edited by JOHN J. ALEXANDER
exam question exchange Correlating Experimental Data with Theory M. Pilar Tarazona Depanamento de O~lmlcaFlslca J n verstdad de A ca A ae henares 28871 AlcalA de Henares Madrid, Spain
Question Values ofmolarheat capacity at constant pressure, C,, at three different temperatures for several substances are given in Table 1. If the substances are (in alphabetical order) C&, CO, COz,He, HzO, and Na, assign them the Cp values taking into account that all data are referred to the gaseous state except those for Na a t 298 and 1000 K, which are for solid and liquid, respectively. Acceptable Solution Table 2 lists the total heat capacity a t constant volume of the gases calculated by adding the contributions from translation, rotation and vibration, estimated using the equipartition principle, together with the C, values calculated using the ideal gas relation Cp = C, + R. However, molecular vibrations contribute to the heat capacity only at temperatures high enough for their excitation; thus, at 298 K, the vibrational contribution must be small. The eighth column of Table 2 shows the calculated value of Cp without a vibrational contribution. Comparing Tables 1 and 2, it is easy to match the experimental and calculated values for gases. A small difficulty may arise in assigning the two diatomic molecules Hz and CO. However, taking Table 1. Values of Cp(J K-' mol-')a
Substance
T(K) 298
a b c d e f 9 'JANAF
20.78 28.15 28.84 29.14 33.59 35.64 37.13
1000 20.78 28.94 30.20 33.18 41.27 71.79 54.31
3000 20.78 21.29 37.09 37.22 55.75 101.39 62.23
Thermochamical Tables, 3rd ed.; Chase Jr.. M. W.; Davies, L. A.;
Downey Jr.. J. R.: Frurip. D. J.; McDonald, R. A.; Syvernd, A. N., Eds.: American Chemical Society:New Yark, 1986.
114
Table 2. Translational, Vibrational and Rotational Contributions to Cvand Calculated Values of C ~ a n d Cp for Ideal Gases
Gas Trans Rot
One of the eoals of a physical chemistr, course is to teach . . students how to correlate cxperimenral results with predicted theoretical data since, most of the time, the experimental values of a given quantity deviate from those calculated using t h e simplest models. Once one or more different theories that permit calculation of a given quantity have been explained, it is important to discuss the experimental values of this quantity. Questions in which experimental data have to be asslgned to different systems i r e useful in evaluating student achievement in this area. Such questions involve intellectual skills a t the applications level.
Journal of Chemical Education
MiMtyofciMati Cincinnati, OH 45221
He Hz CO COz Hz0 CHd
R
R
Vib R
312 312 312 312 312 312
0 1 1 1 312 312
0 1 1 4 3 9
5
Cp
R
R
Cp ~
/
312 512 20.78 712 912 37.41 712 912 37.41 1312 1512 62.36 58.20 6 7 12 13 108.08
CdlowT) Subk ~ ~ Stance ~ 20.78 29.10 29.10 29.10 33.26 33.26
a c d g e f
into account thatthe value of C, at 298 K for compound c is lower than that estimated using the translational and vibrational contributions, it is easy to deduce that these data must correspond to HZfor which the rotational contribution has not yet reached the classical value predicted for the equipartition principle due to its low mass. The unassigned b data must correspond to Na. These data are clearly different since the Cp value at 3000 K is smaller than those for lower temperatures and this fact would be in accord with the discontinuity due to the phase transition to gaseous state. Moreover the value of Cp for solid Na. 28.15 J K-' mol-' (3.39R ) is not far away from the predicted limit ofElnstcm9stheory ofsolids. C, = - 3 ~or, the emo~ricallaw of Dulona and Petit. Caution must he taken in ;sing the first argument since a lower value of C, at high temperatures in a polyatomic gas also can indicate dissociation of the molecule.
Redox Equations with an Infinite Number of Balanced Solutions Jeremy Riley and Thomas G. ~ichmond' University of Utah Salt Lake City, UT 84112
Balancing chemical equations involving oxidationlreduction reactions is a topic covered in virtually all introductory chemistry courses a t both the college and preparatory levels.' Typically the half-reaction (ion-electron)method is introduced as the most convenient method to obtain a balanced chemical equation. However, some mathematically minded students prefer to use various algebraic methods to solve this class of problems. This can lead to a result that, a t first, is surprising: Redox reactions, in which the reactants can disproportionate, actually yield a family of balanced equations rather than a unique solution. Extensive discussions of these methods appeared in this ~ o u r n a l .The ~ - ~following question can be used to lead students to discover that there are multiple ways to balance certain redox equations. 'Author to whom corresoondence should be addressed. to high schools, see Davis, E. J. 2 ~ oar discussion Chern. Educ 1990,67,671 3Kolb,D. J. Chern. Educ. 1978,55,326. 4Kolb,D.J. Chern. Educ. 1979,56,181. 5~olb. D. J Chern. Educ. 1981,58.642. ~~~~~
~
Question a. Using the half-reaction method, balance the reaction of dichromate ion with hydrogen peroxide under acidic conditions, according to eq 1.
[~rz07l~%q) + HzOz(aq)+ OAg) + cr3+(aq)
(1)
b. Keeping in mind that hydrogen peroxide may undergo metal-catalyzed disproportionation to oxygen and water, discuss if your solution to part a is unique. Solution The half-reaction method leads to a straightforward solution to part l.
(
3 H,O,(aq) 6e-+ 1 4 ~ + ( a q+)[cr,0,lS(aq)
8Hi(aq) + [cr20,?(aq1
+ 3H,0z(aq)
-
4
4
O,(aq) + 2HiK +
%j
2cr3+(aq)+ 7H20
302(g)+ 2 c P ( a q ) + 7H20
(2)
An acceptable solution to part b would involve writing a balanced equation for the disproportionation of hydrogen peroxide as given in eq 3. It would also require a statement indicating that eqs 2 and 3 may proceed independently in unknown relative proportions. Thus, no unique solution is obtained. 2HzOz(aq) --,2HzO + Oz(g)
(3)
Alternatively, the algebraic approach could be used revealing more variables than equations.
Hydrogen and chromium are balanced as the variables are assigned. To balance oxygen which reduces to I b + c = 2 d + a
To balance charge 2a - 26 = 6b
which reduces to If b is assigned the value 1,then a = 4, but we are stiU left with 3 + c = 2d, that is one equation with two unknowns. To generate eq 2, c may be chosen as 3. However, any value given c will afford a balanced equation. Thus, an infinite number of solutions are possible. These equations are linear combinations of eq 1and the disproportionation of hydrogen peroxide (eq 3). Although not presented in most general chemistry texts, in this instance the algebraic approach to balancing redox equations provides chemical insight into the reaction of interest. It is well-known and often demonstrated in general chemistry lectures, that many transition metal ions catalyze the disproportionation of hydrogen peroxide. Thus, the infinite family of solutions to this problem may more accurately mirror the reactions that actually take place if such a process were carried out in the laboratory!
Volume 69 Number 2 February 1992
115
