Chapter Nineteen
Copyright © Tyna L. Heise 2001 - 2002
All Rights Reserved
No. 1
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Spontaneous Processes
Understanding and designing chemical
reactions:
How rapidly does the reaction proceed?
- reaction rates
- controlled by a factor related to
energy
- the lower the activation energy,
the faster the reaction proceeds
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Spontaneous Processes
How far toward completion will the
reaction go?
- equilibrium constants
- depends on rates of forward and
reverse reactions
- equilibrium should also be
dependent on energy in some way
due to dependence on reaction
rates
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Spontaneous Processes
Chemical thermodynamics is the relationship
between equilibrium and energy.
First Law of Thermodynamics: for a
reaction that occurs at constant
pressure, the enthalpy change equals
the heat transferred between the system
and its surroundings
Energy is conserved!!
* Enthalpy is important in helping us
determine if a reaction will proceed!
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Spontaneous Processes
Spontaneous Processes:
energy is neither created nor destroyed
in any process, energy can only be
transferred or converted from one form to
another
DE = q + w
a spontaneous process occurs without
any outside intervention as energy is
conserved
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Spontaneous Processes
No. 6
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Spontaneous Processes
Spontaneous Processes:
temperature is going to effect the
spontaneity of a process
if discussing a phase change, at the
temperature of the phase change, the
phases compete for spontaniety, and
neither is said to win over the other
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Spontaneous Processes
Sample exercise: Under 1 atm pressure,
CO2(s) sublimes at -78°C. Is the
transformation of CO2(s) to CO2(g) a
spontaneous process at -100°C?
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Spontaneous Processes
Sample exercise: Under 1 atm pressure,
CO2(s) sublimes at -78°C. Is the
transformation of CO2(s) to CO2(g) a
spontaneous process at -100°C?
*No, if the temp had been higher it
would change phase spontaneously, but lower
than the sublimation point favors the reverse,
so the solid remains a solid.
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Spontaneous Processes
Reversible and Irreversible Processes:
State functions: define a state and do not
depend upon the pathway
temperature
internal energy
enthalpy
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Spontaneous Processes
Reversible and Irreversible Processes:
Reversible processes is a unique way for
a system to change its state, than go back
to its state by following the exact same
path but in the opposite direction
phase changes at constant temp
only one specific value of q (heat)
system in equilibrium
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Spontaneous Processes
Reversible and Irreversible Processes:
Irreversible processes cannot be simply
restored to their original state using the
same path, it may be forced to go back,
but by a different pathway
phase changes at different temps
two q values need to be established
qforward and qreverse
any spontaneous reaction
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Spontaneous Processes
Thermodynamics can tell us
direction of reaction
extent of reaction
NOT
speed of reaction
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Entropy and the 2nd Law
Spontaniety depends upon two factors
Enthalpy (DH): heat of reaction
exothermic normally spontaneous
endothermic normally NOT spontaneous
Entropy (DS): disorder of the system
natural law indicates reactions go in the
direction that leads to more disorder
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Entropy and the 2nd Law
The Spontaneous Expansion of a Gas:
When the stopcock is opened,
the gas will spontaneously
flow to fill the empty half,
but
it WILL NOT flow backward
without work being done on
the system.
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Entropy and the 2nd Law
The Spontaneous Expansion of a Gas:
Gas expands because of the tendency for the
molecules to ‘spread out’ among the different
arrangements that they can take.
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Entropy and the 2nd Law
Entropy: measurement of randomness or
chaos
melting ice
dissolving salts
The more disordered or random a system,
the larger its entropy
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Entropy and the 2nd Law
Sample exercise: Indicate whether each of
the following reactions produces an increase
or decrease in the entropy of the system:
a) CO2(s)  CO2(g)
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Entropy and the 2nd Law
Sample exercise: Indicate whether each of
the following reactions produces an increase
or decrease in the entropy of the system:
a) CO2(s)  CO2(g)
Entropy increases
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Entropy and the 2nd Law
Sample exercise: Indicate whether each of
the following reactions produces an increase
or decrease in the entropy of the system:
b) CaO(s) + CO2(g)  CaCO3(s)
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Entropy and the 2nd Law
Sample exercise: Indicate whether each of
the following reactions produces an increase
or decrease in the entropy of the system:
b) CaO(s) + CO2(g)  CaCO3(s)
Entropy decreases
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Entropy and the 2nd Law
Entropy: measurement of randomness or
chaos
state function
for a process that occurs at constant
temperature, the entropy change is
dependent on the heat transferred during the
reverse of the reaction (qrev)
DS = qrev/T
No. 22
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
No. 23
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
DS = qrev/T
No. 24
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
DS = qrev/T
qrev = DHvap
= 38.56kJ/mol
T = 351.45 K
No. 25
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
DS = qrev/T
38.56kJ 1000 J 1 mol
mol 1 kJ 46 g
T = 351.45 K
No. 26
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
DS = qrev/T
838.26 J/g * 25.8 g =
-21627 J
T = 351.45 K
No. 27
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Entropy and the 2nd Law
Sample exercise: The normal boiling point of
ethanol, C2H5OH, is 78.3°C, and its molar
enthalpy of vaporization is 38.56 kJ/mol.
