Chapter 19 – Principles of
Reactivity: Entropy and
Free Energy
Objectives:
1) Describe terms: entropy and
spontaneity.
2) Predict whether a process will be
spontaneous.
3) Describe: free energy.
4) Describe the relationship between
DG, K, and product favorability.
Thermodynamics
• Thermodynamics is _______________
____________________.
• First Law of Thermodynamics
– The law of conservation of energy: ______
________________________________.
DE = q + w
– The change in internal energy of a system
is the sum of the heat transferred to or
from the system and the work done on or
by the system.
Spontaneous Change
• Chemical changes, physical
changes
• Spontaneous change: occurs
_____________________.
It leads to ____________.
• Example: heat transfers
spontaneously from a hotter
object to a cooler object.
• ____________ is reached in
product-favored and in
reactant-favored processes.
Spontaneous Chemical Reactions
2 H2 + O2  2 H2O
CH4 + 2 O2  CO2 + 2 H2O
2 Na + Cl2  2 NaCl
HCl + NaOH  NaCl + H2O
• Common feature: _____________
• But many processes are ____________
and spontaneous.
• H2 + I2
2 HI (g) _______________
can be approached from either
direction.
Spontaneous Processes
Spontaneous Processes
• Dissolving NH4NO3 in water: DH = +25.7 KJ/mol
• Expansion of a gas into a vacuum: energy
neutral, heat is neither evolved nor required.
• Phase changes: melting of ice requires ~ 6
kJ/mol; but only occurs if T > 0oC.
– ______________ determines whether a process is
spontaneous.
• Heat transfer: The T of a cold substance in a
warm environment will rise until the substance
reaches the ambient T.
– The required heat comes from the _____________.
Entropy
• To predict whether a process will be
spontaneous.
• Entropy, S is a thermodynamic function
– State function: a quantity whose value is determined
only by the initial and final states of a system.
• Second Law of Thermodynamics
– _______________________________________
_____________________________________.
– _______________________________________
_____________________________________.
Dispersal of Energy
• By statistical analysis:
• Energy is distributed of a number of particles
• Most often case is when energy is distributed
over all particles and to a large number of
states.
• As the number of particles and the number of
energy levels grows, one arrangement turns
out to be vastly more probable than all
others.
Dispersal of Energy
• Dispersal of __________ often
contributes to energy dispersal.
Boltzmann Equation
• Ludwig Boltzmann (1844-1906)
• Look at the distribution of energy over
different energy states as a way to calculate
____________.
S = k log W
• K – Boltzmann constant
• W – represent the number of different ways
that the energy can be distributed over the
available energy levels.
• A maximum entropy will be achieved at
_________________ , a state in which W
has the maximum value.
Matter and Energy Dispersal
Matter and Energy Dispersal
Summary: Matter and Energy Dispersal
• A final state of a system can be more probable
than the initial state if:
– The atoms and molecules can be more
____________ and/or
– ___________ can be dispersed over a greater
number of atoms and molecules.
• If energy and matter are both dispersed in a
process, it is _______________.
• If only matter is dispersed, then quantitative
information is needed to decide whether the
process is spontaneous.
• If energy is not dispersed after a process
occurs, then that process will ____________
_____________________.
Entropy
• Entropy is used to __________________
___________ resulting from dispersal of
energy and matter. The greater the _______
in a system, the greater the value of S.
• Third Law of Thermodynamics
• There is no disorder in a perfect crystal at
0K, S=0.
• The entropy of a substance at any T can be
obtained by measuring the heat required to
raise the T from 0K, where the conversion
must be carried by a reversible process (very
slow addition of heat in small amounts).
Entropy
• The entropy of a substance at any T can be
obtained by measuring the ________ required
to raise the T from 0K, where the conversion
must be carried by a reversible process (very
slow addition of heat in small amounts).
• The entropy added by each incremental change
is: DS =
• Adding the entropy changes gives the total
entropy.
• All substances have ___________ entropy
values at temperatures above 0K.
