1. ENVE 501 – Week 3
1

2. Quiz 1
Imagine you can follow a molecule of copper
from the source to the mouth of the Mississippi
River. Assume that the average concentration of
dissolved copper remains constant at 5 µg/L.
The total dissolved solids, however, increases
significantly from source to mouth.
Would you expect the activity to increase,
decrease, or remain the same? Why?
From the perspective of a fish, what is more
important, the activity or the concentration of
copper? Briefly explain your answer.
2

3. Quiz 1 - Solution
• Activity will decrease
• Ionic strength will increase as the total dissolved
solids (TDS) concentration increases
• Activity coefficient will decrease as the ionic
strength increases
ai = γi × [i]
• Activity is more important
• Toxicity involves biochemical reactions
• Reactions depend on chemical activity
3

4. Teams - I
Jennings Diane
Jin Weidong
Knudsen Westen
Gobi Mu Binbin
Temino Boes Regina
Xu Limeimei
Dai Wei
Douce Jordan
Kalahari Glick Nicole
Huang Jinjin
Jiang Yuanzan
4

5. Teams - II
Mirza Muhammad
Sharpe Jessica
Atacama Short Bradley
Wang Chen
Wang Zehua
Gong Qijian
Izadmehr Mahsa
Meyer Kristine
Sahara
Mocny William
She Congwei
Wang Xiaolang
5

6. Teams - III
Azu Laud
Hernandez Alcides
Huo Yechen
Sonoran Pujari Aniket
Ramos Tiffanie
Zhang Yue
Cao Jicong
Lin Kehsun
Mojave Tadeusz Bobak
Wen Jiahong
Young Meghan
6

7. Team assignment for next week
• Use a 9” by 9” slide in Landscape format
• Introduce your consulting firm
• Names
• Photographs(? )
• History
• Specialization
• Company logo….
7

8. Details posted on Blackboard
• No more than four slides
• Email to me by 8 PM on September 14
• Use Blackboard tools to collaborate
• Select one team member to present next week
• Students at remote sites should consider an
electronic presentation
8

9. Chapter 3 –
Thermodynamics
9

10. Key Concepts
Entropy
pp 62-64; 70-74
Gibbs (free) energy
pp 80-93; 98-104
Thermodynamics & chemical systems
pp 104-125
• Page numbers are from Chapter 2 of our text.
• The handout provides a more comprehensive
description.
10

11. Laws of thermodynamics - I
Zeroth law of thermodynamics
• Systems in thermal equilibrium have the same
temperature.
First law of thermodynamics
• Energy is conserved.
11

12. Laws of thermodynamics - II
Second law of thermodynamics
• Nature is not symmetric; the quality of energy
decreases.
• A process will occur spontaneously if it increases
the total entropy of the universe
Third law of thermodynamics
• Entropy = 0 for a perfect crystal at T = 0.
12

13. An alternative look at the laws
We can't win
We are sure to lose
We can't get out of the game
(Garrett Hardin from an unknown source)
13

14. Entropy and life
Life is an open or continuous system that is able
to decrease its internal entropy at the expense
of free energy taken in from the environment
and subsequently rejected in a degraded form.
• Lovelock, J. (2000) GAIA. A new look at life on
Earth. Oxford University Press, New York.
14

15. Entropy and death
Entropy is a measure of a system's thermal
disorder. It implies the predestined and
inevitable run-down and death of the Universe.
• Lovelock, J. (2000) GAIA. A new look at life on
Earth. Oxford University Press, New York.
15

16. Fundamental
thermodynamic properties
Gibbs (free) energy = G
• Energy available to do useful work
Enthalpy = H
• Total energy (molecular + mechanical work)
Entropy = S
• A measure of disorder
16

17. Changes in G, H, and S
are related
∆Gi = ∆H i − T ∆Si
17

18. What do the symbols mean?
Subscript identifies the compound
Gi ≡ Gibbs energy associated with total i
The over-bar means per mol
Gi ≡ energy per mol of i
The superscript ‘o’ refers to the standard state
o
Gi ≡ standard molar Gibbs energy of i, also the
standard Gibbs energy of formation
18

