Chemistry 100 Chapter 19 - Spontaneity of
Chemical and Physical Processes
What Is Thermodynamics?
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Study of the energy changes that accompany chemical and
physical processes.
Based on a set of laws.
In chemistry, a primary application of thermodynamics is
as a tool to predict the spontaneous directions of a
chemical reaction.
What Is Spontaneity?
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Spontaneity refers to the ability of a process to occur on
its own!
Waterfalls
“Though the course may change sometimes, rivers always
reach the sea” Page/Plant ‘Ten Years Gone’.
Ice melts at room temperature!
The First Law of Thermodynamics
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The First Law deals with the conservation of energy
changes.
E = q + w
The First Law tells us nothing about the spontaneous (i.e.,
the natural) direction of a process.
Entropy and Spontaneity
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To help us predict the spontaneous direction of a process,
we will look at the entropy change of the process as well
as its enthalpy change (heat flow).
Entropy – the degree of randomness of a system.
 Solids – highly ordered  low entropy.
 Gases – very disordered  high entropy.
 Liquids – entropy is variable between that of a solid and
a liquid.
Entropy Is a State Variable

Changes in entropy are state functions
S = Sf – Si
Sf = the entropy of the final state
Si = the entropy of the initial state
Entropy Changes for Different Processes
S > 0 entropy increases (melting ice or making steam)
S < 0 entropy decreases (examples freezing water or
condensing steam)
Example - The Solution Process

For the dissolution of NaCl (s) in water
NaCl (s)  Na+(aq) + Cl-(aq)
 We can clearly see that in the solution state the Na+
(aq) and the Cl- (aq) become randomly sidtributed
throughout the solvent. The entropy increases from
the pure solid to the solution state.
The Entropy Change in a Chemical Reaction
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Look at the following example. Burning ethane!
C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l)
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The entropy change is calculated from the difference in
the entropies of the products vs. the reactants.
 rS   np S (products) -  nr S (reactants)
 np and nr represent the number of moles of products and
reactants, respectively.
For the ethane combustion reaction
1 C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (l)
rS   np S(products) -  nr S(reactants)

