1. Ppt04 (PS8), Thermodynamics
energy changes (heat flows)
Very global (“the universe”)
Applies to all fields of science
Led to creation of heat engines/refrigerators/air conditioners
In chemistry, focus is on physical and chemical changes
How can we use “spontaneous processes” to “do work”?
We can use tabulated data to make predictions (and even calculate K’s for possible processes!)
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2. “Spontaneous”  “fast” !!
Spontaneity refers to “directionality”
Does it occur (on its own) in the forward [or stated] direction, or does it not?
Thermodynamics will answer / address this question.
Does not depend on how the process is carried out.
Fast refers to rate at which it occurs (in the spontaneous direction)
How fast will it occur?
Kinetics will answer / address this question
Does depend on how the process occurs (“pathway”; transition state, activation energy, etc.)
Recall that we should be spontaneously combusting right now!!
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3. Figure 18.4 Thermodynamics domain vs. Kinetics domain
Initial and final states, spontaneity
“Path independent”
“Pathway dependent”
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4. Reminder: 1st Law of Thermodynamics
The total amount of energy in the universe never changes (it’s constant)
Euniv=  Esys = -Esurr
“If energy leaves the system, it must go to the surroundings, and vice versa”
Has nothing to do with “spontaneity”
1st Law is consistent with a casserole dish coming out of the oven colder than when it was put in, as long as the oven would get hotter!
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5. Spontaneity and the 2nd Law
A spontaneous process is one that occurs (at a specified set of conditions) “without external intervention”.
Gases expand into the volume of their containers
Ice will melt at 10°C
Water will freeze at -10°C
A chemical reaction will occur in the forward direction if Q < K
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6. Spontaneity and the 2nd Law (continued)
At the same conditions, the reverse process of a spontaneous one is not spontaneous
Gases won’t spontaneously “contract” into a smaller volume (perfume molecules going back into the bottle after opened?)
Water will not freeze at 10°C
Ice will not melt at -10°C
A chemical reaction will not occur in the reverse direction if Q < K
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7. Spontaneity and the 2nd Law (continued)
What determines whether a given process is spontaneous or not?
This question is not addressed by the 1st Law; it is addressed by the 2nd Law.
2nd Law:
For a spontaneous process to occur, the entropy (S) of the universe must increase.
Suniv > 0 for any spontaneous process
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8. **Suniv  Ssys!** Be careful!
The universe is made up of two “parts”—system and surroundings:
ΔSuniv = ΔSsys + ΔSsurr
It doesn’t ultimately matter if ΔSsys is positive, or if ΔSsurr is positive. The key is whether ΔSuniv is positive! (next slide).
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9. Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl)
ΔSsys + ΔSsurr = ΔSuniv
This turns out to be very temperature dependent (See Section 18.5 in Tro). We will come back to this issue later.
plus
equals
See next slide
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10. Bar plots to visualize idea on prior slide (system and surroundings both contribute to DSuniv!)
ΔSsys + ΔSsurr = ΔSuniv
|ΔSsurr| > |ΔSsys|
|ΔSsys| > |ΔSsurr|
NOTE: These two examples both have DSsys < 0 and DSsurr > 0, but all possible variations are possible!
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11. What is entropy? Hard thing to define conceptually!
Something like “energy dispersal”--not necessarily “spatial” dispersal, but energy dispersed over the various motions of atoms & molecules. Reflects quality or usefulness of energy.
Let’s start with individual substances
Look at what affects the amount of entropy in a sample of a single substance (without necessarily understanding “what” it is”).
Then we’ll look at mixtures of substances (in a reactive system, e.g.)
When a change occurs in the system, Ssys changes because there are new substances present (or new physical states)
We’ll consider the surroundings last
Just a “bunch of substances not doing anything” (Prof. Mines’ view)
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12. Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 18.7)
Entropy of a sample of a substance increases with:
Energy added (T increase or phase change)
Solid < Liquid << Gases
Volume (gas or solution species)
Complexity of basic unit of substance
Number of moles of the substance
Entropy, like energy, is a “per mole” kind of quantity
Recall that we’ll typically only specify substances that undergo some change to be part of the system.