19. Thermodynamic properties
are relative
By definition, the Gibbs energy of a pure element
under standard conditions is 0
• For example (T = 25°C, P = 1 bar)
GH2 = 0; pure hydrogen gas
GCu = 0; pure solid copper
For dissolved ions (T = 25°C, P = 1 bar)
GH = 0; dissolved hydrogen ion
+
• Other ions defined relative to this reference
19

20. Corrections for different activity
20

21. 21
+ RT ln ci + RT ln γ i
Activity
Effect on Gibbs energy
coefficient
Concentration
o
Gi = Gi
Standard Gibbs
energy of formation

22. Example
What is the Gibbs energy per
mole of atmospheric oxygen?
Actual Gibbs energy =
Standard Gibbs energy +
Corrections for activity
o
Gi = Gi + RT ln ci + RT ln γ i
22

23. 23

24. Example
What is the Gibbs energy per
mole of atmospheric oxygen?
24

25. Example
Substituting
25

26. So what?
Change in Gibbs energy tells us if a reaction can
occur.
Reaction is possible:
• If the change in the Gibbs energy of reaction is
negative
26

27. Example
Find molar Gibbs energy of reaction for:
1.0 mole hydrogen gas
+ 0.5 mole oxygen gas
= 1.0 mole water as a gas
Assume 1.0 bar and T = 25° C
H2(g) + ½ O2(g) ↔ H2O(g)
27

28. 28

29. Example
Substitute
29

30. -228.57 kJ per mol of what?
Reaction involves:
1 mol H2 + 0.5 mol O2 + 1 mol H2O
Per mole of stoichiometric reaction.
For example:
aA+bB ↔ cC+dD
1.0 mole of stoichiometric reaction:
a mol A + b mol B → c mol C + d mol D
Stoichiometry is important
30

31. For the reaction:
aA+bB↔ cC+dD
G r , products = G r ,C + G r , D
o o
= G r ,C + RT ln aC + G r , D + RT ln aD
G r ,reactants = G r , A + G r , B
o o
= Gr, A + RT ln a A + G r , B + RT ln aB
31

32. o o o o  aC aD 
∆ Gr = GC + GD − GA − GB+ RT ln  
 a A aB 
o  aC aD 
= ∆ Gr + RT ln  
 a A aB 
 γ c [ C ] c γ d [ D] d
 C 

o D
∆Gr = ∆Gr + RT ln  
 γ A [ A] γ B [ B ]
a a b b
 

32

33. For dilute solutions
As γi → 1.0, NOTE: From here on
we often assume
a dilute solution
[ C ] c [ D] d
 

o
∆Gr = ∆Gr + RT ln  
 [ A] [ B ]
a b
 

o
= ∆Gr + RT ln Q
Q = reaction quotient 33

34. Example
H2(g) + ½ O2(g) ↔ H2O (g
• Average atmospheric partial pressures
pH2 = 6×10-7 bar
pO2 = 0.21 bar
pH2O = 0.02 bar
• Gibbs energy under conditions different from
the standard state is :
Gi = G + RT ln ai
i
o
34

35. 35

36. Example
Under standard conditions:
RT = 8.314 (J/mol.K) × 298 (K)
RT = 2.48 (kJ/mol)
ai = γi × pi
36

37. Example
37

38. Substituting for each term
38

39. Quiz 2
• Match the reaction described in the first
column with the dimensions of the rate
constant in the final column
• You might not need all the choices for k
39

40. Interpreting values of ∆Gr
∆Gr < 0
reaction can proceed in forward direction
∆Gr > 0
reaction can proceed in reverse direction
∆Gr = 0
forward and reverse reaction rates are the same;
reaction is at equilibrium
40

41. A special reaction quotient when ∆ Gr = 0
o
∆Gr = − RT ln K eq
Keq = equilibrium constant

  [ C ] [ D]

o
∆Gr
c d
K eq = exp − =
 RT  [ A] [ B ]
a b
 
41

42. Example
How much oxygen should be dissolved in
Lake Michigan when T = 25°C?
Reaction is:
O2(g) ↔ O2(aq)
Gibbs energy for the reaction as written is:
o
 [ O2 ] 
 
∆Gr = ∆Gr + RT ln  
 pO2 
 
What are the dimensions of the term in
brackets?
42

43. 43

44. Example
Recall the “hidden” terms:
 mol 
cO2  
Numerator is:  L 
 mol 
1.0 
 L 
Denominator is:
pO2 ( bar )
1.0( bar )
Change in standard state Gibbs energy for
this reaction is:
o
∆Gr = 16.32 - 0 = 16.32 (kJ/mol)
44

