1. Chapter 18 Chemical Thermodynamics – Entropy and Free Energy

2. 3 Laws of Thermodynamics
1st- Energy is Conserved (Chapter 5)
2nd-Spontaneous Processes and Entropy
3rd-Entropy at Absolute Zero

3. 1st- Energy is Conserved (Chapter 5)
System – portion of the universe that we are concerned with
Surroundings- everything outside the system
ΔE = q + w
q = ΔH at constant pressure
- ΔH = exothermic
+ ΔH = endothermic

4. 1st- Energy is Conserved (Chapter 5)
Energy can be transferred between a system and its surroundings
Energy can be converted from one form to another
ΔE = q + w
“ΔE” is the change in the internal energy of a system
“q” is the heat absorbed or released by the system or from the surroundings
“w” is the work done on the system or by the system

5. 1st- Energy is Conserved (Chapter 5)
Helps us determine how the energy is transferred and if work is done on the system or by the system
If q > 0 means the system is absorbing heat from the surroundings
If w > 0 then the surroundings are doing work on the system

7. 1st- Energy is Conserved (Chapter 5)
State functions- properties that depend on the initial and final state, not on how the change was made.
ΔH and temperature are state functions
Work is not a state function
Calorimetry – process used to measure heat changes, by using ΔT of water in a calorimeter

8. 1st- Energy is Conserved (Chapter 5)
Two ways a system can exchange energy with the surroundings:
Change in heat(q) or work (w)
ΔE = q + w
W= fxd
Doing work on a system increases the potential energy
Work done by the system decreases the potential energy of the system

9. 1st- Energy is Conserved (Chapter 5)
Hess’s Law –total energy change is a sum of all the changes(steps) that occur during a reaction
ΔHf- the energy required to form a mole of a substance from its elements
ΔHf of an element = 0
ΔHreaction = ΔHf products –ΔHf reactants

10. Chapter 18 During a reaction, energy is transferred in 3 directions
Energy lost to universe (entropy- amount of disorder)
During a reaction, energy is transferred in 3 directions
Energy lost or gained (enthalpy)
Energy that can be used to do work (Gibbs free energy)

11. Spontaneous Processes
A process that occurs on its own without any outside assistance
Spontaneity is independent of time it takes for the process to occur (that is kinetics)
Most reactions and processes have a directionality
A spontaneous process occurs in one direction.
The reverse reaction is nonspontaneous
Some processes are spontaneous without being energy-driven; these processes are driven by an increase in disorder

12. Spontaneous Processes

13. Example 1:
Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or at equilibrium
A. Water gets hotter when a piece of metal heated to 150° C is added
Spontaneous – heat is always transferred from the hot object to the cool one
B. Water at room temperature decomposes into hydrogen and oxygen gas
Not spontaneous
C. ice cube melts at room temperature
Spontaneous- heat is transferred from the hot object to the cool one

14. Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0 C, it is spontaneous for ice to melt.
Below 0 C, the reverse process is spontaneous.
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16. Second Law of Thermodynamics
States that for any spontaneous process the total entropy of the universe must increase.
(The universe is becoming more disorderly)
Entropy (S) – degree of disorder.

17. Second Law of Thermodynamics
Maximum entropy, most disorder S > 0
Second Law of Thermodynamics
If S > 0 disorder increases
If S < 0 disorder decreases (more ordered)
Minimal entropy, most order, S < 0

18. What causes a reaction to be spontaneous?
2 Laws of Nature drive spontaneous processes.

19. Law of Nature #1 Systems tend to attain a state of minimum energy, (lose energy, cool off)
large negative ∆H are spontaneous
2Na + 2H2O  NaOH + H2
∆H = ‑281.9 kJ (exothermic)
Reaction is spontaneous forward
2Ag2S + 2H20  Ag + 2H2S + O2
∆H = 595.6kJ (endothermic)
not spontaneous, but the reverse reaction is.
Some endothermic processes may be spontaneous
ex. cloth drying, ice melting

