In the Classroom edited by
Green Chemistry
Mary M. Kirchhoff American Chemical Society Green Chemistry Institute Washington, DC 20036
Some Exercises Reflecting Green Chemistry Concepts
Yu-min Song,* Yong-cheng Wang, and Zhi-yuan Geng Department of Chemistry, Northwest Normal University, Lanzhou, 730070, PRC; *
[email protected] In a recent article (1) Singh et al. have defined green chemistry as the use of chemistry techniques and methodologies that reduce or eliminate the use or generation of feedstock products, by-products, solvents, reagents, etc., that are hazardous to human health or the environment.
2Al + 3H2SO4
While green chemistry is commonly used in industrial applications, the concepts of green chemistry can be incorporated into educational pedagogy. Along with chemistry concepts, knowledge of which processes present the minimum hazard to human health or the environment should be part of the curriculum (2). Chemists trained in this manner will have a significant impact on solving environmental problems. This article presents some exercises to introduce students to the concept of green chemistry. By doing these exercises, students develop an appreciation for the role of green chemistry on feedstock substitution, milder reaction conditions, reduced environmental exposure, and resource conservation. This, coupled with the “maximum atom utilization theory or atom economy concept” (3) serves as the basis for using green chemistry in laboratory pedagogy. Exercise 1 Aluminum hydroxide, Al(OH)3, is an ingredient of a drug that can be used to treat gastritis. The students are asked to use aluminum, dilute sulfuric acid, and a solution of sodium hydroxide to prepare 8 moles of aluminum hydroxide (4). There are two reaction schemes: Al
H2SO4
NaOH
Al
NaOH
Al2(SO4)3 Na[Al(OH)4]
(1)
Al(OH)3 H2SO4
Al(OH)3
For reaction 1, 24 moles of NaOH and 12 moles of H2SO4 are needed to produce 8 moles of aluminum hydroxide:
(2)
Al2(SO4)3 + 6NaOH 2Al + 3H2SO4 + 6NaOH
2. Propose a new reaction that can save more chemicals.
The answer to the first question can be found using the following rationale: when the feedstocks and reaction conditions are the same, a reaction using less feedstock to prepare a mole of products is better. The stepwise solution is as shown below:
Step 1 Because 8 moles of aluminum produces 8 moles of aluminum hydroxide, the same number of moles of aluminum are needed in both reactions. Thus it is necessary to calculate how many moles of H2SO4 and NaOH are needed in each reaction to produce 8 moles of aluminum hydroxide. www.JCE.DivCHED.org
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3Na2SO4 + 2Al(OH)3 3Na2SO4 + 2Al(OH)3
+ 3H2
For reaction 2, 8 moles of NaOH and 4 moles of H2SO4 are needed to produce 8 moles of aluminum hydroxide:
2Al + 2NaOH + 6H2O
2Na[Al(OH)4] + 3H2
2Na[Al(OH)4] + H2SO4
2Al(OH)3 + Na2SO4 + 2H2O
2Al + 2NaOH + H2SO4 + 4H2O
2Al(OH)3 + Na2SO4 + 3H2
Reaction 2 needs fewer moles of NaOH and H2SO4 than reaction 1, so reaction 2 can save more moles of NaOH and H2SO4. Thus reaction 2 is better from a green chemistry perspective.
Step 2 A better reaction than either reaction 1 or reaction 2 is sought. Students are asked to compare the properties of the intermediate product of reaction 1 with that of reaction 2. The Al 2 (SO 4 ) 3 solution is acidic and the solution of Na[Al(OH)4] is basic. In mixing the two solutions, Al(OH)3 is produced by the metathesis of Al2(SO4)3 and Na[Al(OH)4], so students can suggest a new reaction.
The students are asked to answer the question above with the following considerations: 1. Which reaction is better in terms of using less feedstock?
