Download THERMODYNAMICS

Chemistry 102
Chapter 17
THERMODYNAMICS

Thermodynamics is concerned with the energy changes that accompany chemical and physical
processes.

Two conditions must be fulfilled in order to observe a chemical or physical change:
1. The change must be possible (determined by Thermodynamics)
2. The change must occur at a reasonable speed (determined by Kinetics)
Important Terms in Thermodynamics:
1. System and Surroundings:


System : The part of the universe we happen to be studying
Surroundings: Everything outside the system
Three types of systems can be present:



Open: exchange of energy and matter
Closed: exchange of energy but not matter
Isolated: No exchange of energy or matter
The State of the system can be evaluated in terms of the fixed properties of the:


System
Temperature
Pressure
Components of the system
 Number of moles
 Physical State of each system
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Chemistry 102
Chapter 17
2. Adiabatic and Isothermal Changes:
Adiabatic Change
Isothermal Change

Takes place in a system that is insulated from
the surroundings.

Takes place in a system that is not
insulated from the surroundings.

Heat cannot flow between system and
surroundings.


The temperature of the system changes:
Heat can flow between system
and surroundings and as such the
temperature can be maintained
constant.

A change that occurs at constant
temperature.


Temp. increases if rxn is exothermic.
Temp. decreases if rxn is endothermic
3. State Function

A quantity whose value depends only on the current state of the system, and does
not depend on the prior history of the system.
Example: Temperature, Internal Energy
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Chemistry 102
Chapter 17
4. Internal Energy


Internal Energy (U) is a state function.
It is a property of a system that depends only on its present sate, determined by variables such as temperature and pressure
Internal Energy, U
=
Kinetic Energy
of the particles
making the system
Potential Energy
of the particles
making the system
results from the
energy of motion
of
electrons
atoms
results from
molecules
chemical bonding
between atoms
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attractions
molecules
Chemistry 102
Chapter 17
FIRST LAW OF THERMODYNAMICS

The 1st Law of Thermodynamics is the “Law of Conservation of Energy” applied to
thermodynamic systems.

When a system changes from one state to another (ex: warming), its internal energy
changes from one definite energy to another.

Change in Internal Energy : U = Uf  Ui
where: Uf = final value of the internal energy
Ui = initial value of the internal energy

The Changes in Internal Energy, U :
 are more important in thermodynamics than the absolute values of Internal Energy
 are measured by noting the exchanges of energy between the system and the surroundings
Heat = q
Work = w
The energy that moves into or out of the
system because of a temperature
difference between the system and its
surroundings
If heat is lost
U drops
The energy exchange that results when a force F
moves an object through a distance, d
Work = Force x Distance
If work is done by the system
U drops
q is negative
If heat is absorbed
U rises
w is negative
If work is done on the system
U rises
q is positive
w is positive
Examples:
Determine the sign of the work for each of the changes shown below (carried out at constant P):
a)
2 SO2 (g) + O2 (g)  2 SO3 (g)
w=
b)
CaCO3 (s)  CaO (s) + CO2 (g)
w=
c)
CO2 (g) + H2O (l) + CaCO3 (s)  Ca2+ (aq) + 2 HCO3– (aq)
w=
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Chemistry 102
Chapter 17
HEAT (q) AND WORK (w) RELATIONSHIP IN A PHYSICAL SYSTEM
A gas is enclosed in
a vessel equipped
with a movable
piston, momentarily
fixed in position
Heat passes from the
the surroundings to the
vessel and the temperature
of the gas is increased
Assume q = + 165 J
The gas pressure
increases and causes
the gas to expand.
This will cause to lift
the piston and the
weight on top of it
In lifting the
weight, the system
does work = the
energy gained by
piston and weight
Assume w = - 92 J
The system gains Internal Energy, U from the heat absorbed : q = + 165 J
The system loses internal energy via the work done:
The change in
Internal Energy = U = (+165 J) + (- 92 J) = + 73 J
of the System
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w = - 92 J
Chemistry 102
Chapter 17
FIRST LAW OF THERMODYNAMICS

The Change in Internal Energy of a system, U, equals q + w

Consider a system where work is done on the system and heat is lost by the system:
A gas in enclosed in a vessel
equipped with a movable piston,
momentarily fixed in position.
Work is done on the system by
pushing down the piston and thus
compressing the gas.
U = q + w
The gas pressure increases and
causes the gas to warm up.
This will cause heat to flow out
of the system.
Assume w = + 92 J
Assume q = - 165 J

The system gains Internal Energy, U from the work done on it :
w = + 92 J

The system loses internal energy via the heat lost:
q = - 165 J

The change in Internal Energy of the System =
U =
q
+
w
U = (-165 J) + (+ 92 J) = - 73 J
Heat (q) and Work (w) relationship in a chemical system
 Consider the following exothermic chemical reaction carried out in a beaker open to the
atmosphere:
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)
qp = - 152.4 kJ/mol Zn
Zn (s) + 2 H+ (aq) → ZnCl2 (aq) + H2 (g)
qp = - 152.4 kJ/mol Zn
Note: qp = indicates that the process occurs at constant pressure (atmospheric pressure)

The H2 gas that is produced:
 increases the volume of the system
 must push against the atmosphere in order to evolve


It follows that work must be done by the system.
Can the work done by the system be measured and/or calculated?
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Chemistry 102
Chapter 17
PRESSURE-VOLUME WORK
F
Assume that the pressure exerted by the
atmosphere is replaced by a piston and
weights, whose downward force of gravity,
F, creates a pressure on the gas equivalent to
that of the atmosphere
F = force of gravity= atmospheric pressure
V = increase in volume of the system
due to the production of H2 gas
V = (cross sectional area) x( height)
height = h =
V
A
The work done by the system in expanding = (Force of Gravity) x (Distance the piston moves)
Atmospheric
pressure
w = -F x h = - F x

V
F
=x
A
A
V
Negative sign is given because work “w” is done by the system and represents energy lost by it.
It follows:
w = - P V
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Chemistry 102
Chapter 17
Examples:
1. In the reaction shown below, 1.00mol of Zn reacts with excess HCl to produce 1.00 mol of H2 gas.
At 25C and 1.00 atm, this amount of H2 occupies 24.5 L. Calculate the change in Internal Energy
(∆U) for this reaction. (1 Latm = 0.101 kJ)
Zn (s)
+
2H3O+ (aq)
Zn2+ (aq) + 2 H2O (l) + H2 (g)
q = –152.4 kJ/mol
∆U = q + w
q = –152.4 kJ
w=?
w done by system (to push back atmosphere) = –PV
= –(1.00 atm)(24.5 L)(0.101 kJ/1 Latm)
= –2.47 kJ
U = qp + w
= ( –152.4 kJ ) + (–2.47 kJ) = –154.9 kJ
2. What is U when 1.00 mol of liquid water vaporizes at 100C? (H0 vap = 40.66 kJ/mol at 100 C)
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Chemistry 102
Chapter 17
Examples:
3. Consider the combustion (burning) of methane, CH4, in oxygen:
CH4(g) +
2 O2(g)
CO2(g) + 2 H2O(l)
The heat of reaction at 25 C and 1.00 atm is –890.2 kJ.
a) What is the change in volume, V when 1.00 mol CH4 reacts with 2.00 mol O2?
(Ignore the volume of liquid water formed in the reaction)
b) What is w for this change?
c) Calculate U for the change indicated by the chemical equation.
a)
Find the change in volume (V)
Volume occupied by 1 mole of any gas at STP = 22.4 L
Volume occupied by 1 mole of any of the gases in the equation at 25 C:
V = 22.4 L x
CH4(g)
298 K
= 24.45 L
273 K
+ 2 O2(g)
CO2(g) +
3 moles of gas
Gaseous Volume decreases:
2 H2O (l)
H = –890.2 kJ
1 mole of gas
3 moles of gas
1 mole of gas
V = Volume occupied by 3 moles – Volume occupied by 1 mole
V = (3 x 24.45 L) – 24.45 L
= 48.9 L
b)
Calculate the work (w) done on the system
w = + P V = (1.00 atm)(48.9 L)(0.101 kJ/1 Latm) = 4.94 kJ
c)
Calculate change in internal energy, U
= + 4.94 kJ
qp = –890.2 kJ
U = qp + w = (–890.2 kJ) + (+ 4.94 kJ) = –885.3 kJ

Conclusion: Energy leaves the system
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Chemistry 102
Chapter 17
4. What is U for the following reaction at 25 C?
N2 (g) +
3 H2 (g)
2 NH3 (g)
4 moles of gas

2 moles of gas
Volume occupied by 1 mole of any of the gases in the equation at 25 C:
V = 22.4 L x

H = – 91.8 kJ
298 K
= 24.45 L
273 K
The decrease in Volume:
V == 2 x (24.45 L) = 48.9 L

Since the Volume decreases, work is done on the chemical system by the atmosphere:
w
= + P V = (1.00 atm) x (48.9 L)(0.101 kJ/1 Latm) = + 4.94 kJ
U = qp + w = (–91.8 kJ) + (+ 4.94 kJ) = – 86.9 kJ

Conclusion: Energy leaves the system
5. Calculate the amount of work done in each of the following reactions. In each case, is the work
done by or on the system?
a) oxidation of HCl at 200 C
4 HCl (g) + O2 (g)  2 Cl2 (g) + 2 H2O (g)
b) The decomposition of NO2 at 300 C
2 NO (g)  N2(g) + O2 (g)
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Chemistry 102
Chapter 17
ENTHALPY AND ENTHALPY CHANGE

Recall:
Enthalpy = The Heat of Reaction, H, at constant pressure (qp)
Actually: Enthalpy = H = U + PV

U (internal Energy), P (pressure), V (volume) are state functions.

If follows: H (Enthalpy) is also a state function
Consequences:
1. For a given temperature and pressure, a given amount of a substance has a definite enthalpy.
2. If the enthalpies of substances are known, the change of enthalpy for a reaction, H, can be
easily calculated.
H = Final Enthalpy – Initial Enthalpy = Hf – Hi
At constant pressure, P :
Hf = Uf + PVf
and
Hi = Ui + PVi
H = (Uf + PVf) – (Ui + PVI) = (Uf – Ui) + (PVf – PVi ) = U + PV
H = U + PV
and as found earlier:
U = qp – P V

It follows: H = (qp – P V) + P V = qp – P V + P V = qp

Conclusion:

