Lecture Notes Chemical Engineering 28x4j

LECTURE NOTES in

CHEMICAL REACTION ENGINEERING Eng’r Nilo T. Aldon ChE 009692 College of Engineering Colegio San Agustin-Bacolod November 12, 2012

1 1.1

Introduction Chemical Kinetics ___ the study of the rate of reaction and mechanism by which one chemical species is converted to another. Analyzing the influence of different reaction conditions on the reaction rate gives information about the reaction mechanism and the transition state of a chemical reaction. In 1864, Peter Waage, a Norwegian pioneered the development of chemical kinetics by formulating the law of mass action (the speed of a chemical reaction is proportional to the quantity of the reacting substances.) Focal Points Of Chemical Kinetics

1.

Rate of reaction

2.

Mechanisms

Kinetics deals with the experimental determination of reaction rates from which a rate law and reaction rate constant are derived. Essential rate laws exist for zero order reactions (for which reaction rates are independent of initial concentration), first order reactions, and second order reactions, and can be derived for others through calculus. In consecutive reactions the rate-determining step often determines the kinetics. In consecutive first order reactions, a steady state approximation can simplify the rate law. The activation energy for a reaction is experimentally determined through the Arrhenius equation and the Eyring equation. The main factors that influence the reaction rate include: the physical state of the reactants, the concentrations of the reactants, the temperature at which the reaction occurs, and whether or not any catalysts are present in the reaction.

In chemistry, a reaction mechanism is the step by step sequence of elementary reactions by which overall chemical change occurs. Although only the net chemical change is directly observable for most chemical reactions, experiments can often be designed that suggest the possible sequence of steps in a reaction mechanism. An overall description of how a reaction occurs. A mechanism describes in detail exactly what takes place at each stage of a chemical transformation. It describes the transition state and which bonds are broken and in what order, which bonds are formed and in what order, and what the relative rates of the steps are. A complete mechanism must also for all reactants used, the function of a catalyst, stereochemistry, all products formed and the amount each.

Principal Function of Reaction Kinetics from Chemical Engineers Point of View • • • • •

Establishing the chemical reaction mechanism Collecting experimental data Correlating rate data by mathematical equation Deg suitable reactors Specifying operating conditions, methods of control and auxiliary equipment.

1.2 Chemical reaction engineering _is the branch of engineering that is concerned with the exploitation of chemical reactions on a commercial scale for purposes other than the production of power.

2

Fundamentals of Chemical Kinetics

2.1 Classification of Reaction: Page

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2.1.1

Mechanism (Elementary or non-elementary) 1. Irreversible A  B 2. Reversible

A

B

Reversible Reactions_is one which results in the formation of an equilibrium mixture. The concept of a reversible reaction was introduced by Berthollet (1803) after he had observed the formation of sodium carbonate crystals at the edge of a salt lake. 2NaCl + CaCO3 → Na2CO3 + CaCl2 He recognized this as the reverse of the familiar reaction Na2CO3 + CaCl2→ 2NaCl + CaCO3

3. Simultaneous

2.1.2 2.1.3

2.1.4

A B A  C A  BC

4. Consecutive 5. Autocatalytic A+B B+B 6. Homogenous Catalyzed A+C R+C Phases 5. Homogeneous 6. Heterogeneous Operating Conditions 7. Isothermal @ constant volume 8. Isothermal @ constant pressure 9. Adiabatic 10. Non-adiabatic and non-isothermal Molecularity 11. Unimolecular A→ B Example is the decomposition of ozone : O3 → O 2 + O

12. Bimolecular

A +B→P H + O2 → OH + O

13. Trimolecular or Termolecular A + B+C → P A + 2B → P 3A → P 2 NO + O2 → 2 NO2

2.1.5 2.1.6

2.1.7

2.1.8 2.1.9

Order 14. Integral (1st, 2nd, 3rd) 15. Fractional or Zero System 16. Batch 17. Flow 18. Semi-batch or semi-flow Equipment 19. Stirred tank (single or multistage) 20. Tubular (single or multiple) 21. Packed bed (fixed bed, moving bed, fluidized bed-dense phase/dilute phase) Catalyst 22. Catalyzed 23. Uncatalyzed Heat evolved 24. Exothermic Page

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25. Endothermic

2.2

Rate of reaction

(− r ) A

___ the number of units of mass of some participating reactants which is transformed into a product per unit time and per unit volume.

1 dN A =− V dt

The negative sign denotes the disappearance of reactant A . A positive sign on the other hand denotes formation of a product.

2.2.1 Basic Factors Affecting The Rate of Reaction : a. Nature of the Reactants The active and inactive are used in describing the nature of reactants. For instance, we say that sodium is a very active metal which reacts violently with water. The activity of elements are predicted by the use of the periodic table. An example of which is the activity series of metals which ranks potassium as the most active. When covalent bond formation takes place between the molecules and when large molecules are formed, the reactions tend to be very slow. b. Frequency and efficiency of collisions of the reactant particles. It follows that any factor (physical state, composition, temperature, pressure, area of exposure, catalysts, etc) which affects the frequency and efficiency of collisions of the reactant particles will necessarily alter the speed of reaction.

2.

1. Physical State Physical state (solid, liquid, or gas) of a reactant is also an important factor of the rate of change. When reactants are in the same phase, as in aqueous solution, thermal motion brings them into . However, when they are in different phases, the reaction is limited to the interface between the reactants. Reaction can only occur at their area of , in the case of a liquid and a gas, at the surface of the liquid. Vigorous shaking and stirring may be needed to bring the reaction to completion. This means that the more finely divided a solid or liquid reactant, the greater its surface area per unit volume, and the more it makes with the other reactant, thus the faster the reaction. To make an analogy, for example, when you start a fire, first you use wood chips and small branches - you don't start with big logs right away. In organic chemistry On water reactions are the exception to the rule that homogeneous reactions take place faster than heterogeneous reactions. Concentration or composition of the reactant(s) Concentration plays an important role in reactions. According to the collision theory of chemical reactions, this is due to the fact that molecules must collide in order to react together. As the concentration of the reactants increases, the frequency of the molecules colliding increases, striking each other faster by being in closer at any given point in time. Imagine two reactants being in a closed container. All the molecules contained within are colliding constantly. By increasing the amount of one or more of the reactants you cause these collisions to happen more often, increasing the reaction rate. Law of Mass Action _states that the rate of chemical reaction is at each instant proportional to the concentration of the reactant with each raised to a power equal to their coefficient or the actual number of molecules participating in the reaction. This law can be interpreted by several complex mechanisms but it can simply be explained as follows: when two or more molecules react, it must come close to one another or must collide. Therefore, it is expected that the rate of reaction increases if the molecules are crowded closely together, i.e., the concentration is high.

3.

Temperature Temperature usually has a major effect on the speed of a reaction. Molecules at a higher temperature have more thermal energy. When reactants (reactant + reactant → product) in a chemical reaction are heated, the more energetic atoms or molecules have a greater probability to collide with one another. Thus, more collisions occur at a higher temperature, making a product in a chemical reaction. More importantly however, is the fact that at higher temperatures molecules have more vibrational energy, that is, atoms are vibrating much more violently, so raising the temperature not only increases the number of collisions but also collisions that can result in rearrangement of Page

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atoms within the reactant molecules. For example, a refrigerator slows down the speed of the rate of reaction since it cools the molecules. On the other hand, an oven gives heat (energy) to the molecules which in turn speeds up the rate of reaction, cooking the food faster. Chemical kinetics can also be determined using a Temperature Jump. This involves using a sharp rise in temperature and observing the relaxation rate of an equilibrium process. The kinetic energy of particles follows the Maxwell-Boltzmann distribution. An increase in temperature not only increases the average speed of the reactant particles and the number of collisions, but also the fraction of particles having kinetic energy higher than the activation energy. Thus, the effective collision frequency increases. 4.

Pressure By increasing the pressure, you decrease the volume between molecules and will increase the number of collisions between reactants, increasing the rate of reaction. This is because the activity of a gas is directly proportional to the partial pressure of the gas. This is similar to the effect of increasing the concentration of a solution. 5. Catalysts A catalyst is a substance that accelerates the rate of a chemical reaction but remains chemically unchanged afterwards. The catalyst increases rate reaction by providing a different reaction mechanism to occur with a lower activation energy. In autocatalysis a reaction product is itself a catalyst for that reaction leading to positive . Proteins that act as catalysts in biochemical reactions are called enzymes. MichaelisMenten kinetics describe the rate of enzyme mediated reactions. In certain organic molecules specific substituents can have an influence on reaction rate in neighboring group participation. Agitating or mixing a solution will also accelerate the rate of a chemical reaction, as this gives the particles greater kinetic energy, increasing the number of collisions between reactants and therefore the possibility of successful collisions. 

1. 2. 3.

2.2.2 Factors Affecting the Rate of Homogeneous Reactions : Composition rA= f (temperature, pressure, composition) Temperature Pressure These variables are interdependent in that the pressure is fixed, given the temperature and composition of the phase. This we may write without loss of generality.

rA= f (temperature, composition) 2.2.3 Factors Affecting Heterogeneous Reactions :

1. Mass transfer factors (e.g. diffusion characteristics of fluid phases) 2. patterns of phases ( each phase may be in one of two ideal flow patterns i.e. plug or back-mix flow. There are a number of possible combinations of ing patterns) 3. Fluid dynamic factors (e.g. mass velocity, degree of turbulence, etc) 4. Interfacial surface area 5. Geometry of the reaction vessels 6. Chemical kinetic factors (i.e. activation energy, concentration of reactants, etc) 7. Temperature and pressure

Some of these factors are not completely independent and may interact with each other.

