# Chapter 12 Kinetics.

OBJECTIVEAfter completing this chapter you should be able to

* draw a graph of a Michaelis-Menten reaction and identify the critical kinetic parameters.

* linearize the Michaelis-Menten equation.

* write equations for first-order and zero-order processes.

* determine the half-life and mean residence time in first-order and zero-order processes.

Overview

Kinetics deals with factors that determine, measure, and predict the rate of chemical and biochemical processes in soils. Kinetics lets us determine how quickly a process occurs. Knowledge of kinetics and kinetic parameters is essential for a soil scientist to determine such things as the rate of mineralization of materials in soil, and consequently the rate at which nutrients can be delivered to plants. Kinetics is also employed to determine how quickly environmental contaminants will be removed from an environment when a radioactive compound has decayed to the point where it can be safely handled, or how old soil organic matter is. Kinetics will be spread throughout some of the subsequent chapters of this book. For now, we will address the basic kinetic equations that will be employed.

Michaelis-Menten Kinetics

A good place to start looking at kinetics is the classic enzyme kinetics described by the Michaelis-Menten equation.

[upsilon] = [V.sub.max] [S] / [K.sub.m] + [S] = [12-1]

where

[upsilon] = velocity of the reaction

[V.sub.max] = maximum velocity of the reaction when the enzyme is saturated with substrate

[S] = substrate concentration

[K.sub.m] = Michaelis-Menten constant, which is the substrate concentration at [V.sub.max]/2

Example 12-1

For the graph below, determine the [V.sub.max], [K.sub.m], and reaction velocity when the substrate concentration is 3 mM.

[ILLUSTRATION OMITTED]

Solution

[V.sub.max] appears to be 8 [micro]mol [min.sup.-1] because the rate does not increase beyond this level. [K.sub.m] is the substrate concentration at [V.sub.max] / 2, or 4 [micro]mol [min.sup.-1], and is approximately 1 mM based on this plot. The velocity of the reaction at 3 mM could be determined directly from the graph but is more accurately determined by the equation.

[upsilon] = [V.sub.max][S] / [K.sub.m] + [S]

or

[upsilon] (8 [micro]mol [min.sup.-1])(3 mM) / 1 mM + 3 mM = 6 [micro]mol [min.sup.-1]

[FIGURE 12-1 OMITTED]

A linear version of the Michaelis-Menten equation is the Lineweaver-Burke plot (Figure 12-1).

1 / [upsilon] = ([k.sub.m] / [V.sub.max]) (1/[S]) + 1/[V.sub.max] [12-2]

Example 12-2

For the kinetic parameters described in Example 12-1, determine the velocity of the reaction if the substrate concentration was 5 mM by using the Lineweaver-Burke equation.

Solution

It was determined that [V.sub.max] = 8 [micro]mol [min.sup.-1] and [K.sub.m] = 1 mM, therefore

1 / [upsilon] = ([K.sub.m] / [V.sub.max]) (1/[S]) + 1 / [V.sub.max]

1 / [upsilon] = (1 mM / 8[micro]mol [min.sup.-1]) (1 / 5 mM) + 1 / 8 [micro]mol [min.sup.-1]

1 / [upsilon] = 0.15 [micro]mol [min.sup.-1]

[upsilon] = 6.67 [micro]mol [min.sup.-1]

Notice that there are two distinct parts to the curve generated by Michaelis-Menten kinetics. At low substrate concentrations, below the [K.sub.m] ([K.sub.m] >> [S]), the reaction rate increases as the substrate concentration increases. If you discount the [S] term in the denominator, the Michaelis-Menten equation simplifies to

[upsilon] = [V.sub.max] / [K.sub.m] [S] [12-3]

In other words, the reaction rate is proportional to a constant ([V.sub.max] /[K.sub.m]) times the substrate concentration, or as a soil scientist is likely to say, the reaction rate is substrate dependent. Adding substrate increases the reaction rate.

Likewise, when there is a high substrate concentration ([K.sub.m] <<< [S]), the terms for [S] cancel, and the Michaelis-Menten equation simplifies to v = [V.sub.max]. In other words, the reaction velocity is a constant ([V.sub.max]), and the reaction is said to be substrate independent. Adding more substrate has no effect on the reaction rate.