20. Law of Nature #2 –Systems tend to attain a state of maximum disorder (entropy)
Entropy (S) -randomness, or the amount of disorder, of a system.
State function ( doesn’t depend on how you got there)
Entropy increases in spontaneous processes.
Greater the disorder, the higher its entropy.
Every substance has a characteristic entropy
∆S = Sfinal –S initial

21. Calculate change in Entropy (ΔS)
If a process is at constant temperature (isothermal)
ΔS = qrev qrev = heat T T = temp in Kelvin
Units of ΔS are J/K

22. Calculate change in Entropy for phase changes
ΔS for phase changes – remember that during a phase change temp remains constant so substitute ΔH of the phase change for qrev
ΔSphase change = qrev = ΔHphase change
T T
Tabulated value

23. Example 2:
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is °C, and its molar enthalpy of fusion for mercury is ΔHfusion = 2.29 kJ /mol.
What is the entropy change of the system when 50.0 g of Hg (liquid) freezes at the normal freezing point?

24. Example 2:
A few notes : fusion is melting, the question is asking for entropy change when mercury freezes, so reverse the sign of ΔH
ΔH is given to us in kJ / mol, so use stoichiometry to get q into J
Answer = J/K

25. ΔH = = -571J Step 2 –convert -38.9oC to 234.1K
50.0g Hg
1 mole Hg
-2.29 kJ
1000J
200.59g Hg
1 mol Hg
1kj
= -571J x x x
Step 2 –convert -38.9oC to 234.1K
ΔSsys = ΔH = J = -2.44J/K
T K
Step 3
ΔS is negative b/c heat is flowing from the system to the surroundings when it is freezing

26. Total Entropy Reversible : ΔSuniverse = ΔSsystem + ΔSsurroundings = 0
Irreversible (spontaneous) =
ΔSuniverse = ΔSsystem + ΔSsurroundings = 0
The entropy of the universe increases in any spontaneous process
It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases

27. Molecular Entropy
Microstate – single possible arrangement of the positions and kinetic energies of molecules in a particular thermodynamic state
Each thermodynamic state has a characteristic number of microstates, W
Boltzmann developed an equation
S = k ln W
k = Boltzmann constant, 1.38 x J/K

28. Molecular motion
When a substance is heated, the motion of its molecules increases
The higher the temperature the faster the molecules move
Hotter systems have a larger distribution of molecular speeds
3 different types of motion
Translational
Vibrational
rotational

29. invisible motion
Computational chemistry

30. Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
S(g) > S(l) > S(s)
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32. Entropy increases
a. in a reaction that goes from a solid to a liquid or gas
b. in a reaction from fewer moles to more moles
c. simpler molecules to more complex molecules
d. smaller molecules to longer molecules
e. ionic solids with strong attractions to ionic solids with weaker attractions
f. ionic salts dissolve in water (involves disorder of ionic solid but ordering of water molecules – overall
increase in disorder)
g. gas dissolved in water to escaped gas

33. Example 3
Predict whether ΔS is positive or negative or 0 for each process, assuming constant temperature
H2O (l) → H2O (g)
Positive
Ag+ (aq) Cl- (aq) → AgCl (s)
Negative
4 Fe(s) O2 (g) → 2 Fe2O3 (s)
N2 (g) + O2 (g) → 2 NO (g)

34. Entropy changes in chemical reactions
standard molar entropies (S°)
1. standard state – the state of a pure substance at 1atm
2. standard molar entropy of ELEMENTS at 298 K are NOT 0 like enthalpy
3. standard molar entropy of gases are greater than solids and liquids
4. standard molar entropy generally increase with increasing mass of compound

35. The entropy change for a chemical reaction
ΔS° = 𝑛 𝑆 ° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑛 𝑆 ° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