2.3

Mathematical Expression of Rate of Chemical Reaction

2.3.1

Stoichiometry: 1. Stoichiometry may suggest whether we have a single reaction or not. 2. Stoichiometry can suggest whether a single reaction is elementary or not because no elementary reactions with molecularity greater than three have been observed. Page

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The stoichiometric equation is a chemical equation which expresses an overall chemical reaction in of the simplest ratio of reactant and product molecules. Consider the irreversible elementary reaction

aA

bB→

+

rR

+

sS

The stoichiometric relationship of this given chemical reaction is

-rA a rA

-rB b

= =

rR r

=

- dC A dt

=

rS = k A CaA C Bb a

=

rS s

= k A CaA CbB rB =

rR

=

+ dC R dt +dCS dt

- dCB dt

= k BCaA C Bb

= k R CaA CbB = k SCaA CBb

k BCaA C Bb b

=

k R CaA C Bb r

=

k SCaA CbB s

From the preceding equations the values of the specific rate constants kB, kR and kS are solved as a function of kA thus,

b k BCaA CbB kB =   kA ; a b r s k R =   k A ; kS =   k A a a

k A CaA C Bb a

=

Kinetically, the system is at equilibrium if the net rate of change of the forward and backward elementary reactions is zero. Consider the elementary reversible reaction k1 A+B R+S k2 The rate of formation of R by the forward reaction is: Rate of Formation of R : rR = k1C A C B And the rate of disappearance by the reverse reaction is Rate of Disappearance of R : - rR = k 2 C R C S At equilibrium the net rate of formation of R is zero, then

rforward=rbackward=0

rforward = rbackward k1CACB = k2CRCS k1 C R C S = k 2 C AC B

Since for this reaction kC is defined as Page

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kC =

C R CS CA CB

at equilibrium we have kC =

k1 k2

=

C Re CSe CAe CBe

kC, which is equal to k1/k2, is a constant independent of concentration. CAe, CBe, CRe, and CSe are equilibrium concentrations with respect to A, B, R, and S, respectively. Thus, the equation for k c is very specific only for equilibrium condition. The rate of chemical reaction is dependent on temperature, pressure, and concentration. In particular, an important relationship is expressed by the Law of Mass Action. The Law of Mass Action states that the rate of chemical reaction is at each instant proportional to the concentration of the reactant with each raised to a power equal to their coefficient or the actual number of molecules participating in the reaction. This law can be interpreted by several complex mechanisms but it can simply be explained as follows: when two or more molecules react, it must come close to one another or must collide. Therefore, it is expected that the rate of reaction increases if the molecules are crowded closely together, i.e., the concentration is high. The statement of the Law of Mass Action is translated into its mathematical expression using the following defined notations: ri = rate of reaction of any substance i i = any substance in the reaction Ci = concentration of any substance i in the reaction. Suppose the reaction is represented by k A  →P Applying the principle of the Law of Mass Action

( − rA )

α CA

The preceding equation is the mathematical expression of the Law of Mass Action. Given an irreversible chemical reaction k aA + bB + cC+... → dD + eE +...

where:

A, B, C = are the reactants participating in the chemical reaction; D, E = are the products formed during chemical reaction; a, b, c, d and e are the number of molecules of each substance involved in the chemical reaction, then

From the given chemical reaction the rate expression of A is given as:

( − rA ) α C A a C B b C C c ( − rA ) = kC A a C B b C C c Also:

( − rB ) = b kCA a CBbCCc a

( − rC ) = c kCA a CB bCC c

a Consider the rate of change of component i involved in the chemical reaction. If the rate of change in the number of moles of this component is dNi/dt, the rate in the various areas of kinetics is defined as follows Page

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Based on a unit volume of reacting fluid,

dN i Change in No. of moles of any substance i = Vf dt ( unit volume of fluid ) ( time) Based on unit volume of reactor, if different from the rate based on unit volume of fluid, dN i Chnage in the No. of moles of any substance i ri = = Vr dt ( Unit volume of reactor ) ( time ) Based on unit interfacial surface in two fluid systems or based on unit surface of solid in gas-solid systems, ri

=

ri

=

dN i Sdt

-

Change in the No. of moles of any substance i ( Unit surface area) ( time)

Based on unit mass of solid in fluid-solid systems,

ri

=

dN i = Ws dt

Change in

the No. of moles of any substance i ( unit mass of solid ) ( time )

2.3.2 Molecularity and Order of Reaction The molecularity and order of an elementary reaction is the number of molecules involved in the rate determining step of a reaction. Molecularity of reactions has been found to be one, two, and occasionally three. Molecularity refers only to an elementary reaction and can only be whole numbers. Example: Consider the irreversible elementary reaction

aA + bB + cC +..  product where a, b, and c are stoichiometric coefficients. We call the power to which the concentration raised the order of reaction.

are

( − rA ) = kC A pC Bq CC r When the stoichiometric equation truly represents the mechanism of the reaction, the order and molecularity both are n = a + b + c and individually p = a, q = b, r = c

Thus, the reaction is ath order with respect to A bth order with respect to B cth order with respect to C nth order over all n = a + b + c + …. Example: 1. A  P 2. A + B 

P

(n=1) (n=2) Page

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3. 2A + B  P 4. A + B + C  P

(n=3) (n=3)

Rates of disappearance of reactants or formation of products are related to A by the stoichiometric coefficients:

( − rA ) = ( − rB ) = ( − rC ) a

b

c

2.3.3 Rate Constant k :

For Equation : aA  product (s) The rate of disappearance of reactant A:

( − rA ) = k n A 

n

( − rA ) = k n A   V

n

 V  unit mass  1   ( unit time )  unit volume 

k=

( unit volume) n −1 ( unit mass) n −1( time)

k=

unit mass ( unit volume)( time)

for zero-order

k=

1 ( time)

for 1st-order

k=

( unit volume) ( unit mass) ( time)

for 2nd-order

k=

( unit volume) 2 ( unit mass) 2 ( time)

for 3rd-order

n

unit of k for an nth order chemical reaction.

The reaction rate constant could also be expressed in of pressure ( π ) and mole fraction (NA): P πN A CA = A = If we let RT RT (-rA) = kc (CA)n = kp (PA)n ( − rA ) = k ( PA ) n kc = ( CA ) n p ( CA ) n = k P ( RT )

n

= kN

( NA ) n ( CA ) n

 RT  = kN   π 

n

Rate and Order Questions 1. A reaction has the stoichiometric equation 2A → R + S What is the order of reaction? Answer: The reaction is elementary, irreversible, second order 2.

Given the reaction 1 2 NO2 + O2 → N 2O5 2 Page

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What is the relation between the rates of formation and disappearance of three components of the reaction? Answer:

( − rNO ) = k [ NO2 ] 2 [O2 ]0.5 ( − rO ) = 14 k [ NO2 ] 2 [O2 ]0.5 = 14 ( − rNO ) 2

2

2

( rN O ) = 12 k [ NO2 ] 2 [O2 ]0.5 = 12 ( − rNO ) 2

3.

5

2

A reaction with a stoichiometric equation

1 A+ B → R 2

Has the following rate expression: ( − rA ) = 2[ A] 0.5 [ B ] What is the rate expression for this reaction if the stoichiometric equation is written as A + 2 B → 2 R + 2S Answer: ( − rA ) = 2[ A][ B ] 2

4. a. b. c. d. e. f. g.

For the elementary reaction: 2A+B  2S CAo= 8 mols/liter, CBo= 6 mols/liter, CSo= 2mols/liter What is the order of reaction with respect to A? What is the over-all order of reaction? What is the limiting reactant? If 75% of the limiting reactant is reacted after 5 minutes, what will be the values of CA, CB,,and CS ? What is (-rA)? What is (rB)? What is (rS)? SOLUTION: a. Second-order b. Third-order c. A is the limiting reactant. 2 mols of A will react with 1 mol of B, hence, 8 mols A will react with only 4 mols B. 8 ( 0.75 ) 3 mol 2 mol 8 mol ANSWER: C = 8 ( 0.75) + 2 = d. C = 8 ( 1 − 0.75) = ; C = 6− = ; A

liter

B

2 liter dC A 12k mol 2 2 ANSWER: = kC ACB = k ( 2 ) ( 3) = e. ( − rA ) = dt liter•minute r 6 kmol ANSWER: f. −rB = − A = 2 liter•minute

g. rs = ( − rA ) =

5.

12k • mol liter • min ute

S

liter

ANSWER:

The specific reaction rate of a first-order reaction is 2.5 x10-7/s and the initial concentration is 0.1gmol/lit. What is the initial rate expressed in of seconds, liters gmols. SOLUTION: −dC A 2.5 x10−7 0.1gmol 2.5 x10−8 gmol = kC A = x = dt s lit lit

6.

ANSWER:

The initial rate of a second-order reaction is 5.0x10-7 gmol/lit-s when CA is 0.2 gmol/lit. What is the specific reaction rate expressed in units of second, liter, gmol. SOLUTION: Page

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dnA 5.0 x10 −7 gmol  0.2 gmol  = kC A2 + =k  Vdt lit − s  lit  1.25 x10−5 lit ANSWER: k= ( gmol )( s )

2

(− rA ) =

7.

