Download Spontaneous Processes Thermodynamics vs. Kinetics

Spontaneous Processes
Thermodynamics vs. Kinetics
• In thermodynamics, a spontaneous process
is one that proceeds in a given direction
without outside intervention.
• A nonspontaneous process only occurs for
as long as energy is continually added to the
system.
• examples
 rusting nails
 inflating tires
 exploding balloons!
2 H2 + O2  2 H2O
spontaneous
2 H2O  2 H2 + O2
(non-spontaneous)
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Tro: Chemistry: A Molecular Approach, 2/e
Entropy
Thermodynamics
• If you flip a coin 4 times, which outcome
would you bet on?





• Entropy (S) is a measure of the distribution
of energy in a system at a specific
temperature.
all heads
all heads
1) HHHT
2) HHTH
3) HTHH
1) HHTT
4) THTH
2) TTHH
5) HTTH
3) HTHT
6) THHT
all tails
all tails
all tails
• energy distribution and the arrangement of
matter are described by microstates
4 heads  1 out of 16 poss. (1 microstate)
3 heads  4 out of 16 poss. (4 microstates)
2 heads  6 out of 16 poss. (6 microstates)
1 head  4 out of 16 poss. (4 microstates)
all tails  1 out of 16 poss. (1 microstate)
4) THHH
all heads
• The second law of thermodynamics states
that the total entropy of the universe
increases in any spontaneous process.
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Entropy and the 2nd Law
• In a 4-flip coin toss, you should bet on ½
head/½ tails
 If you repeat this experiment, you aren’t likely to
get the same microstate (HTHT, for example) but
you are likely to get one of the equivalent
microstates (HTTH, HHTT, TTHH, THTH, etc.)
• Think about your room
 Does it tend towards messy or orderly?
 Does it always end up in the same mess? Or a
different, equivalent mess?
• The 2nd Law says processes move toward
situations with the most microstates
Thermodynamic Entropy
• Entropy is also studied on the macroscopic scale
• Consider an isothermal process (a process that
takes place at a constant temperature).
Statistical Entropy
• Entropy is related to the number of
microstates by the following equation:
S = k ln(W)
 S is entropy
 W is the number of microstates
 k is the Boltzmann constant (k = 1.38 x 10-23 J/K)
• This equation indicates that entropy increases
as the number microstates increases.
• This is entropy on the microscopic scale
Trends (read on your own)
• Ssolid < Sliquid << Sgas
 H2O(g) is 188.8 J/(mol•K) and H2O(l) is 69.9 J/(mol•K)
S = qrev/T
 qrev means the flow of heat as the process is carried out
reversibly in very small steps.
Suniv = Ssys + Ssurr
• Spontaneous processes must result in an
increase in the entropy of the universe.
(2nd Law of Thermo: Suniv > 0)
• Entropy is the currency of the universe; it is the
price that is paid for anything to happen
Other Trends (read on your own)
• Entropy increases as the
complexity of molecular structure
increases.
• CH4 < CH3CH3 < CH3CH2CH3 <
CH3(CH2)2CH3
• also high MM > low MM
 as MM  there are more electrons,
protons, etc. to store energy
Free Energy
• Free energy (G) is an indication of the
energy available to do useful work.
• Free energy change (G) is the change in
free energy of a process.
 For spontaneous processes at a constant
temperature and pressure G < 0.
Gsys = Hsys  TSsys
For G you only need to consider the system
- not the system + surroundings !
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Gsys = Hsys  TSsys
Prove it!
• Since Hsys = qsys (at constant P) =  qsurr
• and
Then
Hsys = ( qsurr /T)T =  T Ssurr
Plug into:
Gsys = Hsys  TSsys
Gsys =  T Ssurr  TSsys
Gsys =  T (Ssurr Ssys )
or Gsys =  T (Suniverse )
If Suniverse>0, then Gsys <0
(2nd Law!!)
Third Law of Thermodynamics
• Third Law of Thermodynamics - the entropy of a
perfect crystal is zero at absolute zero.
 S is explicitly known (=0) at 0 K, S values at other
temps can be calculated.
• Absolute entropy is the entropy change of a
substance taken from S = 0 (at T = 0 K) to some
other temperature.
• Standard molar entropy (So) is the absolute
entropy of 1 mole of a substance in its standard
state.
Calculation of Free Energies for
Chemical Reactions
(read and practice on your own)
G0rxn
=
nG0f, products
-
mG0f, reactants
The free energy for a chemical reaction
indicates the maximum amount of energy that
is free to do useful work… only at 298K!!
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
50.7
2(0)
394.4 2(228.6)
kJ/mol kJ/mol kJ/mol
kJ/mol
Calculating Entropy
Changes (read on your own)
• Entropy change for a reaction (Srxn) is
Srxn = nSoproducts - mSoreactants
where n and m are the coefficients of the
products and reactants in the balanced
equation.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
186.3 2(205.1) 213.7
2(188.8)
J/K mol J/ K mol J/K mol J/K mol
Srxn =  nSoproducts -  mSoreactants
= 5.2 J/K
G and temperature
(read and practice on your own)
• Go can only be used at 25 oC (298K)
• Horxn and Sorxn are relatively
independent of temperature
• So if you need a Grxn at 200 oC, then
look up Horxn and Sorxn
GTrxn  Horxn  TSorxn
Grxn =  Goproducts -  Goreactants
= 801 kJ
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G and temperature
(read and practice on your own)
G and temperature
(read and practice on your own)
The beauty of GTrxn is that you can now
determine whether a reaction is
spontaneous at any reasonable temp!
4 CuO(s)  2Cu2O(s) + O2(g)
GTrxn
 0 when first spontaneous….
Solve: 0 = Horxn  TSorxn
Horxn / Sorxn
For this reaction: Horxn kJ
Sorxn  J/K
(from Appendix)
Or: kJ  kJ/K)

Or conversely, at what temperature a
reaction will become spontaneous!
Example:
4 CuO  2Cu2O + O2
At what temperature will this
reaction become spontaneous?
A biochemical use for thermo:
• Mammalian metabolism:
ATP + H2O  ADP + H3PO4 G = 31kJ
36 ADP + 36 H3PO4 + 6 O2 + C6H12O6 
36 ATP + 6 CO2 + 42 H2O
Adenosine
triphosphate
(ATP)
energy storage
Coupled reactions
• Mammalian metabolism: necessary
reactions that are non-spontaneous
are made spontaneous by “coupling”
them with ATP
e.g. the
production
of glutamine, an
essential amino
acid
Bioenergetics
The problem:
glutamic acid + NH3  glutamine +
H2O
G = +kJ
non-spontaneous
But…
ATP + H2O  ADP + H3PO4 G = 31kJ
So, if these two systems were coupled
somehow…
Bioenergetics
glutamic acid + NH3  glutamine + H2O G = +kJ
+
ATP + H2O  ADP + H3PO4
G = 31 kJ
G = 17 kJ
glutamic acid + ATP + NH3 
ADP + H3PO4 + glutamine
This coupling is how many biochemical
reactions proceed. It happens in the confines
of an enzyme or protein.
It is an example of Hess’ Law.
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Gibbs Free Energy
• in our oxygen rich atmosphere, many
elements are most stable in their
highest oxidation state
Form
PO43P4
G°f
(kJ/mol)
-1019
-12
Form
NO2
N2
G°f
(kJ/mol)
51.3
Form
SO2
G°f
(kJ/mol)
-300
0.0
SO3
-371
NO3-
-108.7
H2SO4
-745
NO2NH3
-37.2
-16.5
SO42-
-745
H2S
-33
S8
0.0
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