium, to be
spontaneous the T ∆S term must be smaller than ∆H,
which requires T<190C
Free Energy Changes
G0rxn = nG0f, products - mG0f, reactants
The free energy for a chemical reaction
indicates the maximum amount of
energy that is free to do useful work.
Practice
Calculate the free energy change for the
following reaction using the G0f values in
the appendix.
C12H22O11 (s) + 12O2 (g) ---> 12CO2 (g) + 11H2O (l)
Driving the Human Engine
• Exergonic reactions are spontaneous
(G < 0).
• Endergonic reactions are
nonspontaneous (G > 0).
• The laws of thermodynamics describe
the chemical reactions that power the
human engine.
Break Down of Glucose
• Glucose is molecule
that humans use for
energy.
• Glycolysis is a
series of reactions
that converts
glucose into
pyruvate.
Phosphorylation
• A phosphorylation reaction results in the
addition of a phosphate group to an organic
molecule.
ATP
The hydrolysis of ATP to ADP is an exergonic reaction.
ATP
• ATP - adenosine triphosphate
• The ATP produced as result of the
breaking down (metabolizing) of food
can be used to drive endergonic
cellular reactions.
THE END
ChemTour: Entropy
Click to launch animation
PC | Mac
This ChemTour includes an “Entropy Battle” game
that challenges students to maintain order within a
system as the temperature rises and the phase
level moves from solid to gas.
Dissolution of Ammonium Nitrate
Click to launch animation
PC | Mac
Gibbs Free Energy
Click to launch animation
PC | Mac
Students learn to calculate the maximum potential energy
available to do work in a system. An interactive “Gibbs free
energy calculator” allows students to manipulate variables
entropy, enthalpy, and temperature to explore the effect on
DG of a reaction.
Chapter 13 Review
Shown to the left are three
possible configurations (A, B,
and C) for placing 4 atoms in two
boxes. Which of the following
processes is accompanied by the
largest increase in entropy, ΔS?
A) A → B
B) B → C
Entropy of Four Atoms in Two Boxes
C) C → A
From the figures below, we see that A has
only one microstate, B has 4 microstates, and
C has 6 microstates.
A
B
C
B has a change of 3 microstates
C has a change of 2 microstates
A has a change of -5 microstates
1
A
1
2
3
4
1
B
2
4
C
3
1
3
2
1
4
2
2
3
1
24
4
2
1
23
4
3
1
22
3
1
3
2
4
3
2
4
1
3
4
2
4
3
2
1
1
4
From the figures below, we see that A has
only one microstate, B has 3 microstates, and
C has 6 microstates.
A
A
A
B has a change of 3 microstates
B
C has a change of 2 microstates
C
A has a change of -5 microstates
B change is greatest, thus most change in entropy.
1
2
3
4
1
B
2
4
C
3
1
3
2
1
4
2
1
2
3
1
24
4
2
1
23
4
3
1
22
3
1
3
2
4
3
2
4
1
3
4
2
4
3
2
1
1
4
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
Isothermal Expansion of an Ideal Gas
C) ΔSsurr < 0
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
C) ΔSsurr < 0
ΔSuniv
ΔS + ΔSsurr
ΔSuniv =
Reversibly means
almost at equilibrium,
meaning
Isothermal Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
ΔSuniv =
C) ΔSsurr < 0
ΔS + ΔSsurr
Reversibly means almost at equilibrium
Isothermal Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
ΔSuniv =
C) ΔSsurr < 0
ΔS + ΔSsurr
Reversibly means almost at equilibrium ΔSuniv ~ 0
Isothermal Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
ΔSuniv =
C) ΔSsurr < 0
ΔS + ΔSsurr
Reversibly means almost at equilibrium ΔSuniv ~ 0
ΔS > 0, due to expansion
Isothermal Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
ΔSuniv =
ΔS +
C) ΔSsurr < 0
ΔSsurr
Reversibly means almost at equilibrium ΔSuniv ~ 0
ΔS > 0, due to expansion, then ΔSsurr< 0
Isothermal Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally and
reversibly against an external pressure
of 1.0 atm. What can be said of the
change in the entropy of the
surroundings, ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
ΔSuniv =
ΔS +
C) ΔSsurr < 0
ΔSsurr
Reversibly means almost at equilibrium ΔSuniv ~ 0
ΔS > 0, due to expansion, then ΔSsurr< 0
Isothermal Expansion of an Ideal Gas
An ideal gas is expanded
adiabatically (q = 0) into a
vacuum. Which of the following
statements is true for this process?
A) ΔEsys < 0
B) ΔGsys < 0
Gas Expansion into a Vacuum
C) ΔSsys < 0
An ideal gas is expanded
adiabatically (q = 0) into a
vacuum. Which of the following
statements is true for this process?
A) ΔEsys < 0
B) ΔGsys < 0
C) ΔSsys < 0
Because expansion is happening the this is a spontaneous
process.
Gas Expansion into a Vacuum
An ideal gas is expanded
adiabatically (q = 0) into a
vacuum. Which of the following
statements is true for this process?
