edited by: JOHN J. ALEXANDER University d Cincinnati Cincinnati. Ohio 45221
exam que~tionexchange Temperature and Concentration Dependence of Homogeneous Chemical Equilibria C. Brent Smlth North Carolina State University
Raleigh. NC 27695 Conversion efficiency of raw materials into products is an important economic consideration in commercial chemical manufacturing. This efficiency can be studied through reaction rates and positions of equilibria. Such topics are conceptually introduced to students in fundamental courses and later analyzed in detail in junior level undergraduate physical chemistry. An interesting problem for analysis is available1 that allows for the presentation of many concepts in a situation of commercial importance. This problem is useful as an undergraduate exam or homework problem, and can also be used in graduate-level courses. Effects of temperature and molar starting concentration upon positions of equilibria may seem extraordinary a t first glance. Data are available1 for equilibria of dihydroxymethylpyrimidinones in aqueous solution, in particular, equilibrium states of >NCHzOH and CH30H with >NCHzOCH$, i.e., N-hydroxymethylpropyleneurea, plus 21 stochiometric amounts of methanol with N-methoxymethylpropyleneurea in aqueous solutions:
lihrium percent conversions of 28% a t 30 'C and 18%a t 70 OC, when approached from either side of eauilibrium (i.e., eithe; formation of ~ - m e t h o x y m e t h ~ ~ r o ~ ~ leneurea from 0.5 M solution of N-hydroxymethylpropyleneurea and 1.0 M methanol or hydrolysis of 0.5 M N-methoxymethylpropyleneurea). From these data, estimate t h e free energy, enthalpy, and entropy changes (all a t 30 O C and a t 70 'C) for the formation reaction of eq 1.Also estimate the equilibrium position a t 100 OC in otherwise similar circumstances (i.e., starting with 0.5 M aqueous solution of N-hydroxymethylpropyleneurea and 1.0 M methanol, as above). Work in calories, and compute standard free energy of the reaction based on molar concentration units. Acceptable Solutlon a. For chemical equilibrium, the law of mass action states that K = [Products]"' = constant [Reactants]"' where n, and n, are the stoichiometrie coefficientsin the balanced equation. Thus, starting with any concentration Co, of his 2NCH20H in aqueous solution with stoiehiometricamounts (2Co) of CHsOH, they presumably would react according to eq 1farming fractionFof:NCH20CH3. This would give the followingequilibrium concentrations: Starting Concenhation
Equilibrium
Species
Concentration
or simplifying: "Equilibrium position" is the fraction or percent of material on the right of the equilibrium as shown. The question is put in two parts: Puestton a. Describe the "expected" relationship of molar starting concentrations to equilibrium position. Make quantitative com~arisonsof expected equilibrium positions with the foilowing data tb demonsirate whether these observations are consistent with expectations for equilibrium.
Where Co = starting concentration (molar), and R'o = percent conversion a t 30 OC. Also, discuss the "expected" temperature dependence of equilibrium position. b. Figures in the same reference1 (2.1221123) show equi-
'
Peterson. Harro. "Cross-Linking with Formaldehyde-Containing Reactants". Chapter 2 in Functional Finishes, Volume II, Part B; Lewis. M.; Sello. S. B., Eds.; Dekker. New York. 1983; pp 200ff.
To solve this cubic equation quantitatively for F would he difficult, but not impossible, in a general case. However, a quantitative comparison to the given data set can be easily made by calculating a value ofKfor each Coand Fohserved. IfKremains constant, the law of mass action is being obeyed, and the equilibrium position is as expected. This gives: K=
c, (molar)
0.50 1.00 2.00 3.00 4.00
F
-
F (observed)
4C?(1 n3 (caicuiated from Ca and fl
0.280 0.400
0.750 0.463 0.468 0.467 0.512
0.575
0.660 0.720
Except for the values of K for Co = 0.50 M and to a smaller extent 4.00 M, these K values remain remarkably constant. Thus, the variation of F with Co as seen in these data is consistent with expectations,except for the 28%conversion at 30 OC. At this concentration, more conversion than expected was observed. In fact, with K = 0.47 the expected fractional conversionFwould be 0.22 at Co = 0.50 M and 0.71 at 4.00 M. Volume 65 Number 11 November 1988
965
(
- - - - - - - - - - - - - - - - -Equilibrium Is 29% al 30.C
Starting Concentration, 6 (mol/ L)
Time (minutes) Flgure 3. Conversion as a lunction of time at 30 and 70 OC
~igure1. Equliibrlum wition vs. mncenbation.