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13. Differences in entropy of physical states (assume a given T)
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14. Fig. 18.5 Entropy of a substance increases with T, and depends (significantly) on state (s, l, g)
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15. Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse
Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E
System A (4 J)
System B (4 J)
E = 5 J
One “way” to arrange the energy (one [micro]state)
Two “ways” to arrange the energy (two [micro]states)
Both systems have 4 J of energy, but the entropy of System B is greater because it has two ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see Slide 17)!
Chapter 17, Unnumbered Figure 2, Page 775
What if an energy state at 5 J was present? Would the # of microstates change in either system?
No, because there isn’t enough energy (in either system) to access that 5 J state. But…
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16. Two “ways” to arrange the energy (vs. one before); two[micro]states
Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E
System A (6-7 J)
System B (6-7 J)
E = 5 J
E = 5 J E
Two “ways” to arrange the energy (vs. one before); two[micro]states
Three “ways” to arrange the energy (vs two before); three [micro]states
Both systems have 6-7 J of energy, but the entropy of System B is (still) greater because it has three ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see next slide)!
Chapter 17, Unnumbered Figure 2, Page 775
What if 2-3 more units of energy were added (to make a total of 6-7 J)?
In both cases, adding energy (raising T) leads to increased S!
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17. More accessible energy levels in gas!
Chapter 17, Figure 17.6
”Places” for Energy
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18. Adding more accessible energy states
Patterns for Entropies of Substances (revisit earlier slide, with nanoscopic “explanation”)
Entropy of a sample of a substance increases with:
Energy added (T increase or phase change)
Solid < Liquid << Gases
Adding units of energy
Adding more accessible energy states
Volume (gas or solution species)
Complexity of basic unit of substance
Number of moles of the substance
Entropy, like energy, is a “per mole” kind of quantity
(at a given T)
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19. (NOTE: Appendix has lots more data. Use appendix for PS8!)
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20. (From another text [McMurry])
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21. Predicting whether processes are “entropy increasing” or “decreasing” (in the system)
(the most dominant effect in most cases is the number of moles of gases made and lost [Dngas], not the complexity of the substance[s]):
C(s) + 2 H2(g)  CH4(g)
N2(g) H2 (g)  2 NH3(g) [see next slide]
2 CO(g) + O2(g)  2 CO2(g)
CaCO3(s)  CaO(s) + CO2(g)
2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
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22. Can quantify the change in S (at a given T) using standard entropies of substances
C(s) + 2 H2(g)  CH4(g) DSrxn° = – ( x 130.6)
x = J/K (per mol of C reacted)
(consistent with prediction, S decreases)
Although CH4 is more complex than H2, it doesn’t have twice the entropy (per mole) as H2, so the dominant effect here (as usual) is the change in the number of moles of gases on reaction, Dngas)
N2(g) + 3 H2 (g)  2 NH3(g) DSrxn° = 2 x – ( x 130.6)
x x = J/K (per mol of N2)
Again, consistent with prediction (S decreases). Though NH3 more complex, only 2 moles of it form compared to 4 moles of gas “lost”).
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23. Recall (1st semester):
Now: DS S° S°
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24. The entropy of a perfect crystal at 0 K is zero (3rd Law).
W = “the number of microstates” (ways to arrange energy)
Tro, Fig. 17.8
Zumdahl, Fig 16.5
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25. Entropy increases for a substance as energy is added to it (either T increases, or a phase change occurs)
Curve differs for different substances.
This is how standard entropies of substances are determined (at, say, 298 K)
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26. See Section IV of Handout Outline
Relating Entropy of the Surroundings to a Property of the System--What is “free energy” (G)?
See Section IV of Handout Outline
Up through the derivation of the expression:
DG = DH –TDS
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27. Consider only (a) for now.
EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S
Consider the reaction for the decomposition of carbon tetrachloride gas:
(a) Calculate G at 25 C and determine whether the reaction is spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous.
Consider only (a) for now.
SOLUTION
(a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules).
The reaction is not spontaneous.
At the specific conditions / concentrations of the reaction system!
© 2011 Pearson Education, Inc.
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28. Back to Outline Conceptual connections…
Now, return to discuss T-dependence of spontaneity (specifically, DSuniv and DGsys)
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29. Now begin to consider (b)!
EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S
Consider the reaction for the decomposition of carbon tetrachloride gas:
(a) Calculate G at 25 C and determine whether the reaction is spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous.
Earlier did (a)
Now begin to consider (b)!
SOLUTION
(a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules).
The reaction is not spontaneous.
At the specific conditions / concentrations of the reaction system!
