Spontaneity of Reaction
I am a sleepless
Slowfaring eater,
Maker of rust and rot
In your bastioned fastenings,
I am the crumbler:
tomorrow.
– Carl Sandburg
Under
Chapter 16
Key Concepts of Chapter 16
• Identifying
Spontaneous Processes.
• Identifying reversible and irreversible
processes.
• Entropy and its relation to randomness.
• Second Law of Thermodynamics.
• Predicting Entropy Changes of a Process.
• Third Law of Thermodynamics.
• Relate temperature change to entropy change.
• Calculating change in standard entropy.
• Free energy in terms of enthalpy and entropy.
• Relating free energy change to spontaneity.
• Calculating standard free energy change.
• Relationship between free energy and work.
• Calculating free energy Δ under nonstandard
conditions.
From Chapter 8
Chemical thermodynamics is the
study of energy relationships in
chemistry.
The First law of Thermodynamics
- energy cannot be created or
destroyed only converted from one
form to another.
• Enthalpy
– heat transfer between the system and its
surroundings under constant pressure.
– Enthalpy is a guide to whether a reaction is
likely to proceed.
– It is not the only factor that determines
whether a reaction proceeds.
• Occur without outside intervention
• Have a definite direction.
– The reverse process is not spontaneous.
• Temperature has an impact on
spontaneity.
– Ex: Ice melting or forming
– Ex: Hot metal cooling at room temp.
KI (aq) + Pb(NO3)2 (aq)  PbI2(s) + KNO3 (aq)
When mixed  Precipitate forms spontaneously.
*It does not reverse itself and become two clear solutions.
• Reversible:
System changes state and can be restored by
reversing original process.
Ex: Water (s)
Water (l)

• Irreversible:
System changes state and must take a different
path to restore to original state.
Ex: CH4 + O2  CO2 + H2O
• Whenever a system is in equilibrium, the
reaction can go reversibly to reactants or
products (water  water vapor at 100 º C).
• In a Spontaneous process, the path between
reactants and products is irreversible. (Reverse
of spontaneous process is not spontaneous).
*Scrambled eggs don’t unscramble*
- The entropy of the universe
always increases in a spontaneous
process and remains unchanged in
an equilibrium process.
“But ma, it’s not my fault… the universe wants my room like this!”
• A measure of randomness or disorder
• S = entropy in J/K·mole
• Increasing disorder or increasing
randomness is increasing entropy.
• Three types of movement can lead to
an increase in randomness.
Entropy is a state function
• Change in entropy of a system
S= Sfinal- Sinitial
• Depends only on initial and final states, and
not the pathway.
-S indicates a more ordered state
(think: < disorder or - disorder)
Positive (+) S = less ordered state
(think: > disorder or + disorder)
Entropy, S
- a measure of disorder
Ssolid

Sliquid

Sgas
If entropy always increases, how can we account for the fact
that water spontaneously freezes when placed in the freezer?
• Movement of compressor
+
• Evaporation and condensation of refrigerant
+
• Warming of air around container
Net increase in the entropy of the universe
On the AP exam, you will likely be asked to:
1) predict whether a process leads to an increase
in entropy or a decrease in entropy.
2) Determine if ΔS is + or –
3) Determine substances or reactions that have the
highest entropy.
Processes that lead to an Increase in Entropy
1) When a solid melts.
2) When a solid dissolves in solution.
3) When a solid or liquid becomes a gas.
4) When the temperature of a substance increases.
5) When a gaseous reaction produces more
molecules.
6) If no net change in # of gas molecules, can be +
or -, but small.
Predict whether the entropy change is
greater than or less than zero for each
of the following processes:
a) Freezing liquid bromine S<0
b)Evaporating a beaker of
ethanol at room temperature S>0
c) Dissolving sucrose in water S>0
d)Cooling N2 from 80ºC to 20ºC S<0
Predict whether the entropy
change of the system in each of the
following reactions is positive or
negative:
1.) Ag+(aq)+ Cl-(aq)AgCl(s)
1)S –
2.) NH4Cl(s) NH3(g)+ HCl(g) 2)S+
3.) H2(g) + Br2(g)2HBr(g)
3)S?
According to the 2nd law of thermodynamics; the
entropy of the universe always increases.
?
What if the entire senior class assembles in the
auditorium?
Aren’t we decreasing disorder, and therefore
decreasing entropy?
If so, how can the second law of thermodynamics
be true?
If we consider the senior class as the system, the
ΔS of the system would indeed decrease.
ΔS of the system is –
In order for the students to gather, they would:
1) Metabolize food (entropy increase of surroundings)
2) Generate heat (entropy increase of surroundings)
The magnitude of the entropy increase of the
surroundings will
always be greater than the entropy decrease of the
system.
Theoretical values
Suniverse = Ssystem + Ssurroundings
Suniverse = (-10)
+
(+20)
Suniverse = +10
+ means entropy increases
The same can be considered in a chemical
process.
When a piece of metal rusts:
4Fe(s) + O2(g)  2Fe2O3(s)
The entropy of the solid slowly decreases.
Although this is a slow process, it is exothermic, and heat
is released into the surroundings causing an overall
increase in entropy of the universe!
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts above 0 0C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds
spontaneously?
Spontaneous reactions at 25 °C
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H0 = -890.4 kJ
H+ (aq) + OH- (aq)
H2O (l) H0 = -56.2 kJ
H2O (s)
NH4NO3 (s)
H2O (l) H0 = 6.01 kJ
H2O
NH4+(aq) + NO3- (aq) H0 = 25 kJ
TWO Trends in Nature
• Order  Disorder