What is the change in entropy when 25.8 g of
C2H5OH(g) at 1 atm pressure condenses to
liquid at the normal boiling point?
DS = qrev/T
= -21627 J/351.45 K
= -61.5 J/K
No. 28
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Entropy and the 2nd Law
Second law of Thermodynamics: In any
reversible process, DSuniv = 0. In any
irreversible reaction, DSuniv >0.
DSuniv = DSsys + DSsurr
In an isolated system, just the entropy of the
system is considered.
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Molecular Interpretation
On the microscopic level, the number of gas
molecules can be directly related to the
amount of entropy in a system.
The more gas molecules present, the
higher the entropy value
a phase change that increases the
number of gas molecules would increase
entropy
a phase change that decreases the
number of gas molecules would decrease
entropy
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Molecular Interpretation
Three moles of gas combine to form two
moles of gas, thus decreasing the number of
molecules.
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Molecular Interpretation
Degrees of freedom
creating new bonds decreases the
freedom of movement atoms may have
had.
3 degrees
motion in one direction,
translational movement
vibrational movement
spinning, rotational movement
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Molecular Interpretation
Figure 19.12
No. 33
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Molecular Interpretation
Sample exercise: Choose the substance with
the greatest entropy in each case:
1 mol of H2(g) at STP or 1 mol of H2(g) at
100°C and 0.5 atm.
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Molecular Interpretation
Sample exercise: Choose the substance with
the greatest entropy in each case:
1 mol of H2O(s) at 0°C or 1 mol of H2O(l) at
25°C.
No. 35
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Molecular Interpretation
Sample exercise: Choose the substance with
the greatest entropy in each case:
1 mol of H2(g) at STP or 1 mol of SO2(g) at
STP.
No. 36
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Molecular Interpretation
Sample exercise: Choose the substance with
the greatest entropy in each case:
1 mol of N2O4(g) at STP or 2 mol of NO2(g) at
STP.
No. 37
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Molecular Interpretation
Sample exercise: Predict whether is DS is
positive or negative in each of the following
processes:
HCl(g) + NH3(g)  NH4Cl(s)
No. 38
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Molecular Interpretation
Sample exercise: Predict whether is DS is
positive or negative in each of the following
processes:
2SO2(g) + O2(g)  2SO3(s)
No. 39
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Molecular Interpretation
Sample exercise: Predict whether is DS is
positive or negative in each of the following
processes:
cooling of nitrogen gas from 20°C
to -50°C
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Calculation of Entropy Changes
Entropy Calculations:
no easy method for measuring entropy
experimental measurements on the
variation of heat capacity with
temperature can give a value known as
absolute entropy
zero point of reference for perfect
crystalline solids
tabulated as molar quantities, J/mol-K
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Calculation of Entropy Changes
Entropy differs from enthalpy
standard molar entropies are not 0
entropies of gases are greater than those
of liquids and solids
entropies increase with molar mass
entropies increase with number of atoms
in formula
DS° = nS(products) - mS(reactants)
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Gibbs Free Energy
Spontaneity involves both thermodynamic
concepts: entropy and enthalpy
DG = DH – TDS
G = Gibbs Free Energy
H = Enthalpy
T = Temperature (K)
S = Entropy
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Gibbs Free Energy
Gibbs Free Enegry
 If DG is negative, the reaction is spontaneous and
proceeds in the forward direction
 If DG is zero, the reaction is at equilibrium
 If DG is positive, the reaction is nonspontaneous
and proceeds in the reverse direction
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Gibbs Free Energy
Gibbs Free Enegry
 State function
 Table 19.3
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Gibbs Free Energy
Sample exercise: By using data from Appendix C,
Calculate DG at 298 K for the combustion of
methane:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
