Transcript Chapter 19: Entropy and Free Energy

Packet 17:
Free Energy and Thermodynamics
last updated: 7/21/2015
CHE 180 Packet 17 - 1
Concept Area I: Terminology
spontaneous
entropy, S
second law of
thermodynamics
Gibbs free energy, G
standard entropy
change, ΔSºrxn
standard Molar
entropies
third law of
thermodynamics
standard free energy
change, ΔGºrxn
standard free energy
of formation, ΔGfº
reversible reaction
irreversible reaction
free energy change,
ΔGrxn
CHE 180 Packet 17 - 2
Concept Area II: Entropy
a. You should understand that entropy is a
measure of disorder.
b. You should recognize that entropy can be
determined experimentally as the heat
change of a reversible process.
c. You should know how to calculate
entropy changes.
d. You should be able to identify common
processes that are entropy-favored.
CHE 180 Packet 17 - 3
Review:
The First Law of Thermodynamics
“DE = q + w: Energy can neither be created
nor destroyed.”
Consequence
New stuff now…
CHE 180 Packet 17 - 4
Entropy, S
S = is used to quantify the extent of disorder
resulting from dispersal of energy and matter
DS = is the change in entropy
So, for a reaction DS = Sproducts – Sreactants
The bigger and more positive the DS, the more
entropy we have and the more likely the
reaction will go as written – spontaneous.
CHE 180 Packet 17 - 5
Dispersal of Matter
Which of these situations
is the most likely after
we open the valve?
Why?
Tro page 645
CHE 180 Packet 17 - 6
Dispersal of Matter
Probabilities can
help us understand
why disorder is
favored over order.
Do mixtures ever
spontaneously separate into the chemicals they
consist of?
CHE 180 Packet 17 - 7
The Second Law of Thermodynamics
“DSuniv = DSsys + DSsurr.
So, in a reversible process, DSuniv = 0;
in an irreversible (spontaneous) process, DSuniv > 0.”
Consequence
CHE 180 Packet 17 - 8
Experimentally determining ΔS
To determine the entropy change
experimentally, heat transfer
must be measured for a
reversible process.
A reversible process is any process
that exists in equilibrium.
(top picture)
In an irreversible process, the
surroundings must be changed to
get the system back to the original
state. (bottom picture)
CHE 180 Packet 17 - 9
Why must we have a reversible process
to experimentally measure ΔS?
Well, remember DE = q + w?
If the work term, w, is zero (as it is in a
reversible process), then ΔE is equal to q,
the heat!
And to determine the entropy change
experimentally, heat transfer must be
measured for a reversible process.
CHE 180 Packet 17 - 10
The Third Law of Thermodynamics
“The entropy of a pure crystalline
solid at 0 K is zero.”
Consequence
A perfect crystal at 0 K
has only one possible
way to arrange its
components.
Tro page 656
CHE 180 Packet 17 - 11
Standard Molar Entropies can be found in
Table 17.2 or in Appendix II-B
Tro page 657
CHE 180 Packet 17 - 12
So is this reaction favored according to entropy?
Let’s calculate a DS!
Does the following chemical reaction
become more or less ordered?
N2(g) + O2(g) → 2 NO(g)
Entropy, S (J/K·mol)
N2(g) 191.6
O2(g) 205.2
NO(g) 210.8
DSrxn = S Sproducts – S Sreactants
CHE 180 Packet 17 - 13
Identifying entropy-favored processes:
Great! We can calculate the ΔS for a reaction.
But, what if we just need to have a general idea
and don’t need a specific value?
What processes are generally entropy favored?
CHE 180 Packet 17 - 14
Entropy Rules of Thumb
What has more entropy?
gases > liquids > solids
larger molecules/atoms > smaller molecules/atoms
complex molecules > simple molecules
higher temperature > lower temperature
larger volume > smaller volume
more moles > fewer moles
Tro page 659
CHE 180 Packet 17 - 15
Predicting Relative Entropy
For the pairs listed below, determine which one
has the higher entropy. Defend your choice.
