Chapter 17
Spontaneity, Entropy, and Free Energy
• The goal of this chapter is to answer a
basic question: will a given reaction occur
“by itself” at a particular temperature and
pressure, without the exertion of any
outside force?
Thermodynamics
• The science that deals with heat and energy
effects.
• First Law of thermodynamics / The Law of
Conservation of Energy:
– Energy can be neither created nor destroyed
by a process, it is only transformed from one
form to another.
• Any energy lost by a system must be gained by
the surroundings and vice versa
– CH4+2O2(g) CO2(g)+ 2H2O(g) + energy
Here potential energy has been converted to
thermal energy.
Figure 16.1 Methane and Oxygen React
Spontaneous Processes
• A process is said to be spontaneous if it
occurs without outside intervention.
• Thermodynamics lets us predict whether
a process will occur or not.
• Spontaneous processes may be fast or
slow.
• Thermodynamics can tell the direction in
which a process will occur but can say
nothing about the speed of the process.
Figure 16.2 Rate of Reaction
Entropy
• What common characteristic cause the
processes to be spontaneous?
– The driving force for the spontaneous process is
an increase in the Entropy (denoted by s) of the
universe
• What is Entropy?
– Entropy is a measure of randomness or disorder
– State property depends upon the state of a
system
– Thermodynamic function that describe the
number of arrangements
S = Sfinal - Sinitial
Microstate
• Each configuration that gives a particular
arrangement is called a microstate.
• Which arrangement is most likely to
occur?
– One with greatest number of microstate.
• Positional probability (microstates) which
depends on the number of configurations
in space.
– Positional probability increases (entropy
increases) in going from solid to liquid to gas.
Ssolid < Sliquid << Sgas
Figure 16.3 The Expansion of an Ideal Gas into an Evacuated Bulb
Figure 16.4 Three Possible Arrangements (states) of Four Molecules in a Two-Bulbed Flask
Examples
• Choose the compound with the greatest
positional entropy in each case.
a. 1 mol H2 (at STP) or 1 mol H2 (at 100 oC, 0.5
atm) H2 at 100 oC and 0.5 atm; higher
temperature and lower pressure means greater
volume and hence, greater positional entropy.
b. 1 mol N2 (at STP) or 1 mol H2 (at 100 K, 2.0
atm) N2 at STP has the greater volume.
c. 1 mol H2O(s) (at 0 oC) or 1 mol H2O(l) (at 20 oC)
H2O(l) is more disordered than H2O(s)
Second Law of Thermodynamics
• In any spontaneous process there is
always an increase in the entropy of the
universe.
– Universe
• System: Portion of the universe in
which we are interested
• Surroundings: everything else in the
universe besides the system.
Change in entropy of the universe
Suniv = Ssys+ Ssur
• Where Ssys and Ssurr represent the
changes in entropy
• Suniv - positive the process is
spontaneous in the direction written.
• Suniv - negative the process is
spontaneous in the opposite direction.
• Suniv - zero the system is at
equilibrium.
The Effect of Temperature on Spontaneity
Consider, H2O(l)  H2O(g)
1 mole, 18 grams, 18 mL  1 mole, 18
grams, 31 litters (1 atm, 100 oC)
Positional probability increases – entropy of
the system increases, Ssys  Positive
– The sign of Ssurr depends on the
direction of the heat flow
– The magnitude of Ssurr depends on the
temperature
– Ssurr depends directly on the quantity of
heat transferred and inversely on
temperature.
Continued....
Exothermic process: Ssurr = + quantity of
heat (J) / temperature (K)
Endothermic process: Ssurr = - quantity of
heat (J) / temperature (K)
Heat flow (constant P) = Change in
enthalpy = H
Endothermic H positive and exothermic
H negative
Ssurr = - H/T (at constant temp. and
pressure)
Example
Calculate Ssurr for the following reactions at
25 oC and 1 atm.
