Rates of Reaction

Dr. Pedro Julio Villegas [email protected]

2 0 1 6

TEXTBOOK 

Brown, T. L. ; LeMay Jr., H. E.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M. “Chemistry, The Central Science, Chapter 14: Chemical Kinetics”. 13th Edition, Prentice Hall Publisher, ISBN: 978-0321-910-417. USA, January, 2014.

2

INTRODUCTION (1) 



Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. It includes investigations of how different experimental conditions can influence the speed of a chemical reaction and yield information about the reaction’s mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a 3 chemical reaction.




Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time. 4

During the course of this reaction, reactant A is consumed while the concentration of product B increases. The reaction rate can be determined by measuring how fast the concentration of A decreases, or by how fast the concentration of B increases.




The experimental determination of reaction rates involves measuring how the concentrations of reactants or products change over time. For example, the concentration of a reactant can be measured by spectrophotometry at a wavelength where no other reactant or product in the system absorbs light. For reactions which take at least several minutes, it is possible to start the observations after the reactants have been mixed at the temperature of interest. 6

THE OBJECTIVES OF THIS PRESENTATION ARE:  Describe how the reaction rate can be measured.  Detail the conditions that affects the chemical reactions rate.  Show how express the relationship of rate with the variables affecting the rate.  Finally, it will be show what happens on a molecular level during a chemical reaction. 7

REACTION RATES (1) 

Chemical reactions require varying lengths of time for completion. • This reaction rate depends on the characteristics of the reactants and products and the conditions under which the reaction is run. • By understanding how the rate of a reaction is affected by changing conditions, one can learn the details of what is happening at the molecular level. 8

REACTION RATES (2) 

Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction.  What variables affect the reaction rate?  Concentration of reactants.  Concentration of a catalyst.  Temperature at which the reaction occurs.  Surface area of a solid reactant or catalyst. 9

FACTORS AFFECTING REACTION RATES (1) CONCENTRATION OF REACTANTS  More often than not, the reaction rate increases when the concentration of a reactant is increased.  Increasing the population of reactants increases the likelihood of a successful collision.  In some reactions, however, the rate is unaffected by the concentration of a particular reactant, as long as it is present at some concentration. 11

FACTORS AFFECTING REACTION RATES (2) CONCENTRATION OF A CATALYST (1)  A catalyst is a substance that increases the rate of a reaction without being consumed in the overall reaction.  The catalyst generally does not appear in the overall balanced chemical equation (although its presence may be indicated by writing its formula over the arrow).

2H2O2(aq)     2H2O(l)  O2(g) HBr(aq)

12

FACTORS AFFECTING REACTION RATES (3) CONCENTRATION OF A CATALYST (2)  See an example of the HBr decomposition of H2O2 to H2O and O2.

catalyzed

2H2O2( aq)     2H2O(l)  O2(g) HBr ( aq)

 A catalyst speeds up reactions by reducing the “activation energy” needed for successful reaction.  A catalyst may also provide an alternative mechanism, or pathway, that results in a faster rate. 13

FACTORS AFFECTING REACTION RATES (4) TEMPERATURE AT WHICH THE REACTION OCCURS  Usually reactions speed up when the temperature increases.  A good “rule of thumb” is that reactions approximately double in rate with a 10oC rise in temperature. 14

FACTORS AFFECTING REACTION RATES (5) SURFACE AREA OF A SOLID REACTANT OR CATALYST

Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area. 15

DEFINITION OF REACTION RATE (1) 

The reaction rate is the increase in molar concentration of a product of a reaction per time unit.



It can be also expressed as the decrease in molar concentration of a reactant per time unit. 16

DEFINITION OF REACTION RATE (2)  Consider the gas-phase decomposition of nitrogen pent-oxide. 2N2O5 (g)  4NO2 (g)  O2 (g)  If we denote molar concentrations using brackets, then the change in the molarity of O2 would be represented as:

D[ O 2 ]

 Symbol, D means “the change in”. 17

DEFINITION OF REACTION RATE (3)  Then, in a given time interval (Dt), the molar concent. of O2 would increase by D[O2].  The rate of the reaction is given by: D[O2 ] Rate of oxygen formation  Dt • This equation gives the average rate over the time interval, Dt. • If Dt is short, you obtain an instantaneous rate, that is, the rate at a particular instant (See next Figure). 18

INSTANTANEOUS REACTION RATE

For the reaction 2N2O5 (g)  4NO2 (g)  O2 (g) The concentration of O2 increases with the time. Instantaneous rate is obtained from the slope of the tangent at the point of the curve corresponding to that time.


