Chemical Education Today
Letters The Reaction Quotient Is Unnecessary To Solve Equilibrium Problems The Reaction Quotient (Q ) IS Useful After All Paul Matsumoto was absolutely correct in writing “The Reaction Quotient Is Unnecessary To Solve Equilibrium Problems” (1). As he pointed out, algorithms that solve for changes in reactant or product molarity (x) using the equilibrium constant expression, will tell you from the sign of x whether the reaction goes forward or backward to reach equilibrium. Thus calculating the reaction quotient is unnecessary if you plan to solve the problem either by approximation or by algorithm (exactly). Two points are worth considering here however, one regarding the real advantage of the reaction quotient, and one regarding the advisability of using algorithms and K eq to solve for “exact” equilibrium concentrations. The reaction quotient (Q ) has two major advantages in characterizing reaction systems away from equilibrium. First, by calculating Q and comparing it to Keq, one can make an accurate qualitative prediction as to which direction the reaction will proceed in order to reach equilibrium. Second, the maximal free energy available to do work as this system proceeds toward equilibrium can be calculated from the simple equation ∆G = ∆G° + RT lnQ = RT ln (Q/Keq) These two uses of Q generate quick and useful predictions for chemists dealing with real, nonstandard-state, nonequilibrium systems; this is where Q serves its most useful purpose. Furthermore, these conclusions do not require any complicated algorithmic solution regarding the eventual equilibrium situation. This brings me to my second point. It has been stated many times, in this Journal and elsewhere, that asking students to use complex algorithms to solve for “exact” equilibrium concentrations is not a good use of classroom time (2–6). Owing to the effects of both non-ideality (influence of pressure, temperature, and ionic strength on activity coefficients) and competing side reactions, “exact” calculations based on tabulated equilibrium constants may yield concentrations that differ from experimental results by an order of magnitude or more (2–7). These “exact” mathematical predictions regarding equilibrium are inaccurate, and thus not very useful from a chemical standpoint. Class time is better spent on the underlying chemistry, for example, a more rigorous qualitative discussion of equilibrium and non-equilibrium systems. Literature Cited
4. Clark, R. W.; Bonicamp, J. M. J. Chem. Educ. 1998, 75, 1182– 1185. 5. Hawkes, S. J. J. Chem. Educ. 1996, 73, 421–423. 6. Silverstein, T. P. J. Chem. Educ. 2000, 77, 1120. 7. Stolzberg R. J. J. Chem. Educ. 1999, 76, 640–641. Todd P. Silverstein Professor of Chemistry Willamette University Salem, OR 97301
[email protected] No Problems with Q The school day before I read Paul S. Matsumoto’s article (1), I performed a demonstration for my AP chemistry students that they found interesting. I immersed the electrodes of a light-bulb conductivity apparatus into 1.0 M acetic acid. No observable conductivity. I then proceeded to dilute the solution with distilled water, doubling the solution volume. The bulb lit up. My students were perplexed; the observation was counterintuitive to what they expected. After writing the dissociation of the weak acid, and the equilibrium expression, the students plainly see that cutting all the equilibrium concentrations by half will disturb the equilibrium, resulting in a shift to produce more ions in solution. The students realize this because the new reaction quotient, Q, will have a value less than the Ka; the reaction shifts right. Without an understanding of Q—what it really means —this problem cannot be adequately solved. Nor can precipitation problems, where Q and Ksp will determine whether precipitation occurs in a solution where various ion concentrations are mixed. Mr. Matsumoto’s students are to be congratulated for discerning an interesting mathematical procedure. Exclusively utilizing this algorithm, however, short-cuts the understanding of the chemistry involved. Students of chemistry should be challenged to understand why something occurs, and not to be satisfied with how to perform the often mundane calculations. The title of the article is unfortunate and misleading. The reaction quotient is necessary to solve many equilibrium problems. The meaning of Q should always be taught, and the calculation of it utilized, to encourage proper appreciation of chemical equilibria. Literature Cited 1. Matsumoto, P. S. J. Chem. Educ. 2005, 82, 406–407. Robert Lederer
1. Matsumoto, P. S. J. Chem. Educ. 2005, 82, 406–407. 2. Meites, L.; Pode, J. S. F.; Thomas, H. C. J. Chem. Educ. 1966, 43, 667–672. 3. Hawkes, S. J. J. Chem. Educ. 1998, 75, 1179–1181.