45. Example
 ∆ Gro 
 
K eq = exp  − 
 RT 
 
 16.32 × 103 
  −3
= exp  −  = 1.38 × 10
 8.314 × 298 
 
Solving for equilibrium dissolved oxygen:
[O2] = 1.38×10-3 × 0.21
[O2] = 2.9×10-4 (mol/L) ⇒ 9.3 mg/L
Dimensions follow from “hidden” terms.
45

46. Example
Estimate Lake Michigan dissolved oxygen
concentration for T = 1°C.
It looks as if :

 

o
∆Gr
K eq = exp − 
 RT 
 
o
Knowing ∆Gr
Can we substitute T = 1°C?
46

47. Example
No!
Superscript ‘o’ refers to the reference state.
These values apply only for T = 25°C
47

48. Example
Equilibrium constant at T = 1°C estimated
from the van’t Hoff equation.
 ∆H r  1 1  

o

K eq ,T2 = K eq ,T1 exp   − ÷
 R  T1 T2  
 
−3  −11.71 × 103  1
 1  
K eq ,T2 = 1.38 × 10 exp   − ÷
 8.314
  298 274  

= 2.09 ×10−3
[O2] = 2.09×10-3 × 0.21
[O2] = 4.39×10-4 (mol/L) ⇒ 14 mg/L
48

49. Example
Compare available energy
Aerobic versus anaerobic process
Degradation of acetic acid
Aerobic process
C2H4O2(aq) + 2 O2(aq) = 2 H2CO3 (aq)
∆Gr = 2×(-623.2) – (-396.6) - 2×(16.32)
∆Gr = -882.44 kJ/mol
49

50. Example
Anaerobic process
C2H4O2(aq) + H2O(l) = CH4(aq) + H2CO3(aq)
∆Gr = (-34.39)+(-623.2)-(-396.6)-(-237.18)
∆Gr = -23.81 kJ/mol
Relative to anaerobic processes, aerobic process
release more energy
50

51. Example
Temperature affects equilibrium:
Dissolution of O2
∆H r = -11.71 kJ/mol
Precipitation of CaCO3(s)
Ca2+(aq) + CO32-(aq) = CaCO3(s)
∆H r= (-1207.4) - (-542.83) – (-677.1)
∆H r = 12.53 kJ/mol
51

52. Example
Sign of ∆H is important
O2 solubility increases with decreasing
temperature
Implications for fish?
CaCO3(s) solubility decreases with decreasing
temperature
Implications for water softening?
52

53. Example
Precipitation of FeCO3(s)
Find changes in:
Gibbs energy
Enthalpy
Entropy
Fe2+ + CO32- = FeCO3(s)
53

54. 54

55. Example
∆Gr = -666.7 – (-78.87 – 527.9)
= -59.93 kJ/mol
∆H r = -737.0 – (-89.10 – 677.1)
= 29.2 kJ/mol
∆Sr = 105 – (-138 – 56.9)
= 299.90 J/mol.K
Entropy increases for formation of this solid.
Why?
55

56. Summary I
∆Gi = ∆H i − T ∆Si
o
Gi = Gi + RT ln ci + RT ln γ i
∆Gr = Gproducts − Greactants
56

57. Summary II


o
∆Gr { C} { D}

c d
K eq = exp − =

 RT  { A} a { B} b

 ∆H r  1 1  

o

K eq ,T1 = K eq ,T2 exp   − ÷
 R  T1 T2  
 
57

58. Some time for teams
or
Maxwell’s Demon
58

59. 0.5 N2 + O2 ⇔ NO2 + heat
The expression above shows an exothermic
reaction (heat is generated) wherein
nitrogen gas is oxidized to NO2.
What effect would each of the following have
on the reaction?
• Decrease the temperature.
• Increase the volume.
• Decrease the O2 concentration.
• Add a catalyst.
59

60. Solutions…
Decrease the temperature.
Equilibrium shifts to the right to increase the
temperature.
Increase the volume.
Equilibrium shifts to the left to increase the
pressure.
Decrease the O2 concentration.
Equilibrium shifts to the left to increase O2.
Add a catalyst.
A catalyst will affect the rate of the reaction, but not
the equilibrium.
60