Research: Science & Education
How Do Approximations Affect the Solutions to Kinetic Equations? Jonathan M. Goodman* Department of Chemistry, University of Cambridge, Lensfield Road, Cambridge CB2 1EW, U.K.
Rate equations can be confusing to learn, especially if analytical solutions are not available and approximations need to be introduced. What is the effect of each approximation? Does it matter? Rate equations may readily be solved by numerical methods, now that computers are ubiquitous and sufficiently powerful to run the algorithms required to find accurate numerical solutions to these equations. Approximate solutions to the equations were developed so that results could be obtained without powerful computers, at the time when such computers were not available. Computers now make it possible to make visual comparisons of the approximate and the precise solutions. Such a presentation should help give an understanding of the effects of various approximations. A common kinetic scheme is A changing into B, followed by another reaction to form C (eq 1). The rate equations for this reaction can be solved analytically, and so there is no doubt about the correct expression for the variation of the concentrations with time. k1
k2
A → B → C
(1)
A more complex situation is shown in eq 2. A reacts with E to form B, in a reversible reaction, which can then form C and regenerate E in an irreversible reaction. A+E
k1 k᎑1
k2
B → C + E
(2)
All reactions are potentially reversible, so this is an approximation; but many reactions are, for all practical purposes, irreversible. This is commonly used as an approximation to enzyme-catalyzed reactions, with E representing the enzyme, A the substrate, B the enzyme–substrate complex, and C the product of the reaction. The same equations can also be applied to nonenzymic reactions, where E now represents a catalyst and B, an activated intermediate. In this more complex case, the rate equations cannot be solved analytically. How can these rate equations be analyzed? Analytical solutions to the equations for these reactions may not be available, but numerical solutions may readily be found. Approximate solutions may also be found, but there is a choice of approximations that may be made. Approximation I The first approach to an approximate solution to the rate equations for these reactions invokes the steady state ap-
proximation. An expression for the rate of change of concentration of B may be written down:
d B
= k 1 A E – k ᎑1 + k 2 B
dt
If the concentration of B is assumed to be constant (the steady state approximation), the equations become simpler. This assumption is reasonable, provided the concentration of the intermediate B is small, which will be the case if the reactions that consume B are faster than those that create it. The sum of the concentration of E and the concentration of B must be a constant, since E is regenerated as B breaks down. The initial concentration of E can be labeled [E]0, so [E] = [E]0 – [B]. Substituting this expression for [E] and setting the rate of change of B to zero gives
d B dt
= k 1 A E 0 – k 1 A + k ᎑1 + k 2 B = 0
(4)
The rate of formation of C is just k2 times the concentration of B, and eq 4 is an expression for the concentration of B. This gives the following expression for the rate:
d C dt
=
k1 A Km + A
E
(5)
0
Km, which is called the Michaelis constant, is
k 2 + k ᎑1 k1 This expression is the Briggs–Haldane analysis of the reaction (1). The Michaelis–Menten analysis (2) incorporates the additional approximation that k2 is much less than k ᎑1. This leads to a very similar result, but Km has the simpler form k ᎑1/k1. These equations are very useful in the analysis of enzymecatalyzed reactions, and fuller accounts of this are available (3). It is still not possible to analytically integrate this equation. Remembering, however, that [E] is likely to be small, we can introduce a second approximation, and set [A] = [A]0 – [C], giving eq 6. This version of the Briggs–Haldane equation can be integrated, using two standard integrals given as eq 7:
d C dt
*Email:
[email protected].
(3)
=
k2 A 0 – C Km + A 0 – C
E
0
JChemEd.chem.wisc.edu • Vol. 76 No. 2 February 1999 • Journal of Chemical Education
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Research: Science and Education
d x = ln p + q x p + qx q
x dx x p p q x (7) p + q x = q – q 2 ln +
At time zero, the concentration of the product, C, must be zero, so we arrive at eq 8. At last we have an expression relating the concentration of C to time, and so we can plot the concentration of the product, C, against time as the reaction proceeds. For any particular concentration of C we can apply the equation to discover the time at which it occurs. It is not easy to visualize the form of this equation. It is only possible to take logarithms of positive numbers, so [C] ≤ [A]0. As [C] approaches [A]0 the value of the logarithm will become large, so [C] must approach [A]0 asymptotically as time goes on. If [C] is very small, then it is approximately linearly related to time. The curve must therefore bend in the middle.
k 2 E 0 t = C + K m ln
A
0
(8)
A 0– C
Approximation II A different approach to integrating an approximate version of the rate equation may be tried. Returning to the original rate equation, eq 3, substitute [E] = [E]0 – [B], as before. If A is present in excess, as is often the case for enzyme-catalyzed reactions, we can make the approximation [A] = [A] 0. The equations can now be integrated.
d B dt
= k1 A
0
E 0 – k 1 A 0 + k ᎑1 + k 1 B
(9)
This gives an expression for the concentration of B as the reaction proceeds. Applying the boundary condition that [B] is zero when the reaction begins, we obtain:
k2 A
B =
0
E
1 – e᎑
0
k 1 A 0+k ᎑1+k 2 t
(10)
k 1 A 0 + k ᎑1 + k 2 The rate of formation of C is just k2 times the concentration of B, so this equation can be integrated again to give an expression for how the concentration of C varies with time, eq 11. When t is large, the concentration of C increases linearly with time, the opposite behavior to that found for approximation I.