The third-order gas-phase reaction 2NO+O2 2NO2 has a specific reaction rate of kC=2.65 x104 lit2/ (gmol)2(s) at 30oC. Find kP and kN. SOLUTION: kC lit 2 1.7239 ANSWER: = = 2.65 x10 4 a. k = P

( RT ) n

( gmol ) 2 ( s ) 2 (0.08205) 3 (303k ) 3

(lit )( gmol) 2 ( s) 2 (atm) 2

  n 4   2   π 2 . 6 5 x 1 0 ( lit ) atm     b. k N = k C   =   2 2  R T   ( gm o l ) ( s )   ( 0.0 82 05) ( lit) ( atm) ( 30 3 K )      K

(

8.

)

3

=

1.7239

ANSWER:

(lit )( gmol) 2 ( s) 2

The third-order gas-phase reaction 2NO + O2 2NO2 has a specific reaction rate at 30oC and 1 atm. Find kp and kn. Solution; k c = 2.65 x10 4

Solution:

L2 ( g.mol ) 2 .s n

 RT  n k c = ( RT ) k p =   kn  π  kp =

2.65 x10 4 L2

kc

( RT )

n

=

( g.mol ) 2 .s

 0.08206 L.atm ( 303 K )    g.mol.K 

3

= =1.7239

g mol L . atm 3 .s

gmol  g.mol 3 kn = π n k p = ( 1 atm )  1.7239  == 1.7239 3 L . atm . s L.s  

9.

Answer

Answer

Ex. At a given temperature the following data were taken: CH3CHO (g)  CH4 (g) + CO (g) A B C Initial Pressure, A(Torr) Experiment Number 1 300 2 200 a. Write the equation b. Find the order of reaction SOLUTION:

( r ) = dπ = kPAn ; r1 = kPAn1 ; dt

r2 =

kPAn2

kPAn  PA r : 1 = n1 =  1 r2 kPA2  PA  2

    

Initial Rate of Increase in Total pressure (Torr) 0.61 0.27

n

r1 0.61 log r2 0 .27 = 2 n= = 300 PA1 log log 200 PA log

2

a. r = kPA2 b. 2nd-order reaction

Answer:

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2.4

Mechanism __ is the sequence of individual chemical events whose over-all result

produce

the observed reaction. __ the sequence of steps that takes place to complete a chemical change. In most instances the postulated mechanism is a theory devised to explain the end results observed by experiments. Like other theories, mechanisms are subject to change over the years as new data is uncovered or as new concepts regarding chemical interactions are developed. The mechanism of an overall chemical reaction consists of a series of elementary reactions, which together bring about the overall reaction summarized in the stoichiometric equation.

•

A reaction mechanism is a set of steps at the molecular level. Each step involves combinations or rearrangements of individual molecular species. The steps in combination describe the path or route that reactant molecules follow to reach the product molecules. The result of all steps is to produce the overall balanced stoichiometric chemical equation for reactants producing products.

2.4.1 Steps Followed in Establishing Reaction Mechanism : 1. Postulate a simple mechanism with a corresponding stoichiometry. 2. If the stoichiometry appears to indicate the reaction to be one-step and elementary, proceed to acquire kinetic data and analyze them according to the integral method. 3. For non-elementary reactions, assume that the overall reaction consists of several elementary reaction steps with formation of intermediate compounds. 4. Formulate a rate expression for each of the elementary steps and sum up the individual rate expressions to describe the overall rate. 5. If the resulting rate expression agrees with experimental kinetic data, the assumed mechanism is acceptable. Otherwise, continue to assume alternative mechanisms, and repeat step 4 until a desired degree of agreement is obtained. 6. Frequently, it may be simpler to use a purely empirical approach to correlate the data.

2.5

Elementary and Non-Elementary Reactions

2.5.1 Elementary Reactions __ any such reaction in which the rate equation suggested by stoichiometric equation represents the actual mode of action and occurs in a single step. Ex. A + 2B → 2S + P

( − rA ) = ( rP ) = KC A C B2 ( − rB ) = ( rS ) = 2 KC A C B2

2.5.2

Non-elementary Reactions __ are those where there is no direct correspondence between stoichiometric equation and the rate expression and occurs in two or more series of reactions. Ex. A + 2B → 2S + P

If or

( − rA ) = ( rP ) ≠ KC A C B2 ( − rA ) = ( rP ) = KC B2

the reaction is non-elementary. Page

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Two Types of Non-Elementary Reaction Mechanism 1. Non-chain reaction mechanism Reactants 

(intermediates)*

(intermediates)*  Example:

Products

A

P

A

B*

B*

P

Mechanism

2. Chain Reaction Mechanism Reactants  (intermediates)* (intermediates)* + Reactant  (Another intermediates)* + Product (another intermediates)*  Example: 1. A + B  S + P

Product

Mechanism A  I* I* + B  R* + S R*  P 2. 2A + B

P + 2R

Mechanism A+B AB* + A S*

AB* S* + 2R P

Example: Present mechanisms consistent with experimentally found rate equation for the following reaction: 2A2B=2AB + A2 ;

rA2 =

1500 [ A2 B ] 2 [ AB] + 20[ A2 B ]

SOLUTION: A2 B = A* + AB A* + A2 B = A2 + AB rA2 = k 3 [ A* ][ A2 B ] − k 4[ [ A2 ][ AB ]

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rA* = k1[ A2 B] − k 2 [ A* ][ AB] − k3[ A* ][ A2 B] + k4 [ A2 ][ AB ] = 0 A* =

 k [ A B ] + k4 [ A2 ][ AB]  k1[ A2 B ] + k4 [ A2 ][ AB] ; rA2 = k3[ A2 B] 1 2  − k4 [ A2 ][ AB ] k2 [ AB] + k3[ A2 B]  k2 [ AB] + k3[ A2 B] 

=

k1 k 3 [ A2 B] 2 − k k 4 [ A2 ][ AB] 2 2

k 2 [ AB] + k 3 [ A2 B ] Assume k4=0 rA = 2

k1 k 3 [ A2 B ]2 k 2 [ AB] + k 3 [ A2 B]

k1k3

=

1

multiply by

1

k2 k2

2

[ A B] k2 2 1500[ A2 B]2 = k [ AB ] + 3 [ A2 B ] [ AB] + 20[ A2 B ] k2

Therefore, the mechanism will be: A2 B = A * + AB

ANSWER:

k A* + A2 B → A2 + AB 3

2.6. Effect of Temperature on Rate of Reaction: 2.6.1 Arrhenius Equation: The Arrhenius equation is a simple, but remarkably accurate, formula for the temperature dependence of a chemical reaction rate, more correctly, of a rate coefficient, as this coefficient includes all magnitudes that affect reaction rate except for concentration. The equation was first proposed by the Dutch chemist Jacobus Hendricus van’t Hoff in 1884; five years later in 1889, the Swedish chemist Svante Arrhenius provided a physical justification and interpretation for it. Nowadays it is best seen as an empirical relationship. In short, the Arrhenius equation is an expression that shows the dependence of the rate constant k of chemical reactions on the temperature T (in Kelvin) and activation energy Ea, as shown below: where A is the pre-exponential factor or simply the prefactor and R is the gas constant. The units of the pre-exponential factor are identical to those of the rate constant and will vary depending on the order of the reaction. If the reaction is first order it has the units s-1, and for that reason it is often called the frequency factor or attempt frequency of the reaction. When the activation energy is given in molecular units, instead of molar units, e.g. joules per molecule instead of joules per mol, the Boltzmann constant is used instead of the gas constant. It can be seen that either increasing the temperature or decreasing the activation energy (for example through the use of catalysts) will result in an increase in rate of reaction.

k =Ae

−E RT

−E

k = e RT A

k E =− A RT E ln k = − + ln A RT ln

Equation of a straight line: y = mx + b where: y = lnk m = slope = -E/R x = 1/T b = ln A ln

k2 − E  1 1   −  = k1 Rg  T2 T1 

Where: k = reaction rate constant

E = activation energy, cal/gmol A = frequency factor T = absolute temperature, K ΔHo = standard-state enthalpy change, cal/gmol R = 1.987 cal/gmol.K

Slope=-E/R

lnk2 Lnk1 1/T22 Page

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nilotaldon2/7/2013

The Arrhenius Equation

− EA

gives a good empirical description of the temperature dependence of

k = ko e RT

the rate constant of an elementary reaction. Both E A and ko are independent of temperature. This expression fits experimental data well over wide temperature ranges and is strongly suggested from various standpoints as being a reasonable first approximation of the true temperature dependency. 1.