Al2(SO4)3 + 3H2
NaOH
Na[Al(OH)4]
Al
Al(OH)3 Al2(SO4)3
H2SO4
The relevant reactions are: 2Al + 2NaOH + 6H2O 2Al + 3H2SO4 6Na[Al(OH)4] + Al2(SO4)3 8Al + 6NaOH + 3H2SO4 + 18H2O
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2Na[Al(OH)4] + 3H2 Al2(SO4)3 + 3H2 8Al(OH)3
+ 3Na2SO4
8Al(OH)3 + 3Na2SO4 + 12H2
Journal of Chemical Education
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In the Classroom
Three moles of H2SO4 and 6 moles of NaOH are required to prepare 8 moles of Al(OH)3. Based on the above discussion, the new reaction needs the least moles of H2SO4 and NaOH of the three reactions. For reaction 1, the atom utilization ratio (3) is 26%: 26% = [2FWAl(OH)3兾(2FWAl(OH)3 + 3FWNa2SO4 + 3FWH2)] × 100 The atom utilization ratio is 52% and 58% for reaction 2 and reaction 3, respectively. Exercise 2 Copper(II) nitrate, Cu(NO3)2, is an intermediate dye used in textile mills (5). Three students are asked to prepare 3 moles of Cu(NO3)2. Concentrated nitric acid is used by student A, dilute nitric acid is used by student B, and student C changes Cu into CuO, then uses CuO to prepare Cu(NO3)2. (The products are NO2 and NO with concentrated and dilute nitric acid, respectively.) The students are asked to evaluate the proposed scheme in terms of the quantities of the feedstocks and reagents used and the quantities and identities of the byproducts formed (4). The three students prepare Cu(NO3)2 according to the following reactions: Student A Cu + 4HNO3(conc) Cu(NO3)2 + 2NO2
+ 2H2O
3Cu + 8HNO3(dil)
Student B
3Cu(NO3)2 + 2NO + 4H2O 2Cu + O2
Student C
2CuO Cu(NO3)2 + H2O
CuO + 2HNO3
The amount of feedstock used and the products and byproducts produced are listed in Table 1. To prepare 3 moles of Cu(NO3)2, student A needs the most moles of HNO3 and produces the most pollution. Student C needs the fewest moles of HNO3, produces no pollutants, and generates environmentally benign water as the only byproduct. According to the maximum atom utilization and the principle (6) of green chemistry for using feedstock, student C’s plan is the best of these plans. Exercise 3 The pentahydrated salt of sodium thiosulfate, Na2S2O3⭈5H2O, known as “hypo”, is used in photography. The thiosulfate ion is also used in quantitative analysis as a reducing agent for iodine. Na2S2O3⭈5H2O can be prepared Table 1. Feedstock, Product, and Byproducts from Reactions To Produce Copper(II) Nitrate Student
Cu/ mol
HNO3/ Cu(NO3)2/ mol mol
H2O/ mol
NO2 or NO/ mol
A
3
12
3
6
6
B
3
8
3
4
2
C
3
6
3
3
0
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from the following scheme: SO2
Na2SO3
FeS
Na2S2O3·5H2O H2S
S
The relevant reactions are: Na2SO3 + S SO2 + 2NaOH
Na2S2O3 Na2SO3 + H2O
SO2 + 2H2S
3S
+ 2H2O
FeS + 2HCl
FeCl2 + H2S
4FeS + 7O2
2Fe2O3 + 4SO2
Students are asked to give the best stoichiometric ratio (i.e., the number of moles of FeS needed to prepare SO2 divided by the number of moles of FeS needed to prepare H2S) for maximum feedstock utilization. Because the preparation of Na2S2O3 from FeS involves several steps, students have to write all the reactions and balance them. Then they can calculate the best ratio of starting material (4). From the reactions above, to prepare 3 moles of Na2S2O3⭈5H2O, 3 moles of S and 3 moles of Na2SO3 are needed. Then to prepare 3 moles of Na2SO3, 3 moles of SO2 are needed. But to prepare 3 moles of S, 1 mole of SO2 and 2 moles of H 2S are needed. So to prepare 3 moles of Na2S2O3⭈5H2O, 4 moles of SO2 and 2 moles of H2S are needed. To prepare 4 moles of SO2 and 2 moles of H2S, 4 moles of FeS and 2 moles of FeS are needed, respectively. In sum, the number of moles of FeS for preparing SO2 is twice the number of moles of FeS for preparing H2S. The ratio is two for maximum feedstock utilization. The best ratio of stoichiometry map is: 3SO2 4FeS
3Na2SO3
4SO2 SO2
2FeS
3Na2S2O3 ·5H2O
3S
2H2S
Acknowledgment The support received from the World Wildlife Federation Educators’ Environment Initiative is acknowledged. Literature Cited 1. Singh, M. M.; Szafran, Z.; Pike, R. M. J. Chem. Educ. 1999, 76, 1684. 2. Collins, T. J. J. Chem. Educ. 1995, 72, 965. 3. Trost, B. M. Science 1981, 254, 1471–1477. 4. Cheng, Shi-hua; Liu, Xian-qiao. J. Chemical Teaching (Chinese) 2001, 3, 38. 5. Sneed, M. C.; Maynard, J. L.; Brasted, R. C. Comprehensive Inorganic Chemistry; D. Van Nostrand Co.: New York, 1954; Vol. 2, p 85. 6. Anastas, P. T.; Warner, J. C. Green Chemistry Theory and Practice; Oxford University Press: New York, 1998.
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