Next Figure shows the increase in concentration of O2 during the decomposition of N2O5.



Note that the rate decreases as the reaction proceeds.

20

CALCULATION OF THE AVERAGE RATE When the time changes from 600 to 1200s, the average rate is 2.5 x 10-6 mol/(L*s). Later when the time changes from 4200 s to 4800 s, the average rate has slowed to 5 x 10-7 mol/(L.s). Thus, the rate of a reaction decreases as the reaction proceeds.


Because the amounts of products and reactants are related by stoichiometry, any substance in the reaction can be used to express the rate.

D[N2O5 ] Rate of N2O5 decomposition  Dt



Note the negative sign. This results in a positive rate as reactant concentrations decrease. 22


The rate of decomposition of N2O5 and the formation of O2 are easily related.

D[O2 ] 1  D[N2O5 ]   2 *  Dt Dt   

Since two moles of N2O5 decompose for each mole of O2 formed, the rate of the decomposition of N2O5 is twice the rate of the formation of O2. 23

EXPERIMENTAL DETERMINATION OF REACTION RATES (1) 

To obtain the rate of a reaction it must determine the concentration of a reactant or product during the course of the reaction. • One method for slow reactions is to withdraw samples from the reaction vessel at various times and analyze them. • More convenient are techniques that continuously monitor the progress of a reaction based on some physical property of the system. 24

EXPERIMENTAL DETERMINATION OF REACTION RATES (2) GAS-PHASE PARTIAL PRESSURES 

When di-nitrogen pent-oxide crystals are sealed in a vessel equipped with a manometer and heated to 45oC, the crystals vaporize and the N2O5(g) decomposes.

2N2O5 (g)  4NO2 (g)  O2 (g) 

Manometer readings provide the concentration of N2O5 during the course of the reaction based on partial pressures. 25

EXPERIMENTAL DETERMINATION OF REACTION RATES (3) COLORIMETRY  Considering the reaction of the hypochlorite ion with iodide:     ClO ( aq)  I ( aq)  IO ( aq)  Cl ( aq)  The hypo-iodate ion (IO-) absorbs near 400 nm. The intensity of the absorption is proportional to [IO-], and you can use the absorption rate to determine the reaction rate. 26

DEPENDENCE OF RATE ON CONCENTRATION (1) 

Experimentally, it has been found that the rate of a reaction depends on the concentration of certain reactants as well as catalysts. • Let’s look at the reaction of nitrogen dioxide with fluorine to give nitryl-fluoride.

2NO2 (g)  F2 (g)  2NO2F(g)

• The rate of this reaction has been observed to be proportional to the concentration of nitrogen dioxide. 27


When the concentration of nitrogen dioxide is doubled, the reaction rate doubles. • The rate is also proportional to the concentration of fluorine; doubling the concentration of fluorine also doubles the rate. • We need a mathematical expression to relate the rate of the reaction to the 28 concentrations of the reactants.


A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.

Rate  k * [NO2 ] * [F2 ] 

Were k is the rate constant. It is a proportionality constant in the relationship between rate and 29 concentrations.

DEPENDENCE OF RATE ON CONCENTRATION (4) As a more general example, consider the reaction of substances A and B to give D and E.



aA  bB   dD  eE

C  catalyst

C



Then, the rate law could be written as fallows:

Rate  k * [A] * [B] * [C] The exponents m, n, and p are frequently, but not always, integers. They must be determined experimentally and cannot be obtained by simply looking at the balanced equation. 30 m



n

p

DEPENDENCE OF RATE ON CONCENTRATION (5) REACTION ORDER 

The reaction order with respect to a given reactant species is equals to the exponent of the concentration of that species in the rate law, as determined experimentally.  The overall order of the reaction equals the sum of the orders of the reacting species in the rate law. 31

DEPENDENCE OF RATE ON CONCENTRATION (6) REACTION ORDER  Consider

the reaction of nitric oxide with hydrogen according to the following equation. 2NO(g)  2H2 (g)  N2 (g)  2H2O(g)

 The

experimentally determined rate law is: Rate  k * [NO]2 * [H2 ]  Thus, the reaction is second order in NO, first order in H2, and third order overall. 32

DEPENDENCE OF RATE ON CONCENTRATION (7) REACTION ORDER 

Although reaction orders frequently have whole number values (particularly 1 and 2), they can be fractional. • Zero and negative orders are also possible. • The concentration of a reactant with a zero-order dependence has no effect on the rate of the reaction. 33