When only two sets of k and T are given For a given k1 and T1 the equation becomes

k1 = ko e

− EA RT1

Apply the natural logarithm to both sides of the equation and simplify − EA − EA E = ln ko − A RT1 ln k1 = ln ko e = ln ko + ln e RT1 RT1

ln ko = ln k1 +

Solving for lnko will give

EA RT1

Similarly for k2 and T2, Equation 1 becomes − EA

− EA

ln k2 = ln k oe RT2 = ln ko + ln e RT2 E ln ko = ln k 2 + A RT2 Equating the two equations:

ln k1 +

EA E = ln k 2 + A RT1 RT2

Simplifying further : ln k2 − ln k1 = ln

ln

EA E − A RT1 RT2

k2 E A  1 1  =  −  k1 R  T1 T2 

ln k

k2 E A  T2 − T1  =   k1 R  T1T2 

1/T

2. When more than two sets of rate constant (k) and temperature (T) are available:

− EA

k = ko e RT

Applying the natural logarithm to both sides of Equation and simplifying will give −EA E ln k = ln ko − A is an equation of a straight line where ln ko and are the intercept and the slope of RT R the line, respectively. The natural logarithm of k

2.6.2 Collision Theory:

( ln k ) is plotted against the reciprocal of temperature 

1 . T 

1

−E

2

RT

k = T Ae

For a chemical reaction to proceed, molecules must have effective collisions. The two requirements for an effective collisions are; a.The molecules must be suitably reactive and must have enough energy, Page

14 /38

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b. The molecules must be arranged in a proper position. Collision theory, was proposed by Max Trautz and William Lewis in 1916 that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. It assumes that for a reaction to occur the reactant particles must collide, but only a certain fraction of the total collisions, the effective collisions, cause the transformation of reactant molecules into products. This is due to the fact that only a fraction of the molecules have sufficient energy and the right orientation at the moment of impact to break the existing bonds and form new bonds. The minimal amount of energy needed so that the molecule is transformed is called activation energy. Collision theory is closely related to chemical kinetics. Collision theory views the rate to be governed by the number of energetic collisions between reactants. What happens to the unstable intermediate is of no concern. It is simply assumed that this intermediate breaks down rapidly enough into products so as not to influence the rate of the over-all process. − Ec e RT

k = ko T Where: k is the specific rate constant, ko’ is the collision theory constant, T is the temperature (K), R is the universal constant (1.987 cal/g-mol K), and ECT is the collision theory activation energy (cal/g-mol). 1. When only two sets of k and T are available For a given k1 and T1 equation becomes k1 = ko' e T1 ln

k1 T1

= ln ko' −

ln

k2 E = ln ko' − CT RT T2 2 k2 E + CT T2 RT2

ln ko' = ln ln

Equating

ln   ln    

k2 T2 k1 T1

E RT1

k E + CT T1 RT1

ln ko' = ln Similarly for k2 and T2

− ECT RT1

k2 E k E + CT = ln 1 + CT T2 RT2 T1 RT1

k2 k E E − ln 1 = − T2 T1 RT1 RT2

     = E  T2 − T1  ; ln  k2 T1  = ECT  T2 − T1  k T  R  T2 T1   R  T1T2  2   1  

The preceding equation is used to determine the activation energy (ECT) if two sets of rate constants (k) and temperature (T) are given. 2. When more than two sets of k and T are given the equation can be rearranged as shown below

k =k

' o

Te

k = ko' e T

− ECT RT

− ECT RT

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15 /38

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Applying natural logarithm to both sides of the equation, then simplify will result into an equation of a straight line.

ln 1  k  The plot of ln   vs T is  T 

k E = ln ko' − CT RT T

 k ln    T

1/T

Plot of

 k  ln   as a function of (1/T) for the determination of activation energy (ECT) and constant (ko’).  T 

2.6.3 Transition-State Theory:

−E

k = TAe RT

Activated Complex or Transition State Theory: Henry Eyring, an American chemist, postulated an alternative to collision theory. He hypothesized that an intermediate species called an activated complex forms during collision. This intermediate species exist very briefly. It dissociates to form either the product (if reaction occurs) or the original reactants (if reaction does not occur) Most reactions proceed in many steps called elementary processes. The combined effect of all the elementary processes gives the overall reaction. The slow step determines the rate of the chemical reaction and is called the determining step. The observed rate of the overall reaction is the equivalent to the rate of the slow reaction.Transition-state theory views the reaction rate to be governed by the rate of decomposition of Page 16 /38 nilotaldon2/7/2013

intermediate. The rate of formation of intermediate is assumed to be so rapid that it is present in equilibrium concentrations at all times. The governing equation is

k = k o" Te

− E ACT RT

where, k is the specific rate constant, ko” is the activated complex theory constant, T is the temperature (K), R is the universal constant (1.987 cal/g-mol K), and EACT is the activation energy (cal/g-mol). 1. When two sets of k and T are given For k1 and T1

k1 = k T e " o 1

k1 = ko"e T1

− E ACT RT1

− E ACT RT1

k  ln  1  = ln ko" + ln e  T1 

− E ACT RT1

k  E ln ko" = ln  1  + ACT  T1  RT1 similarly for k2 and T2

k  E ln ko" = ln  2  + ACT  T2  RT2

Equating

 k  T  T − T  E ln  2   1  = ACT  2 1  R  T 2T1   k1   T2 

2. When more than 2 sets of k and T are available − E ACT k = ko" e RT T − E ACT k ln = ln ko'' e RT T k ln   T 

E k ln   = ln ko" − ACT RT T  k The Plot ln  T Plot of

 1  vs is  T

1/T k ln   as a function of (1/T) for the determination of activation energy (EACT) and constant (ko”). T 

2.6.4 General Equation: 2.6.5 Van’t Hoff Equation:

k = T m Ae d ( ln k ) ∆H o = dT

−E RT

Rg T 2

Example: 1. If a first-order reaction has an activation energy (E) of 25,000 cal/gmol and in the equation k=Ae-E/RT, A=5x1012/s. At what temperature will the reaction has a half-life of 1-minute? SOLUTION:

k=

1 1

t2

ln 2 =

− 25, 000 −E 1 12 ln 2 = 0.01155 = Ae RT = 5 x 10 e RT 60s

ANSWER:

Page

17 /38

nilotaldon2/7/2013

T = 373.28 K

2.7.

EQUILIBRIUM The system is in the state of equilibrium if the following characteristics are observed: 1. It is a closed system ( the amount of matter in the system does not change) 2. There is no change in the properties of the system as time es. 3. Two processes, which are opposite in direction, simultaneously take place at the same rate. 4. The ratio of the product of the molar concentration of the substances formed to the product of the molar concentration of the reactants is constant.

2.7.1 Factors that Affect Chemical Equilibrium: 1. Temperature 2. Pressure 3. Composition

For Equilibrium: Pressure may be fixed given the temperature and composition of the phase, (due to the interdependency of P and T ) Therefore: r = f ( T, C ) . This assumption could also be applied to non-equilibrium systems in the absence of better supposition (theory).

2.7.2 Le Chatelier’s Principle:

If a stress (disturbance or change) is applied to a system in a state of equilibrium, the system will shift in such a way to neutralize the effect of the stress.

Ex. In the decomposition of CaCO3 =CO2 + CaO, the removal of CO2 gas will trigger the system in a state of equilibrium to produce more CO2 in order to offset the reduction in pressure.

The tendency of a reaction, whether physical or chemical, is determined by a balance between these two factors: 1. The tendency toward a state of minimum energy, or low enthalpy; and 2. Tendency toward a state of maximum disorder, or high entropy. If we represent the change in the heat content of a system as ∆ H, the change in entropy as ∆S, and absolute temperature as T, an equation which is helpful in predicting reaction tendency is ;

∆ G = ∆H - (T) (∆S). (free-energy change) In spontaneous reactions, ∆G is negative; in systems in equilibrium , ∆G is zero.

2.7.3 Hess’s Law : The total change in enthalpy of a system is dependent on the temperature, pressure, state of aggregation, and state of combination at the beginning and at the end; it is independent of the number of intermediate reactions

3

Design of Homogenous Batch Reactors

3.1

Constant-Volume Reactions: Irreversible Reactions − dC dt

− rA ) –=OrderA = k ( C A )A  B a. ( Zero a

-(CA - CAo )=Kt b. First- Order

C − ln A = kt C Ao

A

aA

 bB

dC B b = kC A dt CA = CAo – aKt

( rB ) =

 B

C A = C Ao e − kt Page

18 /38

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c.

Second – Order Type 1

2A  B

1 1 − = kt C A C Ao

CA =

C Ao 1 + ktC Ao

Type 2 A+BR C C 1 ln B Ao = kt ( C Bo − C Ao ) C Bo C A d. Third – Order Type I 3A  B 1 1 − = 2kt C A 2 C Ao 2 Type II

2A + B  R

( 2C Bo − C Ao )( C Ao − C A ) C Ao C A

Type III

( 2C

Ao

+ ln

CA =

C Ao

(1 + 2ktC ) 2

1

2

Ao

C A C Bo ( 2C Bo − C Ao ) 2 kt = C Ao C B 2

A + 2B  R

− C Bo )( C Bo − C B ) C C 2 + ln B Ao = ( 2C A − C Bo ) kt C Bo C B C Bo C A o

Type IV

A+B +C R C 1 C 1 C ln Ao + ln Bo + ln Co = kt ( C Ao − CCo )( C Ao − CBo ) C A ( CBo − CCo )( CBo − C Ao ) CB ( CCo − C Ao )( CCo − CBo ) CC 1

Complex Constant-Volume Reactions a.

k1 A → B

Side or Simultaneous Reactions

k2 A → C

C A = C Ao e −( k1 + k2 ) t

  1 − e -( k1 + k2 ) t  

 k CA CC = C Co +  2 o k +k 2  1

  1 − e - ( k1 + k 2 ) t  

C B = C Bo

b.

[

 k CA + 1 o k +k 2  1

[

CB =

CC =

]

C B − C Bo =

]

k1 k CC − 1 CCo k2 k2

k k A → B → C

Consecutive Reactions C A = C Ao e

rB dC B k1 C B − C Bo = = = rC dCC k 2 C C − C Co

1

2

− kt

tC B max

k1C Ao  − k1 t − e − k 2t  e  k 2 − k1 

 k2    k1  = ln  k 2 − k1     

k2

 k  k 2 − k1 CBMax = C Ao  1   k2  k1 CBequib = C Ao e − k1t k2

C Ao  −k t −k t  k 2 − k1 + k1e 2 − k2e 1   k 2 − k1 

A

k1 k2

B

Page

19 /38

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c.