36. Example 4
Calculate the change in the standard entropy of the system, ΔS° , for the synthesis of ammonia from N2 (g) and H2 (g) at 298 K
N2 (g) + H2 (g) → 2 NH3 (g)
ΔS° = 2 S° (NH3) – [S° (N2) S°(H2) ]
Look up S° values in table and plug in
ΔS° = J/K

37. Entropy changes and the surroundings
. ΔSsurroundings = - qsystem recall that if a reaction is taking place
T at constant pressure that the q = ΔH so
For a chemical reaction,
ΔSsurroundings = - ΔH system T
Can use to find ΔS °universe = ΔS °system ΔS °surroundings

38. Gibbs Free Energy (G)
Another way to determine if a process is spontaneous or not is to look at the energy that’s used to do work…. Gibbs Free energy

39. There are two equations to determine ΔG
One involves reaction in standard conditions ( 298K 0r 25oC and 1 atm)- use Appendix C
Other involves calculating values for ΔH, temperature and ΔS and using them to determine Gibbs free energy
They may differ a little due to errors in experimentation

40. Entropy Change in the Universe
Since
Ssurroundings =
and
qsystem = Hsystem
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T, we get
TSuniverse = Hsystem  TSsystem
qsystem T
Hsystem T
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Gibbs Free Energy
TDSuniverse is defined as the Gibbs free energy, G.
When Suniverse is positive, G is negative.
Therefore, when G is negative, a process is spontaneous.
Start here 4/1/10
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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.
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43. Standard Free Energy Changes
similar to standard enthalpies of formation are standard free energies of formation G: AT STANDARD CONDITIONS f
DG = SnDG (products)  SmG (reactants) f
where n and m are the stoichiometric coefficients.
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Free Energy Changes
At temperatures other than 25 C,
DG = DH  TS
How does G change with temperature?
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45. Determination of Gibbs Free Energy Changes
∆G = ∆H ‑ T (∆S)
Gibbs free energy
Temperature in Kelvin
Change in enthalpy (heat) in kJ
Change in entropy (randomness)
ΔS is change in entropy
∆S is given in joules in Apprendix C so convert to kJ
T is in kelvin

46. Look at Appendix C in textbook
Use Appendix C page to find out
ΔH (kJ/mol)
ΔGf (kJ/mol
S (J/mol) * be careful to convert to kJ
Now let’s try a problem…..

47. Example 5: Calculate ΔG for the reaction below at 25oC (298K)
Example 5: Calculate ΔG for the reaction below at 25oC (298K) N2(g) H2(g)  2NH3(g)
Appendix C ΔHf
ΔGf
Convert to kJ ΔS
ΔHf = 2mol(-46.19) – (0) +3(0) = kJ
ΔS = 2 mol(.1925) – [1(.19150) + 3(.13058)] =
ΔG = ΔH –TΔS
ΔG = ( )
ΔG = kJ

48. Example 6: Using Appendix C, calculate ΔG at 298K for the following reaction using both equations.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Ans kJ vs. (-853kJ)
Why the difference? Experimental data can be off

49. Interpreting ΔG
if ∆G is negative, the reaction is spontaneous in the forward direction
if ∆G is positive, the reaction is spontaneous in reverse
If ∆G is zero, the reaction is at equilibrium

51. Simplified ∆G = -RT ln Keq
The equilibrium constant is related to both ∆H and the ∆S in the following way: ln Keq = ∆H/(RT) + ∆S/R
Or……….
lnKeq = -∆H ∆S
RT R
Since ∆G =∆H - T∆S then……
Simplified ∆G = -RT ln Keq

52. Free Energy and Equilibrium
At equilibrium, Q = K, and G = 0.
The equation becomes
0 = G + RT ln K
Rearranging, this becomes
G = RT ln K or
K = e
G/RT

53. Example 7:
The standard free-energy change for the formation of NH3 from N2 + H2 is
-33,000 J/mol. Calculate the equilibrium constant for the process at 25°C.
Answer : K = 7 x 105