The Enthalpy or Reaction equals the Heat of Reaction at constant pressureApplications:
H = qp
(Chapter 6)
Heat absorbed (endothermic reaction) or
given off (exothermic reaction) by a
chemical reaction
H0 = Standard Enthalpy Change
H0 = Standard Enthalpy Change for a Reaction can be calculated if the
Standard Enthalpies of formation, H0f for both reactants and products are known.
H0f
= Standard Enthalpy of Formation of a Substance
=
Enthalpy change for the formation of one mole of substance in its standard
state from its elements in their reference form ( most stable form at 250C and
1.00 atm) and in their standard states. (Given in Table 6.2 and Appendix C)
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Chemistry 102
Chapter 17
Examples:
1. Calculate the heat absorbed or evolved in the reaction between NH3 gas and CO2 gas that yields
aqueous urea (NH2CONH2) and liquid water.
2 NH3 (g)
+ CO2 (g)
From Appendix C:
H0f (NH3)
H0f (CO2)
H0f [CO(NH2)2]
H0f (H2O)
2 NH3 (g)
+
2 x (– 45.9 kJ/mol)
H0
CO(NH2)2 (aq) +
gas
gas
aqueous
liquid
CO2 (g)
(– 393.5 kJ/mol)
H2O (l)
= – 45.9 kJ/mol
= – 393.5 kJ/mol
= – 319.2 kJ/mol
= – 285.8 kJ/mol
CO(NH2)2 (aq) +
(– 319.2 kJ/mol)
H2O (l)
(– 285.8 kJ/mol)
=  n H0f (products) –  n H0f (reactants)
H0 = [(– 319.2) + (– 285.8 kJ)] – [2 x (– 45.9 ) + (– 393.5)] = – 119.7 kJ

Since H0 is negative, we conclude that the reaction is exothermic.
2. Use data in appendix C to find H0 for the reactions shown below:
a)
NH4NO3 (s)  NO2 (g) + 2 H2O (l)
b)
SiO2 (s) + 3 C (graphite)  SiC (s) + 2 CO (g)
12
H0 = ?
Chemistry 102
Chapter 17
THE SECOND LAW OF THERMODYNAMICS

Spontaneous Processes: Physical or chemical changes that occurs by themselves, without outside
assistance
Spontaneous Physical Change: A ball rolling from the top of a hill
High Potential Energy
Low Potential Energy
Spontaneous Chemical Change: The reaction of solid K(potassium) with liquid water
spontaneous
2 K(s) + 2 H2O (l)

2 KOH(aq) + H2(g)
In order for spontaneous changes to go in the opposite direction work would have to be expended, and
the changes would be called non-spontaneous.
Non-spontaneous Physical Change: Rolling a ball uphill
Non-spontaneous Chemical Change Change: Obtaining K(s) from KOH(aq)
non-spontaneous
2 KOH (aq) + H2 (g)

2 K(s) + 2 H2O(l)
This process requires several chemical reactions.
NOTE:
 The Second Law of Thermodynamics determines if a reaction is spontaneous or not.
 Exothermic Reactions (H  0) are not necessarily spontaneous and Endothermic Reactions (H  0)
are not necessarily non-spontaneous
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Chemistry 102
Chapter 17
ENTROPY (S)




Entropy is the thermodynamic quantity that is a measure of the randomness or disorder in a system.
Entropy is a state function: - the quantity of entropy in a substance depends only on variables that
determine the state of the substance (temperature and pressure)
Entropy is measured in J/K (SI unit)
Entropy increases when the disorder in a sample of the substance increases:
Example: Consider the melting of ice at O0C and 1 atm.
MELTING
ICE
H2O molecules occupy regular fixed positions
(crystal lattice)
Ordered crystalline structure




LIQUID WATER
H2O molecules move about freely, giving a
disordered structure
Less-ordered structure
Lower Entropy
Higher Entropy
Si = Initial Entropy
= Entropy of Ice (41 J/K)
Sif= Final Entropy
= Entropy of liquid water (63 J/K)
melting at 00C and 1 atm
H2O(s)
Si = 41 J/K


H2O(l)
Sf = 63 J/K
S = 22 J/K
In any natural process the degree of disorder (Entropy) increases.
Natural processes are changes that occur by themselves (spontaneous changes).
Second Law of Thermodynamics in reference to the system and its surroundings
The Total Entropy of a System and its Surroundings always increases for a Spontaneous Process.


ENTROPY
is created during a spontaneous change
14

ENERGY
can be neither created or destroyed
during a spontaneous change
Chemistry 102
Chapter 17
SECOND LAW OF THERMODYNAMICS IN REFERENCE TO THE SYSTEM ONLY

Consider a spontaneous change and the changes that occur in the system at a given temperature:
1. Entropy is created
2.
Heat (q) flows into or out of the system
Entropy accompanies the heat flow
Entropy Change
q
associated with

heat flow (q)
T
Entropy created
S =
during
the spontaneous process
+
Change in entropy
associated with
heat flow
S
+
Entropy Flow
S
=
=
Entropy Created
Entropy Created
q

T
+
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Chemistry 102
Chapter 17
S = Entropy Created +
q

T
For a Spontaneous change:
S  0
q

T
Entropy Created  0
may be positive or negative
q
It follows: S  
T
For a Spontaneous Process at a Given Temperature, the Change in Entropy of the System is greater than the
Heat divided by the Absolute Temperature
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Chemistry 102
Chapter 17
ENTROPY CHANGE FOR A PHASE TRANSITION

The entropy of a sample of matter increases as it changes state from a solid to a liquid or from a
liquid to a gas. This is due to the increase in disorder of molecules in liquid compared to solid and
gas compared to liquid.

We can therefore predict the sign of ∆S for processes involving changes of state (or phase). In
general, entropy increases (∆S > 0) for each of the following:




the phase transition from a solid to a liquid
the phase transition from a solid to a gas
the phase transition from a liquid to a gas
an increase in the number of moles of a gas during a chemical reaction
Examples:
1. Predict the sign of ∆S for each of the following processes:
a) the boiling of water
∆S =
b) I2 (g) → I2 (s)
∆S =
c) CaCO3 (s) → CaO (s) + CO2 (g)
∆S =
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Chemistry 102
Chapter 17
ENTROPY CHANGE FOR A PHASE TRANSITION

The criterion for spontaneity is an increase in the entropy of the universe. However, there are
several spontaneous processes that include a decrease in entropy. For example, when water freezes
below 0C, the entropy of the water decreases, yet the process is spontaneous.

