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Chapter 17
Thermodynamics:
Spontaneity, Entropy
& Free Energy
Hill, Petrucci, McCreary & Perry 4th Ed.
Thermodynamics & Equilibrium
First Law of Thermodynamics
∆u = q + w
Change in the
Internal Energy
of a System
=
Heat absorbed
or lost to a
system
+
Work done on or
by a system on its
surroundings.
The internal energy,"u", is a "state function". This
means that it is a property that depends only on the
present state (T, P, phase) of the system.
Another State Function: Enthalpy
The “Heat Content” of a System
∆H = Hfinal - Hinitial
In practice only ∆H's are known!
o
∆H =
Σn∆Hf (products)
o
Σm∆Hf (reactants)
o
-
The small zeros "o" indicate the system is at "Standard State".
"Standard State"
The system state (phase)
at 25 C; 1 atm Pressure.
Note: System phase may not be the normal phase!
o
o
∆Hf H2O(l) = ∆Hf H2O(g)
Spontaneous Processes and Entropy
•A spontaneous process is a chemical or physical
change that occurs without outside agency.
•Entropy – A measure of a system’s disorder or
degree of randomness.
∆S = S final - S initial
Second Law of Thermodynamics - For any
spontaneous process the total Entropy of the
system and its surroundings always increases.
Entropy is created when heat flows into
a System
∆S =
Entropy
Created
Entropy created
cannot usually be
measured
A mathematical way to define the
Second Law of Thermodynamics
∆S >
q
T
+
q
T
For a change at equilibrium:
∆S =
Kelvin
Temp.
At constant T & P
∆H
>
T
q
T
heat
∆H
=
T
Gibb’s Free Energy - ∆G
∆S >
rearranging:
∆H
T
For a spontaneous process at
constant temperature and pressure.
∆H ∆S < 0
T
A negative quantity!
Rearranging again:
∆H
- T∆S < 0
Defining a quantity!
∆H
- T∆S
This quantity is the energy
available to a system to
do useful work.
∆G
The "Gibb's Free Energy"
or just the "Free Energy".
Quantifying Spontaneity
∆H – T∆S / ∆G
∆G is negative Y Process is spontaneous
∆G is zero Y Process is at equilibrium
∆G is positive Y Process is non-spontaneous
In Practice:
∆G# -10 kJ/mole Y complete reaction
∆G $ +10 kJ/mole Y no reaction
∆G = K10 kJ/mole Y equilibrium mixture
Thermodynamic Quantities
o
∆Hf
Standard Enthalpy of Formation
The Heat liberated (or consumed) when forming one
mole of a substance from the elements.
o
∆Gf
Standard Free Energy of Formation
Defined analogously for formation of one mole of a
substance from the elements. See Appendix C1 p A13.
o
∆G =
Σn∆Gf (products)
o
-
Σm∆Gf (reactants)
o
The Third Law of Thermodynamics
A perfectly crystalline substance at Absolute
Zero (0 oK) has an entropy of zero.
So
Standard Entropy
A measure of the disorder of one mole of a substance at
25 C and 1 atm pressure. See Appendix C1 p A13.
For a chemical reaction:
o
∆S =
ΣnS(products)
o
Note: these are absolute
entropies.
-
ΣmS(reactants)
o
A Rule of Thumb in Predicting Entropies
S0gases >>
S0liquids >
S0solids
Processes producing gases tend to have large,
positive entropy changes.
Commonly, not always, ∆S is positive when
gases are formed.
∆S is positive when solids go to liquids or when
solids go into solution
Predict the Sign of Then Calculate ∆S
Predicted
Sign of ∆S
Reaction:
CaCO3(s)
Remarks
Calc. ∆S
CaO(s) + CO2(g)
CS2(g)
CS2(l)
2 Hg(l) + O2(g)
2 HgO(s)
2 Na2O2(s) + 2 H2O(l)
o
∆S =
ΣnS(products)
o
4 NaOH(aq)
+ O2(g)
-
ΣmS(reactants)
o
Calculating an Entropy Change
2 Hg(l) + O2(g)
o
∆S =
ΣnS(products)
o
o
o
2 HgO(s)
-
ΣmS(reactants)
o
o
o
∆S = [2 SHgO(s)] - [2 SHg(l) + SO2(g) ]
o
∆S = [ 2mol(70.29 J/molK)] [2mol(76.02 J/molK) +1mol(205.0 J/molK)]
o
∆S = [ 140.58 J/K] - [152.04 J/K + 205.0 J/K]
o
∆S = [ 140.58 J/K] - [ 357.0 J/K]
o
∆S = -216.5 J/K
Can a Process with Negative Entropy be
Spontaneous?