Standard Molar Entropy Values
Standard Molar Entropy Values
Thermodynamics
• First Law: The total energy of the universe
is a constant.
• Second Law: The total entropy of the
universe is always increasing.
• Third Law: The entropy of a pure, perfectly
formed crystalline substance at 0K is zero.
- A local decrease in entropy (the assembly of
large molecules) is offset by an increase in
entropy in the rest of the universe -.
Standard Entropy
• So, is the entropy gained by converting it
from a perfect crystal at 0K to standard
state conditions (1 bar, 1 molal solution).
• Units: J/Kmol
• Entropies of gases are ____________than
those for liquids, entropies of liquids are
____________ than those for solids.
• Larger molecules have a _________ entropy
than smaller molecules, molecules with more
complex structures have ________entropies
than simpler molecules.
Entropy
• The entropy of
liquid water is
___________ than
the entropy of solid
water (ice) at 0˚ C.
S˚(H2O sol) < S˚(H2O liq)
Entropy
Entropies of ionic solids depend on
___________________________.
So (J/K•mol)
Mg2+ & O2-
MgO
26.9
NaF
51.5
Na+ & F-
The larger coulombic attraction on MgO than NaF leads to a lower entropy.
Which substance has the higher
entropy, why?
• O2 (g) or 03 (g)
• SnCl4 (l) or SnCl4 (g)
Arrange the substances in order of increasing entropy.
Assume 1 mole of each at standard conditions.
HCOOH(l)
CO2(g)
Al(s)
CH3COOH(l)
Predict whether S for each reaction would be greater
than zero, less than zero, or too close to zero to decide.
CO(g) + 3 H2(g)  CH4(g) + H2O(g)
2 H2O(l)  2 H2(g) + O2(g)
I2(g) + Cl2(g)  2 ICl(g)
Entropy Change
S increases
slightly with T
S increases a
large amount
with phase
changes
Entropy Change
• Entropy usually increases when a pure
liquid or solid ______________ in a
solvent.
• Entropy of a substance ____________
with temperature.
Entropy Change
• The entropy change is the sum of the
entropies of the products minus the sum
of the entropies of reactants:
DS0system = S S0
0
–
S
S
(products)
(reactants)
You will find DSo values in the Appendix L of your book.
Calculate the standard entropy changes for the
evaporation of 1.0 mol of liquid ethanol to ethanol vapor.
C2H5OH(l)  C2H5OH(g)
Calculate the standard entropy change for
forming 2.0 mol of NH3(g) from N2(g) and H2(g)
N2(g) + 3 H2(g)  2 NH3 (g)
Using standard absolute entropies at 298K, calculate
the entropy change for the system when 2.35 moles of
NO(g) react at standard conditions.
2 NO(g) + O2(g)  2 NO2(g)
Calculate the standard entropy change for the
oxidation of ethanol vapor (CH2H5OH (g)).
Entropy in the Universe
DS0univ = DS0sys + DS0surr
• 2nd Law of Thermodynamics: DSuniv is
positive for a spontaneous process.
• For a nonspontaneous process:
DS0univ < 0 (negative)
• If DSuniv = 0 the system is at equilibrium.
• Calculate first the DS0sys, then DS0surr.
DS0surr = qsurr/T = -DH0sys/T
DH0sys = S H0 (products) – S H0 (reactants)
Show that DS0univ is positive (>0) for
dissolving NaCl in water
DS0univ = DS0sys + DS0surr
1) Determine DS0sys
2) Determine DS0surr
NaCl(s) NaCl (aq)
Show that DS0univ is positive (>0) for
dissolving NaCl in water
Predicting whether a Process will
be Spontaneous – Table 19.2
Based on the values of DH0sys and DS0sys there are 4 types:
1) DH0sys < 0 Exothermic & DS0sys > 0 Less order
DS0univ > 0 ________________ under all conditions.
2) DH0sys < 0 Exothermic & DS0sys < 0 More order Depends on values,
more favorable at __________ temperatures.
