# 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 - â1997 - âRelated a...

Chemical Education Today

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!