Transcript Thermodynamics

Thermodynamics
Spontaneity, Entropy, and Free
Energy
First Law of Thermodynamics
• Law of Conservation of Energy
– Energy can change forms
– Not “lost”, but changed
– Discuss things like
• How much energy is exchanged?
• Where does the energy go? (calorimeter)
• What form is the energy?
Spontaneous Processes
• A process is spontaneous if it occurs without outside
intervention.
– We discuss the direction of the reaction
– Says nothing of the kinetics or rate
• For example:
– A ball rolls down hill, but never spontaneously rolls uphill.
– Iron exposed to water rusts. Rust does not spontaneously turn
into iron
– A container will fill uniformly with a gas; the gas does not
spontaneously pool at one end.
Spontaneous Processes
• Spontaneous processes are
those that can proceed
without any outside
intervention.
• The gas in vessel B will
spontaneously effuse into
vessel A, but once the gas is
in both vessels, it will not
spontaneously return to
vessel B.
Kinetics
The reaction pathway
Thermodynamics
the initial and final states
nd
2
Law of Thermo
• Entropy in the universe is increasing
• The driving force for spontaneous processes is an
increase in Entropy
– Natural tendency is to go from ordered to disordered
– Take a deck of cards. Throw them into air. When you
put them back, what are the chances they are all in
order?
• But there is a chance, however unlikely.
2nd Law
• Entropy is a function that describes the
number of possible arrangements
– Available to a particular system
– Nature proceeds toward the states that have the
highest probability of existing
– The driving force is “probability”
Let’s Look at a Simple System
• Four atoms of an ideal gas
• Three possible arrangements
• How many ways can each
state be achieved?
Examine All Possibilities (Pg 795)
Possibilities
• The arrangement with two on each side is
most likely to occur
By the ratio of 6:4:1
Probability of finding all the Molecules in the Left Bulb as a
function of the total number of molecules
Unlikely to Occur
1 in 10
2 x 1023
or not likely to occur
But it is possible!
Positional Entropy
• A gas expands into a vacuum
– Because the expanded state has the highest
positional probability or entropy of all the states
available to the gas
• Illustrated by changes of state
– The larger the intermolecular distances, the
more states available
• The more states, the more entropy
Coffee Cup
• Explain on a molecular level how a hot cup of
coffee cools to room temperature
What is the possibility of this whole process going in reverse?
But it is possible! Next time your coffee is cold, just wait for
it to get hot.
Entropy on the Molecular Scale
• Ludwig Boltzmann described the concept of
entropy on the molecular level.
• Temperature is a measure of the average kinetic
energy of the molecules in a sample.
Entropy on the Molecular Scale
• Molecules exhibit several types of motion:
– Translational: Movement of the entire molecule from one
place to another.
– Vibrational: Periodic motion of atoms within a molecule.
– Rotational: Rotation of the molecule on about an axis or
rotation about  bonds.
– All of these are considered microstates of a system.
Entropy on the Molecular Scale
• Each molecule has a specific number of microstates, W,
associated with it.
• Entropy is
S = k lnW
where k is the Boltzmann constant, 1.38  1023 J/K.
Entropy on the Molecular Scale
• The change in entropy for a process, then, is
S = k lnWfinal  k lnWinitial
S = k ln
Wfinal
Winitial
• Entropy increases with the number of
microstates in the system.
Standard Entropies
Larger and more complex molecules have
greater entropies.
Entropy
• Kinetic-molecular view
•
For an ideal gas at one
atmosphere of pressure, as the
temperature is lowered, the
volume will be reduced.
•
At 0 K, the molecules will
have no energy of motion.
•
There is only one possible
arrangement for the molecules.
Ideal gas at one
atm and 0 K.
Entropy and temperature
• The entropy of an ideal gas at constant
pressure increases with increasing
temperature.
•
This is because the volume increases.
0K
T1
T2
T3
Entropy and temperature
• There are other reasons for
entropy to increase with increasing
temperature.
Increased temperature will
result in a greater distribution of
T1 speeds.
molecular
number
•
T2
T3
speed
T1 < T2 < T3
Entropy and temperature
• Increased temperature also results in more
energy levels in atoms and molecules being
occupied.
