1. Chemical Thermodynamics
Chapter 19

2. Chemical Reactions: Why ?
Why are some reactions reversible and others are not?
HC2H3O2 < > C2H3O2 + H
What does it mean that the Ka = 1.8 x 105 ?
Not all reactions that you can write on paper will occur. CO2 (g) ---> C (s) + O2 (g) does not, but the reverse does.
What determines if a reaction will occur?
Most chemical reactions are exothermic. Why?
Thermodynamics answers these questions.

3. Spontaneous Processes
Certain processes will always occur.
ice melts on a table at 20 C
an iron nail left outside will rust
Spontaneous Processes occur without outside intervention
May be fast or slow (combustion vs. conversion of coal to a diamond)

4. A spontaneous process has a definite direction, while energy is conserved in both forward and reverse directions.
Temperature plays a role
Exothermic processes tend to occur spontaneously.
Spontaneous processes produce “Free Energy”.

5. Reversible vs. Irreversible Processes
A reversible process:
the system can be restored to it’s exact state by simply reversing the process which led to the change +Q
Example: ice ----> liquid water Q
ice < liquid water

6. Reversible vs. Irreversible Processes
An irreversible process:
the system cannot be restored by simply
reversing the procedure.
It must be restored to its original state by a new pathway which changes the surroundings.
Example: Freezing water at 30C,
it takes work to keep it from freezing
between 0 and 30 C.

7. Reversible vs. Irreversible Processes
When a chemical system is in equilibrium, reactants and products can move back and forth reversibly.
For any spontaneous process, the path between reactants and products is irreversible.
Keep in Mind . . .
a spontaneous reaction always has a certain direction, while energy is conserved in both forward and reverse directions.
Some spontaneous reactions are VERY slow.

8. First Law of Thermodynamics
Energy is neither created or destroyed
The energy of the universe is constant
Energy can be transferred between the system and the surroundings
DE = q + w
DE = energy absorbed by the system
q = heat absorbed by system
w = work don on the system by the surroundings

9. Second Law of Thermodynamics
The Entropy of the universe is increasing.
Entropy, S, is disorder or randomness.
The more disorder, the more randomness
Units: J/K
∆S = Sf  Si
+S = final state more disordered
S = initial state more disordered

10. Entropy (S)
What would drive spontaneous forces? An increase or decrease in disorder?
An increase in spontaneous forces
Unlike energy, entropy is NOT conserved.
The Suniv is continually increasing.
No process that produces order within the system can proceed without producing an equal or greater disorder in its surroundings.
Entropy is defined in terms of probability.
Substances take the arrangement that is most likely.
The most likely is the most random

11. 2 possible arrangements 50 % chance of finding the left empty
Calculate the number of arrangements for a system.
1 molecule
2 possible arrangements
50 % chance of finding the left empty

12. 2 molecules
4 possible arrangements
25% chance of finding the left empty
50 % chance of them being evenly dispersed

13. 12 possible arrangements 8% chance of finding the left empty
4 molecules
12 possible arrangements
8% chance of finding the left empty
50 % chance of them being evenly dispersed
Substances take
the most random
arrangement

14. Ssolid < Sliquid <<< Sgas
Entropy
Which State of matter will have the most entropy? The least? Explain.
Ssolid < Sliquid <<< Sgas
there are many more ways for gas molecules to be arranged than as a liquid than a solid.
Gases have a huge number of positions possible.
Solids are very orderly.

15. Entropy increases when:
Temperature increases.
Gases are formed from solids or liquids.
The number of molecules of gas increases during a reaction.
Solutions are formed from solids or liquids.
Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate.

16. Reversible: DSuniv = DSsys + DSsurr = 0
Entropy
DSuniv = DSsys + DSsurr
Reversible: DSuniv = DSsys + DSsurr = 0
Irreversible: DSuniv = DSsys + DSsurr >0

17. Entropy DSuniv = DSsys + DSsurr
If DSuniv is positive the process is spontaneous.
If DSuniv is negative the process is spontaneous in the opposite direction.
For exothermic processes:
DSsurr is +.
For endothermic processes:
DSsurr is .
Consider this process: H2O(l)® H2O(g)
DSsys is positive
DSsurr is negative
DSuniv depends on temperature.

18. Temperature and Spontaneity
Entropy changes in the surroundings are determined by the heat flow.
An exothermic process is favored because by giving up heat the entropy of the surroundings increases.
The size of DSsurr depends on temperature
DSsurr = -DH/T

19. DSuniv = DSsys + DSsurr Yes No, Reverse ? At High Temp At Low Temp ? +
Spontaneous? + + +
Yes - -
No, Reverse - + - ?
At High Temp
At Low Temp - + ?
DSuniv = DSsys + DSsurr

20. Entropy Changes in Chemical Reactions
Standard Molar Entropies, So: the molar entropy values of substances in their standard states.
Where standard enthalpies of elements are 0, standard entropies are not 0.
What state would have the highest standard molar entropies? The lowest?
Gases = Highest, Solids = Lowest
So increase or decrease with molar mass?
Increases
So increase or decrease with increasing number of molecules involved in the formula?

21. Third Law of Thermodynamics
3rd Law- The entropy of a pure crystal at absolute zero ( 0 K) is 0.
Gives us a starting point. All others must be > 0.