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Gibbs Free Energy
Sample exercise: By using data from Appendix C,
Calculate DG at 298 K for the combustion of
methane:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
-50.8
0
-394.4 2(-228.57)
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Gibbs Free Energy
Sample exercise: By using data from Appendix C,
Calculate DG at 298 K for the combustion of
methane:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
-50.8
0
-394.4
2(-228.57)
DG = products - reactants
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Gibbs Free Energy
Sample exercise: By using data from Appendix C,
Calculate DG at 298 K for the combustion of
methane:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
-50.8
0
-394.4
2(-228.57)
DG = products – reactants
(-394.4 + 2(-228.57)) – (-50.8 + 0)
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Gibbs Free Energy
Sample exercise: By using data from Appendix C,
Calculate DG at 298 K for the combustion of
methane:
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
-50.8
0
-394.4 2(-228.57)
DG = products – reactants
(-394.4 + 2(-228.57)) – (-50.8 + 0)
-800.7
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Gibbs Free Energy
Sample exercise: Consider the combustion of
propane to form CO2(g) and H2O(g) at 298K.
Would you expect DG to be more negative or less
negative than DH?
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Gibbs Free Energy
Sample exercise: Consider the combustion of
propane to form CO2(g) and H2O(g) at 298K.
Would you expect DG to be more negative or less
negative than DH?
more negative, using Gibbs Free Enegry
formula, more moles of gas being produced would
be increasing entropy, +DS,
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Free Energy and Temperature
Table 19.4
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Free Energy and Temperature
Sample exercise: Using standard enthalpies of
formation and standard entropies in Appendix C,
calculate DH and DS at 298 K for the following
reaction:
2SO2(g) + O2(g)  2SO3(g)
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Free Energy and Temperature
Sample exercise: Using standard enthalpies of
formation and standard entropies in Appendix C,
calculate DH and DS at 298 K for the following
reaction:
2SO2(g) + O2(g)  2SO3(g)
2(-296.9) 0
2(-395.2)
DH = 2(-395.2) - 2(-296.9) = -196.6 kJ
No. 55
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Free Energy and Temperature
Sample exercise: Using standard enthalpies of
formation and standard entropies in Appendix C,
calculate DH and DS at 298 K for the following
reaction:
2SO2(g) + O2(g)  2SO3(g)
2(248.5) 205.0
2(256.2)
DS = 2(256.2) – (2(248.5)+205.0) = -189.6 J
No. 56
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Free Energy and Temperature
Sample exercise: Using the values obtained,
estimate DG at 400 K
DG = DH – TDS
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Free Energy and Temperature
Sample exercise: Using the values obtained,
estimate DG at 400 K
DG = DH – TDS
= -196.6 kJ – 400(-0.1896 kJ)
= -120.8
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Free Energy and the
Equilibrium Constant
2 other important ways free energy is a powerful
tool
 Tabulate free energy under nonstandard
conditions
 Directly relate free energy to equilibrium
constants
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Free Energy and the
Equilibrium Constant
Tabulate free energy under nonstandard
conditions
DG = DG° + RTlnQ
R = 8.314 J/mol-K
T = temperature K
Q = reaction quotient
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Free Energy and the
Equilibrium Constant
Sample exercise: Calculate DG at 298 K for
the reaction of nitrogen and hydrogen to form
ammonia if the reaction mixture consists of 0.50
atm N2, 0.75 atm H2, and 2.0 atm NH3.
DG = DG° + RTlnQ
No. 61
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Free Energy and the
Equilibrium Constant
Sample exercise: Calculate DG at 298 K for
the reaction of nitrogen and hydrogen to form
ammonia if the reaction mixture consists of 0.50
atm N2, 0.75 atm H2, and 2.0 atm NH3.
DG = DG° + RTlnQ
2(-16.6) + 0.008314(298)ln(2.02/0.50(0.753))
-26.0 kJ
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Free Energy and the
Equilibrium Constant
At equilibrium, DG is equal to 0 so…
DG° = -RTlnKeq
DG° negative: K > 1
DG° zero : K = 1
DG° positive: K < 1
Keq = e-DG°/RT
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