In the Classroom edited by
Second-Year and AP Chemistry
John Fischer Ashwaubenon High School Green Bay, WI 54303-5093
The Reaction Quotient Is Unnecessary To Solve Equilibrium Problems Paul S. Matsumoto Galileo Academy of Science and Technology, 1150 Francisco Street, San Francisco, CA 94109;
[email protected] The impetus for this article is the algorithm that a number of my advanced placement (AP) chemistry students used in an exam on equilibrium. The specific problem was to find the equilibrium concentration of chemicals in a reaction, given the equilibrium constant (Keq) and the initial concentrations of all chemicals in a reaction. The algorithm was similar to the traditional algorithm (1, 2) used to solve such problems without evaluating the reaction quotient (Q). The purpose of this article is to describe the algorithm and its benefits. The study of chemical equilibrium has important applications (1, 2). Many students have difficulty in solving problems involving equilibrium (3, 4). A reason for student difficulty in solving equilibrium problems may be that such problems involve many concepts, such as concentration, stoichiometry, and the abstract nature of using mathematics to describe the system (5). Description of the Traditional Algorithm The first step in the traditional method to determine the equilibrium concentration of chemicals in a reaction, given Keq and the initial concentration of chemicals, involves the determination of Q. For the general chemical reaction,
aA + bB
determine Q and to arbitrarily choose either expressions in Table 1 or Table 2 to substitute into eq 1. The comparison between Q and Keq to determine the sign of the change in the initial concentration of the chemicals in a reaction requires an abstract mathematical analysis that students may lack; therefore its elimination would greatly simplify the algorithm to solve such problems. Notice that if x = ᎑y, then the expression in Table 1 and Table 2 are identical. The consequence of this situation is that identical results would be obtained by using either expressions in Table 1 or Table 2 in eq 1. Identical to the traditional algorithm, the equilibrium concentration of the chemicals in the reaction is obtained by substituting the roots of the equilibrium equation into the appropriate equilibrium expressions in Table 1 or Table 2, where a negative concentration of any chemical makes that root invalid. This situation is analogous to solving physics problems involving Kirchhoff ’s loop rule, where the direction of the current is arbitrarily chosen to be positive, yet the correct answer would be obtained even if the initial choice is invalid (7). The lack of concern to determine the direction of the current simplifies solving such problems. A similar situation is present to determine the equilibrium concentration of all
cC + dD
Keq is
Table 1. ICE Values at Q < Keq c
K eq =
Concentration
d
[C ] [ D ] [ A ]a [B]b
(1)
[Change]
where the equilibrium concentration of the chemicals are used in the Keq expression. The value of Q is obtained by substituting the initial concentration of the chemicals into the Keq expression. Use Q to determine which direction the equilibrium shifts, set up the ICE table (Table 1 or Table 2), solve the mathematical equation (6), and then assign concentrations. Modification of the Traditional Algorithm The problem with the traditional algorithm is the large number of steps in the algorithm and the abstract nature of using mathematics to describe the system (5). The modification to the traditional method by my students was not to
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[Initial]
•
[Equilibrium]
A
B
C
D
[A]o
[B]o
[C]o
[D]o
−ax
−bx
+cx
+dx
[A]o − ax
[B]o − bx
[C]o + cx
[D]o + dx
NOTE: The concentrations of reactants decrease, while the concentrations of the products increase.
Table 2. ICE Values at Q > Keq Concentration
A
B
C
D
[Initial]
[A]o
[B]o
[C]o
[D]o
[Change]
+ay
+by
−cy
−dy
[A]o + ay
[B]o + by
[C]o − cy
[D]o − dy
[Equilibrium]
NOTE: The concentrations of reactants increase, while the concentrations of the products decrease.
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In the Classroom
chemicals in a reaction, given Keq and the initial concentrations of all chemicals in the reaction. That is, the student need not be concerned whether the change in the initial concentration of the chemicals is an increase or decrease in concentration. Conclusion A pedagogical consequence of this algorithm is the simplification of the traditional method, while maintaining its validity. Such a reduction in the complexity of the algorithm may increase the number of students that would be able to solve such equilibrium problems. Based upon the simplicity and validity of the modified algorithm presented in this article, I would recommend that chemistry teachers and textbook authors utilize this algorithm. A sample problem is given in the Appendix. In addition, this algorithm should take less time than the traditional algorithm, which is advantageous when students are taking a timed exam, such as the AP chemistry exam. This experience has also demonstrated that the alternative algorithm that students use in solving problems may be a source of a more efficient algorithm. I would praise such students to encourage their originality and raise their selfconfidence.
Acknowledgments I thank the students in my AP Chemistry class that came up and correctly used this method: Chyanne Xiao Yan Chen, Hai Jian Guan, Rocky Jin Yong Huang, Thai Quoc Tran, Annie Tsang, Rebecca Lijia Wang, and Leonard Lianghui Zhen. I also thank the following educators for reviewing an earlier version of the manuscript: Leighton Izu (University of Maryland), Jonathan Ring, and Gary Tong (Galileo Academy of Science & Technology). Literature Cited 1. Brown, T. L.; LeMay, H. E.; Bursten, B. E. Chemistry, The Central Science, 7th ed.; Prentice Hall; Upper Saddle River, NJ, 1997. 2. Chang, R. Chemistry, 6th ed.; McGraw-Hill; Boston, MA. 1998. 3. Tyson, L.; Triages, D. F.; Ducat, R. B. J. Chem. Educ. 1999, 76, 554–558. 4. Banerjee, A. C. J. Chem. Educ. 1995, 72, 879–881. 5. Kean, E; Middlecamp, C. How To Survive and Even Excel in General Chemistry; McGraw Hill; New York, 1994. 6. Donato, H. J. Chem. Educ. 1999, 76, 632–634. 7. Cutnell, J. D.; Johnson, K. W. Physics, 4th ed.; John Wiley and Sons; New York, 1998.
Appendix The following example, at 375 ⬚C, N2(g) + 3H2(g)
2NH3(g) Keq = 1.2 (2)
will be solved using the expressions in Table 1 and Table 2 to demonstrate that the proper choice of the expression for the equilibrium concentration of the chemicals is not important. Using the expression in Table 1
while the first root is valid, since [N 2 ] e , [H 2 ] e , and [NH3]e > 0. Using the expression in Table 2 Table 4. ICE Values Employing Style of Table 2 Concentration [Initial]
Table 3. ICE Values Employing Style of Table 1 Concentration [Initial] [Change] [Equilibrium]
N2
H2
NH3
2.00 M
1.00 M
3.00 M
−x
−3x
+2x
2−x
1 − 3x
3 + 2x
substituting the expressions for the equilibrium concentration of chemicals in the reaction into eq 1 and solving for x (6), yields x = ᎑0.1407 and 2.247. The second root is invalid, since [N2]e = (2 − x) < 0 and [H2]e = (1 − 3x) < 0,
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[Change] [Equilibrium]
N2
H2
NH3
2.00 M
1.00 M
3.00 M
+y
+3y
−2y
2+y
1 + 3y
3 − 2y
substituting the expressions for the equilibrium concentration of chemicals in the reaction into eq 1 and solving for y (6), yields y = 0.1407 and ᎑2.247. The second root is invalid, since [N2]eq = (2 + y) < 0 and [H2]eq = (1 + 3y) < 0, while the first root is valid, since [N2]eq, [H2]eq, and [NH3]eq > 0. Notice that x = ᎑y. In both situations, the equilibrium concentration of all chemicals are identical: [N2]eq = 2.14 M, [H2]eq = 1.42 M, and [NH3]eq = 2.72 M.
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