• High energy  Low energy

The Driving Forces
• Energy Factor (ΔH)
• Randomness
Factor (ΔS) or
more lately the
“dispersal” factor.
Entropy (S) is a measure of the randomness or disorder of a
system…the tendency to spread energy out – disperse energy.
order
disorder
S
S
S = Sf - Si
If the change from initial to final results in an increase in randomness
S > 0
Sf > Si
For any substance, the solid state is more ordered than the liquid
state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
S > 0
First Law of Thermodynamics
Energy can be converted from one form to another but energy
cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process
and remains unchanged in an equilibrium process.
Spontaneous process:
Suniv = Ssys + Ssurr > 0
Equilibrium process:
Suniv = Ssys + Ssurr = 0
The Second Law of Thermodynamics
- The entropy of the universe
increases in a spontaneous process
and remains unchanged in an
equilibrium process.
Suniverse = Ssystem + Ssurroundings
Suniverse > 0 for spontaneous rxn
Suniverse = 0 at equilibrium
In terms of temperature, how would you describe
an object that has an entropy value of 0?
0 K
Perfect solid crystal with no motion
Only Theoretical
It is not possible to reach absolute 0!
Entropy of universe is always increasing!
3rd Law of Thermodynamics
the entropy of a perfect crystalline substance is zero at absolute zero
*Based on 0 entropy as a reference point, and
calculations involving calculus beyond the scope of
this course, data has been tabulated for
Standard Molar Entropies
ΔSº
Pure substances, 1 atm pressure, 298 K
Standard Molar Entropies
ΔSº
1)
Standard molar entropies of elements are not 0 (unlike ΔHºf).
(0 entropy is only theoretical; not really possible)
2)
S.M.E of gases > S.M.E of liquids and solids.
(gases move faster than liquids)
3)
S.M.E. increase with increasing molar mass.
(more potential vibrational freedom with more mass)
4)
S.M.E. increase as the number of atoms in a formula increase.
(same as above)
Calculating the Entropy Change
Sorxn = n So(products) - m So(reactants)
Units for S
S=J/mol•K
Since we are considering ΔS°
J/K are often used because moles are
assumed and cancel in the calculations
when considering standard states.
Calculate the standard entropy change
(Sº) for the following reaction at 298K
Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)
Substance
Al
Al2O3
H2O(g)
H2(g)
Sº(J/mol-K) at 298K
28.32
51.00
188.8
130.58
Sorxn = n So(products) - m So(reactants)
Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)
So = [2Sº(Al) + 3Sº(H2O)] [Sº(Al2O3) + 3Sº(H2)]
 2(28.32 J ) 3(188.8 J )   51.00 J 3(130.58 J ) 
S   





mol

K
mol

K
mol

K
mol

K

 

