1. Spontaneity, Entropy, and Free Energy
Thermodynamics
Spontaneity, Entropy, and Free Energy

2. 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?

3. 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.

4. 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.

5. Kinetics
The reaction pathway
Thermodynamics
the initial and final states

6. 2nd 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.

7. 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”

8. Let’s Look at a Simple System
Four atoms of an ideal gas
Three possible arrangements
How many ways can each state be achieved?

9. Examine All Possibilities (Pg 795)

10. Possibilities
The arrangement with two on each side is most likely to occur
By the ratio of 6:4:1

11. Probability of finding all the Molecules in the Left Bulb as a function of the total number of molecules

12. Unlikely to Occur 1 in 10 or not likely to occur But it is possible!
2 x 1023
But it is possible!

13. 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

14. 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.

15. 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.

16. 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. 

17. 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.

18. Entropy on the Molecular Scale
The change in entropy for a process, then, is
S = k lnWfinal  k lnWinitial
Wfinal
Winitial
S = k ln
Entropy increases with the number of microstates in the system.

19. Standard Entropies
Larger and more complex molecules have greater entropies.

20. 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.

21. Entropy and temperature
The entropy of an ideal gas at constant pressure increases with increasing temperature.
This is because the volume increases.
0 K T T T3

22. Entropy and temperature
There are other reasons for entropy to increase with increasing temperature.
Increased temperature will result in a greater distribution of molecular speeds.
speed
number T3 T2 T1
T1 < T2 < T3

23. 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.

24. 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

25. 2nd 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.

26. 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!

27. 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.

28. Star

29. Star
Further away

30. 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

31. 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.

32. 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)

33. 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)

34. 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.
Change in enthalpy
exotherm = neg
endotherm = pos
J = - H = Ssurr
K T

35. Spontaneity S system Surrounding Universe Spontaneous? + Yes -
No (process in opposite direction ?
Yes if
Ssys > Ssurr
Ssys < Ssurr

36. 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?

37. G = H - T S 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!

38. 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

39. Free Energy G = H - T Ssys If we divide by –T
-G = - H + Ssys H = Ssurr
T T T
-G = Ssurr + Ssys = Suniv at constant T, P T
At what value of G , is Suniv > 0 or “spontaneous”

40. Spontaneity Again Processes are spontaneous
H2O(s)  H2O (l)
H = 6.03 x 103 J/mol
S = 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

41. 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 = J/K • mol
G = H - T Ssys

42. Calculate Sunv and G G -10 263 -22.9 -0.8 5.81 + 2.2 x102 273 -22.1
H = 6.03 x 103 J/mol
S = J/K • mol
G = H - T Ssys
T (°C) T K
-H = Ssurr
S + Ssurr=Sunv
TS
X 103 G
-10
263
-22.9
-0.8
5.81
+ 2.2 x102
273
-22.1
6.03 10
283
-21.3
+0.8
6.25
- 2.2 x102

43. The spontaneity of the process depends on the temp
Result
Positive
Negative
Spontaneous at All temps
Spontaneous at High Temps
(exotherm not important)
Spontaneous at Low Temps
(Exotherm is important)
Not Spontaneous Process
Reverse spontaneous at all temps

44. Gibbs Free Energy
If DG is negative, the forward reaction is spontaneous.
If DG is 0, the system is at equilibrium.
If G is positive, the reaction is spontaneous in the reverse direction.

45. G < 0 for spontaneous process G = 0 for equilibrium process
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.

46. What About Reactions?
Chemistry is all about the changes that occur. How can we use thermo and entropy to evaluate the changes around us?

47. Which has greater positional entropy?

48. @ 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?

49. 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.

50. And this? Cdiamond → Cgraphite ∆Go = -3kJ
So how come we still have diamonds?

51. 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.

52. 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.

53. 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:
Where R is the ideal gas constant, J/mol•K

54. Free Energy & the Equilibrium Constant
At equilibrium, Q = K and G = 0, so
From the above we can conclude:
If G < 0, then K > 1.
If G = 0, then K = 1.
If G > 0, then K < 1.

55. G = - RT lnK Free Energy & the Equilibrium Constant
Solving for the equilibrium constant, K ,
G = - RT lnK

56. 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?

57. Summary of Thermo 1st law says you can’t win, only break even.
2nd law says you can’t break even.
Explains energy crisis!