3) DH0sys > 0 Endothermic & DS0sys > 0 Less order Depends on values,
more favorable at __________ temperatures.
4) DH0sys > 0 Endothermic & DS0sys < 0 More order
DS0univ < 0 ___________________ under any conditions.
Remember that –∆H˚sys is proportional to ∆S˚surr
An exothermic process has ∆S˚surr > 0.
Classify the following as one of
the four types of Table 19.2
CH4 (g) + 2 O2 (g)
2 H2O (l) + CO2 (g)
2 FeO3(s) + 3 C (graphite)
4 Fe(s) + 3 CO2 (g)
DH0 (kJ)
-890
+467
DS0 (J/K)
-242.8
+560.7
Calculate the entropy change of the UNIVERSE when
1.890 moles of CO2(g) react under standard conditions
at 298.15 K.
Consider the reaction
6 CO2(g) + 6 H2O(l)  C6H12O6 + 6O2(g)
for which DHo = 2801 kJ and DSo = -259.0 J/K at 298.15 K.
• Is this reaction reactant or product favored under standard conditions?
Gibbs Free Energy
DSuniv = DSsurr + DSsys
DSsurr = -DHsys/T
DSuniv = -DHsys/T + DSsys
Multiply equation by –T
-T DSuniv = DHsys –TDSsys
J. Willard Gibbs (1839-1903)
DGsys = -T DSuniv
DGsys = DHsys –TDSsys
DGsys < 0, a reaction is ____________
DGsys = 0, a reaction is _____________
DGsys > 0, the reaction is ____________
Gibbs Free Energy and
Spontaneity
• J. Willard Gibbs (1839-1903)
• Gibbs free energy, G, “free energy”, a thermodynamic
function associated with the ________________.
G = H –TS
H- Enthalpy
T- Kelvin temperature
S- Entropy
• Changes during a process: DG
• Use to determine whether a reaction is __________.
 DG is ___________related to the value of the
_____________________________ , and hence to
product favorability.
“Free” Energy
DG = w max
• The free energy represents the maximum energy
____________________________.
Example: C(graphite) + 2 H2 (g)
CH4 (g)
DH0rx = -74.9 kJ; DS0rx = -80.7 J/K
DG0rx = DH0 – TDS0
= -74.9 kJ – (298)(-80.7)/1000 kJ
= -74.9 kJ + 24.05 kJ
DG0rx = - 50.85 kJ
• Some of the energy liberated by the reaction is needed
to “order” the system. The energy left is energy
available energy to do_________, “free” energy.
 DG < 0, the reaction is _______________.
Calculate DGo for the reaction below at 25.0 C.
P4(s) + 6 H2O(l) → 4 H3PO4(l)
Species D H (kJ/mol) S (J/K·mol)
P4(s)
0
22.80
H2O(l) -285.8
69.95
H3PO4(l) -1279.0
110.5
f
f
DG0rx = DH0 – TDS0
Standard Molar Free Energy of
Formation
• The standard free energy of formation of a compound, DG0f, is
the free energy change when forming __________of the
compound from the __________________, with products and
reactants in their __________________.
• Then, DG0f of an element in its standard states is _________.
Gibbs Free Energy
DG0rxn is the increase or decrease in
free energy as the reactants in their
standard states are converted
completely to the products in their
standard states.
* Complete reaction is not always
________________.
* Reactions reach an _____________.
DG0system = S G0 (products) – S G0 (reactants)
Calculating DG0rxn from DG0f
 DG0system = S G0 (products) – S G0 (reactants)
Calculate the standard free energy change for the
oxidation of 1.0 mol of SO2 (g) to form SO3 (g).
DGf0 (kJ/)
SO2(g) -300.13
SO3(g) -371.04
DG0system =
Free Energy and Temperature
• G = H – TS
• G is a function of T, DG will change as T
changes.
• Entropy-favored and enthalpy-disfavored
• Entropy-disfavored and enthalpy-favored
Changes in DG0 with T
Consider the reaction below. What is DG0 at 341.4 K and will
this reaction be product-favored spontaneously at this T?