Reversible Reaction

d.

t=

( k + k1 ) C Ao − k2C Ao 1 ln 2 k 2 + k1 ( k2 +k1 ) C A − k2C Ao

t=

1 C − C Ae ln Ao k 2 + k1 C A − C Ae

Homogeneous Catalyzed Reactions

k A → R k A + C → R + C 1

2

Ln

e.

CA

o

CA

= − Ln(1 − x A ) = ( k1 + k 2C c ) t = ( k observed ) t A + R → R + R k

Autocatalytic Reactions

Ln

C Ao ( Co − C A )

(

= Ln

)

C A Co − C Ao

CA o CR

(

)

= Co kt = CAo + CRo kt

C ACRo

Where : Co = C A + C R = C Ao + C R o = Constan t

3.2 Variable Volume Reactions: − dn A n  = k A  Vdt V 

a.

n

Zero-Order

C Ao ln(1 + ε A x A ) = kt εA b. First Order  ∆V − ln1 −  ε AV o c.

  = − ln (1 − x A ) = kt 

Second-Order (1 + ε A ) x + ε ln(1 − x ) = kC t A Ao (1 − x A ) A A

d. Third-Order

(1 + ε ) ( 2 x − x 2(1 − x ) 2

A

A

2 A

2

A

Where:

) − 2 x (ε

)

+ εA 2 2 − ε A ln(1 − xA ) = kCAo t (1 − x A )

A

2

A

x A = fractional conversion ε A = fractional cahnge in volume=

Vx A =1 −Vx A= 0 VxA =0

Sample Problems:

Page

20 /38

nilotaldon2/7/2013

1. For the reaction A + B  S, equal volumes of 1 molar A and 2 molar B are mixed and allowed to react for 1 hr which half of A had been reacted. If the reaction is 1st-order with respect to A and 1st-order with respect to B, how much time in hours is required for 75% of A to react? a. 2.714 b. 2.26 c. 4.351 d. 6.732 Initial co ncentratio n of A and B after mixing: (1)(1) = 0.5; C = ( 2)(1) = 1 C Ao = Bo 2 2 For sec ond-order reaction

t=

1 C C Ln B Ao ; C A = C Ao (1 − x A ) = ( 0.5)(1 − 0.5) = 0.25 k ( CBo − C Ao ) CBo C A

CBo = (1) − ( 0.5)( 0.50 ) = 0.75 k=

( 0.75)( 0.5) = 0.8109 1 C C 1 Ln B Ao = Ln t ( CBo − C Ao ) CBo C A (1) (1 − 0.5) (1)( 0.25)

C A = C Ao (1 − x A ) = ( 0.5)(1 − 0.75) = 0.125 CBo = (1) − ( 0.5)( 0.75) = 0.625

t=

2.

1 C C 1 ( 0.625)( 0.5) = 2.26 h Ln B Ao = Ln k ( CBo − C Ao ) CBo C A .( 0.8109 )(1-0.5) (1)( 0.125)

The reaction A +B  S is of second-order and k at 0oC is 39.1 liter/(mole)(minute). An aqueous solution is made of 0.004 molar A and 0.005 molar B. How long will it take for 90% of A to react? a. 30 minutes b. 26 minutes c. 50 minutes d. 20 minutes For second - order reaction C A = C Ao ( 1 − x A ) = ( 0.004 )( 1 − 0.9 ) = 0.0004 C B = ( 0.005 ) − ( 0.004 )( 0.90 ) = 0.0014 C C 1 1 ( 0.0014 )( 0.004 ) = 26.33 minutes Ln B Ao = Ln k ( C Bo − C Ao ) C Bo C A 39.1 ( 0.005 − 0.004 ) ( 0.005 )( 0.0004 )

t=

3. The reaction 2A  2R + S takes place isothermally in a constant-volume experimental reactor. Starting with a mixture of 80% A and 20% inserts, the initial pressure of 10 atm increases by 25% in 8 minutes. What conversion is attained? a. 37.5% b. 62.5% c. 35.5% d. 64.5% 2A

i

+ I → 2R + S

8

∆ - x'A

2 0

0 x

' A

f 10 + 0.5 x 'A = 12.5; PA = 8 − 5 = 3 :

xA =

0 0.5 x 'A x A' = 5 PAo − PA 8 − 3 = = 6.52% PAo 8

4. If the same feed in the preceding problem is introduced in an isothermal variable-volume batch reactor, what is the time required for the same conversion? a. 9.3 minutes b. 8.5 minutes c. 6.5 minutes d. 10.5 minutes @ constant-volume 1 1 1 k=  − t  C A C Ao

 11 1  =  −  = 0.02604  83 8 

Variable Volume: Page

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nilotaldon2/7/2013

εA = t=

3.3.

Vx A =1 − Vx A = 0 Vx A = 0

3− 2 = ( 0.8) = 0.4;  2 

  (1 + 0.4 )( 0.625)  1  (1 + ε A ) x A 1 + ε A Ln(1 − x A )  = t = + ( 0.4) Ln(1 − 0.625)  = 9.32 min  ( 0.02604)( 8)  (1 − 0.625) kPAo  (1 − x A )  

REACTOR DESIGN

3.3.1 Material Balance: Reactant Leaves

υfCA dt REACTOR

Reactant Enters

υoCAo dt (-rA)Vdt

Reactant Disappears by reaction

Reactant Accumulates

Types of Reactors 1. b. c. d. 2. a. b.

d (VCA)

Single Batch Reactors (Continuously-Stirred Tanks) a. Feed is introduced before the start of reaction and product is drawn out after reaction is completed or terminated. No feed is introduced and no product is drawn out during reaction time Concentration of reactants and products varies as a function of time Concentration of reactants and products is uniform throughout the reactor volume

Back-Mix Reactors (Continuously-Stirred Tanks, steady-state) Feed is introduced simultaneously as product is drawn out The volumetric feed rate is equal to the product draw-out rate Page

22 /38

nilotaldon2/7/2013

d.

c. Reactor volume and reactant concentration remain constant and do not vary as function of time. Concentration of reactants and products is uniform throughout the reactor volume.

Plug-Flow or Tubular Reactors

3.

a. Feed is introduced simultaneously as product is drawn out b. Concentration of the reactants and products varies as a function of location or length of the reactor c. The concentration of the reactants and products at the specific length of the reactor does not vary as function of time d. The volumetric feed rate may not necessarily be equal to the volumetric product draw-out rate.

REACTOR COMPARISON 1. When fluid density is constant

t batch π = k = t plugflow

2. For all conditions

Where:

t batch π = k = t batch v = k = t plugflow = τ plugflow

3. For Batch

τ = t + ts

4. For Back-Mix

τ = t (1 + ε A x A )

τ

= space time or cycle time; the time required to process one reactor volume

t = holding time or reaction time (batch and plug-low)

t = holding time, mean residence time (for back-mix)

ts = shutdown time; time required for loading and unloading of reactants and products

respectively, cleaning etc. Applicable for batch reactors s= space velocity; volume of entering feed at a specified conditions per unit time per void volume of reactor. =1/τ

3.3.2 Single Batch Reactors (Continuously-Stirred Tanks)

In a batch reactor, neither the reactants nor the products flow into or leave the system when the reaction is carried out. They are either the constant-volume or constant-pressure reactors. The operation of a batch reactor is unsteady-state. Although the composition throughout the reactor is uniform at a given instant, it changes with time.

Material Balance

Reactant disappears by reaction

(-rA)V dt

Page

BATCH REACTOR

d (VCA)

23 /38

nilotaldon2/7/2013

Reactant Accumulates

0 = (-rA)V dt + d (VCA) dt = −

d (VC A ) V ( − rA )

where: (-rA) = kCAn

dC A =− ( − rA )

Holding or Reaction time, t_ retention time of reactants inside the reactor volume Constant Volume a.

Zero-Order

C Ao − C A

t=

b.

k

C Ao (1 − x A ) k

First-order

t= c.

=

CA 1 −1 Ln o = Ln(1 − x A ) k CA k

Second-Order t=

1  1 1  1  xA    − =   k  C A C A  kC A  1 − x A  o

d.