22. Entropy Changes in Chemical Reactions
DSo = SnSo (products) – SmSo (reactants)
S = “sum of”
n and m are the coefficients in the chemical equation.
Calculate the DSo for the synthesis of ammonia from N2 (g) and H2 (g) at 298 K
N2 (g) + 3H2 (g)  2NH3 (g)
DSo for N2 = J/mol K
DSo for H2 = J/mol K
DSo for NH3 = J/mol K
DSo = 2 (192.5) – [(191.6) + 3(130.7)
DSo = J/K … spontaneous or not?

23. Gibb's Free Energy
Free Energy= energy available to do work, (however, much is in the form of heat)
All spontaneous processes produce free energy.
Whether a process is spontaneous or not depends on 2 thermodynamic concepts:
1. Enthalpy, ∆H and 2. Entropy, ∆S

24. Gibbs Free Energy DG = DH -TDS at constant temperature (use Kelvin!)
Formula: G = H-TS
Never used this way. But rather as:
DG = DH -TDS at constant temperature (use Kelvin!)
Divide by -T
-DG/T = -DH/T-DS
-DG/T = DSsurr + DS
-DG/T = DSuniv
If DG is negative at constant T and P, the process is spontaneous.
If ∆G is positive, the process is nonspontaneous.
If ∆G is 0, the process is at equilibrium.

25. - + - ? + + - - ? - + + DG = DH -TDS DH DS Spontaneous? ∆G
At all Temperatures
At high temperatures,
“entropy driven” ? + +
At low temperatures,
“enthalpy driven” - - ?
Not at any temperature,
Reverse is spontaneous - + +

26. Is this process spontaneous?
For the reaction H2O(s) ® H2O(l)
DSº = 22.1 J/K mol DHº =6030 J/mol
Calculate DG at 10ºC and -10ºC
DG=DH-TDS
∆G = 6030 J/mol - (283K)(22.1J/K mol)
∆G =
∆G = yes, at 10C
DG= 6030 J/mol - (263K)(22.1J/K mol)
∆G =
∆G = no, at -10C

27. Free Energy and Work Free energy is that energy free to do work.
The maximum amount of work possible at a given temperature and pressure.
Never really achieved because some of the free energy is changed to heat during a change, so it can’t all be used to do work.

28. If ∆G = , reaction is spontaneous
In Summary
If ∆G = , reaction is spontaneous
If ∆G = +, reaction is Non-spontaneous;
work MUST be done to reverse
the reaction
If ∆G = 0, reaction is at equilibrium
If ∆G = very large (-) #, then a lot of energy is released to surr. and is capable of doing much work. (burning of gasoline)

29. Entropy in Reactions
Enthalpy, ∆H, is measured experimentally
No easy way to measure Entropy, ∆S
Use absolute Entropy, S, values from tables in back of textbook
These values are based on difficult calculations using absolute zero data.

30. Free Energy in Reactions
DGº = standard free energy change.
DGº = DHº - TDSº
Can’t be measured directly, must be calculated from other measurements.
Free energy change that will occur if reactants in their standard state turn to products in their standard state.
∆G = ∑ n Gf products  ∑ m Gf reactants

31. Free energy in Reactions
DGo = DHo –TDSo
Example:
The Haber process for the production of ammonia involves the following equillibrium: N2 + 3H2 2NH3
Assume DHo and DSo don’t change with temperature.
Predict the direction in which DGo changes for this reaction with increasing temperature.
Expect DSo to be (–) because number of moles of gas is smaller on product side, thus –TDS term will be positive and get bigger as temperature increase.

32. Calculate the values of DGo for the reaction at 25oC and 500oC.
First we need to Calculate the DHo and DSo.
DHo values = N2 (0); H2 (0); NH3 ( kJ/mol)
DSo values = N2 (191.6 J/mol K); H2 (130.7 J/mol K); NH3 (192.5 J/mol K)
At 298K, DGo = kJ
At 773K, DGo = 61 kJ

33. Free Energy and Equilibrium
For a system at equilibrium, ∆G = 0.
we can find the ∆G for any conditions, not only ∆G, with this formula:
DG = DGº +RT ln (Q)
where Q is the reaction quotients (pressure of the products / pressure of the reactants), and R is the ideal gas law constant, 8.31 J/mol K
At equilibrium, Q = Keq
DG = DGº +RT ln (Q)
0 = DGº + RT ln (Keq)
DGº =  RT ln (Keq)

34. Free Energy and Equilibrium
DGº =  RT ln (Keq)
If ∆Gis negative, Keq >1
If ∆Gis zero, Keq =1
If ∆Gis positive, Keq <1

35. How far will an equilibrium reaction continue to go?
During the course of a reaction, the free energy always decreases.
DG tells us spontaneity at current conditions. When will it stop?
It will go to the lowest possible free energy which may be an equilibrium.
Analogy of boulder rolling down a hill

36. Equilibrium/Free Energy analogy
Reactants
Free Energy
Products PE
Equilibrium
position in
valley
Mixture at
Equilibrium
Course of Reaction
Position
Equilibrium Position = Minimum free energy available to the system

37. Free Energy and Work “Free Energy” is the energy “free” to do “work”
The amount of work obtained is always less than the max.
Henry Bent’s First two Laws of Thermodynamics:
“You can’t win, you can only break even”
“ You can’t even break even”
The amount of free energy of our system (Earth) is decreasing