C =
k2 A
0
E
0
k 1 A 0 + k ᎑1 + k 2
2
e᎑
k 1 A 0+k ᎑1+k 2 t
–1 +
k2 A 0 E
0
k 1 A 0 + k ᎑1 + k 2
t
(11)
Discussion These two apparently reasonable approaches have given two very different results. Which is the better approximation? 276
This can be a source of confusion in understanding the kinetics of such systems. The problem can be resolved by plotting graphs of the results from these equations against a numerical solution to the problem, which shows how the system actually behaves (4). The numerical solutions illustrated here were generated by a Fortran program, based on an algorithm from Numerical Recipes (5), which can be downloaded from http://www.ch.cam.ac.uk/MMRG/software/ as a compiled executable for SGI workstations. Figure 1 shows the results of plotting the concentration of the product, C, against time for the three methods. The initial concentrations of both A and E are set to 0.5. The rate constants k1 and k ᎑1 are set to 1.0 and 0.1, respectively, so this equilibrium will favor the formation of B. However, k2 is set to 1.1, so the concentration of B will not build up very much. Approximation I (䉭) works well, mirroring the shape of the numerical solution (ⵧ ), except at the beginning of the reaction, when [C] and [B] are comparable in size. The approximations that were introduced were first, that the concentration of B is constant, and second, that the concentration of A is much greater than the concentration of B. The first of these is not true at the beginning of the reaction, as the concentration of B must increase from zero. The second may well be true, as the reactions that destroy B should go faster than the reactions that form it. Approximation II looks very bad, at first glance. The initial concentration of A is 0.5, so it is not possible to obtain a concentration of C greater than 0.5, but the approximation happily predicts this after time = 3. The strength of this approximation is at the beginning of the reaction, when it follows the shape of the numerical curve rather precisely. In this region, the approximation [A] = [A]0 is reasonable. Approximation I does badly before time 1 because the concentration of the intermediate B is building up. Figure 2 shows how the concentration of the product changes when k1 = 1, k ᎑1 = 0.1, k 2 = 1.1, [E]0 = 0.2, [A]0 = 0.8. The rate constants are the same as before, but the initial concentrations of A and E are changed to give an excess of A. At first glance, it seems that approximation I is far superior to approximation II. However, this is only the case as the reaction proceeds. At the very beginning of the reaction, before time = 1, approximation II follows the course of the numerical solution much more precisely. Conclusion Two solutions to the rate equations are derived, which give rather different expressions for the rate of formation of the product of the reaction. A program has been written to allows students to compare the results of these approximations with a numerical solution to the rate equations, for any choice of rate constants and initial concentrations. This enables students to experiment with the effects of the approximations in different situations and to find conditions for which they may be reliable. This program may be downloaded from the Internet; an animated version of the equations is available on: http://www.ch.cam.ac.uk/SGTL/teach/kinetics1.html. Each of the two approximations to the kinetic equations works well. The first gets the general shape of the line correct and the second gets the initial stage of the reaction rather precisely. Using computers in this way may help students to
Journal of Chemical Education • Vol. 76 No. 2 February 1999 • JChemEd.chem.wisc.edu
Research: Science & Education 1.0
of C
1.0
Approximation II 0.8
Concentration of C
Concentration
0.8
0.6
Approximation I
0.4
Approximation II
0.6
0.4
Approximation I
Numerical Solution 0.2
0.2
0.0 0
2
4
6
8
10
Time Figure 1. k1 = 1; k᎑1 = 0.1; k 2 = 1.1; [E] 0 = 0.5; [A] 0 = 0.5; ⵧ : numeric solution; 䉭: approximation I; 䊊 : approximation II.
Numerical Solution
0.0 0
2
4
6
8
10
Time Figure 2. k 1 = 1; k ᎑ 1 = 0.1; k 2 = 1.1; [E] 0 = 0.2; [A] 0 = 0.8; ⵧ : numeric solution; 䉭: approximation I; 䊊 : approximation II.
better understand the results of various kinetic schemes. Acknowledgments I thank the Royal Society for support and Alan Fersht for helpful discussions. Literature Cited 1. Briggs, G. E.; Haldane, J. B. S. Biochem. J. 1925, 19, 338–339. 2. Michaelis, L.; Menten, M. L. Biochem. Z. 1913, 49, 333–369. 3. See, for example, Fersht, A. R. Enzyme Structure and Mechanism; Freeman: New York, 1988; pp 98–101. 4. The mathematical requirements of this approach have recently been outlined: Pavlis, R. R. J. Chem. Educ. 1997, 74, 1139– 1140. 5. Press, W. H.; Teukolsky, S. A.; Vetterling, W. T.; Flannery, B. P. Numerical Recipes in FORTRAN; Cambridge University Press: New York, 1992.
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