Why Does an Equilibrium Constant Not Appear Constant? - Journal of


Why Does an Equilibrium Constant Not Appear Constant? - Journal of...

1 downloads 221 Views 68KB Size

Information • Textbooks • Media • Resources

Why Does an Equilibrium Constant Not Appear Constant? R. Bruce Martin Chemistry Department, University of Virginia, Charlottesville, VA 22903

Students and researchers commonly find it difficult to obtain a constant value for an equilibrium constant over different sets of initial conditions. In some student laboratory experiments the equilibrium constant for a single reaction appears so inconstant over the course of a reaction that students wonder whether there even is an equilibrium constant, and instructors may drop the experiment from the curriculum. There are several reasons for inconstancy. Often the experimental results take such a form that the calculation for the equilibrium constant involves small differences of large numbers, especially at low and high percentage completion of a reaction. This article focuses on the role of errors in the values of a property of pure reactant and pure product that is reused in calculations, causing variations in the calculated equilibrium constant. Unless precise data are obtained for the terminal values, serious errors are introduced into the value for the calculated equilibrium constant. In an approach to equilibrium situation, the value of the pure final product may be grossly underestimated (1). Not considered in this article are other sources of inconstancy, such as variations in ionic strength in reactions between ions (2). For the conversion of reactant R to product P, R P, we define the equilibrium quotient according to Q = [P]/[R] = f /(1 – f )

(1)

where the brackets represent molar concentrations and f is the mole fraction of product P at equilibrium, f = [P]/([R] + [P]). This mole fraction varies from 0 < f < 1, and it expresses the extent of a reaction. For a property proportional to the concentration, such as the absorbance at a fixed wavelength, we have AR = εR [R]

and

Thus the equilibrium quotient may be determined from three absorbances: those of pure reactant, pure product, and the system at equilibrium. For concreteness the above analysis employs absorbance as the measured property to follow the reaction extent. The principle and equations are of the same form for other properties, such as the magnitude of optical rotation, intensity in NMR spectra of systems in slow exchange, the chemical shift in NMR spectroscopy of systems in fast exchange, and even the amount of titrant used in laboratory determination of an equilibrium constant for ester hydrolysis. Thus the analysis presented here applies widely. The purpose of this article is to evaluate the role of errant terminal absorbance values A R and AP in causing drifts or extrema in calculated equilibrium constants over a range of initial conditions. In evaluating the equilibrium quotient from eq 2, the terminal values are reused for each determination. Thus for reliable results it is essential that AR and AP be precisely known. On many spectrophotometers it is difficult to obtain reliable absorbance values when the absorbance is greater than one. We evaluate the effects of errant terminal values by dividing the calculated Q value from eq 2 by the true value from eq 1. The results appear in Table 1, where

AP = εP [P]

where εR and εP are molar absorptivities of pure substances R and P, respectively. At equilibrium we have for the observed absorbance A = (1 – f ) εR [R] + f εP[P] = (1 – f )AR + f AP

Table 1. Ratio of Calculated to True Equilibrium Constants for Several Terminal Absorbances and Extents of Reaction Absorbance

f (fraction of reactant reacting)

AR

AP

0.20

0.33

0.50

0.67

0.80

0.00

1.00

1.00

1.00

1.00

1.00

1.00

0.10

1.00

0.50

0.70

0.80

0.85

0.88

᎑0.10

1.00

1.50

1.30

1.20

1.15

1.13

0.00

1.10

0.89

0.87

0.83

0.77

0.67

0.00

0.90

1.14

1.18

1.25

1.43

2.00

0.05

0.95

0.80

0.92

1.00

1.09

1.25

from which f = (A – A R )/(A P – A R ), and the equilibrium quotient becomes

᎑0.05

1.05

1.18

1.07

1.00

0.93

0.85

0.05

1.05

0.71

0.79

0.82

0.80

0.75

Q = (A – A R )/(A P – A)

᎑0.05

0.95

1.33

1.24

1.22

1.27

1.42

(2)

Continued on page 1498

JChemEd.chem.wisc.edu • Vol. 75 No. 11 November 1998 • Journal of Chemical Education

