1. Ch. 17 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 (position and/or energy levels) that are available to a system existing in a given state
Nature proceeds toward the states that have the highest probabilities of existing (½)^n

4. p. 775

5. Figure 17.4 Arrangements & Microstates
Also see Table 17.1

6. 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
The tendency to mix is due to the increased volume available to the particles of each component of the mixture.

7. 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 (∆S > 0)
Suniv = Ssys + Ssurr

8. Relationship between ∆H (kJ) and ∆S (J/K)
- ∆H => + ∆S of surroundings, -∆Ssys
+∆H => - ∆S of surroundings, +∆Ssys
The ∆S’s are opposing, so who controls the ∆Suniv? In other words, which ultimately says whether a process is spontaneous?
Temperature
The impact of energy transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures.

9. What about the entropy in the universe?!
∆Ssurr = -∆H/T, at constant pressure
Why is there a negative in the equation?
The minus sign changes the point of view from the enthalpy in the system to the entropy of the surroundings.
Table 17.3 to determine the sign ∆Suniv
Classwork: p. 807 #13, 15, 22, 23, 28, 30

10. H, S, G and Spontaneity
G = H - TS
G is free energy, H is enthalpy, T is Kelvin temperature, S is entropy
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 temp); |∆H| > |T∆S|
Spontaneous if the absolute value of TS is greater than the absolute value of H (high temp); |T∆S| > |∆H|

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. 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 free energy of a rxn system changes as the reaction proceeds
The more negative the value for G0, the farther to the right the reaction will proceed in order to achieve equilibrium
Equilibrium is the lowest possible free energy position for a reaction

13. For reactions at constant T and P:
Calculating Free Energy Method #1
For reactions at constant T and P:
G0 = H0 - TS0 < 0 to be spontaneous at a given T
Ex: Is it a given reaction spontaneous at 400 K?
H0 : 79.3 kJ/mol
S0 : J/K mol
Plug into equation:
=79.3 kJ/mol – (400K)(0.1621kJ/Kmol)
G0 = kJ/mol
<- Both are positive…
which one dominates?
<- not very spontaneous at K.

14. 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
Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ
CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ
Cdiamond(s)  Cgraphite(s)
G0 =
-3 kJ

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

16. Classwork Appendix 4 begins on p.A19
P.808 #37, 41, 46, 49 choose 2 of 4, 51, 56, 57

17. Sec. 17.7 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

18. Sec. 17.8 Free Energy and Equilibrium
A given process will be spontaneous until the free energy is minimized and G =0
Equilibrium point occurs at the lowest value of free energy available to the reaction system
Not necessarily to completion
At equilibrium, G = 0 and Q = K
(See figure 17.8)

19. G K Shift G = 0 K = 1 None G < 0 K > 1 Right G > 0
To find k: G = -RT ln(K) G K
Shift
G = 0
K = 1
None
G < 0
K > 1
Right
G > 0
K < 1
Left

20. Temperature Dependence of K
Notice: Y=mx + b
So, ln(K)  1/T

21. This agrees well with the low K at 400 K
Remember this example?
A given rxn has the following T-dependent values of K. Is it spontaneous at 400 K?
H0 : 79.3 kJ/mol
S0 : J/K mol
Plug into equation:
=79.3 kJ/mol – (400K)(0.1621kJ/Kmol)
G0 = kJ/mol
This agrees well with the low K at 400 K T
350
400
450 K
3.98 x 10^-4
1.41 x 10⁻²
1.86 x 10⁻¹

22. Sec. 17.9 Free Energy and Work
The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy
The amount of work obtained is always less than the maximum
Henry Bent's First Two Laws of Thermodynamics
First law: You can't win, you can only break even
Second law: You can't break even