The 2nd Law of Thermodynamics states that for a spontaneous process, the entropy of universe
increases (∆Suniv > 0). In the example above, even though the entropy of water decreases, the
entropy of the universe must increase in order for the process to be spontaneous.

Similar to the distinction of a system and surroundings in thermodynamics, we can distinguish
between the entropy of the system and the surroundings. In the example of the freezing water, ∆Ssys
is the entropy change of water, and the ∆Ssurr is the entropy change of the surroundings.

The entropy change of the universe is then the sum of the entropy changes of the system and the
surrounding.
∆S univ = ∆Ssys + ∆Ssurr

If a spontaneous process includes a decrease in ∆Ssys , then it would follow that there must be a
greater increase in ∆Ssurr in order for the ∆Suniv to increase.

Since freezing of water is an exothermic process, it would follow that the heat given off by the
process increases the entropy of the surroundings by a greater amount than the decrease of entropy
of the system, making it a spontaneous process.

To summarize:
 An exothermic process increase the entropy of the surroundings.
 An endothermic process decreases the entropy of the surroundings.
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Chemistry 102
Chapter 17
QUANTIFYING ENTROPY CHANGE OF SURROUNDINGS & SYSTEM

We know that when a system exchanges heat with surroundings, it changes the entropy of the
surroundings. At constant pressure, qsys can be used to quantify the entropy change of the
surroundings.

In general,
 A process that emits heat into the surroundings (qsys = negative) increases the entropy of
the surroundings (positive ∆Ssurr)
 A process that absorbs heat from the surroundings (qsys = positive) decreases the entropy
of the surroundings (negative ∆Ssurr)

The magnitude of the change in entropy of the surroundings is proportional to the magnitude of
the qsys
∆Ssurr α – qsys

It can also be shown that for a given amount of heat exchanged with the surroundings, the
magnitude of ∆Ssurr is inversely proportional to the temperature
∆Ssurr α 1/T
- qsys
- Hsys

Combining the above relationships:

Since the sum of ∆Ssys and ∆Ssurr must be positive for a spontaneous process, it would then
follow that for a spontaneous process
Ssurr =
Ssys >

T
=
T
(constant P, T)
qsys
T
In an equilibrium process, such as a phase change, no significant amount of entropy is created by
molecular disorder, and almost all the change in entropy is due to the absorption of heat.
Therefore,
Ssys =
qsys
(equilibrium process)
T
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Chemistry 102
Chapter 17
Examples:
1. Calculate S for the melting of ice (Heat absorbed = Heat of Fusion = Hfus = 6.0 kJ/mol)
Ssurr =
Hfus
6.0x103 J
=
= 22 J/K
T
273 K
2. Liquid ethanol, C2H6O (l), at 250C has an entropy of 161 J/(mol x K). If the heat of vaporization,
Hvap at 25 C is 42.3 kJ/mol, what is the entropy of the vapor in equilibrium with the liquid at 25 C?
When the liquid evaporates, it absorbs heat: Hvap = 42.3 kJ/mol at 250C
q
Hvap
42.3 kJ/mol
Entropy Change = S =  =  =  = 141.9 J/K x mol
T
(25 + 273) K
298 K
The entropy of 1 mole of vapor is calculated using the entropy of 1 mole of liquid (161 J/K) to which the
entropy change resulting from the heat absorption (141.9 J/K) is added:
Entropy of Vapor = 161 J/K + 141.9 J/K = 303 J/K
3. Consider the combustion of propane gas:
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)
∆Hsys = – 2044 kJ
a) Calculate the ∆Ssurr associated with this reaction occurring at 25C.
b) Determine the sign of ∆Ssys
c) Determine the sign of ∆Suniv. Will this reaction be spontaneous?
20
Chemistry 102
Chapter 17
2ND LAW OF THERMODYNAMICS AND SPONTANEITY OF REACTIONS

Consider the preparation of urea at constant temperature and pressure:
2 NH3 (g) + CO2(g)

CO(NH2)2 (aq) + H2O (l)
Is this a spontaneous reaction? (Does it go from left to right as written?)
 For a spontaneous reaction:
(constant T and P)
 For a spontaneous reaction:
(constant T and P)
qp
H
S   = 
T
T
H
S  
T
H
 – S  0
T
smaller larger
 Multiplying each term by the positive quantity T yields:
For a spontaneous reaction: H – T S  0
(constant T and P)

If S is available and H is calculated (Table 6.2), one can predict if the reaction is spontaneous
or not:
 H – T S  0
The reaction is spontaneous
 H – T S  0
The reaction is non-spontaneous (
)
The reaction is spontaneous
(
)
The reaction is at equilibrium (
)
 H – T S = 0

For a spontaneous reaction:
H – T S  0
21
(
(constant T and P)
)
Chemistry 102
Chapter 17
INTERPRETING SPONTANEITY

Two factors determine the spontaneity of a chemical change:
1. Nature’s tendency for the potential energy to be at a minimum.
 Any change that tends to lower the potential energy tends to occur spontaneously
 The quantity that is related to the lowering of the potential energy of changes that take
place at constant pressure is H
 For exothermic reactions H  0
 Exothermic reactions “tend” to occur spontaneously
2. Nature’s tendency toward disorder
 Any change that tends to increase the disorder of a system tends to occur spontaneously
 The quantity that is related to the increase in the degree of disorder for changes that take
place at constant pressure is S
 The larger the entropy of a system, the greater its statistical probability
 Reactions that go in the direction of greater entropy “tend” to occur spontaneously.
S = Sfinal – Sinitial

When S  0
Spontaneity is favored
The entropy (S) of a system depends on several factors:
(a) Physical state of the components of the system
S (solid)  S (liquid)  S (gas)
(b) Temperature of the system
Entropy (S) increases with temperature
(c) Mixing of components
 If the volume of the mixture is larger than the volumes of the components the Entropy (S)
increases
 If two liquids are mixed to form a solution, S increases
 If two gases are mixed, S increases only if the total volume increases
22
Chemistry 102
Chapter 17
THIRD LAW OF THERMODYNAMICS

A substance that is perfectly crystalline at 0 K has an entropy of zero

This law helps determine the Entropy of different substances at different temperatures, if one
considers the following facts:

When the temperature of a substance is raised from 0 K,
 the substance absorbs heat, and
 the substance becomes more disordered

The entropy change that occurs when heat is absorbed,
S=
q
T
Examples:
1. What is S for a substance if it is heated from near 0.0 K to 2.0 K and the heat absorbed is 0.19 J?
0.19 J
0.19 J
S =  =  = 0.19 J/K (at 1.0 K)
T(average)
1.0 K
2. What is S for the same substance if it is heated from near 2.0 K to 4.0 K and the heat absorbed
is 0.88 J?
0.88 J
0.88 J
S =  =  = 0.29 J/K (at 3.0 K)
T(average)
3.0 K

Proceeding this way, the entropy at any temperature can be determined.
Standard entropy of methyl chloride, CH3Cl, at various temperatures
Notes:
1. The entropy increases gradually as
the temperature increases.
2. The entropy increases sharply
when a phase change occurs.
23
Chemistry 102
Chapter 17
Standard Entropy of a Substance or Ion (Absolute Entropy), S0

S0 is the entropy value for the standard state of the species.
 For a substance:
Standard State is the pure substance at 1 atm
 For a species in solution: Standard state is the 1 M solution

S0 is temperature dependent; it is usually given for 25 C
Entropy Changes (S) for a Reaction
A. S can sometimes be predicted
 The entropy usually increases in the following situations:
(a) A reaction in which a molecule is broken into two or more smaller molecules.
Example: fermentation of glucose (grape sugar) to alcohol:
C6H12O6 (s)
2 C2H5OH (l) + 2 CO2 (g)
1 large molecule
4 smaller molecules
(b) A reaction in which there is an increase in moles of gas.
This may result from a molecules breaking up, in which case Rule (a) and Rule (b) are related.
Example: the burning of acetylene in oxygen:
2 C2H2 (g)
+
3 O2(g)
4 CO2 (g) + 2 H2O (g)
5 moles of gas
6 moles of gas
n gas = 6 – 5 = +1
(c) A process in which a solid changes to a liquid or gas or a liquid changes to a gas
melting
Ice at 0 C
Liquid
Liquid Water at 0 C
vaporization
Ethyl alcohol vapor at 25 C
Ethyl alcohol
at 25 C
sublimation
Solid CO2 at 25 C
CO2 gas at 25 C
24
Chemistry 102
Chapter 17
B. S can be calculated if the values of standard entropies are available
Examples:
1. Calculate the change in entropy, S0 for the fermentation of glucose.
The standard entropy of glucose is C6H12O6(s) is:
212 J/mol x K
The standard entropy for C2H5OH (l) is:
(ethyl alcohol)
161 J/mol x K
The standard entropy for CO2(g) is:
213.7 J/mol x K
C6H12O6 (s)
S : 1 x (212)
2 C2H5OH (l)
2 x (161)
S0 =
-
0
 nS0(products)
S0 = [(2 x 161) + 2 x 213.7)]
+
2 CO2 (g)
2 x (213.7)
 nS0(reactants)
-
[1 x 212] = 537 J/K
2. Predict the sign of S0 for each reaction shown below. If you cannot predict the sign for
any reaction, state why.
a) C2H2 (g) + 2 H2 (g)  C2H6 (g)
S0 =
b) N2 (g) + O2 (g)  2 NO (g)
S0 =
c) 2 C (s) + O2 (g)  2 CO (g)
S0 =
3. Using S0 values in your textbook, calculate S0 for the synthesis of ammonia.
N2 (g) + 3 H2 (g)  2 NH3 (g)
25
Chemistry 102
Chapter 17
PREDICTING SPONTANEITY
Recall:
For a spontaneous reaction: H – T S  0
(constant T and P)

The value of H – T S can be used as a criterion for spontaneity:
If:
 H – T S  0
 H – T S  0
 H – T S = 0
The reaction is spontaneous from left to right
(
The reaction is non-spontaneous from right to left (
)
The reaction is non-spontaneous from left to right (
)
The reaction is spontaneous from right to left
)
The reaction is at equilibrium (
(
)
Examples:
1. Consider the reaction by which urea is prepared from NH3 and CO2 , at 25 C
2 NH3 (g) + CO2 (g)
CO(NH2)2 (aq)
+
H2O (l)
H0 = – 119.7 kJ
Determine if the reaction is spontaneous from left to right as written.
The value of H – T S must be determined:
H0 = – 119.7 kJ
T = 25 + 273 = 298 K
S = ?
From Table 18.1:
S0(NH3)
gas
S0 (CO2)
gas
2 NH3 (g)
2 x 193
S0
= 193 J/mol x K
= 214 J/mol x K
+
S0 [CO(NH2)2] aqueous = 174 J/mol x K
S0(H2O)
liquid = 70. J/mol x K
CO2 (g)
214
CO(NH2)2 (aq)
174
+
S0= n S0(products) – n S0(reactants) = [(174 + 70] – [2 x (193 ) + (214)] =
H2O (l)
70
– 0.356 kJ/K
H – T S = (119.7 kJ)  (298 K) ( 0.356 kJ/K) = 13.6 kJ