Third-Order

t=

1 2k

o

 1 1  1 =  2 − 2  C A  2kC A  C A o

o

2

 2x A − x A 2   2   (1 − x A ) 

Variable -Volume Reactors a. Zero-Order t=

CA

o

ε Ak

Ln (1 + ε A x A )

b. First Order 1 1  ∆V  t = − Ln(1 − x A ) = − Ln 1 −  k k  ε AVo 

c. Second-Order

t=

 1  (1 + ε A ) x A + ε A Ln(1 − x A )   kC A  (1 − x A )  o

d. Third-Order Page

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t=

(

)

(

)

2 2 2  1  (1 + ε A ) 2 x A − x A 2 ε A + ε A xA 2 − − ε A Ln(1 − x A )  2  2 (1 − x A ) kC A  2(1 − x A )  o

Note: For Batch, Space Time or cycle Time,

τ = t + ts

where:

ts= shutdown time

3.3.3 Back-Mix Reactors (Continuously-Stirred Tanks, steady-state) Material Balance

Reactant Enters

Reactant Leaves

υfCAodt

υoCAo dt

BACK-MIX REACTOR

Reactant Disappears by Reaction

(-rA)Vdt

ν oC Ao dt = ν f C Adt + ( − rA ) V dt

@ steady-state: υo = υf ; = CA, V are constant

υo (CAo-CA)dt = (-rA)Vdt V C −C C x = τ = Ao A = A o A ( − rA ) ( − rA ) νo For single reactor

( − rA ) = − C A τ

+

CA

o

_

equation of a straight line

τ

(-rA) Page

CAn

25 /38

C Ao = y - intercept τ nilotaldon2/7/2013

For reactors in series:

CAn

( − rA ) = −

+

τ

C An −1

τ

Back-Mix Constant Volume: (Steady-State) 1. Zero-Order

(− r ) = k A

τ=

CA − CA o

k

=

C A = C Ao − kτ

CA xA o

k

For Reactors in Series:

CA = CA

n −1

n

− kτ

2. First-Order

( − r ) = kC A

τ=

C Ao − C A kC A

=

A

CA 1 = C Ao 1 + kτ

xA

k (1 − x A )

For Uniform Volume Reactors in series:

C An =

C Ao (1 + kτ ) n

3. Second-Order

( − r ) = kC A

τ=

2 A

CA − CA o

kC A

2

=

xA

kC A (1 − x A )

2

o

CA =

−1± 1+ 4kτ C A o 2k τ

For Reactors in series: C An =

− 1 ± 1 + 4kτ C An − 1 2kτ n

4. Third-Order

( − r ) = kC A

τ=

3 A

CA − CA o

kC3A

=

xA

kC 2Ao

(1 − x A ) 3 Page

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Back-Mix Variable Volume: (Steady-State) General Equation:

CA − CA

τ=

o

( − rA )

=

1 x A (1 + ε A x A ) kC nAo−1 (1 − x A ) n

n

Space Time, τ 10.

Zero-Order τ=

11.

CA − CA o

k

=

CA x A o

t=

k

o

CA x A o

=

k(1+ ε A x A )

k ( 1+ ε A x A )

1 x A (1 + ε A x A ) k (1 − x A )

t=

x 1 k (1 −x A

)

A

Second-Order τ=

1 x A (1 + ε A x A ) kC A (1 − x A ) 2

2

t=

o

13.

CA − CA

First-Order

τ =

12.

−

Mean Holding Time, t

1 x (1 + ε x (1 − x ) kC A

A

)

A

2

Ao

A

Third-Order 1 x A (1 + ε A x A ) τ= kC 2Ao (1 − x A ) 3

3

Note: a. When density is constant or ε

t=

A

= 0,

1 kC 2A o

x (1 + ε x (1 − x ) A

A

A

)

2

3

A

τ=t

b. When density varies with conversion,

τ = t (1+ ε A x A )

10.5.4 For Back-Mix Reactors, (Unsteady State)

υoCAo dt = υfCAdt + (-rA)Vdt + VdCA + CAdV

Where: V = Vo + (υo-υf)t

10.6. Plug-Flow or Tubular Reactors Page

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10.6.1 Material Balance υoCAo

dV

FAo

FA

Reactant

υoCAf

FA + dFA

xA (-rA)dV

Enters

FAf xA + dxA xA

Reactant

Leaves

Reactant Disappears by Reaction

Material Balance for Incremental Volume: F A = F A + dF A + ( − rA dV ) − dF A = ( − rA ) dV FA = FAo (1 − x A ) dF A = − F Ao dx A

but:

and

Substituting values: F Ao dx A = ( − rA ) dV dx A

1 FA

∫ dV = ∫ (−r

A

o

also :

)

FA = ν o C A o

o

V τ = νo τ dx A = C Ao (−rA )

∫

General equation:

dX A A)

τ = CA

∫ (− r

τ =

1

o

kC A

o

10.6.2

n −1

∫

(1 + ε A x A ) n dx A (1 − x A ) n

Space Time and Holding Time Space Time, τ

Holding Time, t Page

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1. Zero-Order

CA − CA

τ=

o

k

=

CA xA

t=

o

k

C Ao kε A

Ln(1 + ε A x A )

2. First-Order

τ=

 1 1 − ε A xA  (1 + ε A ) ln ( k 1 − xA ) 

t=−

3. Second-Order (1 + ε A ) 2 x A  1  2 τ= 2 ε ( 1 + ε ) ln ( 1 − x ) + ε x +  A A A A A (1 − x A )  kC A  o

t=

1 Ln(1 − x A ) k

 1  (1 + ε A ) x A + ε A Ln(1 − x A )   kC A  (1 − x A )  o

4. Third-Order τ=

1 kC 2Ao

(

)(

)

(

)

 2x A − x A 2 1 + 3ε A + 6ε A 2 − 2ε A 3 6ε A 2 + 3ε A x A  2 3 ( ) − 3 ε ln 1 − x − ε x −   A A A A (1 − x A )  2(1 − x A ) 2 

t=

1 kC 2Ao

(

) (

)

 (1 + ε A ) 2 2x A − x A 2 2 ε A 2 + ε A x A  2 − − ε A Ln(1 − x A )   2 (1 − x A ) 2(1 − x A )  

Sample Problems: 1.

The reaction 2A  2R + S takes place isothermally in a constant-volume experimental reactor. Starting with a mixture of 80% A and 20% inert, the initial pressure of 10 atm increases by 25% in 8 minutes. What conversion is attained?

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2A + I → 2R + S initial

8

Δ

− xA

2

0

0

x A 0.5x A

8 − x A + 2 + x A + 0.5x A = π = ( 1.25 )( 10 )

final

x' A = 5 = PAo − PA PA = 8 − 5 = 3 atm %conversion =

2.

PAo − PA 5 = x100% = 62.5% PAo 8

Answer

If the same feed in the preceding problem is introduced in an isothermal variable-volume batch reactor, what is the time required for the same conversion? Solving for k

: 2nd − order constant volume reactor 1 1 1 1 − = kt = − = k ( 8 ) C A C Ao 3 8 k = 0.026

for 2nd − order variable volume reactor

(1+ ε A ) x + ε ln(1− x ) = kC t A Ao (1− x A ) A A

Solving for fractional change in volume :εA εA=

VxA =1 -Vx A =0 Vx A =0

Substituting values: ( 1+ε A )

( 1-x )

 3-2  = 0.8=0.4  2 

xA+εA ln ( 1-xA )=kCAo t

A

( 1 + 0.4 )

( 1 − 0.625 )

0.625 +0.4ln (1 − 0.625 )= (0.026 ) 8 ( t)

t = 9.3 minutes

3.

Answer

For a gas reaction 2A R+2S taking isothermally in a constant volume reactor. Starting with 3 atm A and 1 atm inerts, the pressure rises to 4.5 atm in 60 minutes. What space time, space velocity and holding time is required to effect this conversion in a) plug flow b) back-mix reactor. I + 2A  R + 2S initial 1 3 Δ -xA’ xA’ 2xA’ Final = π =1+3 -xA’+ xA’+ 2xA’=4.5 xA’= 1.0 PA = PAo - xA’ = 3-1= 2 Fractional conversion P −P 1 = = 33.33 % = P 3 Ao

A

Aa

For 2 -order constant volume reactor: nd

1 1 − = kt PA PA 0

 1   1 1 1 k =    −  = Answer  6 0  2 3  3 6 0

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a. For 2nd-order variable volume plug-flow reactor:

τ=

(1 + ε A ) 2 x A  1  2 ( ) ( ) 2 ε 1 + ε ln 1 − x + ε x +  A A A A A (1 − x A )  kC A  o

3−23  = 0.375 ε A =   2 4

360  (1 + 0.375 ) 2 ( 0.333 )  = 68.77 minut es 2( 0.375 )(1 + 0.375 ) ln (1 − 0.333 ) + ( 0.375 ) 2  0.333  + 3  (1 − 0.333 )  

τ=

Answer

 1  (1 + ε A ) x A + ε A Ln(1 − x A )   kC A  (1 − x A ) 

t=

o

 360  (1 + 0.375) 0.333 + 0.375Ln(1 − 0.333)  = 64.15 minutes Answer  3  (1 − 0.333) 

t=

b. for back-mix 2 1 xA ( 1 + ε A xA ) 360 ( 0.333) ( 1 + (0.375) (0.333)) τ= = 2

( 1 − xA )

kC Ao

t=

3

( 1 − 0.333)

2

1) ( (0.375)(0.333) + ( 1 xA ( 1ε+ x A A ) 3600.333 == 2 2 kC Ao ( 1 − x A ) 3 ( 1 − 0.333 )

2

113.66 = min utes

) = 101.66

min utes

Answer

Answer

or: t=

4.

τ

=

113.66

(1 + ε A x A ) (1 + (0.375)(0.33))

= 101.66 minutes

Answer

The isothermal irreversible aqueous phase A+B E @ 100oF obeys rE=KCACB; k=15ft3/lbmol.h. Using a 1000ft3 stirred tank reactor with aqueous feed of 2000 ft3/h, solve for the active concentration of E if the inlet concentration of A and B are both 0.25 lbmol/ft3. Solution: τ=

V 1000 ft 3 = = 0.5h υ 2000 ft 3 / h

For 2nd-order Back-mix reactor: τ =

CA =

− 1 ± 1 + 4kτC Ao 2kτ

=

C Ao − C A kC A 2

− 1 ± 1 + 4(15)( 0.5)( 0.25) = 0.1277 2(15)( 0.5)

C E = C Ao − C A = 0.25 − 0.1277 = 0.1233 lbmol / ft 3 Answer

5.