0
1
I
0.0 0.1 0.2 0.3 0.4 0.5 0.6 Equilibrium Pasition, F
0.7
0.8 0.9
1.0 Temperature, C '
Figure 2. Relative error in Kvs. equilibrium pasition, F.
Figwe 4. Fraction conversion in 1 h vs. temperahne of reaction.
As a practical matter, determination of equilibria that lie far to one side tend to he not very accurate, so errors of this type are not totally unexpected. To put this in perspective, the "expected" vs. observed values are plotted in Figure 1. To illustrate the effect of eouilihrium shifts uDon the relative accuracy of determining K values, one can look at the functional relatiomhip oi K and F given in eq 4. M,the error in K , can be estimated by differentiating eq 4 to give
and
~
~
~~~~~~~~
~~
where 1 F i s the error in measuring F(percent conversion) and AK is the resultant error in the equilibrium constant K . The relative error in K would he AKIK or
then -=-In
R
If AHo < 0, then Kdecreases with increasing T (as in this case). The value of ryIO can he calculated to be -4.3 k d m o l (calculation shown in part b). This reaction involves forming 2 mol each of HOH and C-OC bonds and breaking 2 mol each of CO-H and C-OH bonds. Specific knowledge of the G O and H-0 bond enthalpies of these environments could lead to estimation of AHo. Certainly the value -4.3 kcal/mol is not unreasonable. h. To calculate AHo, using the method above, the following values are obtained: If3, = 0.750
If a constant error in measuring F exists (AF constant), then the relative error in K would be proportional to (1 2F)l(F(1 - F)). This is shown in Figure 2, and it is clear that equilibria which lie far to the left (values of near 0) or right (values of F near 1.0) have higher relative errors than values in the range of 0.3 < F < 0.6. As to the temperature dependence of the equilibrium, one would expect that the values of K from the law of mass action above would depend on temperature through the principle of Le Chatelier. In particular, since
AH' = a constant
d (llQ
@
30 'C (from part a)
K,, = 0.326 @ 70 'C (by a similar calculation)
+
Q66
Journal of Chemical Education
Then AHo = -4.3 k d m o l . This is the value a t either (both) temperature. To estimate AGO a t each temperature, we use eq 7, which gives 173.2 caVmol a t 30 OC and 763.4 callmol a t 70 'C. The entropy change is given by
A,y = AHo - AGO
T
- AHo T
AGO
T
To find the equilibrium position at 100 "C,and starting with Co = 0.50 M,first find Klwat 100 "C by using eq 9
Sequencing QuestionsRevisited by Using a Computer Program Kenton 8. Abel and William M. Hemmerlln Pacific Union College Angwln. CA 94508
The resultant value is Klw = 0.197. Then find F to be 0.30 by trial and error, substituting various values of F into the equation and fiding the value of K. When a value of F is found such that K = 0.197,then that value ofFis the desired solution which satisfies the equation. A typical example of such trial and error estimates is Guess (0
0.1
0.13
Thus the percent conversion is 13.05%.An alternative method would maoh. he t o d o t K vs. F and read the desired value from the . It is also inntructive tc,pnint out that this ir a typical example ofa proress where raisrngthe temperaturecauses an undesirahleshiff to a lower product yreld (because AH' < 0).and that a manutacturer'n desire for high conversion is best facilitated by low temperature conditions. However, the rate of reaction is slower at low temperatures. Therefore, if there is a time constraint on how long the reaction may he allowed to run, as there uaually is in commercial production, the maximum yield within a giuen time may occur at some optimum temperature, as indicated in Figures 3 and 4. ~~
~~
.~~~ ~
~~
~
We recently published in this column (June 1987) an article describing the use of sequencing questions in chemistry examinations. Our partial-credit scheme dealt only with four-item sequences, and credit was based on the number of correct relationships the student deduced. We have expanded the partial credit grading scheme to allow four-, five-, six-, and seven-item sequences. The grading grid for any of these choices is easy to generate by use of a computer program. Once the correct sequence is typed into the computer, all the possible combinations are printed along with the partial credit allowed for each. The format is straiehtforward and easy to use. Although the number of possibie combinations for a six-item question (720) makes grading a bit tedious, four- and five-item questions are quite easy to grade. The program was written in Microsoft Basic for the Apple Macintosh. If you would like a printout of the program, the authors will be happy to send it to you. If you send a Macintosh-compatible disk, the program will be copied onto your disk and the disk returned to you.
Volume 65
Number 11 November 1988
967
