Spontaneity of Reaction. 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.
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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.
Chemical thermodynamicsis 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.
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.
System changes state and can be restored by reversing original process.
Ex: Water (s) Water (l)
System changes state and must take a different path to restore to original state.
Ex: CH4 + O2 CO2 + H2O
*Scrambled eggs don’t unscramble*
of Thermodynamics
 The entropy of the universe always increases in a spontaneous process and remains unchanged in an equilibrium process.
S indicates a more ordered state(think: < disorder or  disorder)
Positive (+) S = less ordered state
(think: > disorder or + disorder)
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:
Processes that lead to an Increase in Entropy
Predict whether the entropy change is greater than or less than zero for each of the following processes:
S<0
S>0
S>0
S<0
Predict whether the entropy change of the system in each of the following reactions is positive or negative:
1)S –
2)S+
3)S?
1.) Ag+(aq)+ Cl(aq)AgCl(s)
2.) NH4Cl(s) NH3(g)+ HCl(g)
3.) H2(g) + Br2(g)2HBr(g)
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
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
Spontaneous Physical and Chemical Processes
nonspontaneous
CH4(g) + 2O2(g) CO2(g) + 2H2O (l)DH0 = 890.4 kJ
H+(aq) + OH(aq) H2O (l)DH0 = 56.2 kJ
H2O (s) H2O (l)DH0 = 6.01 kJ
H2O
NH4NO3(s) NH4+(aq) + NO3(aq)DH0 = 25 kJ
Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions at 25 °C
order
disorder
S
H2O (s) H2O (l)
Entropy (S) is a measure of the randomness or disorder of a system…the tendency to spread energy out – disperse energy.
DS = Sf  Si
If the change from initial to final results in an increase in randomness
DS > 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
DS > 0
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.
DSuniv = DSsys + DSsurr > 0
Spontaneous process:
DSuniv = DSsys + DSsurr = 0
Equilibrium process:
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!
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
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.
Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)
So = [2Sº(Al) + 3Sº(H2O)]  [Sº(Al2O3) + 3Sº(H2)]
Al2O3(s) + 3H2(g)2Al(s) + 3H2O(g)
= 180.4 J/K
Predict the sign of ΔSº of the following reaction.
2SO2(g) + O2(g) 2SO3(g)
Entropy decreases, 
Lets’ Calculate
2SO2(g) + O2(g) 2SO3(g)
ΔSº = 187.8 J K1
Predicting spontaneous reactions
(G)
Provides a way to predict the spontaneity of a reaction using a combination of enthalpy and entropy of a reaction.
(sum of standard free energies of formation of products)
minus
(sum of standard free energies of formation of reactants)
Coefficients from equation
the sum of
The values of ΔGºf of elements in their most stable form is 0, just as with enthalpy of formations.
818.0 KJ Spontaneous
835.5 KJ spontaneous
ΔG° = ΔH° – TΔS°(Given on AP Exam)
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 high temp.
+ +
Pos. (nonspontaneous) at low temp
Neg. (spontaneous) at high temp.
ΔS
ΔG
 +
Different sign, not temperature dependent.
+ 
 
Freezing
Same sign, temperature dependent.
+ +
Melting
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 – TS
G = 65 – 298(.030)
= 73.94
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 K1) = +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 K1)
T = 13.65 KJ 0.0758 KJ K1
T = 180 K
Reaction is spontaneous below 180 K, not spontaneous above 180 K
The standard entropy of reaction (DS0rxn) is the entropy change for a reaction carried out at 1 atm and 250C.
aS0(A)
bS0(B)

[
+
]
cS0(C)
dS0(D)
[
+
]
=
aA + bB cC + dD

S
mS0(reactants)
S
nS0(products)
=
DS0
DS0
DS0
DS0
rxn
rxn
rxn
rxn
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g) 2CO2(g)
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
= 427.2 – [395.8 + 205.0] = 173.6 J/K•mol
Entropy Changes in the System (DSsys)
S0(CO) = 197.9 J/K•mol
S0(CO2) = 213.6 J/K•mol
S0(O2) = 205.0 J/K•mol
What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s)
Entropy Changes in the System (DSsys)
When gases are produced (or consumed)
The total number of gas molecules goes down, DS is negative.
Entropy Changes in the Surroundings (DSsurr)
Endothermic Process
DSsurr < 0
Exothermic Process
DSsurr > 0
The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.
ΔGrxn = ΣΔGfo(products) – ΣDGfo(reactants)
DSuniv = DSsys + DSsurr > 0
Spontaneous process:
DSuniv = DSsys + DSsurr = 0
Equilibrium process:
For a constanttemperature process:
Gibbs free energy (G)
DG = DHrxn TDSrxn
DG < 0 The reaction is spontaneous in the forward direction.
DG > 0 The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = 0 The reaction is at equilibrium.

[
+
]
[
+
]
=

mDG0 (reactants)
S
S
=
f
Standard free energy of formation (DGfo) is the freeenergy change that occurs when 1 mole of the compound is formed from its elements in their standard states.
DG0
DG0
rxn
rxn
DGo of any element in its stable form is zero.
f
nDG0 (products)
aDG0 (A)
dDG0 (D)
bDG0 (B)
cDG0 (C)
f
f
f
f
f
The standard freeenergy of reaction (DGorxn ) is the freeenergy change for a reaction when it occurs under standardstate conditions.
mDG0 (reactants)
S
S
=
f
2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l)
DG0
DG0
DG0

[
]
[
+
]
=
rxn
rxn
rxn
[ 12(–394.4) + 6(–237.2) ] – [ 2(124.5) ] = 6405 kJ
=
Is the reaction spontaneous at 25 0C?
12DG0 (CO2)
2DG0 (C6H6)
f
f
6DG0 (H2O)
f
nDG0 (products)
f
What is the standard freeenergy change for the following reaction at 25 0C?
DG0 = 6405 kJ
< 0
spontaneous
Temperature and Spontaneity of Chemical Reactions
DH0 = 177.8 kJ
DS0 = 160.5 J/K
DG0 = DH0 – TDS0
At 25 0C, DG0 = 130.0 kJ
DG0 = 0 at 835 0C
40.79 kJ
=
DS =
373 K
DH
T
Gibbs Free Energy and Phase Transitions
T = ΔH/ ΔS
DG0 = 0
= DH0 – TDS0
= 109 J/K
Gibbs Free Energy and Chemical Equilibrium
Nonstandard Conditions:
DG = DGo + 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
DG = 0
Q = K
0 = DGo + RT lnK
DGo = RT lnK