In the Classroom
Visualizing Entropy Joseph H. Lechner Department of Chemistry, Mount Vernon Nazarene College, 800 Martinsburg Road, Mount Vernon, OH 43050;
[email protected] Entropy is discussed in most introductory chemistry courses as a possible explanation for the spontaneity of chemical reactions. A process is said to be spontaneous if it occurs without outside intervention (1). The second law of thermodynamics attempts to identify which kinds of processes can be spontaneous. The version of it attributed to Rudolf Clausius states that “the entropy of the universe tends toward a maximum” (2). Introductory chemistry texts define entropy using terms like disorder (1, 3–7 ), randomness (1, 3, 5, 8), dispersal (9), energy storage (10), or the number of microstates (11). Sometimes the random arrangement of molecules is emphasized, sometimes random molecular motion is stressed, but entropy clearly encompasses both of these (12, 13). In Ludwig Boltzmann’s formula (eq 1), W represents the number Table 1. Materials for a Rainbow Tube Size
Inner Containers Outer Container (6 each)
Small
5-mL beakers
Closure
25 × 150-mm test tube
#5 rubber stopper
Medium 10-mL beakers
250-mL poly(methylpentene) graduated cylinder a
#7 rubber stopper
Large
1000-mL poly(methylpentene) graduated cylinder a
#12 rubber stopper
50-mL beakers
aTop
section with pour spout should be cut off so that the stopper will make a watertight seal.
Table 2. Solutions for the Rainbow Tube No. Color
Species Responsible Method of Preparationa ,b
1
red
phenolphthalein
combine 1 mL 0.1% (w/v) phenolphthalein (in ethanol) with 100 mL 0.1 M NaOH
2
yellow
mostly Fe(OH)(H2 O)5 2 +
dissolve 2.4 g Fe(NO3 )3 in 100 mL H2 O
3
blue
Cu(NH3 )4 2 +
dissolve 0.19 g Cu(NO3 )2 in 100 mL 1 M NH3
4
orange
probably FeSCN(H2 O)5 2 +
add 1 mL 0.1 M Fe(NO3 )3 and 1 mL 0.1 M KSCN to 98 mL H2 O
5
green
bromcresol green
dissolve 1.4 g NaC2 H3 O2 ?3H2 O in 100 mL 0.1 M HC2 H3 O2 and add 10 drops 1% bromcresol green solution
6
violet
Mn O 4 {
dissolve 0.016 g KMnO4 in 100 mL H2 O
NOTE: We prepare enough material for several repetitions of the demonstration and store each solution in a capped bottle. Solutions should not be placed in the rainbow tube until shortly before class time; otherwise NH3 may diffuse from No. 3 and cause a precipitate in No. 2. aThese formulas yield solutions that have pleasing colors when viewed in 50-mL beakers. A solution’s optical density is directly proportional to its thickness. If smaller beakers are used, more concentrated solutions may be needed to achieve the same color intensity. bThese formulas provide the following final concentrations of the indicated reagent: solution 2, Fe(NO3)3 0.1 M; solution 3, Cu(NO3)2 0.01 M; solution 5, NaC2H3O2 0.1 M; solution 6, KMnO4 0.001 M.
1382
of distinct but equivalent ways to distribute a specified quantity of energy among a collection of molecules, and k has the value 1.38 × 10{23 J K{1. S = k ln W
(1)
A number of analogies have been proposed for helping students to visualize entropy. Toy blocks have more entropy when scattered on the floor than when neatly stacked (14 ). Trash has more entropy if scattered over the countryside than if collected in a wastebasket (7 ). A pile of loose bricks has more entropy than an intact brick wall (8). A jar of mixed nuts has more entropy than a jar containing all peanuts or all cashews (15). A deck of playing cards has more entropy after it has been shuffled (3, 16 ) or thrown in a heap (1, 6, 11, 17) than did the original, unopened deck whose cards were arranged in a predictable order. Umland and Bellama (7 ) object to the latter analogy because each card in a deck is distinguishable from the others, whereas the molecules of a substance are identical. I describe two classroom activities that help students visualize the concept of entropy and appreciate that entropy tends to increase spontaneously. Part 1 (Qualitative): The Rainbow Tube A rainbow tube consists of six beakers (Table 1), each containing a different brightly colored solution (Table 2), stacked inside a large, stoppered test tube or graduated cylinder (Fig. 1A). If it is shown indoors, the rainbow tube can be illuminated by floodlights to emphasize the attractive colors. I first point out to students that the substances inside the rainbow tube are highly organized: all the red solution is in the first beaker; all the yellow solution is in the second beaker; all
A
Figure 1. The rainbow tube (A) before and (B) after inverting.