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Dr. E. P. Scarlett High School Calgary, Alberta Canada
[email protected] Vol. 82 No. 8 August 2005
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Journal of Chemical Education
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Chemical Education Today
Letters while
The author replies: I wish to thank Todd Silverstein and Robert Lederer for their letters regarding my paper about solving equilibrium problems (1) and wish to respond to their points. While my paper states that the reaction quotient (Q) is not needed to solve equilibrium problems, it does not imply that Q is not valuable. In fact, when I teach this topic to my AP chemistry class, I initially use Q to solve the problem, then mention the alternative method described in the paper. Q is valuable in solving problems involving Le Châtelier’s principle (2), as well as how a reaction with a positive ∆G° value with an appropriate value of Q may be spontaneous (3), which is based on the equation:
I =
HA
+ − H + A
HA
where H+ Ka =
e
HA
1150
A− e
e
H+ = HA
Journal of Chemical Education
i
2
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× 100%
(1)
+ e
(2)
i
HA I 2 + K a I − K a = 0 i
which may be solved using the quadratic formula
I =
−K a ±
K a 2 + 4 K a HA 2 HA
i
(3) i
Using L’Hôpital’s rule to evaluate eq 3, lim [ HA ]i → 0 I = 100% shows that as the concentration of HA decreases, there is an increase in percent ionization. Literature Cited 1. Matsumoto, P. S. J. Chem. Educ. 2005, 82, 406–407. 2. Brown, T. L.; LeMay, H. E.; Bursten, B. E. Chemistry, The Central Science, 7th ed.; Prentice Hall: Upper Saddle River, NJ, 1997. 3. Lenhinger, A. L. Biochemistry, 2nd ed.; Worth Publ.: New York, NY, 1975. 4. Clark, R. W.; Bonicamp, J. M. J. Chem. Educ. 1998, 75, 1182– 1185. Hawkes, S. J. J. Chem. Educ. 1998, 75, 1179–1181. Silverstein, T. P. J. Chem. Educ. 2000, 77, 1120. Stolzberg, R. J. J. Chem. Educ. 1999, 76, 640–641. 5. Po, H. N.; Senozan, N. M. J. Chem. Educ. 2001, 78, 1499– 1503. 6. Donato, H. J. Chem. Educ. 1999, 76, 632–634. 7. Levine, I. N. Physical Chemistry; McGraw-Hill: New York, NY 1978. Paul S. Matsumoto
e
− H
e
Solving for [H+]e in eq 2 and substituting into eq 1 followed by rearrangement yields,
∆G = ∆G° + RT ln Q. That is, I agree with Lederer and Silverstein that Q is a valuable concept. The value of not using Q to solve equilibrium problems was to simplify its solution and not to imply that Q is not valuable. I agree with Silverstein’s comment that solving for “exact” equilibrium concentrations is not a good use of class time owing to the complexity of the calculations (4). While my paper simplifies the calculations to determine the equilibrium concentrations of chemicals in a reaction compared to the traditional algorithm, it does not affect the “accuracy” of the results. I agree with Silverstein’s point that class time is better spend on learning chemistry on a semi-rigorous (or more qualitative) level rather than at a more rigorous (or quantitative or tedious) level, especially at the high-school level. For example, (i) solving buffer problems using the Henderson– Hasselbalch equation (2, 3), despite its limitations (5), is preferable to using the traditional “ICE” table (2), (ii) using a graphing calculator (6) is preferable to solve equilibrium problems than to do calculations “by hand”, and (iii) solving for pH as “᎑log [H+]” (2) is preferable to using “᎑log a(H+)” (7). Lederer’s interesting demonstration shows that an understanding of Q may be used to explain an unexpected observation. An alternative explanation of the demonstration would be that a decrease in the (initial) concentration of a weak acid, increases the percent ionization (I ) of the acid (2), which may be rationalize as follows. The dissociation of a weak acid is
H+
Galileo Academy of Science & Technology San Francisco, CA 94109
[email protected] Vol. 82 No. 8 August 2005
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