)
The reaction is spontaneous under standard conditions from left to right, as written.
26
Chemistry 102
Chapter 17
GIBBS FREE ENERGY

Willard Gibbs (American Physicist) brought together the H (Enthalpy Change) and S (Entropy
Change) into a single thermodynamic quantity G:
G = Gibbs Free Energy = H  TS

As a reaction proceeds at a given temperature and pressure:
REACTANTS
PRODUCTS
n Hf (reactants)
m Hf (products)
n S (reactants)
H= n Hf (products) - m Hf(reactants)
mS (products)
S = nS (products) - mS(reactants)

Result: For a change at constant T and P :
G = H  TS

For any spontaneous change that takes place in a system:
 Free Energy G must be lowered (Gproducts  Greactants)
 G must be negative: G
= n G (products) - m G(reactants)  0
–
0
+
G < 0



G > 0



Reaction is spontaneous
Product-favored reaction
Forward reaction is favored
Reaction is non-spontaneous
Reactant-favored reaction
Reverse reaction is favored
Conclusion:
 G gives a composite of the two factors that contribute to spontaneity (H and S)
27
Chemistry 102
Chapter 17
SPONTANEITY & FREE ENERGY

G = H  TS
Consider the following situations:
G = (–) – T(+) = negative
1. H is negative and S is positive
(exothermic)
(increase in entropy)
G = (+) – T(–) = positive
2. H is positive and S is negative
(endothermic)

G = (+) – T(+) = ?
(increase in entropy)
The temperature plays the determining role in controlling spontaneity:
(a) When T is large:
TS  H
G = (+) – Tlarge(+) = negative
spontaneous
(b) When T is small:
TS  H
G = (+) – Tsmall(+) = positive
non-spontaneous
G = (–) – T(–) = ?
4. H is negative and S is negative
(exothermic)

(decrease in entropy)
The temperature plays the determining role in controlling spontaneity:
TS > H
G = (–) – Tlarge(–) = positive
(b) When T is small: TS < H
G = (–) – Tsmall(–) = negative
(a) When T is large:

non-spontaneous
(decrease in entropy)
3. H is positive and S is positive
(endothermic)
spontaneous
non-spontaneous
spontaneous
The effects of the algebraic signs of H and S and the effect of temperature on spontaneity can be
summarized as:
H
S
G = H – TS
Outcome
(–)
(+)
negative
Spontaneous at all temp.
(+)
(–)
positive
Non-spontaneous regardless of temp.
(+)
(+)
negative only at high T
Spontaneous only at high temp.
(–)
(–)
negative only at low T
Spontaneous only at low temp.
28
Chemistry 102
Chapter 17
FREE ENERGY AT STANDARD STATE

For interpreting thermodynamic data, standard states are chosen.
Standard States:
 Standard states are indicated by a superscript degree sign on the symbol of the quantity:
 G0 = standard free-energy change
 H0f = standard enthalpy change = n H0f (products) - m H0f (reactants)
 S0 = standard enthropy change = nS0 (products) - mS0(reactants)

Standard states are:
Pressure
1 atm pressure (for pure liquids and solids)
1 atm partial pressure (for gases)
Concentration
Temperature
1 M (for solutions
25 ºC (298 K)
Examples:
1. Calculate the standard free-energy change (G0) for the following reaction at 25 C, and determine
whether it is spontaneous or not.
CH4(g) +
H0f : − 74.9
S0 :
186.1
2 O2
CO2(g) +
− 393.5
213.7
2x0
(2) x (205.0)
2 H2O(g)
(2) x ( − 241.8)
(2) x (188.7)
H0 = [- 395.5 + (2) x (-241.8)] kJ – [- 74.9 + 0] kJ =
S0 = [213.7 + (2) x (188.7)] J/K – [186.1 + (2) x (205.0)] =
− 802.2 kJ
− 5.0 J/K
G0 = H0 - TS0 = - 802.2 kJ – (298 K) (- 5.0 x 103 kJ/K) = – 800.7 kJ (spontaneous)
2. Calculate the standard free-energy change (G0) for the following reaction at 25 C, and determine
whether it is spontaneous or not.
N2 (g)
H0f (kJ/mol)
S0 (J/mol.K)
0
191.5
+
O2 (g)
0
205.0
2 NO (g)
90.25
210.7
29
Chemistry 102
Chapter 17
STANDARD FREE ENERGY OF FORMATION (G0f)


G0f is the free-energy change that occurs when 1 mol of substance is formed from its elements in
their most stable states at 1 atm and 25 C
G0f has units of kJ/mol of substance

G0f is a constant, with values available in textbooks.
NOTE: Standard free energies of formation of elements in their most stable form = 0