Journal of Chemical Education • Vol. 76 No. 10 October 1999 • JChemEd.chem.wisc.edu
B
In the Classroom
the blue solution is in the third beaker; etc. In other words, the entropy (disorder) of this system is low. I also point out that the rainbow tube did not generate itself spontaneously; careful assembly was required. Each solution was prepared individually from pure chemical compounds, poured into a separate beaker, and carefully lowered into the tube using tongs. After everyone has had an opportunity to appreciate the beauty and orderliness of the rainbow tube, I introduce the second law of thermodynamics: the entropy in an isolated system tends to spontaneously increase. The contents of a rainbow tube are isolated because the tube is sealed. The stopper prevents it from exchanging energy or materials with the outside world. It would be technically more accurate to say that the rainbow tube is closed (i.e., its contents can still exchange energy with the surroundings), but actual energy flow is negligible under the conditions of this experiment. I ask, what would happen if the rainbow tube were inverted? After students have had opportunity to make predictions, I slowly and dramatically invert the rainbow tube. The solutions mix together, chemical reactions occur, and a nondescript (and unattractive) brown sludge is formed (Fig. 1B). This sludge is more random (higher in entropy) than the original solutions were, because the six pure chemical substances are now mixed and scattered throughout the tube. I next ask the class whether it would be possible to refine the brown sludge (high entropy), recover the six compounds, and return each substance to its original beaker (low entropy). If students hesitate, I suggest that someone could dissolve the sludge in concentrated hydrochloric acid; perform ion-exchange chromatography to separate copper, iron, and manganese; and then re-create the original solutions. Admittedly this would be a tedious and time-consuming process. Students are not eager to attempt it themselves, but they seem willing to acknowledge that a technically competent instructor could perform the separation. Yes, it is possible to reduce the entropy of the rainbow tube. However, the restoration method outlined above requires an open system. One must remove materials from the tube, use additional substances that were not part of the original system, and expend energy from an external source. Furthermore, wastes would be generated. The environment’s entropy would increase if the rainbow tube’s entropy were reduced. Finally, I ask, would it be possible to restore the rainbow tube to its original condition without removing the stopper? Most students answer confidently in the negative. To verify their prediction, I leave the tube in the lecture room until the next class day. Of course, the contents remain a brown sludge. This allows me to make an important distinction between what the second law says and what many people think it says. A popular misconception holds that no system’s entropy can ever spontaneously decrease. Thus, for example, it has been argued that evolution is thermodynamically impossible because highly developed organisms would have less entropy than their simpler predecessors. This argument presupposes that the biosphere is an isolated system. It isn’t, because the earth receives energy from the sun. The second law of thermodynamics does not claim that a decrease in entropy is impossible; it states that a decrease in entropy in an isolated system is improbable. When the rainbow tube is inverted and the solutions in Table 2 are mixed, the acetic acid in no. 5 is neutralized by excess NaOH in no. 1; the MnO4{ in no. 6 is reduced to
insoluble MnO2 by the ethanol in no. 1; and the Cu2+ in no. 3 and the Fe3+ in no. 2 precipitate as hydroxides. The resulting mixture’s pH is approximately 11. Addition of 6 M hydrochloric acid (1 mL of acid per 30 mL of mixture) yields a clear yellow solution whose pH is 2–3. Disposal of this solution via sanitary sewer is permissible in most localities. Part 2 (Quantitative): Money by the Pound Office workers shred confidential papers before discarding them because this decreases the likelihood of someone reassembling a document and reading it. For similar reasons, the Federal Reserve Bank shreds paper currency that has been withdrawn from circulation. The average life expectancy for a dollar bill is 18 months; higher denominations tend to last longer because they are handled less frequently (18). Retired bills are destroyed by cutting them lengthwise into strips approximately 1/16′′ wide. The resulting fragments are much more disorderly than an original, intact bill. Cutting a dollar bill into strips is quick and easy. (I suggest demonstrating this with play money!) Reassembling a shredded bill is possible, but is time consuming and tedious. The shredding process is analogous to a chemical reaction in which a large reactant molecule is converted to many small product molecules. Examples of such reactions include the aerobic metabolism of glucose, the detonation of TNT, and the combustion of gasoline. These reactions are highly spontaneous, both because they involve large entropy increases and also because they are exothermic. Shredded U.S. currency is available from American Science & Surplus, Skokie, Illinois 60076. Item 26470, “Money by the Pound”, costs $2 and contains 454 g of shredded bills of mixed denominations—enough to fill a large (16′′ diameter) stainless-steel kitchen bowl (Fig. 2). I have challenged younger students (grades 7–12) to find the fragments of a single bill and to reconstruct it. To facilitate reassembly, each student is given a sheet of cardstock with two 65-mm strips of double-sided cellophane tape placed 15 cm apart. Fragments of currency can be stretched out between the tapes and lightly adhered to the card. To succeed, not only must one find a complete set of strips all from the same denomination, but the serial numbers in all four corners of the bill must match. So far, no one has recovered a complete bill, but one person (working approximately an hour) found adjacent strips comprising 40% of a bill. I have assumed that all of a bill’s fragments are, in fact, present in the bowl, although this is not guaranteed (some students believe that the Treasury Department transports the top and bottom halves of retired bills to different geographical locations to eliminate any possibility of reassembly). Figure 2. Money by the pound.
JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education
1383
In the Classroom
I have asked college freshman chemistry students to estimate the probability that someone could find all the fragments of a single bill by randomly drawing strips from the bowl. Samples of currency were weighed on an analytical balance, yielding the following relevant information: 1. The average mass of an intact, circulated U.S. bill is 0.985 g. 2. The average mass of one strip of shredded currency is 0.0253 g. 3. Each shredded bill was cut into 0.985/0.0253 = 39 strips. 4. A pound of shredded currency contains approximately 454/0.0253 = 16535 strips. 5. A pound of shredded currency is equivalent to 454/ 0.985 = 461 complete bills. NOTE: the Bureau of Engraving and Printing states that uncirculated currency contains 490 bills per pound (19); however, one would expect previously circulated bills to weigh more because they have adsorbed soil and moisture.
Table 3. Probability of Drawing All n Fragments of the Same Bill in n Successive Attempts from a Collection of 461 Shredded Bills
(2)
For the general case where each bill has been split into n fragments, the probability of finding all the pieces of a single bill in n successive draws is given by P = 1 × (n – 1)/(461n – 1) × (n – 2)/(461n – 2) × … × 1/(461n – [n – 1])
(3)
Spreadsheet evaluation of eq 3 yielded the representative results shown in Table 3. Clearly, the probability of success drops precipitously as n increases. Direct computation failed for values of n greater than 33 because my spreadsheet program could not display a number smaller than 1 × 10{100; however, log10(P) could still be calculated for large n using eq 4: n–1
n–1
i=1
i=1
log10 P = Σ log10 n – i – Σ log10 461n – i
1/P
S / J K{1 a
1
1
0
2
1.09 × 10 {3
9.21 × 102
9.42 × 10 {23
3
1.05 × 10
{6
9.54 × 10
5
1.90 × 10 {22
4
9.60 × 10
{10
1.04 × 10
9
2.87 × 10 {22
5
8.54 × 10
{13
1.17 × 10
12
3.84 × 10 {22
6
7.45 × 10
{16
1.34 × 10
15
4.81 × 10 {22
7
6.42 × 10
{19
1.56 × 10
18
5.78 × 10 {22
8
5.47 × 10
{22
1.83 × 10
21
6.76 × 10 {22
9
4.63 × 10
{25
2.16 × 10
24
7.73 × 10 {22
10
3.90 × 10
{28
2.57 × 10
27
8.71 × 10 {22
20
5.81 × 10
{59
1.72 × 10
58
1.85 × 10 {21
30
7.52 × 10
{90
1.33 × 10
89
2.83 × 10 {21
40
9.19 × 10 {121
1.09 × 10120
3.81 × 10 {21
50
1.09 × 10 {151
9.20 × 10150
4.80 × 10 {21
60
1.26 × 10 {182
7.93 × 10181
5.78 × 10 {21
70
1.44 × 10 {213
6.94 × 10212
6.76 × 10 {21
80
1.63 × 10 {244
6.13 × 10243
7.75 × 10 {21
90
1.83 × 10
{275
5.46 × 10
274
8.73 × 10 {21
100
2.04 × 10
{306
4.89 × 10
305
9.72 × 10 {21
aCalculated
using the formula S = k ln (1/P ).
