The steady state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step 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.

## Introduction

Before discussing the steady state approximation, it must be understood that the approximation is derived to simplify the kinetic expression for product concentration, [product]. Consider the following sequential reaction:

$A \xrightarrow[]{k_1} B \xrightarrow {k_2} C$

Calculating the [product] 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

$[C] = [A]_0 \left(1+ \dfrac{k_2e^{-k_1t}-k_1e^{-k_2t})}{k_1 - k_2}\right) \label{1}$

With a more complicated mechanisms, 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. This article concerns the steady state approximation.

The steady state approximation is applies to a consecutive reaction with a slow first step and a fast second step ($$k_1<<k_2$$). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product, such as product B in the above example.

$\dfrac{d[B]}{dt} = 0 = k_1[A] - k_2[B] \label{2}$

Thus

$[B] = \dfrac{k_1[A]}{k_2} \label{3}$

From the mechanism:

$\dfrac{d[C]}{dt} = k_2[B] = \dfrac{k_2k_1[A]}{k_2} = k_1[A] \label{4}$

Solving for $$[C]$$:

$[C] = [A]_0 (1- e^{-k_1t}) \label{5}$

Equation $$\ref{2}$$ is much simpler to derive than Equation $$\ref{1}$$ , especially with a more complicated reaction mechanisms.

Example $$\PageIndex{1}$$

Consider the reaction:

$$A + 2B \xrightarrow[]{} C$$

A: What is the expected rate law according to the following proposed multi-step mechanism under the steady state approximation with $$k_2 >> k_{-1}$$) for the following mechanism:

• $$A + B \xrightarrow[k_{-1}]{k_1} I$$ Slow
• $$I + B \xrightarrow[]{k_2} C$$ 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 values of x, y, and n?
SOLUTION

A:

$\dfrac{d[I]}{dt} = k_1[A][B] - k_{-1}[I] - k_2[I][B] = 0$

$[I] = \dfrac{k_1[A][B]}{ k_{-1} +k_2[B]}$

because $$k_2>>k_{-1}$$ then $$k_{-1} = 0$$. Therefore, $$[I] = \dfrac{k_1[A]}{k_2}$$

$\dfrac{d[C]}{dt} = k_2[I][B]$

$\dfrac{d[C]}{dt} = \dfrac{k_1k_2[A][B]}{k_2}$

$\dfrac{d[C]}{dt} = k_1[A][B]$

B:

• $$x = 1$$
• $$y = 1$$
• $$n = 2$$

## Use of the 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 made:

1. The rate constant of the first step must be slower than the rate constant of the second step ($$k_1 << k_2$$), hence $\dfrac{d[ES]}{dt} = 0$
2. Enzyme concentration must be significantly lower than the substrate concentration to keep the first step slower than the second step.

Figure 1: Steady state dynamicsin enzymes

This gives the following:

$\dfrac{d[P]}{dt} = k_2[ES] \label{6}$

where

$\dfrac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES] \label{7}$

Because

$[S] >> [E] \label{8}$

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

$[E] = [E]_o - [ES] \label{9}$

The following equation is obtained:

$k_1[E]_o[S] = k_{-1}[ES] + k_2[ES] + k_1[ES][S] \label{10}$

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

$[ES] = \dfrac{k_1[E]_o[S]}{(k_{-1} + k_2) + k_1[S]} \label{11}$

Substitute this into Equation $$\ref{3}$$ gives,

$\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{[(k_1+k_2)/k_1]+[S]} = \dfrac{k_2[E]_0[S]}{(K_M+[S]} \label{12}$

where

$K_M = \dfrac{k_{-1}+k_{2}}{k_1} \label{13}$

Because in most of the cases, only the initial d[P]/dt is measured to determine the rate of product formation, (4) can be rewritten as:

$v_0 = \dfrac{d[P]_0}{dt} = \dfrac{k_2[E]_0[S]}{K_M+[S]} \label{14}$

Because [E]0 = Vmax/k2. Equation $$\ref{5}$$ becomes the following:

$v_0 = \dfrac{d[P]_0}{dt} = \dfrac{k_2/k_2)v_{max}[S]}{(K_M+[S]} \label{15}$

$\\ = \dfrac{V_{max}[S]}{K_M+[S]} \label{16}$

This equation is a very useful tool to in calculating vmax and KM (the Michaelis constant), of an enzyme by using the Lineweaver-Burk plot (1/[S] vs. 1/v0) or the Eadie-Hofstee plot (v0/[S] vs. v0).

## Problems

Given the reaction $$A \xrightarrow[]{k_1} B \xrightarrow[]{k_2} C$$

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) Because k1 is much 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

Substitute 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