•
•
•
•
•
For molecules,
this means that
they will be able
rotate and their
bonds can vibrate.
•
•
This further
increases entropy.
Examples of Entropy
• What has more entropy
– Gas or liquid?
– Solid or liquid?
– Homogeneous solution or separate mixture
• sugar dissolved in water or sugar and water
• The more random or lack of order
– The more entropy
• Do you have it?
– Iodine vapor condensing on cold glass?
– Gas at 1 atm or 1 x 10-2 atm?
S = Sfinal – Sintial
nd
2
Law Restated
• In any spontaneous process, there is always an
increase in the entropy of the universe
Suniverse = Ssystem + Ssurroundings
If S univ > 0, process is spontaneous.
If S univ < 0, process is non-spontaneous. The process
is spontaneous in the other direction.
If S univ = 0, process has no tendency to occur or is at
equilibrium.
How can complex molecules
assemble in a bacteria?
• The created order is in the bacteria. The energy
needed for this activity is supplied from an external
source.
– The Universe gains entropy while the cell is organized.
• Most of our energy comes from the sun. The
constant influx of energy supplies the energy to
overcome entropy…. for the time being!
The Sun is Entropic!
• Stars produce light in all directions
• This energy is spread through the universe
– Sounds entropic
• Think about a star that is 1 million light
years away.
Star
Star
Further away
Chaos Theory
• Chaotic events tend to organize themselves
• Best example is a whirlpool. (toilet)
– The particles organize themselves in order to
become disorganized more efficiently
How can we determine if a
process is spontaneous?
Suniverse = Ssystem + Ssurroundings
The sign of Ssurr depends on direction of heat flow
exothermic process adds energy to the universe
The universe now has more random motion
So the universe experiences an increase in entropy
Suniverse > 0 or positive.
Ssurrounding
Magnitude of Ssurr depends on the temperature
If the surroundings have a low temp, additional
energy makes a big difference
If the surroundings have a high temperature,
additional heat does not add much more energy
(entropy) it has little effect.
(little change, small Ssurr)
Entropy Continued
• The tendency for a system to lower its
energy becomes more important at lower
temperatures.
Driving Force
Provided by
energy flow
Magnitude of the
Entropy change of
The surroundings
Quantity of heat (J)
temperature (K)
Entropy depends on Enthalpy
• The change in Enthalpy, H, which is the
direction and magnitude of heat exchanged
• Energy of system is proportional to its temp
in kelvin in an isothermal system.
J = - H = Ssurr
K
T
Change in enthalpy
exotherm = neg
endotherm = pos
Spontaneity
S
system
+
S
S
Surrounding Universe
+
+
Spontaneous?
-
-
-
+
-
?
No (process in
opposite direction
Yes if
Yes
Ssys > Ssurr
-
+
?
Yes if
Ssys < Ssurr
Gibbs Free Energy
• There is a “war” between
– order and disorder
– Enthalpy and Entropy
• The sun is the source of our energy
– It drives our enthalpic world
– If the sun were to stop, how long would live still exist.
– In a million years would things still look the same?
G = H -TS
• This war can be described mathematically
• G is Gibbs Free Energy
– Gibbs Free Energy is the energy “free” to do work
– We will use this to determine the “force” behind
reactions
• Remember the second law!
Free Energy
• G = Gibbs Free Energy
G = H – TS
In processes where temp is constant
G = H - T S
• We are referring to the system
– No subscripts needed
Free Energy
G = H - T Ssys
-G = - H + Ssys
T
T
If we divide by –T
- H = Ssurr
T
-G = Ssurr + Ssys = Suniv at constant T, P
T
At what value of G , is Suniv > 0 or “spontaneous”
Spontaneity Again
• Processes are spontaneous
H2O(s)  H2O (l)
H = 6.03 x 103 J/mol
S = 22.1 J/K • mol
– If they have a positive Suniv
– If they have a negative G , at constant P,T
G = H - T Ssys
Spontaneous processes have negative G
Is Water Melting Spontaneous?
• Will this be spontaneous at -10, 0, or 10oC?