= 180.4 J/K
Predict the sign of ΔSº of the
following reaction.
2SO2(g) + O2(g) 2SO3(g)
Entropy decreases, -
Lets’ Calculate
Calculate the standard entropy change
(Sº) for the following reaction at 298K
2SO2(g) + O2(g) 2SO3(g)
Substance
SO2(g)
SO3(g)
O2(g)
Sº(J/mol-K) at 298K
248.1
256.7
205.0
ΔSº = -187.8 J K-1
• Spontaneous reactions result in an
increase in entropy in the universe.
• Rx’s that have a large and negative 
tend to occur spontaneously.
• Spontaneity depends on enthalpy,
entropy, and temperature.
Provides a way to predict the spontaneity of a
reaction using a combination of enthalpy and
entropy of a reaction.
If Both T and P are constant, the relationship
between G and spontaneity is:
1) G is (-), forward rxn is
spontaneous.
2) G is 0, rxn is at equilibrium.
3) G is (+) forward rxn is not
spontaneous (requires work)
reverse rxn is spontaneous.
Grxn 
(sum of standard free energies of formation of products)
minus
(sum of standard free energies of formation of reactants)
Grxn  nG f (products)  mG f (reactants)
the sum of
Coefficients from
equation
The values of ΔGºf of elements in their most stable
form is 0, just as with enthalpy of formations.
Calculate the ΔG°rxn for the combustion of
methane at 298K and determine if the reaction is
spontaneous.
Substance
CH4
CO2
H2O(l)
Gº(KJ/mol) at 298K
-50.8
-394.4
-237.2
-818.0 KJ Spontaneous
Calculate the (Gº) for the thermite reaction
(aluminum with iron(III) oxide).
Substance
Al2O3 (s)
Fe2O3 (s)
Gº(KJ/mol) at 298K
-1576.5
-740.98
-835.5 KJ spontaneous
• This equation allows us to determine if a process provides energy to do work. A
spontaneous reaction in the forward direction provides energy for work.
• If not spontaneous, ΔG equals the amount of energy needed to initiate the reaction.
• Allows us to calculate the value of ΔG as temperature changes.
• Gibbs free energy (G°) is a state function defined as:
ΔG° = ΔH° – TΔS°
(Given on AP Exam)
 T is the absolute temperature
 ΔG° = ΔH° – TΔS°  ΔG = ΔH – TΔS (when nonstandard)
 If given a temperature change and asked to determine spontaneity or value
of ΔG, this is the equation you would use.
 Value of ΔG tells us if a reaction is spontaneous.
G = H – T(S)
If we know the conditions of ΔH and ΔS,
we can predict the sign of G.
We will see that:
Two conditions always produce the same result, and
two conditions depend on temperature.
Predicting Sign of ΔG in Relation to Enthalpy and Entropy
ΔH
ΔS
-
+
+
-
+
ΔG
Always negative (spontaneous)
Always positive (nonspontaneous)
-
Neg. (spontaneous) at low temp
+
Pos. (nonspontaneous) at low temp
Pos. (nonspontaneous) at high temp.
Neg. (spontaneous) at high temp.
ΔH
ΔS
-
+
+
-
-
Freezing
ΔG
Different sign, not temperature dependent.
Same sign, temperature dependent.
+
Melting
+
G = H – T(S)
Some reactions are spontaneous because they
give off energy in the form of heat (ΔH < 0).
Others are spontaneous because they lead to an
increase in the disorder of the system (ΔS > 0).
Calculations of ΔH and ΔS can be used to probe
the driving force behind a particular reaction.
Example – The entropy change of the
system is negative for the precipitation
reaction:
Ag+(aq) + Cl-(aq) 
AgCl(s)
Ho = -65 kJ
Since S decreases rather than increases
in this reaction, why is this reaction
spontaneous?
Yes, entropy increases, which goes against the
second law. However, in this case, the entropy
decrease is minimal compared to the magnitude of
change in enthalpy. Therefore, the release of heat
drives the reaction to stability, which is why it is
spontaneous.
Theoretical Values
G = H – TS
G = -65 – 298(-.030)
= -73.94
Problem
For a certain reaction, ΔHº = -13.65 KJ and ΔSº = -75.8 J K-1.
A) What is ΔGº at 298 K?
B) Will increasing or decreasing the temperature make the
reaction become spontaneous? If so, at what temperature
will it become spontaneous?
Given:
ΔHº = -13.65 KJ