© 2011 Pearson Education, Inc.
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30. RECALL: Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl)
ΔSsys + ΔSsurr = ΔSuniv
This turns out to be very temperature dependent (See Section 18.5 in Tro). We will come back to this issue later.
plus
equals
See next slide
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31. Revisiting Earlier Example
If both of these plots represent the same exact process under the same exact conditions EXCEPT THAT ONE IS AT A HIGHER TEMPERATURE, then which plot represents the higher T and which the lower T? How do you know?
What would the bar plot look like if the T went to infinity?
To zero?
(see next slides →)
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32. As T → , |DSsurr| → 0! (so DSuniv → DSsys)
T increasing
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33. As T → 0, |DSsurr| →  ! (so DSuniv → DSsurr)
(Y-axis scale much bigger (zoomed out); DSsys same as on right.)
T decreasing
T very small
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34. Summary, T-dependence of DSuniv
At very high T, DSuniv approaches DSsys
Because DSsurr becomes tiny (and DSsys remains essentially unchanged)
At very low T, DSuniv approaches DSsurr
Because DSsurr becomes huge, while DSsys remains essentially unchanged
**A more detailed summary table of this will be discussed a bit later.
NOTE: Many people (most, actually) end up taking a different approach to this “T-dependence” issue: They look at DGsys rather than DSuniv. We’ll explore this next!
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35. -TDS (DSsys dominates)
DG = DH –TDS
Thus: As T goes to…
…infinity, DG →
-TDS (DSsys dominates)
…zero, DG →
DH (DSsurr dominates)
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36. Tro’s “version” of this
DG = DH –TDS
Thus: As T goes to…
…infinity, DG →
-TDS
…zero, DG → DH
DGsys “view” DH
-TDS
opposites
NOTE: You could also look at each of these from a DSuniv “view”!
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37. Add the linear “version” of DG vs T. Slope related to sign of DS
Add the linear “version” of DG vs T. Slope related to sign of DS. Intercept is DH
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38. “Global” (entropy) view (Review of earlier slide)
DSuniv = DSsys + DSsurr;
Thus: As T goes to…
…infinity, DSuniv →
DSsys
…zero, DSuniv →
DSsurr
DSuniv “view”
DSsys
DSsurr
DSsurr
DSsys + +
(DSsurr > 0)
(DSsys > 0) - -
(DSsurr < 0)
(DSsys < 0) - +
(DSsurr > 0)
(DSsys < 0) + -
(DSsurr < 0)
(DSsys > 0)
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39. T-Dependence Explored Further
Table 16.4 (Zumdahl) Results of the Calculation of ΔSuniv and ΔG° for the Process H2O(s)  H2O(l) at -10°C, 0°C, and 10°C
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40. C2H4(g) + H2(g)  C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K
EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S
Now consider part (b)
Consider the reaction for the decomposition of carbon tetrachloride gas:
(a) Calculate G at 25 C and determine whether the reaction is spontaneous.
(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous.
(a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules).
SOLUTION
The reaction is not spontaneous.
At the specific conditions / concentrations of the reaction system!
(b) Since S is positive, G will become more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 17.9 to find the temperature at which G changes from positive to negative (set G = 0 and solve for T). The reaction is spontaneous above this temperature.
FOR PRACTICE 17.3
Consider the reaction:
C2H4(g) + H2(g)  C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K
Calculate G at 25 C and determine whether the reaction is spontaneous. Does G become more negative or more positive as the temperature increases?
© 2011 Pearson Education, Inc.
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41. Relationship of DG to Q (and DG)
See Outline (and board)
ΔG = ΔG° + RT ln Q
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42. Relationship of DG to K
See Outline—Derive from prior slide’s relationship!
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43. Table 16.6(Zumdahl) Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction
DG = -RT lnK
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44. Ppt04_Thermo Chapter 17, Figure 17.14
Free Energy and the Equilibrium Constant
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45. From another text
These data provide a more direct way to calculate the DG for a chemical reaction equation (i.e., you don’t need to calculate DH and DS if you know DGf values!
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46. Fig. 17.10 (Tro). Why “free” energy?
DH = kJ
DS = J/K
DG = kJ
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47. Table 16.1 The Microstates That Give a Particular Arrangement (State)
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48. Table Probability of Finding All the Molecules in the Left Bulb as a Function of the Total Number of Molecules
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