G0 can be calculated directly for any reaction by using:
G0
=  nG0f (products) -  mG0f (products)
Examples:
1. Calculate the standard free energy of the following reaction at 25 C, using standard free energies of
formation given below:
Na2CO3 (s) +
G0f :
– 1048.1
H+ (aq)
2 Na+ (aq) +
0
(2) x (– 261.9)
HCO3(aq)
– 587.1
G0f = [(2) x (–261.9) + (–587.1)] – [(–1048.1) + 0] kJ = – 62.8 kJ
3. Calculate the standard free-energy change (G0) for the reaction shown below, and determine whether it
is spontaneous or not. (Use G0f values in your textbook).
2 NOCl (g)
2 NO (g)
30
+
Cl2 (g)
Chemistry 102
Chapter 17
G0 AS A CRITERION FOR SPONTANEITY
G < –10 kJ
Large negative number
Reaction is spontaneous
Reactants
Products
Reactants transform entirely
to products
–10 kJ <G < +10 kJ
Small negative or positive
number
Reaction gives an
equilibrium mixture
Reactants
Products
Significant amount of both
reactants and products
present
G > +10 kJ
Large positive number
Forward reaction is non-spontaneous
Reverse reaction is spontaneous
Reactants
Products
Mostly reactants present
Examples:
1. Which of the following reactions are spontaneous in the direction written? (See table in your
textbook for G0f values.
(A)
C(graphite)
0
Gf :
0
+
2 H2 (g)
0
G0 = [(–50.80 – (0)] kJ = –50.8 kJ
Ag+ (aq)
(B)
Gf 0 :
77.1
+
CH4 (g)
– 50.8
(spontaneous reaction)
I(aq)
–51.7
AgI (s)
–66.3
G0 =
2. Consider the reaction of nitrogen, N2 and oxygen, O2 to form nitric oxide, NO:
N2 (g)
+
O2 (g)
2 NO (g)
The standard free energy of formation of NO is + 86.60 kJ/mol. Do you expect the reaction to be
spontaneous as written?
Recall:
Standard free energies of formation of elements in their stablest form = 0
G0 = 2 (+ 86.60 kJ) = + 173.2 kJ
31
(non-spontaneous reaction)
Chemistry 102
Chapter 17
FREE ENERGY and USEFUL WORK

In a spontaneous reaction:
 the “free” energy is lowered as reactants change to products,
 the change in “free” energy is released as free-energy change( G)
 this “G” can be harnessed to perform useful work
A spontaneous reaction can be used to obtain useful work
Examples:
1. The combustion of gasoline is used to move a car
2. The reaction in a battery generates electricity that can drive a motor
The free – energy change (G) is the maximum energy available (“free”) to do useful work
Maximum useful work =
 In an ideal situation:
wmax = G
G
completely
Useful work
converted into
 In real situations:
G
converted
Some Useful Work + Some Entropy
into
Conclusions:
1. In theory, all of the free-energy decrease liberated during a spontaneous chemical change can be
used to do work (this would be wmax)
2. In practice, less work is obtained and the difference appears as an increase in entropy
Example:
 Living systems are able to convert only about 40% of the free energy available in the oxidation
of glucose to other forms of stored chemical energy (for example , ATP).

The rest (60%) appears as heat, which ensures the body’s effective temperature control system.
32
Chemistry 102
Chapter 17
FREE-ENERGY CHANGE DURING A SPONTANEOUS REACTION
(combustion of gasoline)
Gasoline + O2
CO2(g) + H2O(g)
Free energy decreases as the reaction proceeds (G  0)
At equilibrium, free energy is at a minimum
(equilibrium mixture is mostly products)
FREE-ENERGY CHANGE DURING A NON-SPONTANEOUS REACTION
(synthesis of NO from elements)
N2(g) +
O2(g)
2 NO(g)
Free energy increases as the reaction proceeds (G  0)
There is a small decrease in free energy as the system goes to
equilibrium (some reaction occurs to give the equilibrium
mixture which consists mostly of reactants)
N2(g) + O2(g)
NO(g)
33
Chemistry 102
Chapter 17
G0 & THE EQUILIBRIUM CONSTANT

G determines the maximum amount of energy that is available to perform useful work as a system
passes from one state to another (G = Gproducts – Greactants)

As a reaction proceeds:
 the system’s capacity to perform useful work decreases (G diminishes),
 the system eventually reaches equilibrium and G  0 (Free Energy ceases to change)

The value of G determines where the system stands with respect to equilibrium:
When G < 0
The reaction is spontaneous
and proceeds in the forward
direction towards
equilibrium (G decreases)
When G = 0
When G > 0
The system is in a state of
dynamic equilibrium
The reaction is not
spontaneous in the forward
direction.
The reaction is spontaneous
in the reverse direction.
Qualitative Relationship between G0 and the Position of Equilibrium
 The direction in which a reaction proceeds toward equilibrium is determined by where the system lies
with respect to the free-energy minimum.
G0
G
Direction of
reaction
Reactants
Products
Position of
equilibrium
NOTE:
 The reaction proceeds spontaneously only in a direction that gives rise to a decrease in free
energy (G is negative)
34
Chemistry 102
Chapter 17
RELATIONSHIP BETWEEN G0 AND EQUILIBRIUM CONSTANT
G = G0 + RT lnQ
Q = Thermodynamic Reaction Quotient
 for reactions involving gases Q is obtained from partial pressures.
 for reactions in solutions, Q is obtained from molar concentrations
R = Gas Law Constant in units of energy = 8.314 J/mol K
G0 = Standard Free-Energy Change
G = Free-Energy Change when Reactants in nonstandard states are changed to products in
non-standard states
At equilibrium:
1. The free energy ceases to change:
G = 0
2. The Thermodynamic Reaction Quotient, Q becomes equal
to the Thermodynamic Equilibrium Constant (K):
G = G0 + RT lnQ
By rearrangement:
becomes:
G0 = – RT ln K
Q=K
0 = G0 + RT lnK
or
G0 = – 2.303 RT log K
Basic Equations relating the standard
free-energy change to the equilibrium constant.