• H2O(s)  H2O (l)
H = 6.03 x 103 J/mol
S = 22.1 J/K • mol
G = H - T Ssys
Calculate Sunv and G
H = 6.03 x 103 J/mol
S = 22.1 J/K • mol
G = H - T Ssys
T
(°C)
T
K
-H = Ssurr
T
S + Ssurr=Sunv
TS
X 103
G
-10
263
-22.9
-0.8
5.81
+ 2.2 x102
0
273
-22.1
0
6.03
0
10
283
-21.3
+0.8
6.25
- 2.2 x102
The spontaneity of the process depends on the temp
S
H
Result
Positive
Negative
Spontaneous at All temps
Positive
Positive
Negative
Negative
Negative
Positive
Spontaneous at High Temps
(exotherm not important)
Spontaneous at Low Temps
(Exotherm is important)
Not Spontaneous Process
Reverse spontaneous at all
temps
Gibbs Free Energy
1. If G is negative, the
forward reaction is
spontaneous.
2. If G is 0, the system is
at equilibrium.
3. If G is positive, the
reaction is spontaneous
in the reverse direction.
Br2(l)  Br2(g)
At what temp is the following process spontaneous at 1 atm?
What is the normal boiling point of liquid Br2?
H = 31.0 kJ/mol S = 93.0 J / K • mol
• G < 0 for spontaneous process
• G = 0 for equilibrium process
G = H - T Ssys
0 = H - T Ssys = 31.0 x 103 – T (93.0)
T = 333K
T > 333 K Ssys is dominant. Liquid vaporizes
T = 333 K G = 0, liquid and vapor coexist (normal BP)
(exothermic processes dominant)
T < 333 K H is dominant. Liquid forms.
What About Reactions?
• Chemistry is all about the changes that
occur. How can we use thermo and entropy
to evaluate the changes around us?
Which has greater positional entropy?
@ Constant Temperature and
Pressure
• Why would we use this as a constraint on a
thermodynamic system?
– 2nd law Suniv = Ssys + Ssurr
– No temp change means no Ssurr
4NH3(g) + 5O2 (g)  4NO(g) + 6H2O(g)
Is this process thermodynamically favored?
How about this?
•
•
•
•
•
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
Same amount of gas on both sides.
Entropy would appear equal.
Its actually +179J/K. Why?
Water is more complex a molecule than
hydrogen.
• More ways it can move = more entropy.
And this?
• Cdiamond → Cgraphite
∆Go = -3kJ
• So how come we still have diamonds?
Third Law of Thermodynamics
• When can perfect order be achieved?
– What conditions would have to be necessary to
first achieve it, and the keep it that way?
• The only time the entropy is zero is when
you have a perfect crystal at 0K
• Any rise in temperature will create
movement and therefore raise entropy.
Other information
• As with enthalpy which is a state function,
• Ho = np Hf products - nr Hf reactants
• So too with entropy and free energy.
– They are both state functions
– So = np S products - nr S reactants
– Go = np Gf products - nr Gf reactants
• free energy of formations for an element in its
standard state is zero.
• Also free energy and entropy for reactions
can be added like Hess’s Law.
Free Energy & the Equilibrium Constant
Recall that G and K (equilibrium constant) apply
to standard conditions.
However, G and Q (reaction quotient) apply to any
conditions.
It is useful to determine whether substances under
any conditions will react:
G  G  RT ln Q
Where R is the ideal gas constant, 8.314 J/mol•K
Free Energy & the Equilibrium Constant
At equilibrium, Q = K and G = 0, so
G  G   RT ln Q
0  G   RT ln K .
 G    RT ln K .
From the above we can conclude:
If G < 0, then K > 1.
If G = 0, then K = 1.
If G > 0, then K < 1.
Free Energy & the Equilibrium Constant
Solving for the equilibrium constant, K ,
G = - RT lnK
K = e
- Gº
RT
G and work
• G is the value of all free energy from a
reaction.
• Therefore its value is equal to the maximum
work possible from a reaction. (if -)
• If G is positive, what does it tell us?
• Used for efficiency.
• Will never be 100%, why?
Summary of Thermo
• 1st law says you can’t win, only break even.
• 2nd law says you can’t break even.
• Explains energy crisis!