= 3 S [H2O (l)] + 2 S [CO2 (g)] - (7/2 S [O2(g)] + 1 S
[C2H6 (g)] )
 Appendix C in your textbook has absolute entropy values
for a wide variety of species.
 Units for entropy values  J / (K mole). Temperature and
pressure for the tabulated values are 298.2 K and 1.00
atm. Note – the elements have NON-ZERO entropy
values!
e.g., for H2 (g)
fH = 0 kJ/mole (by def’n)
S = 130.58 J/(K mole)
 For the combustion reaction above
rS = 3 S [H2O (l)] + 2 S [CO2 (g)] - (7/2 S [O2(g)] + 1
S [C2H6 (g)] )
= 3 mole H2O x 69.91 J/(K mole) + 2 mole CO2 x 213.6 J/(K
mole) – 7/2 mole O2 x 205.0 J/(K mole) –1 mole C2H6 x
229.5 J/(K mole) = -310.1 J/(K mole)
Some Generalizations
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For any gaseous reaction (or a reaction involving gases).
ng > 0, rS > 0 J/(K mole).
ng < 0, rS < 0 J/(K mole).
ng = 0, rS  0 J/(K mole).
Look at the ethane combustion reaction above! ng < 0 –
the entropy change is negative!!
For reactions involving only solids and liquids – depends
on the entropy values of the substances.
The Second Law of Thermodynamics
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The entropy of the universe (univS) increases in a
spontaneous process. univS is unchanged in an
equilibrium process
The Definition of univS?
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univS = sysS + surrS
sysS = the entropy change of the system.
surrS = the entropy change of the surroundings.
In order to obtain univS, we need to obtain estimates for
both the sysS and the surrS.
As an example, let’s examine the following chemical
reaction.
C(s) + 2H2 (g)  CH4(g)
The entropy change for the systems is the reaction
entropy change, rS (calculated as shown above).
How do we calculate surrS?
Calculating surrS
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Note that for an exothermic process, an amount of thermal
energy is released to the surroundings!
 A small part of the surroundings is warmed (kinetic
energy increases). The entropy increases!
Heat
System
surroundings
Insulation
Note that for an endothermic process, thermal energy is
absorbed from the surroundings!
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A small part of the surroundings is cooled (kinetic energy
decreases).
Heat
System
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surroundings
The entropy decreases!
 For a constant pressure process
qp = H
surrS  surrH
surrS  -sysH
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The entropy of the surroundings is calculated as follows.
surrS = -sysH / T
For a chemical reaction
sysH = rH
surrS = -rH/ T
Let’s carry out the calculation for the above reaction.
rS   np S(products) -  nr S(reactants)
= S [CH4 (g)] - 2 S [H2 (g)] - S [C (s)] )
(Values are from Appendix C in your textbook)
rS = 1 mole x 186.96 J/(K mole) – 2 mole x 131.0 J/(K
mole) – 1 mole x 5.69 J/(K mole)
= -81.5 J/K
Next to calculate univS - need rH.
rH   np fH (products) -  nr fH (reactants)
= fH [CH4 (g)] - 2 fH [H2 (g)] - fH [C (s)] )
= fH [CH4 (g)] (i.e., fH [elements] = 0 by definition).
rH = 1 mole x –74.85 kJ/mole
= -74.85 kJ/mole
surrS = -rH / Tsurr  assume Tsurr = 298.2 K
surrS = 74.85 x 103 J / 298.2 K = 251 J/K
univS = surrS +sysS =251 J/K + -81.5 J/K = 170 J/K
 Note that the univS is positive. This indicates that the process
is spontaneous under standard conditions.
The Third Law of Thermodynamics
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Entropy is related to the degree of randomness of a
substance.
Entropy is directly proportional to the absolute
temperature.
Cooling the system decreases the disorder.
At a very low temperature, the disorder decreases to 0
(i.e., 0 J/(K mole) value for S).
The most ordered arrangement of any substance is a
perfect crystal!
The Third Law of Thermodynamics
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The Third Law - the entropy of any perfect crystal is 0 J
/(K mole) at 0 K (absolute 0!)
Due to the Third Law, we are able to calculate absolute
entropy values.
The Gibbs Energy of the System
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For a spontaneous process
univS = sysS + surrS > 0 J/(K mole)
For chemical reactions
sysH = rH
sysS = rS
surrS = -rH/ T
univS = rS -rH/ T
The Gibbs Energy
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TunivS = TrS - rH
Or
-TunivS = -TrS + rH
If we look at the criteria for spontaneity
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We note that the calculation of TunivS requires two system
parameters
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-TunivS < 0  spontaneous process
-TunivS > 0  non-spontaneous process
rS
rH
Why don’t we define a system parameter that allows us
to determine if a given process will be spontaneous?
The Definition of the Gibbs Energy
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The Gibbs energy of the system
G = H – TS
For a spontaneous process
sysG = Gf – G i
Gf = the Gibbs energy of the final state
Gi = the Gibbs energy of the initial state
 We note the following relationships that exist between
the Gibbs Energy and the spontaneity of a process.
sysG < 0 - spontaneous process
sysG > 0 - non-spontaneous process (note that this process
would be spontaneous in the reverse direction)
sysG = 0 - system is in equilibrium
Note that these are the Gibbs energies of the system
under non-standard conditions
Applications of the Gibbs Energy
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The Gibbs energy is used to determine the spontaneous
direction of a process.
Two contributions to the Gibbs energy change (G)
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Entropy (S)
Enthalpy (H)
G = H - TS
H
+
+
-
S
+
+
-
G
< 0 at high temperatures
> 0 at all temperatures
< 0 at all temperatures
< 0 at low temperatures
Spontaneity and Temperature
Standard Gibbs Energy Changes
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The Gibbs energy change for a chemical reaction?
Combustion of methane.
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
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Define
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rG =  np fG (products) -  nr fG (reactants)
fG = the formation Gibbs energy of the substance
The Gibbs Energy Change (cont’d)
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For the methane combustion reaction
1 CH4(g) + 2 O2(g)  1 CO2(g) + 2 H2O(l)
rG =  np fG (products) -  nr fG (reactants)
= 2 fG [H2O(l)] + 1 fG [CO2(g)] - (7/2 fG [O2(g)] + 1 fG
[CH4(g)] )
fG (elements) = 0 kJ / mole  fG [O2 (g)] = 0
kJ/mole.
 Use tabulated values of the Gibbs formation energies to
calculate the Gibbs energy changes for chemical reactions.
Example – for the methane combustion reaction above
From appendix C in the textbook
fG [CO2 (g)] = -394.5 kJ/mole
fG [H2O (l)] = -237.2 kJ/mole
fG [CH4 (g)] = -50.7 kJ/mole
rG = 2 mole x –237.2 kJ/mole + 1 mole x –393.5
kJ/mole - 1 mole x (-50.7 kJ/mole)
= -817.2 kJ - note since the Gibbs energy change is < 0
(under standard conditions), the reaction is spontaneous in
the forward direction under standard conditions.
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Gibbs Energies and Equilibrium Constants
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rG < 0 - spontaneous under standard conditions
rG > 0 - non-spontaneous under standard conditions
The Reaction Quotient
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Relationship between QJ and KJ
Q < K - reaction moves in the forward direction
Q > K - reaction moves in the reverse direction
Q = K - reaction is at equilibrium
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rG° refers to standard conditions only!
For non-standard conditions - rG
rG < 0 - reaction moves in the forward direction
rG > 0 - reaction moves in the reverse direction
rG = 0 - reaction is at equilibrium
Relating K to rG
rG = rG +RT ln Q
rG = 0  system is at equilibrium
rG = -RT ln Qeq
rG = -RT ln K
Gaseous reaction - K = Kp
Solution reaction - K = Kc
Example – Calculate the equilibrium constant for the
reaction
½ N2 (g) + 3/2 N2 (g)  NH3 (g)
given that the enthalpy and entropy changes at 298 K are –
46.2 kJ and –99.2 J/K, respectively.
Solution
Calculate the Gibbs energy change at 298 K
rG = rH - TrS
= -46.2 kJ – (-.0992 J/K x 298 K) = -16.6 kJ
rG = -RT ln Kp
-16.6 x 103 J = 8.314 J/K x 298 K x ln Kp
ln Kp = 6.70  Kp = e 6.70 = 812
Phase Equilibria
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At the transition (phase-change) temperature only - trG
= 0 kJ
tr = transition type (melting, vapourization, etc.)
trS = trH / Ttr
Example – Calculate the entropy of fusion of ethanol given
that the enthalpy of fusion is 6.90 kJ/mole at -117C.
Note  the process we are dealing with is
C2H5OH (s)  C2H5OH (l)
Tfus = -117 C = 156 K
 fusS = fusH / Tfus = (6.90 kJ/mole) / (156 K) = 0.0442
kJ/(K mole) = 44.2 J/(K mole)
As expected, there is an increase in the entropy of the system as
we melt the solid (i.e., a liquid is more disordered than a solid).