1497

Information • Textbooks • Media • Resources

the correct values of the initial and final absorbances are scaled from AR = 0.00 to AP = 1.00 (without loss of generality). The first row of the table shows that for the correct values of the absorbances the ratio of the calculated to the true value of an equilibrium constant is unity for all extents of reaction ( f = 0.20 to 0.80) across the table. The second through fifth rows of the table show the results when first AR and then AP is increased or decreased by 0.10 units, corresponding to 10% in the range of AP–AR. When the range is reduced to 0.90 the ratios in Table 1 increase across the table and when the range is increased to 1.10 the ratios decrease. All the ratios in the second and fourth rows are less than unity and all the ratios in the third and fifth rows are greater than unity. Note the ratios of 0.50 near the beginning of the second row and 2.00 at the end of the fifth row, indicating that the ratio of calculated to true value of the equilibrium constant differs by a factor of 2.00. This pronounced difference results from only a 10% error in a terminal absorbance. In the last four rows of Table 1 the 10% variation in the range of AP–AR is split between AR and AP. In the fourth and third-to-last rows the range is reduced to 0.90 and increased to 1.10 but still centered at 0.50, with the result that the calculated-to-true ratios increase and decrease so that they pass through 1.00 at f = 0.50. In the last two rows the range AP–AR remains at 1.00, but the scale slides up or down 0.05 units. Now the ratios bow with a maximum or minimum at f = 0.50, but the ratio of 1.00 and hence a correct value of the equilibrium constant is never obtained. The results across Table 1 may be used as a helpful guide to faults when drifts are observed in calculated equilibrium constants for increasing extents of reaction. Upward drifts of the calculated equilibrium constant suggest too narrow a range in the terminal values; downward drifts, too wide a range; and bowing, a displaced range with both terminal values in error. Conversion of reactant R to product P is driven by other reactants. For example, if addition of strong base deprotonates weak acid R to give weak base P, the acidity constant K a = (H+)[P]/[R] may be evaluated from absorbance measurements from pK a = pH – log Q (3) The value of Q depends upon the pH and the amount of strong base added. During the course of the titration we start with an absorbance corresponding to that of reactant R when f = 0, and finish with an absorbance corresponding to that of product P when f = 1. When applying eq 3 for increasing values of 0 < f < 1 investigators notice drifts and even a minimum or maximum in pK a values. Such deviations may be due to acid–base impurities. They may also be due to incorrect terminal absorbances and comparison of the drifts with those in Table 1 may suggest the parameter wherein error lies. It is common practice, especially in teaching laboratories, to estimate an equilibrium constant for each of several sets of conditions and then report an average value over several determinations. With even small errors in the reused terminal points, estimated equilibrium constants will be in error and vary over a considerable range as illustrated in Table 1. A more satisfactory procedure is to cast the relevant equations into a linear form and estimate the equilibrium constant from the slope and intercept. 1498

Since the conversion of reactants to products is driven by presence of another substance, S, we may write a more complete reaction and its equilibrium constant as R+S

P

K = [P]/[R][S] = Q/[S]

Solving the equation for the observed absorbance we obtain two equivalent expressions.

A=

K [S]A P + A R 1 + K [S]

and

A – AR =

K [S](A P – A R) 1 + K [S]

Often a form of the second equation is plotted as a linear double reciprocal plot of 1/(A – AR) versus 1/[S] corresponding to a Benesi–Hildebrand plot for evaluation of an equilibrium constant and a Lineweaver–Burke plot for a rate constant in enzyme kinetics. However, such double reciprocal plots weight points at one end of the plot hundreds of times more heavily than at the other, and should never be used (1). It is impossible to allow for such very different weightings when attempting to draw a straight line through scattered points. Preferred is a linear plot of the form [S]/(A – AR) versus [S] from which the difference AP – AR may found from the reciprocal of the slope and the equilibrium constant from the slope over the intercept. This Scott or Hanes plot weights points much more equally, allowing a reasonable straight line to be drawn, and is also amenable to treatment by simple unweighted linear least squares (1). For a system with random error, the most appropriate and sophisticated method for evaluating the unknown parameters is a nonlinear least squares treatment, in which variations in the most error-prone observable are minimized in an equation where it alone appears on the left side and all other variables and parameters to be fitted appear on the right side. The first of the above pair of equations meets this criterion: the most error-prone observable, the absorbance, appears alone on the left side. Note that in contrast to eq 2, a difference of large numbers does not occur in this equation. This equation was employed in a nonlinear least squares analysis of the cases in Table 1. In each case the calculated equilibrium constant is close to the value listed in the table for f = 0.50. For example, in the last case with 5% error in each terminal absorbance, the fitted ratio is 1.25(5), within one standard deviation in the last digit in parentheses to the value of 1.22 in Table 1. The fitted terminal absorbances closely approach those listed in the table: for the last case, AR = ᎑0.048(9) and AP = 0.955(8). With five “correct” intermediate points and two terminal absorbances with 5% error the nonlinear least squares treatment is also unable to bring the ratio near to 1.00 and the terminal absorbances near to 0.00 and 1.00. Errant values in the terminal points are not corrected by any procedure with five correct intermediate points. A greater number of correct intermediate points would help. But the lesson of this article is that no procedure overcomes the limitation of incorrect terminal points in estimating reliable values for an equilibrium constant. An especial effort should be made to obtain reliable terminal points. Literature Cited 1. Martin, R. B. J. Chem. Educ. 1997, 74, 1238–1240. 2. Martin, R. B. J. Chem. Educ. 1986, 63, 471–472.

Journal of Chemical Education • Vol. 75 No. 11 November 1998 • JChemEd.chem.wisc.edu