ΔSº = -75.8 J /K
T = 298 K
A) What is ΔGº at 298 K?
At 298 K the free energy is
ΔG° = ΔH° – TΔS°
ΔG° = -13.65 KJ – 298(-.0758 KJ K-1) = +8.94 KJ
(Reaction is not spontaneous at 298 K)
B) Will increasing or decreasing the temperature make the
reaction become spontaneous? If so, at what temperature
will it become spontaneous?
Because enthalpy and entropy have the same signs,
spontaneity is indeed temperature dependent.
*Since going from not spontaneous to spontaneous crosses
the point of equilibrium, and ΔG° = 0 at equilibrium, we can
make ΔG° = to 0 to find the temperature at which equilibrium
is crossed.
0 = ΔH° – TΔS°
0 = -13.65 KJ – T(-.0758 KJ K-1)
T = 13.65 KJ
0.0758 KJ K-1
T = 180 K
Reaction is spontaneous below 180 K, not spontaneous above 180 K
Entropy Changes in the System (Ssys)
The standard entropy of reaction (S0rxn) is the entropy change for
a reaction carried out at 1 atm and 250C.
aA + bB
S0rxn =
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
S0rxn =  nS0(products)
-  mS0(reactants)
What is the standard entropy change for the following reaction at
250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
Entropy Changes in the System (Ssys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules than it
consumes, S0 > 0.
•
If the total number of gas molecules diminishes, S0 < 0.
•
If there is no net change in the total number of gas
molecules, then S0 may be positive or negative BUT
S0 will be a small number.
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down, S is negative.
Entropy Changes in the Surroundings (Ssurr)
Exothermic Process
Ssurr > 0
Endothermic Process
Ssurr < 0
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero of temperature.
Gibb’s Free Energy --The energy of a
system that is available to do work.
• ΔGorxn = ΔHorxn – TΔSorxn .
• Hess’ Law also applies:
ΔGrxn = ΣΔGfo(products) – ΣGfo (reactants)
• ΔGfo has a similar definition as ΔHfo
• If ΔGorxn is positive, the rxn is nonspontaneous at that
temperature.
• If ΔGorxn is negative, the rxn is spontaneous at that
temperature.
• If ΔGrxn is zero, the rxn is at equillibrium.
Gibbs Free Energy
Spontaneous process:
Suniv = Ssys + Ssurr > 0
Equilibrium process:
Suniv = Ssys + Ssurr = 0
For a constant-temperature process:
Gibbs free
energy (G)
G = Hrxn -TSrxn
G < 0
The reaction is spontaneous in the forward direction.
G > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
G = 0
The reaction is at equilibrium.
The standard free-energy of reaction (Gorxn ) is the free-energy
change for a reaction when it occurs under standard-state conditions.
aA + bB
cC + dD
G0rxn = [ cG0f (C) + dG0f (D) ] - [ aG0f (A) + bG0f (B) ]
G0rxn =  nG0f (products) -  mG0f (reactants)
Standard free energy of formation
(Gfo) is the free-energy change that
occurs when 1 mole of the compound
is formed from its elements in their
standard states.
Gof of any element in its stable form
is zero.
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
G0rxn =  nG0f (products) -  mG0f (reactants)
G0rxn = [ 12Gf0 (CO2) + 6Gf0 (H2O)] - [ 2Gf0 (C6H6)]
G0rxn = [ 12(–394.4) + 6(–237.2) ] – [ 2(124.5) ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
G0 = -6405 kJ < 0
spontaneous
ΔGfo = ΔHfo - TΔSfo
ΔGf
-?
-?
o
@ high T
@ low T
+?
-?
@ all T
@ all T
ΔHf
+
+
-
o
TΔSf
+
+
o
G = H - TS
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
H0 = 177.8 kJ
S0 = 160.5 J/K
G0 = H0 – TS0
At 25 0C, G0 = 130.0 kJ
G0 = 0 at 835 0C
CaO (s) + CO2 (g)
Gibbs Free Energy and Phase Transitions
T = ΔH/ ΔS
G0 = 0 = H0 – TS0
H2O (l)
H2O (g)
H
40.79 kJ
S =
=
T
373 K
= 109 J/K
Gibbs Free Energy and Chemical Equilibrium
Non-standard Conditions:
G = Go + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
G = 0
Q=K
0 = Go + RT lnK
Go =  RT lnK
o
G
=  RT lnK
Chemistry In Action: The Thermodynamics of a Rubber Band
TS = H - G
High Entropy
Low Entropy