2 Hg(l) + O2(g)
o
∆Hr =
Σn∆Hf(products)
o
2 HgO(s)
-
Σm∆Hf(reactants)
o
∆Gr = ∆Hr - T∆Sr
o
∆Sr = -216.5 J/K
o
Use tabulated ∆Hf values to calculate the Enthalpy
and use the Gibbs Eqn. to calculate the Free Energy.
Calculate the ∆G for the Reaction Directly
2 Hg(l) + O2(g)
o
∆Gr =
Σn∆Gf(products)
o
o
2 HgO(s)
-
Σm∆Gf(reactants)
o
Use tabulated values of ∆G f to calculate the Free Energy
directly for the reaction.
Thermodynamic Quantities
o
∆Hf
Zero for an element in its standard state*.
*The normal phase of the element at T = 25 C and 1 atm. P.
∆Gfo
Zero for an element in its standard state*.
*These data will NOT be found in Appendix C1 p A13
because they are defined values.
Like Enthalpies, Entropy and Free Energy values for chemical
compounds must ultimately be determined by experiment!
o
∆Hrxn
-
o
T∆Srxn
o
∆Grxn
The Chemical Coupling of Reactions
The rusting of iron is a very spontaneous reaction!
4 Fe(s) + 3 O2(g)
2 Fe2O3(s)
o
The reverse reaction, the refining ∆Grxn
= -1487 kJ
of iron is very non-spontaneous!
2 Fe2O3(s)
4 Fe(s) + 3 O2(g)
How does one enable a nonspontaneous reaction?
o
= +1487 kJ
∆Grxn
Couple it with a spontaneous reaction!
2 CO(g) + O2(g)
Spontaneous
Reaction
+
2 CO2(g)
non-Spontaneous
Reaction
∆Grxn = -514.4 kJ
o
Spontaneous
Reaction Couple
Couple the two reactions so the couple is spontaneous!
Free Energy & the Equilibrium Constant
For Non-Standard States:
∆G = ∆Go + RT ln Q
But at Equilibrium ∆G = zero & Q = K
0 = ∆Go + RT ln K
Rearranging:
∆Go = - RT ln K
∆Go = -2.303 RT log K
or
or
log K =
-∆Go
2.303 RT
What are the Values of the Equilibrium
Constant for various ∆Go?
log K =
-∆Go
2.303 RT
~
K =
[products]
[reactants]
If ∆Go => Spontaneous:
If ∆Go => Non-Spontaneous:
Variation of ∆G with Temperature
From the Gibbs Equation :
o
∆G =
o
∆H
o
- T∆S o
o
But ∆H and ∆S don't vary much
with a change in Temperature.
o
Since ∆H and ∆S o are nearly constants
o
∆G varys mostly with T .
∆GT
o
o
~
= ∆H - T∆S
The above equation is a good approximation
of ∆GT at other temperatures.
Predicting Spontaneity w.r.t Temperature
∆H
o
negative
negative
positive
positive
Color code:
∆S
o
positive
negative
negative
positive
∆G
always negative
negative low Temp
positive high Temp
always positive
positive low Temp
negative high Temp
A Non-spontaneous effect
A Spontaneous effect
Predict the Reaction Spontaneity
N2(g) + 3 H2(g)
∆H is negative
CaCO3(s)
∆H is positive
2 NH3(g)
∆S is negative
∆G = ?
High Temp
Low Temp
CaO(s) + CO2(g)
∆S is positive
∆G = ?
High Temp
Low Temp