• 1 mol of SO2(g) or 1 mol of SO3(g)
• 1 mol of CO2(g) or 1 mol of CO2(s)
• 3 mol of O2(g) or 2 mol of O3(g)
• 1 mol of KBr(s) or 1 mol of KBr(aq)
• seawater at 2ºC or seawater at 23ºC
• 1 mol of CF4(g) or 1 mol of CCl4(g)
CHE 180 Packet 17 - 16
Your turn!
Balance the following equations, then predict
the sign of ΔSºrxn if possible. Finally,
calculate its value at 25ºC using the entropy
values in Appendix II-B.
1. NaOH(s) + CO2(g)  Na2CO3(s) + H2O(ℓ)
2. Fe(s) + H2O(g)  Fe2O3(s) + H2(g)
CHE 180 Packet 17 - 17
1.2 NaOH(s) + CO2(g)  Na2CO3(s) + H2O(ℓ)
2.2 Fe(s) +3 H2O(g)  Fe2O3(s) +3 H2(g)
CHE 180 Packet 17 - 18
Concept Area III:
Will a process be spontaneous?
a. You should be able to use entropy and
enthalpy changes to predict whether a
reaction is spontaneous or not.
b. You should understand the roll of
temperature on whether a reaction is
spontaneous or not.
CHE 180 Packet 17 - 19
Quick Review!
How do we decide the extent of a reaction?
Equilibrium! (already covered)
How do we predict how fast a reaction will
go?
Kinetics! (haven’t covered yet)
How do we predict if a reaction will go or
is spontaneous if given enough time?
Thermodynamics! (finish covering now)
CHE 180 Packet 17 - 20
Thermodynamics & Kinetics
We will finish learning how to calculate if a
reaction is spontaneous or not.
Usually to be spontaneous, a reaction must
be thermodynamically favored or
____________.
However, just because a reaction is
spontaneous does not mean it happens
quickly! For that we need kinetics.
CHE 180 Packet 17 - 21
Thermodynamics & Kinetics
Let’s look at the kinetics for this spontaneous
reaction for instance…does it happen very
quickly? _________
C(diamond)  C(graphite)
Tro page 426
DHrxn= –1.8 kJ/mol
CHE 180 Packet 17 - 22
Thermodynamics & Kinetics
Can we convert a nonspontaneous reaction to a
spontaneous one using a catalyst or other means?
Can we speed up a really slow reaction with a
catalyst or other means?
So if something is nonspontaneous , does that
mean it will never
happen?
Tro page 643
CHE 180 Packet 17 - 23
Spontaneous?
We’ve been seeing the word “spontaneous”
an awful lot this chapter. What does it
mean?
CHE 180 Packet 17 - 24
When is a reaction thermodynamically spontaneous?
Type
DHºrxn
1 Exothermic
DHºrxn< 0
2 Exothermic
DHºrxn< 0
3
4
DSºrxn
Less Order
DSºrxn> 0
More Order
DSºrxn< 0
Spontaneous?
Always
DSºuniverse> 0
Depends
more favorable at
lower temperatures
Endothermic Less Order Depends
more favorable at
DHºrxn> 0
DSºrxn> 0
higher temperatures
Endothermic More Order Never
DHºrxn> 0
DSºrxn< 0
DSºuniverse< 0
CHE 180 Packet 17 - 25
Review!
If ΔHrxn = +182 kJ, so is that exo- or endothermic?
CHE 180 Packet 17 - 26
So, is the reaction thermodynamically favored?
For the reaction:
N2(g) + O2(g) → 2 NO(g)
The ΔHrxn = +182 kJ (read off graph on previous
slide), which is not spontaneous for enthalpy.
The DSrxn = +24.8 J/K (we calculated on slide 13),
which is spontaneous for entropy.
So, is this reaction spontaneous overall?
CHE 180 Packet 17 - 27
Thermodynamics 2 Handout
Do you get it? Let’s see!
Calculate ΔH°rxn and ΔSºrxn for part “a” on
the Thermodynamics 2 worksheet.