• C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Ho
= -2221kJ
Ssurr = - H/T = -(-2221 kJ) / (25+273)K =
7.45 kJ/K = 7.45 X 103 J/K
• 2NO2(g)  2NO(g) + O2(g) H = 112 kJ
Ssurr = - H/T = -112 kJ/298 K = -0.376 kJ/K
= -376 J/K
Free Energy
 Free energy is a thermodynamic function
related to spontaneity and is useful in
dealing with temperature dependence of
spontaneity, defined by the relationship:
G = H – TS where, G is free energy, H is
enthalpy, T is Kelvin temperature, S is
entropy
For a process that occurs at constant
temperature, the changes in free energy
(G) is given by the equation,
G = H – TS
G = H – TS
Lets see how this equation relates to
spontaneity,
 -G/T = - H/T – TS/-(T) [Divide both
side by –T]
-G/T = - H/T + S = Ssurr+ S = Suniv
[Recall, Ssurr = - H/T]
 Suniv = - G/T at constant T and P
 G negative – process is spontaneous
at constant T and P
Example
• Example: The boiling point of chloroform
(CHCl3) is 61.7 oC. The enthalpy of
vaporization is 31.4 kJ/mol. Calculate the
entropy of vaporization.
G = H - TS
At the boiling point, G = 0, so TS = H
S = H/ T = 31.4 kJ/mol ÷ (273.2+61.7)K
=9.38 X 10-2 kJ/K.mol = 93.8 J/K.mol
• Example: For mercury, the enthalpy of
vaporization is 58.51 kJ/mole and the
entropy of vaporization is 92.92 J/K.mole.
What is the normal boiling point of
mercury?
At the boiling point , G = 0, so H = TS
T=H/S=58.51X103J/mol92.92 J/K.mol
= 629.7 K
 Example: For ammonia (NH3), the enthalpy of
fusion is 5.65 kJ/mol and the entropy of fusion is
28.9 J/K.mol
a. Will NH3(s) spontaneously melt at 200 K?
b. What is the approximate melting point of
ammonia?
a. NH3(s)  NH3(l)
G = H - TS = 5650 J/mol – 200 K.(28.9
J/K.mol)
= 5650 J/mol - 5780 J/mol = - 130 J/mol
Yes, NH3 will melt since G < 0 (negative) at
this temperature
b. At the melting point G = 0
H = TS
T =  H / S = 5650 J/mol  28.9 J/K.mol
= 196 K
Entropy Changes in Chemical Reactions
• The entropy changes in the system (the
reactants and the products of the
reaction) are determined by positional
probability. eg. In the ammonia synthesis
reaction
N2(g) + 3H2(g) 2NH3(g)
– Four reactant molecules becomes two
product molecules.
– Fewer molecules mean fewer possible
configuration.
Entropy Changes in Chemical Reactions
• Does positional entropy increases or
decreases for the following reaction
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
9 gaseous molecule  10 gaseous
molecule  Positional entropy increases
If the number of molecules of the
gaseous products is greater than the
number of molecules of the gaseous
reactants, positional entropy increases,
and S will be positive for the reaction.
