1. Spontaneity, Entropy and Free Energy

2. Spontaneous Processes and Entropy
First Law
“Energy can neither be created nor destroyed"
The energy of the universe is constant
Spontaneous Processes
Processes that occur without outside intervention
Spontaneous processes may be fast or slow
Many forms of combustion are fast
Conversion of diamond to graphite is slow

3. Entropy (S) A measure of the randomness or disorder
The driving force for a spontaneous process is an increase in the entropy of the universe
Entropy is a thermodynamic function describing the number of arrangements that are available to a system
Nature proceeds toward the states that have the highest probabilities of existing

4. Positional Entropy
The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved
Ssolid < Sliquid << Sgas

5. Example Which has higher positional S? Solid CO2 or gaseous CO2?
N2 gas at 1 atm or N2 gas at 0.01 atm?
Predict the sign of DS
Solid sugar is added to water
+ because larger volume
Iodine vapor condenses on cold surface to form crystals
- because gs, less volume

6. Second Law of Thermodynamics
"In any spontaneous process there is always an increase in the entropy of the universe"
"The entropy of the universe is increasing"
For a given change to be spontaneous, Suniverse must be positive
Suniv = Ssys + Ssurr

7. 2nd Law of Thermodynamics
In any spontaneous reactions, there is always an increase in entropy of the universe
∆Suniv = ∆Ssys + ∆Ssurr
If ∆Ssys is negative, it can still be spont. as long as the ∆Ssurr is larger and positive
∆Suniv > 0 : spontaneous
∆Suniv = 0 : no tendency, at equilibrium
∆Suniv < 0 : not spontaneous

8. Ch. 16: Spontaneity, Entropy, and Free Energy
16.3 The Effect of Temperature on Spontaneity

9. H2O(l)  H2O(g) ∆Ssys has + sign b/c of the increase in # of positions
∆Ssurr is determined mostly by heat flow
Vaporization is endothermic so it removes heat from the surroundings
So decreases random motion of surroundings
Negative ∆Ssurr

10. Temperature’s Effects
If the ∆Ssys and ∆Ssurr have different signs, the temperature determines the ∆Suniv
For vaporization of water
Above 100°C, ∆Suniv is positive
Below 100°C, ∆Suniv is negative
Impact of the transfer of heat will be greater at lower temperatures

11. Calculating Entropy Change in a Reaction
Entropy is an extensive property (a function of the number of moles)
Generally, the more complex the molecule, the higher the standard entropy value

12. Gibb’s Free Energy D Suniv= D Ssystem + D Ssur.
Entropy of something will increase as it gets more heat.
The particles will move more. The effect this heat has
on the particles is dependant upon the temp. A small
amount of heat at a high temp will not make much of an
impact, while a small amount of heat at a small temp. will.

13. This leads to this relationship.
S= D H T

14. Thus the D Ssurr. Can be expressed as
- DHsys T
The sign should be opposite that of the systems, so it is negative because from the systems perspective, the surrounding have an opposite sign.

15. Can be changed to D Suniv= D Ssys + - DHsys
So D Suniv= D Ssys + D Ssur
Can be changed to
D Suniv= D Ssys + - DHsys T

16. T D Suniv= T D Ssys + - DHsys
Multiply the whole thing be T and this gives:
T D Suniv= T D Ssys DHsys
Multiple by –1 and you get :

17. T D Suniv= -T D Ssys + DHsys
Rearranging this gives:
T D Suniv= DHsys -T D Ssys

18. -T D Suniv So: D G= DHsys - T D Ssys
DG is what we call the relationship
-T D Suniv
So:
D G= DHsys - T D Ssys

19. Standard Free Energy Change
G0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
G0 cannot be measured directly
The more negative the value for G0, the farther to the right the reaction will proceed

20. G0 = H0 - TS0 Calculating Free Energy Method #1
For reactions at constant temperature:
G0 = H0 - TS0

21. H, S, G and Spontaneity H is enthalpy, T is Kelvin temperature
G = H - TS
H is enthalpy, T is Kelvin temperature
Value of H
Value of TS
Value of G
Spontaneity
Negative
Positive
Spontaneous
Nonspontaneous
???
Spontaneous if the absolute value of H is greater than the absolute value of TS (low temperature)
Spontaneous if the absolute value of TS is greater than the absolute value of H (high temperature)

22. An adaptation of Hess's Law:
Calculating Free Energy Method #2
An adaptation of Hess's Law:
Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ
Cgraphite(s) + O2(g)  CO2(g) G0 = -394 kJ
CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ
Cdiamond(s)  Cgraphite(s)
G0 = -3 kJ

23. Using standard free energy of formation (Gf0):
Calculating Free Energy Method #3
Using standard free energy of formation (Gf0):
Gf0 of an element in its standard state is zero

24. Sample problem using free energy to find the boiling point of a substance.
At what temperature is the following process spontaneous at 1 atm?
Br2 (l)  Br2(g)
DH= 31.0 Kj/mol & DSo = 93.0 j/k.mol
G0 = H0 - TS0
make G0 = 0, so 0 = H0 - TS0
0 = j/mol - T (93.0 j/k.mol), solve for T….
T = 333 K is boiling point

25. The Dependence of Free Energy on Pressure
Enthalpy, H, is not pressure dependent
Entropy, S
entropy depends on volume, so it also depends on pressure
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure