How Do We Keep the Journal So Affordable?


How Do We Keep the Journal So Affordable?pubs.acs.org/doi/pdfplus/10.1021/ed074p8.1by K Emerson - ‎1997 - ‎Related a...

1 downloads 82 Views 114KB Size

Chemical Education Today

Letters How Do We Keep the Journal So Affordable? Paul Poskozim asked in a letter in the November issue how we keep the Journal so affordable. This seems a golden opportunity to speak briefly to the readership about the Journal’s economic facts of life. Yes, we are very affordable compared to most similar periodicals in this country and in the entire world. The Journal staff are proud of this and hope that we can keep it that way. There are three crucial factors in keeping subscription rates low. First, the Journal is the property of a nonprofit organization, and all of us who work for it take that seriously. We do a quality job with minimum expenditure—all the time. We are a stand-alone operation; Journal revenues support Journal activities, but they don’t support anything else. Second, we sell advertising space each month, and advertising produces about a quarter of our total revenue. Readers can help increase advertising revenue by letting the advertisers they deal with know that they read and use the ads in the Journal. Using the Reader Inquiry Card regularly helps too—reader service response is a very common way for advertisers to evaluate the return on their ad dollars. Our recent survey shows that readers use the advertising, which is good, because advertising is a major part of what keeps the cost of the Journal low. The third important cost factor is the number of readers—we now have just about 15,000 subscribers, and this is a large enough base so that the printing cost per issue is significantly lower than it would be if the circulation were, say, about 5000. You are important! Maintaining or expanding our subscriber base plays a very important role in keeping the Journal affordable. We need to attract more subscribers, and you can help by advising colleagues, and even students who are interested in teaching as a career, to subscribe. To summarize my response to Paul Poskozim, the Journal is a real bargain because it has • a dedicated, effective staff who believe in it • advertising revenues • many loyal and thoughtful subscribers You, and all other readers, can help maintain the Journal’s affordability by • volunteering to work with the staff as a reviewer, a column editor, or in another capacity • using our Reader Inquiry Card and telling advertisers you saw their product in the Journal • urging others to subscribe and maintaining your subscription.

Computation of Vapor Pressure I read with great interest Rainder Abrol’s piece entitled “Computation of Vapor Pressure” (J. Chem. Educ. 1995, 72, 1083) which can predict vapor pressure based on the van der Waals equation. Such computations may also interest people without FORTRAN and whose computer languages cannot solve cubic equations. It is the solving of cubic equations that permits the user to find the local maximum and local minimum in graph of pressure versus volume for an isotherm, as well as the three volumes, V1, V2, and V3 in order of magnitude, for which the pressure is the same. Abrol’s algorithm compares the ratio of the areas between the trial vapor pressure line and the isotherm between V1 and V2 and between V2 and V3. For those without cubic equation–solving capability, an alternative method would be faster. After selecting a V1 for a trial vapor pressure, V3 (in Abrol’s terminology) may be determined by programmed trial and error. When V3 is found, one need only see how well this trial pressure fulfills the condition RTlog[(V3 – b)/(V1 – b)] + a(1/V3 – 1/V1) – P(V3–V1) = 0 finding the V1, P, and V3 that best fulfill the condition by programmed trial and error. The volumes representing the local minimum, Vmin, and local maximum, Vmax, can be approximated by estimation based on the absolute temperature and the van der Waals constants of the gas. The estimates can be refined by programmed trial and error. Vmax ≈ 8a/9RT + 1.08(a2 – 27RTab/8) 1/2/RT Vmin can be approximated as follows F = 27RTb/8a x ≈ 0.2887641(1 – F0.03096279295)–0.50495929259 (Where the exact value of x is one that fulfills the condition F1–2x + 6F1–x + 12F + 8Fx+1 – 27 = 0) Vmin ≈ b(1 + 2Fx) I readily acknowledge that there may be better methods of estimation or refinements on these approximations, but they should suffice for student use where solving the cubic equation is not an option. A “derivation” of these approximations is available on request. Ben Ruekberg University of Rhode Island Kingston, RI 02881-0809

Kenneth Emerson Journal Publications Coordinator Department of Chemistry and Biochemistry Montana State University Bozeman, MT 59717-0340

8

Journal of Chemical Education • Vol. 74 No. 1 January 1997

Letters continued on page 22

Chemical Education Today

Letters Redox Rap Kids sometimes “hit the wall” when they hit the language and notation required by redox chemistry. Although we teach this topic at secondary school, I wrote the following rap song and videotaped intermediate students performing it in the tradition of their hero, Michael Jackson. Is it possible that redox is a fun topic? My students are now convinced it is! Perhaps other teachers would like to make their own music video! Here are the words: Refrain (To the tune of “Rock me Gently” by Neil Diamond) “Redox gently, redox slowly, Take it easy, don’t you know We ain’t never done redox reactions before!” (× 2)

Rap Beat Here’s what you do, just listen to me. I’ll make it as easy as A, B, C. Above every element in the whole equation Write that number called “oxidation”. Assign a zero to elements that are free Like O2 , He, Zn, and C. Oxygen in compounds is minus two! (For peroxides only minus one is used.) Hydrogen in compounds is always plus one! (For metal hydrides minus one is done.) In metal compounds the following is true: Group I, plus one, … Group II, plus two. If the molecule is neutral, then you know, The sum of the numbers is zero.

Free Energy The recent paper by Treptow (J. Chem. Educ. 1996, 73, 51) on “Free Energy versus Extent of Reaction” covers much the same ground as did my paper in Education in Chemistry, published in 1988. Inevitably, it makes the same point that an essential difference between a reaction involving the interconversion of two solids and one involving two species in a homogeneous phase is that, in the case of the second, there is an important role for the entropy of mixing. However, with regard to his subsidiary point about the overuse of the symbol ∆, it is not at all clear that Treptow’s urgings are well considered. Surely it is enough to say, in regard to either of the reactions mentioned above, that the point of equilibrium will be that where G is a minimum. Thus the first reaction will go to completion but the second will not. Those who are proficient in the calculus will know that, for the second reaction, the equilibrium state can be identified as that where ∂G/∂ξ is zero, but that this parameter plays no useful role in regard to the first reaction. S. R. Logan University of Ulster Coleraine, N. Ireland BT52 1SA

If the molecule is an ion without a charge, The sum of the numbers is zero.

Literature Cited

If the molecule is an ion with a charge, The sum of the numbers is just that large.

Treptow Replies:

So, step number one, as we have said, Write oxidation numbers above each head!

Refrain (× 2) Rap Beat Oxidation–Reduction is the name of the game. Either one or the other is half of the same. So each HALF equation is what you will need On each half of the page before you proceed. For all elements we first balance mass, Save oxygen and hydrogen for the last. Oxygen we balance by adding water, Then at H+ as you know you “otta”. Now balance charge in each half equation By adding electrons with the proper notation. Adjust half-equations by cross-multiplying through To get equal electrons (any number will do). Now add the equations, there’s not much more to do. The electrons must cancel if you’ve carried them through. Write the final equation as neat as a pin. Charge and mass must both balance for the fun to begin!

Refrain × 2, fadeout

22

H+ is read as “H plus”. The students who recorded this video wore dark glasses, baseball caps on backwards, etc. We got the rap beat from a portable keyboard. Certainly a new image for redox chemistry! Laura Kolonie Canterbury University Christchurch, New Zealand

1. Logan, S. R. Educ. Chem. 1988, 25, 44.

I appreciate Professor Logan calling attention to his well-crafted article of several years ago. Had I been aware of it, I would have cited it in my paper. I welcome the opportunity to reiterate my case against overuse of the ∆ symbol. Logan and I both find utility in graphs of G, the free energy of a system, versus ξ, the extent of a reaction occurring in the system. The slope of the graph at any point can be calculated from ∆G° + RT lnQ. It determines if the reaction is at equilibrium or if it will proceed in the forward or backward direction. But, what shall we call this slope? The mathematically correct symbol is ∂G/∂ξ. For students not versed in calculus we can simply refer to it as “the slope of the tangent” or “the rate of change of free energy with respect to advancement of the reaction”. It can even be symbolized ∆G/∆ξ. I believe the current practice of simply calling it ∆G misleads the student. The ∆G symbol should be limited to differences between initial and final states. Instructors who want to avoid the entire issue can continue to determine the reaction direction by comparing Q with the equilibrium constant, K. Richard S. Treptow Chicago State University Chicago, IL 60628

Journal of Chemical Education • Vol. 74 No. 1 January 1997