(4)
When n = 39, P = 1.14 × 10{117, a vanishingly small probability. If this were a lottery, we could say that the odds of winning are one chance in (1/P) = 8.8 × 10116. A person’s chances of hitting the jackpot in a state lottery are roughly 10110 times better than this. For example, the Maryland State Lottery Agency estimates that the odds of winning its Classic Lotto jackpot are one chance in 6.99 million (20). Put still another way, the value of (1/P) represents the number of different possible ways of selecting n strips from the larger collection of strips. Thus (1/P) is analogous to the number of equivalent thermodynamic states that are possible for a system of molecules. I have calculated the “entropy” in a collection of shredded bills using Boltzmann’s principle (eq 1), taking 1/P as the value of W (Table 3). This yields the not-unexpected result that 1384
P
1
n
What is the likelihood of getting all n fragments of the same bill (any bill) by randomly drawing n strips in succession? If the bills have not been shredded at all (n = 1) then, of course, the probability is 1.0 that a complete bill will be found. If each bill has been cut in half (n = 2), the first strip taken will necessarily be a part of one of them (P = 1). There remains one other piece of that particular bill (and 921 pieces total) in the bowl, so the probability of drawing the second strip from the same bill is (P = 1/921). If each bill has been cut into thirds, the probability of drawing all three fragments from the same bill is P = 1 × 2/1382 × 1/1381 = 1.05 × 10{6
entropy is directly proportional to n – 1, where n is the number of strips into which each bill is shredded (Fig. 3). Compared to shredded currency, uncut bills (n = 1) have zero entropy. It seems clear that these calculations yield relative, not absolute, results. A pound of unshredded money must still possess some entropy. Students may be able to suggest several reasons: (i) at a macroscopic level, the bills are of mixed denominations, each bill has a different serial number, the bills are arranged randomly in the bowl, and some of them may be marked or torn in distinctive ways; (ii) at a microscopic level, cellulose molecules in the paper are of varying lengths, they are oriented irregularly, and they vibrate randomly. The foregoing discussion can help students understand why thermo-
Figure 3. Entropy in a collection of 461 shredded bills as a function of the number of fragments per bill.
Journal of Chemical Education • Vol. 76 No. 10 October 1999 • JChemEd.chem.wisc.edu
In the Classroom
dynamic quantities such as G °, H °, and S ° must be calculated relative to a carefully defined standard state. Acknowledgment I thank Tiffany Lippert (Fig. 2) for technical assistance. Literature Cited 1. Zumdahl, S. S. Chemical Principles; Heath: Lexington, MA, 1992; p 372. 2. Barón, M. J. Chem. Educ. 1989, 66, 1001. 3. Ebbing, D. D. General Chemistry, 5th ed.; Houghton Mifflin: Boston, 1996; p 752. 4. Masterton, W. L.; Hurley, C. N. Chemistry: Principles and Reactions, 3rd ed.; Saunders: Philadelphia, 1997; p 471. 5. McMurry, J.; Fay, R. C. Chemistry; Prentice-Hall: Englewood Cliffs, NJ, 1995; p 668. 6. Silberberg, M. Chemistry: The Molecular Nature of Matter and Change; Mosby-Year Book: St. Louis, 1996; p 843. 7. Umland, J. B.; Bellama, J. M. General Chemistry, 2nd ed.; West: Saint Paul, MN, 1996; p 189. 8. Brady, J. E.; Holum, J. R. Chemistry: The Study of Matter and Its Changes; Wiley: New York, 1993; p 563.
9. Moore, J. W.; Stanitski, C. L.; Wood, J. L.; Kotz, J. C.; Joesten, M. D. The Chemical World: Concepts and Applications, 2nd ed.; Saunders: Philadelphia, 1998; p 272. 10. Atkins, P.; Jones, L. Chemistry: Molecules, Matter, and Change, 3rd ed.; Freeman: New York, 1997; p 596. 11. Oxtoby, D. W.; Freeman, W. A.; Block, T. F. Chemistry: Science of Change, 3rd ed.; Saunders: Philadelphia, 1998; p 454. 12. Bickford, F. R. J. Chem. Educ. 1982, 59, 317. 13. Lowe, J. P. J. Chem. Educ. 1988, 65, 403. 14. Fortman, J. J. J. Chem. Educ. 1993, 70, 102. 15. Swanson, R. M. J. Chem. Educ. 1990, 67, 206. 16. Chang, R. Chemistry, 5th ed.; McGraw-Hill: New York, 1994; p 738. 17. Hill, J. W.; Petrucci, R. H. General Chemistry; Prentice-Hall: Upper Saddle River, NJ, 1996; p 735. 18. U. S. Department of the Treasury. Frequently Asked Questions about U.S. Paper Currency; Fact Sheet OPC-34; http://www.treas.gov/opc/ opc0034.html (accessed June 1999). 19. U. S. Bureau of Engraving and Printing. Collector’s Forum; http:// www.bep.treas.gov/forum.cfm (accessed June 1999). 20. Maryland Lottery On-Line; http://www.msla.state.md.us/msla/ index.html (accessed June 1999).
JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education
1385