Figure 16.5 Entropy
Predicting the Sign of So
 Predict the sign of So for each of the following changes
a. AgCl(s)  Ag+(aq) + Cl-(aq)
Increase in disorder; So (+)
b. 2H2(g) + O2(g)  2H2O(l)
Decrease in disorder; n < 0; So (-)
c. H2O(l)  H2O(g)
Increase in disorder; n > 0; So (+)
d. Na(s) + 1/2Cl2(g)  NaCl(s)
Decrease in disorder; n < 0; So (-)
e. 2SO2(g) + O2(g)  2SO3(g)
Decrease in disorder; n < 0; So (-)
Third Law of Thermodynamics
• The entropy of a perfect crystal at 0 K is
zero
• Entropy is a state function of the system
• Entropy change for a given chemical
reaction can be calculated by taking the
difference between the standard entropy
values of products and those of the
reactants:
Soreaction = npSoproducts- nrSoreactants
• Entropy is an extensive property
(depends on amount)
Example
Calculate So at 25oC for the reaction
2NiS(s) + 3O2(g)  2SO2(g) + 2NiO(s)
Soreaction = npSoproducts- nrSoreactants
= 2SoSO2(g) + 2SoNiO(s) – (2SoNiS(s) + 3SoO2(g))
= 2 mol(248 J/K.mol) + 2mol(38 J/K.mol) 2mol(53 J/K.mol) - 3mol(205 J/K.mol)
= 496 J/K + 76 J/K - 106 J/K - 615 J/K
= -149 J/K
We would expect So to be negative because
the number of gaseous molecules decreases.
Free Energy and chemical Reactions
• Go, the change in free energy that will
occur if the reactants in their standard
states are converted to the products in
their standard states.
N2(g) + 3H2(g)  2NH3(g) Go = -33.3 KJ
Go = Ho - TSo
• Example:
C(s) + O2(g)  CO2(g)
The values of Ho and So are -393.5 KJ
and 3.05 J/K, calculate Go at 298 K.
Go = H - TSo
= -3.935 x 105J – (298)(3.05 J/K)
= -3.944 x 105J = -394.4 KJ
(per mol of CO2)
Free Energy Change and
Chemical Reactions
• Go = standard free energy change that
accompanies the formation of 1 mole of
that substance from its constituent
elements with all reactants and products
in their standard states.
Go = npGof (products)
– nrGof (reactants)
• The standard free energy of formation of
an element in its standard state is zero.
Free Energy and Pressure
• G = Go + RTln(P) where,
Go = the free energy of the gas at a pressure of 1 atm
G = the free energy of the gas at a pressure of P atm
R = the universal gas constant
T = the Kelvin temperature
G = npG products – nrG reactants
Where, Gproducts = Go products + RTln(Pproducts)
Greactants= Go reactants + RTln(Preactants)
G = Go + RTln(Q) where,
Q = reaction quotient from the law of mass action
Free Energy and Equilibrium
G = Go + RTln(Q)
At equilibrium,
G = 0 (Gproducts = Greactants)
and Q = K (equilibrium constant)
So, G = 0 = Go + RTln(K)
Go = -RTln(K)
Figure 16.8 The Dependence of Free Energy on Partial Pressure
Continued….
Case 1: Go = 0, the system is at equilibrium when
the pressures of all reactants and products are 1
atm, which means that K = 1
Case 2: Go < 0 , in this case Go products < Go reactants.
The system will adjust to the right to reach
equilibrium, K will be greater than 1, since the
pressure of the products at equilibrium will be
greater than 1 atm and the pressure of the reactants
will be less than 1 atm.
Case 3: Go > O, in this case Go reactants < Go products.
The system will adjust to the left to reach
equilibrium. The value of K will be less than 1.
Temperature Dependence of K
Go = -RTln(K) = Ho - TSo
ln(K) = - Ho/RT + So/R
ln(K) = -[Ho/R][1/T] + So/R
This is a linear equation of the form
y = mx + b
where, y = ln(K), m = -Ho/R = slope,
x = 1/T, and b = So/R = intercept
(Ho and So independent of temperature
over a small temperature range)
Summary
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First law of thermodynamics
Spontaneous process
Entropy
S = Sfinal – Sinitial
Microstate
Second law of thermodynamics
Effect of temperature
System
Surroundings
Summary
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Suniv = Ssys + Ssurr
Free energy G = H – TS
Entropy change in chemical reactions
Soreaction = npSoproducts- nrSoreactants
Go = npGof(products) = nrGof(reactants)
Go = Go + RTln(p) = Go + RTln(Q)
Go = -RTln(K)
ln(K) = -[Ho/R][1/T] + So/R