A gas decomposes @ 900oC according to the reaction: 2A(g)  2R(g) + S(g) with a rate constant of 1000 cm3/gmol.s. Solve for the time required in minutes for 80% of reactant A in a batch reactor @ 900oC and 1 atm. Solution: For 2nd-order variable-volume batch reactor: (1+ ε A ) x + ε ln (1− x ) = kC t A Ao (1− x A ) A A Solving for fractional change in volume :εA Page

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Vx A =1 − Vx A = 0

3− 2 =  = 0.5 Vx A = 0  2  x A = 0.80 1000 cm3  L 1L  k=  = gmol.s  1000cm 3  gmol .s εA =

C Ao=

n Ao  P   T1   1 gmol  273 1  =  =     =0.01039 V  T   PV L 1 1   273+900  1  22.4 

Substituting known values in the equation: ( 1+ε A ) x +ε ln 1-x =kC t ( A) Ao ( 1-xA ) A A t=

 ( 1+0.5 )  1 ( 0.8 ) + ( 0.5 ) ln ( 1-0.8 ) =500 s  ( 1) ( 0.01039 )  ( 1-0.8 ) 

 minutes  =500  =8.33 minutes  60 s 

Answer

11. Glossary Activation energy_ the critical amount of energy required for a reaction to take place. The amount of energy in excess of the average energy level which the reactants must have in order for the reaction to proceed. A kinetic energy greater than a certain minimum, Antimatter_ matter composed of elementary particles that are, in a special sense, mirror images of the particles that make up ordinary matter as it is known on earth. Antiparticles have the same mass as their corresponding particles but have opposite electric charges or other properties related to electromagnetism. For example, the antimatter electron, or positron, has opposite electric charge and magnetic moment (a property that determines how it behaves in a magnetic field), but is identical in all other respects to the electron. The antimatter equivalent of the chargeless neutron, on the other hand, differs in having a magnetic moment of opposite sign (magnetic moment is another electromagnetic property). In all of the other parameters involved in the dynamical properties of elementary particles, such as mass, spin, and partial decay, antiparticles are identical with their corresponding particles.The existence of antiparticles Page

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was first proposed by the British physicist Paul Adrien Maurice Dirac, arising from his attempt to apply the techniques of relativistic mechanics. Catalyst_ acts by lowering the activation energy requirement of a particular reaction or shortening of the reaction pathway. It either hastens or retards the rate of a chemical reaction without itself undergoing a permanent change. a. A substance that changes the rate of chemical reaction b. A substance which changes the rate of reaction without itself undergoing a permanent chemical change. c. A substance which lowers the activation energy requirement of a reaction. Chemical Kinetics ___ the study of the rate of reaction and mechanism by which one chemical species is converted to another. chemical kinetics or reaction kinetics is the study of reaction rates in a chemical reaction. Analyzing the influence of different reaction conditions on the reaction rate gives information about the reaction mechanism and the transition state of a chemical reaction. In 1864, Peter Waage pioneered the development of chemical kinetics by formulating the law of mass action, which states that the speed of a chemical reaction is proportional to the quantity of the reacting substances. Chemical equation_ is a symbolic representation of a chemical reaction. The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers. The first-ever chemical equation was diagrammed by Jean Beguin in 1615. Chemical reaction_ is a process that results in the interconversion of chemical substances.The substance or substances initially involved in a chemical reaction are called reactants. Chemical reactions are usually characterized by a chemical change, and they yield one or more products which are, in general, different from the reactants. Chemical reaction engineering _is the branch of engineering that is concerned with the exploitation of chemical reactions on a commercial scale for purposes other than the production of power. Collision theory of chemical reaction_ rate is directly proportional to the number of collisions per second a. Only certain collisions between particles results in the formation of products b. The molecules must have proper orientation c. A more concentrated solution contains a greater number of particles Collision Theory of a Chemical reaction: For a chemical reaction to proceed, molecules must have effective collisions. The two requirements for an effective collisions are; a.The molecules must be suitably reactive and must have enough energy, b. The molecules must be arranged in a proper position. a. The molecules must be suitably reactive and must have enough energy. The energy that the molecules must possess is known as activation energy. The critical amount of energy required for a reaction to take place. The amount of energy in excess of the average energy level which the reactants must have in order for the reaction to proceed. Collision theory is a theory, proposed by Max Trautz and William Lewis in 1916 that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. It assumes that for a reaction to occur the reactant particles must collide, but only a certain fraction of the total collisions, the effective collisions, cause the transformation of reactant molecules into products. This is due to the fact that only a fraction of the molecules have sufficient energy and the right orientation at the moment of impact to break the existing bonds and form new bonds. Activated Complex or Transition Theory: Henry Eyring, an American chemist, postulated an alternative to collision theory. He hypothesized that an intermediate species called an activated complex forms during collision. This intermediate species exist very briefly. It dissociates to form either the product (if reaction occurs) or the original reactants (if reaction does not occur) Most reactions proceed in many steps called elementary processes. The combined effect of all the elementary processes gives the overall reaction. The slow step determines the rate of the chemical reaction and is called the determining step. The observed rate of the overall reaction is the equivalent to the rate of the slow reaction b. The molecules must be arranged in a proper position. This affects the chance for the reaction to occur.

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Dutch chemist Jacobus van’t Hoff won the 1901 Nobel Prize in chemistry. van’t Hoff investigated the structure of organic compounds and came to be known as the “father of chemical kinetics.”

Jacobus van’t Hoff

Complex chemical reactions_ reactions where actual mechanisms are not represented by simple stoichiometric equations. These are usually non-elementary reactions and involve more than one step to accomplish. Elementary Particles, in physics, particles that cannot be broken down into any other particles. The term elementary particles also is used more loosely to include some subatomic particles that are composed of other particles. Particles that cannot be broken further are sometimes called fundamental particles to avoid confusion. These fundamental particles provide the basic units that make up all matter and energy in the universe.

Structure of Matter Modern physics has revealed successively deeper layers of structure in ordinary matter. Matter is composed, on a tiny scale, of particles called atoms. Atoms are in turn made up of minuscule nuclei surrounded by a cloud of particles called electrons. Nuclei are composed of particles called protons and neutrons, which are themselves made up of even smaller particles called quarks. Quarks are believed to be fundamental, meaning that they cannot be broken up into smaller particles. Elementary Reactions __ any such reaction in which the rate equation suggested by stoichiometric equation represents the actual mode of action and occurs in a single step. Equilibrium_ the system is in the state of equilibrium if the following characteristics are observed: 1. It is a closed system ( the amount of matter in the system does not change) 2. There is no change in the properties of the system as time es. 3. Two processes, which are opposite in direction, simultaneously take place at the same rate. For chemical equilibrium: The ratio of the product of the molar concentration of the substances formed to the product of the molar concentration of the reactants is constant. Equilibrium constant _ the ratio of the forward and reverse rate constant for a reversible reaction. Fermion_ one of the two main classes of fundamental particles that make up matter and energy. Fermions play an important role in the structure of matter. The particles that make up atoms (electrons, protons, and neutrons) are all fermions. Fermions are named for Italian-born American physicist Enrico Fermi. In the 1920s Fermi calculated a set of rules that define the behavior of fermions. Fractional conversion ( xA )_ the amount of the reactant converted to the desired product divided by the original amount of reactants. Fractional change in volume (εA)_ the resulting change in volume when all of the reactants are converted divided by the original volume occupied by the reactants before the start of the reaction . Vx A =1 − Vx A = 0 Vx A = 0 εA = General Theory of Relativity_ The principle of equivalence holds that forces produced by gravity are in every way equivalent to forces produced by acceleration, so that it is theoretically impossible to distinguish between Page

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gravitational and accelerational forces by experiment. In 1915 Einstein developed the general theory of relativity in which he considered objects accelerated with respect to one another. He developed this theory to explain apparent conflicts between the laws of relativity and the law of gravity. To resolve these conflicts he developed an entirely new approach to the concept of gravity, based on the principle of equivalence. Holding time or reaction time (t) = (batch and plug-low); t = holding time, mean residence time (for back-mix)_ the average time required by the reactants to stay inside the rector in order to attain the desired conversion to the desired product Hadron, family of elementary particles. All known particles of matter and energy are classified as hadrons, leptons, or force carriers. Scientists believe leptons and force carriers are fundamental particles, meaning they cannot be divided into smaller particles. Hadrons, on the other hand, have an underlying structure of smaller particles called quarks. The most common hadrons are protons and neutrons, the building blocks of the nucleus in an atom. Hess’s Law _ The total change in enthalpy of a system is dependent on the temperature, pressure, state of aggregation, and state of combination at the beginning and at the end; it is independent of the number of intermediate reactions. Irreversible reactions_ reactions that proceed only in one direction Law of chemical equilibrium_ the ratio of the products of the molar concentration of the substances formed to the product of the molar concentration of the reactants is constant. Law of Mass Action _states that the rate of chemical reaction is at each instant proportional to the concentration of the reactant with each raised to a power equal to their coefficient or the actual number of molecules participating in the reaction. This law can be interpreted by several complex mechanisms but it can simply be explained as follows: when two or more molecules react, it must come close to one another or must collide. Therefore, it is expected that the rate of reaction increases if the molecules are crowded closely together, i.e., the concentration is high. Le Chatelier’s Principle_ If a stress (disturbance or change) is applied to a system in a state of equilibrium, the system will shift in such a way to neutralize the effect of the stress . Lepton_ type of elementary particle, the most basic building block of matter, that does not experience the strong force. The strong force holds particles in the nucleus of an atom together. Physicists have discovered the following leptons: electrons, muons, tau particles, and three corresponding types of neutrino (electron neutrinos, muon neutrinos, and tau neutrinos). Limiting reactant_ the reactant that determines the rate of the chemical reaction. The chemical that determines how far the reaction will go before the chemical in question gets used up, causing the reaction to stop. The chemical of which there are fewer mols than the proportion requires is the limiting reagent. Mechanism __ is the sequence of individual chemical events whose over-all result produce the observed reaction. _ the sequence of steps that takes place to complete a chemical change. Metabolic pathway_ is a series of chemical reactions occurring within a cell. In each pathway a principal chemical is modified by chemical reactions. These reactions are accelerated, more accurately catalyzed, by enzymes. Dietary minerals, vitamins and other cofactors are often needed by the enzyme to perform its task. Many pathways are elaborate. Various metabolic pathways within each cell form that cell's metabolic network. Pathways are needed by an organism to keep its homeostasis. Neutrino_ very small particles with no electric charge and little or no mass. Neutrinos are elementary particles— that is, they cannot be broken into smaller particles. Neutrinos are so small that they right through most material. One important kind of neutrino is created in the nuclear reactions that give the Sun its energy. Neutrinos are of a group of elementary particles called leptons Non-elementary Reactions __ are those where there is no direct correspondence between stoichiometric equation and the rate expression and occurs in two or more series of reactions. Quantum theory_ based on the idea that energy exist in the form of tiny, discrete units called quanta. Quarks_ Quark, smallest known building block of matter. Quarks never occur alone; they always are found in combination with other quarks in larger particles of matter. An elementary particle with an electric charge equal to one-third or two-thirds that of the electron. Quarks are believed to be the constituents of baryons and mesons Quarks and Their Properties Name Symbol up