A) ΔEsys < 0
B) ΔGsys < 0
C) ΔSsys < 0
Because expansion is happening the this is a spontaneous
process. Since the system is spontaneous then ΔGsys < 0
Gas Expansion into a Vacuum
An ideal gas is expanded
adiabatically (q = 0) into a
vacuum. Which of the following
statements is true for this process?
A) ΔEsys < 0
B) ΔGsys < 0
C) ΔSsys < 0
Because expansion is happening the this is a spontaneous
process. Since the system is spontaneous then ΔGsys < 0
Gas Expansion into a Vacuum
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds broken?
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed?
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = ?
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Is ΔS>0 or ΔS<0?
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Is ΔS>0 or ΔS<0
ΔS>0
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Is ΔS>0 or ΔS<0
ΔS>0
ΔG ?
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Is ΔS>0 or ΔS<0
ΔS>0
ΔG<0
Formation of CH Cl from CH and CCl
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
C) ΔG > 0
ΔH involves bonds breaking and bonds forming. How many B-G
bonds and W-G bonds are broken? 2 of each
How many of each are formed? 2 of each
What does ΔH = 0
Is ΔS>0 or ΔS<0
ΔS>0
ΔG<0
Formation of CH Cl from CH and CCl
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O(l),
at 25 °C if the partial pressure of
H2O(g) is 1.0 atm?
A) ΔG° > 0
B) ΔG° = 0
C) ΔG° < 0
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O(l),
at 25 °C if the partial pressure of
H2O(g) is 1.0 atm?
-237.2
-228.6
H2O(g)
→
A) ΔG° > 0
H2O(l),
ΔG°= -237.2-(-228.8)
ΔG°= -8.4 kj
B) ΔG° = 0
C) ΔG° < 0
ΔG degree for Condensation of Water at 25 degree C
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O(l),
at 25 °C if the partial pressure of
H2O(g) is 1.0 atm?
-237.2
-228.6
H2O(g)
→
A) ΔG° > 0
H2O(l),
ΔG°= -237.2-(-228.6 )
ΔG°= -8.6 kj
B) ΔG° = 0
C) ΔG° < 0
ΔG degree for Condensation of Water at 25 degree C
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O(l),
at 25 °C if the partial pressure of
H2O(g) is 1.0 atm?
-237.2
-228.6
H2O(g)
→
A) ΔG° > 0
H2O(l),
ΔG°= -237.2-(-228.6)
ΔG°= -8.8 kj
B) ΔG° = 0
C) ΔG° < 0
ΔG degree for Condensation of Water at 25 degree C
To the left is a plot of vapor
pressure versus temperature for the
vaporization of ethanol,
C2H5OH(λ)
→
C2H5OH(g).
At which temperature is ΔG° = 0
for the vaporization of ethanol at
1.0 atm?
A) > 100 °C B) 100 °C C) < 100 °C
ΔG degree of Vaporization of Ethanol
To the left is a plot of vapor
pressure versus temperature for the
vaporization of ethanol,
C2H5OH(λ)
→
C2H5OH(g).
At which temperature is ΔG° = 0
for the vaporization of ethanol at
1.0 atm?
A) > 100 °C
B) 100 °C
C) < 100 °C
Remember at the boiling point of a liquid the vapor pressure
of the liquid is equal to the atmospheric pressure. Phase
diagrams tell us at the BP evaporation/condensation are in
equilibrium
ΔG degree of Vaporization of Ethanol
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
A)
B)
ΔG degree Versus T for the Sublimation of I (s)
C)
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic?
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG =?
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
ΔG
0
T
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
At 10 K is how does ΔG compare to ΔH
ΔG
T
0
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
At 10 K is how does ΔG compare to ΔH
ΔG < ΔH
ΔG
T
0
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
At 10 K is how does ΔG compare to ΔH
ΔG < ΔH
ΔG
T
0
Which of the following plots shows the
correct relationship between ΔG° (yaxis) and temperature (x-axis) for the
sublimation of solid iodine to iodine
vapor at 1.0 atm?
I2 (s)
I2 (g)
ΔG = ΔH - TΔS
Is sublimation endothermic or exothermic? Endothermic
At absolute zero ΔG = ΔH; ΔG >0, right?
At 10 K is how does ΔG compare to ΔH
ΔG < ΔH
ΔG
T
0
Now A, B, or C?
A)
B)
ΔG degree Versus T for the Sublimation of I (s)
C)
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
2 O3(g)
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
2 O3(g)
ΔG = ΔH - TΔS
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
+
ΔG = ΔH - TΔS
2 O3(g)
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
+
ΔG = ΔH - TΔS
Can ΔG ever be
negative?
2 O3(g)
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
+
ΔG = ΔH - TΔS
Can ΔG ever be
negative? NO
2 O3(g)
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The formation of ozone, O3(g),
from molecular oxygen is an
endothermic process, with
ΔH° = 85 J/mole.
3 O2(g)
+
ΔG = ΔH - TΔS
Can ΔG ever be
negative? NO
2 O3(g)
At what temperatures will the
reaction proceed spontaneously if
PO2 = PO3 = 1.0 atm?
A) High temperatures B) Low temperatures C) No temperatures
Spontaneity of Ozone Formation
The End Ch #14