CHE 180 Packet 17 - 28
Concept Area IV: Gibbs Free Energy
a. You should know how entropy changes, enthalpy
changes and temperature can be used to calculate
the Gibbs free energy for a process.
b. You should understand how a change in
temperature changes the Gibbs free energy of a
process.
c. You should be able to calculate the standard free
energy of formation from standard entropy and
enthalpy values or from standard free energy
values.
CHE 180 Packet 17 - 29
How can we tell if a reaction will go
when “it depends”?
Gibbs Free Energy
DGrxn = DHrxn – TDSrxn
If DGrxn is …
negative, the reaction is spontaneous.
zero, the reaction is at equilibrium.
positive, the reaction is not spontaneous.
CHE 180 Packet 17 - 30
Spotlight on J. Willard Gibbs
A little info on Gibbs:
lived 1839-1903
earned Ph.D. from Yale in 1863, first Ph.D. in
science awarded in U.S.
pretty much founded the science of chemical
thermodynamics and also made large
contributions to equilibria, and electrochemistry
elected to the Hall of Fame of Distinguished
Americans in 1950 (took that long to receive
enough votes!)
CHE 180 Packet 17 - 31
What is “free energy”?
Free energy simply means the energy
available to do work.
It might seem that all energy in a chemical
process is able to do work.
However, if the products are more ordered
than the reactants, some of the energy must
go to ordering the products.
Thus, ΔG is the total energy given off or
needed for a chemical reaction.
CHE 180 Packet 17 - 32
Let’s calculate a DG!
Will the following reaction go
spontaneously at 298 K?
N2(g) + O2(g) → 2 NO(g)
Previously calculated values:
DH
+182 kJ
DS
+24.8 J/K
DGrxn = DHrxn – TDSrxn
CHE 180 Packet 17 - 33
Another DG calculation!
Approximately what temperature will
this reaction proceed spontaneously?
N2(g) + O2(g) → 2 NO(g)
Previously calculated values:
DH
+182 kJ
DS
+24.8 J/K
DGrxn = DHrxn – TDSrxn
Make sense?
CHE 180 Packet 17 - 34
Your turn!
Potassium chlorate, a common
oxidizing agent in fireworks
and match heads, undergoes
a solid-state disproportionation reaction when heated:
4 KClO3(s)  3 KClO4(s) + KCl(s)
Use ΔHfº and Sº values to calculate
ΔGºrxn at 25ºC for this reaction.
CHE 180 Packet 17 - 35
4 KClO3(s)  3 KClO4(s) + KCl(s)
First, calculate ΔHºrxn
at 25ºC
Next, calculate ΔSºrxn
Finally, calculate ΔGºrxn at 25º
CHE 180 Packet 17 - 36
Calculating ΔGºrxn another way…
We can also calculate ΔGºrxn from ΔGºf values.
What are ΔGºf values?
Thus it can be calculated with the following
equation:
CHE 180 Packet 17 - 37
ΔGºf Values
(also found in Appendix II-B in our text)
Tro page 663
CHE 180 Packet 17 - 38
Let’s calculate a DG!
Use ΔGfº values to calculate
ΔGºrxn at 25ºC for this reaction.
N2(g) + O2(g) → 2 NO(g)
ΔGfº (kJ/mol)
N2(g) 0
O2(g) 0
NO(g) 87.6
Δ Gºrxn = S ΔGfº(products) – S ΔGfº(reactants)
Let’s compare the answer to what we got before using
DGrxn=DHrxn–TDSrxn which was 175 kJ.
Are our answers similar?
CHE 180 Packet 17 - 39
Your turn!
Potassium chlorate, a common
oxidizing agent in fireworks
and match heads, undergoes
a solid-state disproportionation reaction when heated:
4 KClO3(s)  3 KClO4(s) + KCl(s)
Use ΔGfº values to calculate
ΔGºrxn at 25ºC for this reaction.
CHE 180 Packet 17 - 40
4 KClO3(s)  3 KClO4(s) + KCl(s)
So, calculate ΔGºrxn
Let’s compare the answer to what we
got before using DGrxn=DHrxn–TDSrxn
which was – 133.1 kJ.