Natural logarithms (ln) and logarithms to the base 10 (log) are related by: ln x = 2.303 log x
G0 = - 2.303 RT log K


Thermodynamic Equilibrium Constant
The equilibrium Constant in which:
 the concentration of gases are expressed in partial pressures in atmospheres.
 the concentration of solutes are expressed in molarities
It follows that:
1. For reactions involving only gases:
K = Kp
2. For reactions involving only solutes in liquid solutions:
K = Kc
3. For net ionic equations:
K = Ksp
35
Chemistry 102
Chapter 17
RELATIONSHIP OF G0, K, AND SPONTANEITY
G0 = –2.303 RT log K
When K > 1
When K  1
When K <1
Reactants transform almost
entirely to products
Significant amounts of both
reactants and products are
present
Mostly reactants present
Reactants
Reactants
Products
Products
Reactants
Products
log K > 0
log K  0
log K < 0
G0 < 0
G0 =  10 kJ
K = (0.018 – 57)
G0 > 0
Reaction is spontaneous
Reaction gives an
equilibrium mixture
Forward reaction is non-spontaneous
Reverse reaction is spontaneous
Examples:
1. Use the standard free energies of formation given to determine the equilibrium constant (K)
for the following reaction at 25 C:
N2 (g)
G0f (kJ/mol)
0
+ 3 H2 (g) 
2 NH3 (g)
–16.66
0
G0 = [ 2 x G0f (NH3)] – [ G0f (N2) + 3 x G0f (H2)] = 2 x (–16.66) – 0 = –33.32 kJ
G = –RT ln K
0
ln K =
ΔG 0
( 33, 320 J)
=
= 13.45
RT
(8.314 J/K)(298 K)
G0
RT
K=e
G 0 /RT
K = e13.45 = 6.9 x 105
2. Use the standard free energies in your text to calculate G0 and K for the following reaction
at 25 C:
2 H2 (g) + O2 (g)  2 H2O (g)
36
Chemistry 102
Chapter 17
CHANGE OF FREE ENERGY WITH TEMPERATURE

How can G0 be found for temperatures other than standard temperatures (25 C)?

An approximate method used to calculate G0T is based on the assumption that both H0 and
S0 are constant with respect to temperature (only approximately true)

Then:
NOTE:
G0T = H0 - TS0
(a convenient approximation for G0T )
G0T is strongly temperature dependent
G0T = Change in free Energy for a substance:
 at 1 atm of pressure (standard pressure) and
 at the specified temperature, T (nonstandard temperature)
depends on the value of
G
SPONTANEITY

determined by
0
TEMPERATURE
T
It follows that SPONTANEITY IS TEMPERATURE DEPENDENT!
Meaning:
 Some chemical changes may be non-spontaneous at one temperature but spontaneous at
another temperature.
Effect of Temperature on the Spontaneity of Reactions
G0T = H0 - TS0
For a Spontaneous Reaction: G0T  0
H0
S0
Temperature
(–)
(+)
no effect
(+)
(–)
no effect
(–)
(+)
G0
Negative
regardless of temp.
Positive
regardless of temp.
Low
Negative
High
Positive
Low
Positive
High
Negative
(–)
(+)
37
Spontaneity
Spontaneous
at any temp.
Non-spontaneous
at any temp.
Spontaneous
at low temp.
Non-spontaneous
at high temp.
Non-spontaneous
at low temp.
Spontaneous
at high temp.
Chemistry 102
Chapter 17
Examples:
1. When glucose ferments, it produces ethyl alcohol and carbon dioxide. The reaction is exothermic.
Under what temperature conditions is this reaction spontaneous?
C6H12O6 (s)
glucose
1 molecule
of
solid
2 C2H5OH (l) +
ethyl alcohol
breaks
2 molecules of liquid
and
2 molecules of gas
into
G0T = H0 - TS0
G0T = H0 - TS0
H0  0
2 CO2(g)
= (Negative Number)
=
S0  0
–
[(Temperature) x (Positive Number)]
–
(–)
[
(+)
]
NOTE: T is absolute temperature in K cannot be negative

G0T is negative (regardless of temperature)

Reaction is spontaneous at any temperature
2. Sodium Carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3.
Estimate the temperature at which NaHCO3 decomposes to products at 1 atm.
2 NaHCO3 (s)

Na2CO3 + H2O (g)
+
CO2 (g)
First calculate the H0 and S0 , using the given Hf 0 and S0 values:
Hf :
S0 :
0
2 NaHCO3 (s)
2 x (–947.7)
2 x 102
Na2CO3
–1130.8
139
+ H2O (g) +
–241.826
188.72
CO2 (g)
–393.5
213.7
H0 = [(–1130.8) + (–241.826) + (–393.5)] – [ (2) x (–947.7)]
=
129.274 kJ
S0 = [(139) + (188.72) + (213.7)] – [ 92) x (102)]= 337.4 J/K
=
0.3374 kJ/K

Let
G0 = 0
Substitute these values into G0 = H0 – TS0 ;
H0 – TS0 = 0
H0
129.274 kJ
T =  =  = 383 K (110 C)
S0
0.3374 kJ/K
[T  383 K (110 C)]
The temperature at which NaHCO3 decomposes to products at 1 atm. should be at least = 383 K (110 C)
38
Chemistry 102
Chapter 17
3. Oxygen was first prepared by heating mercury(II) oxide, HgO. Estimate the temperature at which
HgO decomposes to O2 at 1 atm.
2 HgO (s)

2 Hg (g) + O2(g)
First calculate the H0 and S0 , using the given Hf 0 and S0 values:
2 HgO(s)
Hf : –90.8
S0 :
70.3
0
2 Hg (g)
61.3
174.9
+
O2 (g)
0
205.0
H0 =
S0 =

Substitute these values into G0 = H0 - TS0 ;
H0 - TS0 = 0
H0
T =  =
S0
39
Let
G0 = 0