Mass1 (GeV) u

.002 - .005

Charge2 (e) +2/3

Generation First Page

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down

d

.003 - .009

-1/3

First

charm

c

1.1 - 1.4

+2/3

second

strange

s

.06 - .17

-1/3

second

top

t

163 -179

+2/3

third

b

4.1 - 4.4

-1/3

third

bottom 1

One GeV is equal to 1.78 x 10

-17

kg (3.92 x 10-17 lb).

The charge 'e' is equal to the charge on one proton, or 1.602 x 10-19 coulombs. 2

Rate constant_ experimentally determined constant of proportionality between the reaction rate and the concentrations of reactants that appear in the rate law. Rate Law_ equation or mathematical expression showing the relationship between reactant concentrations and rate of reaction. It has the form: rate = k [A]a[B]b[c]c Rate of reaction ___ the number of units of mass of some participating reactants which is transformed into a product per unit time and per unit volume. Reactant or reagent _ a chemical substance which takes part in a chemical reaction is a substance consumed during a chemical reaction. Solvents and catalysts, although they are involved in the reaction, are usually not referred to as reactants. Reaction mechanism is the step by step sequence of elementary reactions by which overall chemical change occurs. Series of successive elementary steps by which reactants are converted to products. Reducing agent_ substance that produces reaction in another substance. Reduction_ removal of oxygen; combination with hydrogen; involves gain in electrons. Resonance structures_ two or more possible Lewis structures used to describe the bonding in a molecule or ion. Reversible Reactions_is one which results in the formation of an equilibrium mixture. The concept of a reversible reaction was introduced by Berthollet (1803) after he had observed the formation of sodium carbonate crystals at the edge of a salt lake. 2NaCl + CaCO3 → Na2CO3 + CaCl2 He recognized this as the reverse of the familiar reaction Na2CO3 + CaCl2→ 2NaCl + CaCO3 Reversible process, or reversible cycle_ in thermodynamics, is a process that can be "reversed" by means of infinitesimal changes in some property of the system without loss or dissipation of energy. Due to these infinitesimal changes, the system is at rest during the whole process. Since it would take an infinite amount of time for the process to finish, perfectly reversible processes are impossible. Shielding effect_ electrons in filled energy levels that fairly but effectively shield the outer electrons in the outermost energy level from the effect of an equal number of protons in the nucleus. Shutdown time ( ts)_ time required for loading and unloading of reactants and products respectively, cleaning Single replacement reaction_ element replaces another element in a compound. Solid_ state of matter that has a definite shape and volume. Solubility_ measure of the amount of solute that can dissolve in a given amount of solvent at a certain temperature Standard molar volume_ volume occupied by one mole of an ideal gas at STP=22.4 L Stoichiometry_ calculation of quantities of reactants and products in a chemical reaction. Sub-atomic particles_ electrons, protons and neutrons contained in an atom. Solution_ homogeneous mixture of two or more substances. Specific gravity_ ratio of the density of a substance to the density of water at the same temperature. Space time(τ) or cycle time_ the time required to process one reactor volume Space velocity ( s)_volume of entering feed at a specified conditions per unit time per void volume of reactor. =1/τ = the number of reactor volumes of feed at specified conditions which can be processed in unit time Special Theory of Relativity_ According to Einstein, no particular object in the universe is suitable as an absolute frame of reference that is at rest with respect to space. Any object (such as the center of the solar system) is a suitable frame of reference, and the motion of any object can be referred to that frame. Thus, it is equally correct to say that a train moves past the station, or that the station moves past the train. This example is not as unreasonable as it seems at first sight, for the station is also moving, due to the motion of the earth on its axis and its revolution around the sun. All motion is relative, according to Einstein. None of Einstein's basic assumptions was revolutionary; Newton had previously stated “absolute rest Page 36 /38 nilotaldon2/7/2013

cannot be determined from the position of bodies in our regions.” Einstein stated the relative rate of motion between any observer and any ray of light is always the same, 300,000 km/sec (186,000 mi/sec), and thus two observers, moving relative to one another even at a speed of 160,000 km/sec (100,000 mi/sec), each measuring the velocity of the same ray of light, would both find it to be moving at 300,000 km/sec (186,000 mi/sec), and this apparently anomalous result was proved by the Michelson-Morley experiment. Spin _ in physics, intrinsic angular momentum of a sub-atomic particle. In particle and atomic physics, there are two types of angular momentum: spin and orbital angular momentum. Spin is a fundamental property of all elementary particles, and is present even if the particle is not moving; orbital angular momentum results from the motion of a particle. For example, an electron in an atom has orbital angular momentum, which results from the electron's motion about the nucleus, and spin angular momentum. The total angular momentum of a particle is a combination of spin and orbital angular momentum. Stoichiometric equation_ is a chemical equation which expresses an overall chemical reaction in of the simplest ratio of reactant and product molecules. Sub-atomic particles_ electrons, protons and neutrons combined in an atom Temperature_ Measure of the average kinetic energy of the molecules. Temperature jump_ is a technique used in the study of chemical kinetics. It usually involves the discharging of a capacitor (in the kV range) through a small volume (<ml) of a conducting solution containing the molecule/reaction to be studied. In some versions of the apparatus used, the solution is heated instead by the output of a pulsed laser which emits in the near infra-red. When laser heating is employed, the solution need not be conducting. In both cases, the temperature of the solution is caused to rise by a small amount in microseconds (or less in the case of laser heating). This allows the study of the shift in equilibrium of reactions that equilibrate in milliseconds (or microseconds with laser temperature jump), these changes most commonly being observed using absorption spectroscopy. Due to the small volumes involved the temperature of the solution returns to that of its surroundings in minutes. Termolecular reaction_ when elementary step involves the simultaneous collision of three molecules. Ternary compounds_ contain atoms of three elements. Theoretical yield_ calculated quantity of product in chemical reaction. Torr_ metric unit of pressure equal to 1 mm Hg. 760 Torr = 1 atm. Triple bond_ formed by the sharing of three pairs of electrons between atoms. Uncertainty principle_ states that it is impossible to accurately determine the exact position and velocity of an electron at the same time. Unimolecular reaction_ when elementary process involves only one molecule. Unit cell_ smallest repeating unit showing the structural characteristic of a crystal. Universal gas constant_ (R) proportionality constant in the Ideal Gas Law, PV= nRT Valence electrons_ outermost electrons of atoms Valence-shell-electron-pair repulsion (VSEPR)_ theory assumes that electron pairs are arranged around the central atom of a molecule or polyatomic ion so that there is maximum separation and minimum repulsion among electron pairs. Vapor pressure of the liquid_ pressure exerted by the vapor particles on the liquid at equilibrium Viscosity_ tendency of a liquid to resist flow Volatility_ refers to the case with which a liquid vaporizes . 12. References: Chemical Reaction Engineering, Octave Levenspiel Reaction Kinetics for Chemical Engineers, Stanley M. Walas Chemical Engineering Kinetics, J.M. Smith

13. Sample Problems

1.

Derivation of Formula Michaeli’s Menten equation: Page

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E+S

= ⇄ ES → E + P k1

k3

k2

d [ ES ]

= 0 = k1 ( [ ET ] − [ ES ] ) [ S ] − k 2 [ ES ] − k 3 [ ES ] dt k1 [ ET ] [ S ] − k1 [ ES ] [ S ] − k 2 [ ES ] − k 3 [ ES ] = 0

[ ES ] ( k1 [ S ] + k 2 + k 3 ) = k1 [ ET ] [ S ] = 0 k [ E ][ S] [ ET ] [ S ] = [ ES ] = 1 T ( k2 + k3 ) + k1 [ S ]  k2 + k3  + k1  

[ ES ] = [ ET ] [ S ] km + [ S ]

k1

 

k1

[ S]

=

[ ET] [S ] km+ [ S ]

Michaeli’s Menten

equation (1)

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