When describing reaction rates in soil, these two regions define the type of process or equations that can be used to evaluate the reactions. When the reaction rate depends on the substrate concentration, we call this a first-order process. When the reaction rate is constant, we call this a zero-order process. The mathematics of each is described in the next sections.

First-Order Processes

First-Order Rates

In a first-order process (Segal, 1976)

[upsilon] = - [DELTA][S] / [DELTA]t = k[S] [12-4]

where

[upsilon] = velocity or reaction rate

[DELTA][S]/[DELTA]t = substrate consumed per increment of time

k = a constant fraction

[S] = the substrate concentration present at that time

The difference in substrate concentration between when the reaction started ([S.sub.0]) and when the substrate concentration was measured again (St) can be written in the following exponential form:

[S.sub.t] = [S.sub.0] [e.sup.-kt] [12-5]

A plot of this reaction will look like Figure 12-2.

First-order processes proceed slower as time progresses and the substrate runs out. The exponential equation can be written in linear form by taking the natural logarithm (In) of both sides of the equation.

ln [S.sub.t] = ln [S.sub.0] - kt [12-6]

and a plot of the reaction will look like Figure 12-3.

[FIGURE 12-2 OMITTED]

[FIGURE 12-3 OMITTED]

Example 12-3

What percent of the substrate will remain in a first-order process after 7 days if the first-order rate constant is 0.12 [day.sup.-1]?

Solution

Initially [S.sub.0] = 100%, k = 0.12 [day.sup.-1], and the time interval (t) is given as 7 days. Solving by the exponential form gives

[S.sub.t] = [S.sub.0][e.sup.-kt]

[S.sub.t] = (100%) [e.sup.-(7days)(0.12/day)]

[S.sub.t] = 43%

Solving by the linear form gives

ln [S.sub.t] = ln [S.sub.0] - kt

ln [S.sub.t] = ln 100 - (7 days)(0.12 [day.sup.-1])

ln [S.sub.t] = 4.605 - 0.84

ln [S.sub.t] = 3.765 = 43%

Half-life Determinations

The time it takes for half the substrate to disappear ([S.sub.0]/2) is called the half- life ([t.sub.1/2]) and is a very important kinetic property. For a first-order process, the halflife can be calculated by the following equation:

[t.sub.1/2] = ln 2 / k = 0.693 / k [12-7]

Example 12-4

How long will it take for 50 percent of a substrate to disappear in a first-order reaction if it has a reaction rate constant of 0.4 [week.sup.-1]?

Solution

[t.sub.1/2] = ln 2 / k = 0.693 / 0.4 [week.sup.-1] = 1.7 weeks

Mean Residence Time

The mean residence time ([t.sub.mrt]) is the mean time it takes for 100 percent of the substrate to disappear. For a first-order reaction, the mean residence time is

[t.sub.mrt] = 1/k [12-8]

Example 12-5

What is the mean residence time of the reaction described in Example 12-4?

Solution

[t.sub.mrt] = 1 / k = 1 / 0.4 [week.sup.-1] = 2.5 weeks

Zero-Order Processes

Zero-Order Rates

In a zero-order reaction, the reaction rate is unaffected by the substrate concentration, or

[upsilon] = - [DELTA][S] / [DELTA]t = k [12-9]

In this equation, as before, we write -[DELTA][S]/[DELTA]t (a negative rate) because the substrate is disappearing. The linear form of this equation is [S.sub.t] = [S.sub.0] - kt (Figure 12-4).

[FIGURE 12-4 OMITTED]

Example 12-6

If a hydrocarbon is decomposing by a zero-order process, how long will it take for 75 percent of the material to disappear if it has a reaction rate constant of 0.1 [week.sup.-1]?

Solution

[S.sub.t] = [S.sub.0] -kt

25% = 100% - (0.1 [week.sup.-1])(t)

t = (100% - 25%) / 0.1 [week.sup.-1]

t = 750 weeks

Half-life Determinations

The half-life in a zero-order reaction is defined as it was for a first-order reaction, the time it takes for one-half the substrate to disappear. For a zero-order reaction, the equation is

[t.sub.1/2] = [S.sub.0] /2k [12-10]

Example 12-7

What is the half-life in a zero-order reaction if the starting concentration was 150 ppm and the reaction rate constant was 0.25 [week.sup.-1]?

Solution

[t.sub.1/2] = [S.sub.0] / 2k

[t.sub.1/2] = 150 ppm / (2) (0.25 [week.sup.-1])

[t.sub.1/2] = 300 weeks

Mean Residence Time

The mean residence time in a zero-order reaction is the average time it takes for 100 percent of the material to disappear and is calculated as

[t.sub.mrt] = [S.sub.0] / k [12-11]

Example 12-8

If the reaction rate constant for a zero-order reaction is 0.33 [day.sup.-1], how long does it take for a starting substrate concentration of 400 ppm to disappear?

Solution

[t.sub.mrt] = [S.sub.0] / k

[t.sub.mrt] = 400 / 0.33 [day.sup.-1]

[t.sub.mrt] = 1212 days

Reference

Segal, I. H. (1976). Biochemical calculations (2nd ed.). New York: John Wiley & Sons.

Sample Problems

Michaelis-Menten Kinetics

1. If the [V.sub.max] of a reaction is 15 [micro]mol [min.sup.-1], what is the velocity of the reaction at its [K.sub.m]?

2. Given that Michaelis-Menten kinetics apply, what is the velocity of a reaction if [S] = 0.5 mM, [V.sub.max] = 10 [micro]mol [min.sup.-1], and the [K.sub.m] = 0.003 mM?

3. If the velocity of a reaction is 5 [micro]mol [min.sup.-1] at its [K.sub.m] and the [K.sub.m] of the reaction is 4 mM, at what substrate concentration would you expect [V.sub.max] to occur?

4. Use the Lineweaver-Burke equation to calculate the velocity of a reaction given that the y intercept is 0.2 [micro]mol [min.sub.-1], the x intercept is - 0.04 mM, and the substrate concentration is 5 mM.

First-Order Processes

1. If the first-order rate constant is 0.5 [week.sup.-1], what will be the concentration of 10 g of substrate after 2 weeks if it decomposes by a first-order process?

2. After 3 weeks the substrate concentration was 25 percent. What is the first-order decomposition rate constant for this reaction?

3. If the initial substrate concentration in a first-order reaction was 200 ppm and the reaction rate constant was 0.33 [h.sup.-1], what will the substrate concentration be after 15 h?

4. If 1500 ppm of a compound decomposes with a first-order reaction rate constant of 0.25 [year.sup.-1], how long will it take before the concentration is 750 ppm?

5. What is the mean residence time of oxygen in soil if it is consumed with a k of 0.5[h.sup.-1]?

Zero-Order Processes

1. For hydrocarbons that decompose via a zero-order process, what will be the concentration if 3000 mg [kg.sup.-1] decompose for 25 weeks with a reaction rate of 0.1 [week.sup.-1]?

2. What is the reaction rate constant for a substrate that was 3500 ppm on day zero and 1200 ppm on day 16?

3. What is the half-life of the material in the question above?

4. There is a high concentration of N[O.sub.3.sup.-] in a soil sample at the start of a flooding period (10 ppm) and the N[O.sub.3.sup.-] disappears as described in the following table:

Time (h) N[O.sub.3.sup.-] concentration (ppm) 0 10 3 8 6 6 9 4 11 ?

What type of reaction is occurring? What is the reaction rate constant? Will any N[O.sub.3.sup.-] be left to measure after an additional 2 h? If so, how much?

5. A soil sample originally has 120 ppm of an organic substrate. After 8 h it has 100 ppm and after 16 h it has 85 ppm. What type of reaction rate is likely to be represented? What is its reaction constant? How much substrate will be left after an additional 8 h?

6. What is the mean residence time of 1500 ppm benzene if it has a zero-order decomposition rate constant of 0.01 [week.sup.-1]?

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Title Annotation: | Section III: Problem Solving in Soil Biochemistry |
---|---|

Publication: | Math for Soil Scientists |

Geographic Code: | 1USA |

Date: | Jan 1, 2006 |

Words: | 2033 |

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