CaCO3(s)  CaO(s) + CO2(g)
Thermodynamic values:
DHf0 (kJ/mol) S0 (kJ/Kmol)
CaCO3(s) -1206.9
+0.0929
CaO(s)
-635.1
+ 0.0398
CO2(g)
-393.5
+ 0.2136
Estimate the temperature required to
decompose CaSO4(s) into CaO(s) and SO3(g).
CaSO4(s)
CaO(s) + SO3(g)
DH0sys = S H0 (products) – S H0 (reactants)
DH0sys = S S0 (products) – S S0 (reactants)
Thermodynamic values:
DHf0 (kJ/mol) S0 (J/Kmol)
CaSO4(s) -1434.52
+106.50
CaO(s)
-635.09
+ 38.20
SO3(g)
-395.77
+ 256.77
For the reaction: 2H2O(l)  2H2(g) + O2(g)
DGo = 460.8 kJ and DHo = 571.6 kJ at 339 K and 1 atm.
• This reaction is (reactant,product) _____________
favored under standard conditions at 339 K.
• The entropy change for the reaction of 2.44 moles of
H2O(l) at this temperature would be _________J/K.
 DGorxn = DHorxn - DT Sorxn
DSo = (DHo - DGo)/T
DG0, K, and Product Favorability
• Large K – ____________ favored
• Small K – ____________favored
• At any point along the
reaction, the reactants are
not under standard
conditions.
• To calculate DG at these
points:
DG = DG0 + RT ln Q
R – Universal gas constant
T - Temperature (kelvins)
Q - Reaction quotient
DG0, K, and Product Favorability
DG = DG0 + RT ln Q
For a A + b B
cC+dD
Q = [C]c [D]d
[A]a [B]b
DG of a mixture of reactants and products is
determined by DG0 and Q.
When DG is _____________ (“descending”) the
reaction is spontaneous . At ______________
(no more change in concentrations), DG = 0.
0 = DG0 + RT ln K (at equilibrium)
DG0 to be negative, K must be larger
DG0 = - RT ln K For
than 1 and the reation is product favored.
Summary DG0 and K
• The free energy at equilibrium is ________ than the free
energy of the pure reactants and of the pure products.
DG0 rxn can be calculated from:



•
DG0rxn = S G0 (products) – S G0 (reactants)
DGorxn = DHorxn - DT Sorxn
DGorxn = - RT ln K
DGrxn describes the direction in which a reaction proceeds
to reach ___________, it can be calculated from:
DGrxn = DG0rxn + RT ln Q
– When DGrxn < 0, Q < K, reaction proceeds spontaneously to convert
______________________ until equilibrium is reached.
– When DGrxn > 0, Q > K, reaction proceeds spontaneously to convert
______________________ until equilibrium is reached.
– When DGrxn = 0 , Q = K, reaction is ___________________.
The formation constant for [Ag(NH3)2]+ is
1.6 x107. Calculate DG0 for the reaction below.
Ag+ (aq) + 2 NH3 (aq)  [Ag(NH3)2]+ (aq)
DG0 = -RTlnK
The reaction below has a DG0 = -16.37 kJ/mol.
Calculate the equilibrium constant.
1/2 N2 (g) + 3/2 H2 (g)  NH3 (g)
DG0rxn = DG0f NH3 (g)
DG0 = -RTlnK
The value of Ksp for AgCl (s) at 25oC is 1.8 x 10-10.
Determine DGo for the process:
Ag+ (aq) + Cl- (aq)  AgCl (s) at 25oC.
The standard free energy change for a chemical
reaction is -18.3 kJ/mole. What is the equilibrium
constant for the reaction at 87 C? (R = 8.314 J/K·mol)
End of Chapter
• Go over all the contents of your
textbook.
• Practice with examples and with
problems at the end of the chapter.
• Practice with OWL tutor.
• Practice with the quiz on CD of
Chemistry Now.
• Work on your OWL assignment for
Chapter 19.