Are our answers similar?
CHE 180 Packet 17 - 41
Concept Area V: Application of
Thermodynamics to Equilibrium
a. You should be able to describe the
relationship between the free energy
change and equilibrium constants.
b. You should be able to calculate
equilibrium constants from DGºrxn.
c. You should be able to calculate DGrxn at
nonstandard conditions.
CHE 180 Packet 17 - 42
Free Energy & Equilibrium
Remember, if:
K > Q (Q/K < 1) – reaction proceeds to the right
K = Q (Q/K = 1) – reaction is at equilibrium
K < Q (Q/K > 1) – reaction proceeds to the left
And,
ΔG < 0 – reaction spontaneous towards the right
ΔG = 0 – reaction is at equilibrium
ΔG > 0 – reaction spontaneous towards the left
What if we take the natural log of Q/K?
Tro page 653
CHE 180 Packet 17 - 43
Free Energy & Equilibrium
Remember, if:
ln Q/K < 0 – reaction proceeds to the right
ln Q/K = 0 – reaction is at equilibrium
ln Q/K > 0 – reaction proceeds to the left
And,
ΔG < 0 – reaction spontaneous towards the right
ΔG = 0 – reaction is at equilibrium
ΔG > 0 – reaction spontaneous towards the left
Notice some similarities?
CHE 180 Packet 17 - 44
Free Energy & Equilibrium
So, we can take this information and
eventually derive the following useful
equations:
ΔG = ΔGº + RT ln Q
ΔGº = – RT ln K
We can rearrange the second equation to:
K = e–(ΔGº/RT)
to solve for an equilibrium constant!
Now, does everyone remember what R is?
R = 8.3145 J/mol·K or R = 0.082057 L·atm/mol·K
CHE 180 Packet 17 - 45
The relationship Between ΔGº and K at 298 K
Significance
200
9×10–36
100
3×10–18
Essentially no forward reaction; reverse
reaction goes to completion
50
2×10–9
10
2×10–2
1
7×10–1
0
1
–1
1.5
– 10
5×101
– 50
6×108
– 100
3×1017
– 200
1×1035
Forward and reverse reactions proceed
to same extent
reverse reaction
K
forward reaction
ΔGº (kJ)
Forward reaction goes to completion;
essentially no reverse reaction
CHE 180 Packet 17 - 46
Let’s calculate a K using ΔG!
What is the equilibrium constant at
25ºC for the following reaction?
N2(g) + O2(g) → 2 NO(g)
ΔGº = – RT ln K
Previously calculated values:
DHºrxn +182 kJ
DSºrxn
+24.8 J/K
DGºrxn +175kJ
rearrange to
CHE 180 Packet 17 - 47
Why not the same?!?!
We just got:
N2(g) + O2(g) → 2 NO(g)
Keq= 2.11×10–31
But, the previous text reported:
½ N2(g) + ½ O2(g) → NO(g) Keq= 4.59×10–16
Why the difference? Or, are they really?
Your mission, for next class or later in this
class, is to prove that these answers are really
not different from each other!
CHE 180 Packet 17 - 48
Worksheet Time, again!
Let’s finish the Thermodynamics II worksheet.
a.
b.
CHE 180 Packet 17 - 49
Worksheet Time, again!
Let’s finish the Thermodynamics II worksheet.
c.
d.
CHE 180 Packet 17 - 50
Free Energy, Equilibrium and Nonstandard States
Consider the following change:
H2O(ℓ)
H2O(g)
ΔGºrxn = +8.59 kJ/mol
If this process is not spontaneous, why does
spilled water spontaneously
evaporate?
What do we do then?
Tro page 666
They are the SAME!
Your mission was to prove that these answers
are really not different from each other!
We got:
N2(g) + O2(g) → 2 NO(g)
But, the previous text reported:
½ N2(g) + ½ O2(g) → NO(g)
K´eq= 2.11×10–31
K´´eq= 4.59×10–16
CHE 180 Packet 17 - 52
The End of Packet 17!
Questions?
CHE 180 Packet 17 - 53