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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Kinetics > Rate Laws > Reaction Mechanisms > Steady State Approximation

Steady State Approximation

The steady state approximation is a method used to estimate the overall reaction rate of a multistep reaction. It assumes that the rate of change of intermediates in a multistep reaction are constant. This method can only be applied when the first step of the reaction is significantly slower than subsequent step in an intermediate forming consecutive reaction, where the product of the first reaction is the reactant in the subsequent reaction.

Introduction

Before discussing the Steady-State Approximation, it must be understand that the approximation is derived to simplify the kinetic expression for product concentration, [product]. When we have a sequential reaction,

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Calculating the [product] can be rather complicated since it depends on all the rate constants in each step. For example, if the kinetic method was used to find the concentration of C, [C], at time t in the above reaction, the expression would be

[C] = [A]0 (1+( k2e-k1t-k1e-k2t)/ (k1 - k2))              (1)

With a more complicated chain reaction, the kinetic expression becomes harder to derive. To simplify this calculation, scientists developed the Steady-State Approximation and the Pre-equilibrium approximation for determining the overall reaction rates of consecutive reactions. In this module, the discussion will be solely on the Steady-State Approximation.

Steady-State Approximation

The Steady-State Approximation is useful to apply to consecutive reaction with a slow first step and a fast second step (k1<<<k2). If the first step is very slow in comparison to the second step, there will not be any accumulation of intermediate product, such as product B in the above example.

d[B]/dt = 0 = k1[A] - k2[B]

or we can say [B] = k1[A]/k2

From the chain reaction, we have

d[C]/dt = k2[B] = k2k1[A]/k2 =  k1[A]

Solving for [C], we will get

[C] = [A]0 (1- e-k1t)             (2)

Compared to equation (1), equation (2) is much simplier to derive, especially when we have to solve for a more complicated chain reaction.

Example

Consider the reaction:

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A: What is the expected rate law according to the following proposed multistep mechanism under the steady-state approximation (d[I]/dt = 0 and k2 >> k-1)?

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   Slow

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   Fast

 
B: If x is the order of the reaction with respect to A, y is the order of the reaction with respect to B, and n is the overall reaction order. What are the balues of x, y, and n?

Solution

A:

d[I]/dt = k1[A][B] - k-1[I] - k2[I][B] = 0

[I] = k1[A][B] / k-1 +k2[B] , since k2>>k-1, k-1 = 0

Therefore, [I] = k1[A] / k2

d[C]/dt = k2[I][B]

d[C]/dt = k1k2[A][B] / k2

d[C]/dt = k1[A][B]

B: x = 1, y = 1, n = 2

Application of Steady-State Approximation in Enzyme Kinetics

In 1925, George E. Briggs and John B. S. Haldane applied the Steady-State Approximation method to determine the rate law of the enzyme-catalyzed reaction (figure 1). The following assumptions were used:

  1. The rate constant of the first step must be slower than the rate constant of the second step (k1 <<< k2), hence
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  2. Enzyme concentration must be significantly lower than the substrate concentration to keep the first step slower than the second step.

Reaction2.bmp

Figure1

So we will have

d[P]/dt = k2[ES]              (3)

where

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Since

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Using the second assumption and the fact that enzyme concentration equals the initial concentration of enzyme minus the concentration of the enzyme-substrate intermediate, or

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We obtain the following equation:

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From this equation the concentration of the ES intermediate can be found:

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Substitute this into (3), we will have

d[P]/dt = k2[E]0[S]/{[(k1+k2)/k1]+[S]} = k2[E]0[S]/(KM+[S])              (4)

Where KM = (k-1+k2)/k1

Since in most of the cases, people only measure the initial d[P]/dt to determine the rate of product formation, we can rewrite (4) as

v0 = d[P]0/dt = k2[E]0[S]/(KM+[S])              (5)

Since [E]= Vmax/k2

We can rewrite the equation (5) as

v= d[P]0/dt = (k2/k2)vmax[S]/(KM+[S])

                    = Vmax[S]/(KM+[S])

This equation is a very powerful tool to look for vmax and KM (Michaelis constant), of an enzyme by using the Lineweaver-Burk plot (1/[S] vs. 1/v0) or the (Missing file: File:User:Sandy_Algaze/Eadie_Hofstee_Plot.jpg)Eadie-Hofstee plot (v0/[S] vs. v0).

Practice problems

We have a reaction

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where k1 = 0.2 M-1s-1 , k2 = 2000 s-1

1) Write the reaction rates for A, B, and C.

2) Is this a steady-state reaction?

3) Write the expression for d[C]/dt using the Steady State Approximation

4) Calculate d[C]/dt if [A] = 1M

5) Calculate [C] at t = 3 s and [A]0 = 2M

Solutions

1) d[A]/dt = -k1[A]; d[B]/dt = k1[A] - k2[B]; d[C]/dt = k2[B]

2) Since k1 is a lot larger than k2, this is a steady state reaction.

3) d[C]/dt = k2[B]

where d[B]/dt = k1[A] - k2[B] = 0

so, [B] = k1[A]/k2

Replace this into d[C]/dt

d[C]/dt = k1[A]

4) d[C]/dt = 0.2M-1s-1(1M) = 0.2 s-1

5) [C] = [A]0 (1-e-k1t) = 2M(1-e-0.2(3)) = 0.9 M

References:

  1. Chang, Raymond. Physical Chemistry for The Biosciences. Sausalito: University Science Books, 2005. 368-370.
  2. Garrett, Reginald H, Charles M. Grisham. Biochemistry. 4th ed. Boston: Brooks/Cole Cengage Learning, 2010. 389-397.
  3. Segel, Irwin H. Biochemical Calculations. 2nd ed. New Jersey: John Wiley and Sons, inc., 1976. 216-218.

Contributors

  • Melanie Miner, Tu Quach, Eva Tan, Michael Cheung

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Last Modified
09:28, 2 Oct 2013

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