DEPENDENCE OF RATE ON CONCENTRATION (8) DETERMINING THE RATE LAW 

One method for determining the order of a reaction with respect to each reactant is the method of initial rates. • It involves running the experiment multiple times, each time varying the concentration of only one reactant and measuring its initial rate. • The resulting change in rate indicates the order with respect to that reactant. 34

DEPENDENCE OF RATE ON CONCENTRATION (9) DETERMINING THE RATE LAW • If doubling the concentration of a reactant has a doubling effect on the rate, then one would deduce it was a first-order dependence. • If doubling the concentration had a quadrupling effect on the rate, one would deduce it was a second-order dependence. • A doubling of concentration that results in an eight-fold increase in the rate would be a thirdorder dependence. 35

A PROBLEM TO CONSIDER (1) 

Iodide ion is oxidized in acidic solution to tri-iodide ion, I3- , by hydrogen peroxide.

H2O2 ( aq)  3I



( aq)

 2H



( aq)

 I3



( aq)

 2H2O(l)

A series of four experiments was run at different concentrations, and the initial rates of I3- formation were determined. a) From the following data, obtain the reaction orders with respect to H2O2, I-, and H+. b) Calculate the numerical value of the rate constant. 36

A PROBLEM TO CONSIDER (2)

Exp. 1 Exp. 2 Exp. 3 Exp. 4

Initial Concentrations (mol/L) H2O2 IH+ Initial Rate [mol/(L.s)] 0.010 0.010 0.00050 1.15 x 10-6 0.020 0.010 0.00050 2.30 x 10-6 0.010 0.020 0.00050 2.30 x 10-6 0.010 0.010 0.00100 1.15 x 10-6

• Comparing Experiment 1 and Experiment 2, it can be observed that when the H2O2 concentration doubles (with other concentrations constant), the rate doubles. • This implies a first-order dependence with respect to H2O2. 37

A PROBLEM TO CONSIDER (3) Exp. 1 Exp. 2 Exp. 3 Exp. 4


• Comparing Experiment 1 and Experiment 3, it can be observed that when the I- concentration doubles (with other concentrations constant), the rate doubles. • This implies a first-order dependence with respect to I-. 38

A PROBLEM TO CONSIDER (4) Exp. 1 Exp. 2 Exp. 3 Exp. 4


• Comparing Experiment 1 and Experiment 4, it can be observed that when the H+ concentration doubles (with other conc. constant), the rate is unchanged. • This implies a zero-order dependence with 39 respect to H+.




• Because [H+]0 = 1, the rate law is: 

Rate  k * [H2O2 ] * [I ] • The reaction orders with respect to H2O2, I-, and H+, are 1, 1, and 0, respectively. 40




• It can be now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 it is obtained:

mol mol mol 1.15 * 10  k * 0.010 * 0.010 Ls L L 6

41




• It can be now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1 it is obtained:

1.15 * 10 6 s 1 2 k  1.2 * 10 L /(mol  s) 0.010 * 0.010mol / L

42

CHANGE OF CONCENTRATION WITH TIME (1)  A rate law simply tells you how the rate of reaction changes as reactant concentrations change. • A more useful mathematical relationship would show how a reactant concentration changes over a period of time. 43

CHANGE OF CONCENTRATION WITH TIME (2) A

rate law simply indicates how the rate of reaction changes as reactant concentrations change. • Using calculus it can be transformed a rate law into a mathematical relationship between concentration and time. • This provides a graphical method for determining rate laws. 44

CONCENTRATION-TIME EQUATIONS (1) FIRST-ORDER RATE LAW  The rate law can be written in the form:

D[ A ] Rate    k * [A] Dt

45

CONCENTRATION-TIME EQUATIONS (2) FIRST ORDER RATE LAW 

Using calculus, it can be getting the following equation. [ A ]t ln  -k * t [ A ]o

• Where [A]t is the concentration of reactant A at time (t) and [A]o is the initial concentration. • The ratio [A]t / [A]o is the fraction of A remaining at time t. 46

A PROBLEM TO CONSIDER (1) 



The decomposition of N2O5 to NO2 and O2 is a first order reaction with a rate constant of 4.8*10-4 s-1. If the initial concentration of N2O5 is 1.65*10-2 mol/L, what is the concentration of N2O5 after 825 seconds? The first order time-concentration equation for this reaction would be: [N2O5 ]t ln  -k * t [N2O5 ]o 47




The decomposition of N2O5 to NO2 and O2 is a first order reaction with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? Substituting the given information we obtain:

[N2O5 ]t - 4 -1 ln  (4.80 * 10 s ) * ( 825 s) 2 1.65 * 10 mol / L 48




The decomposition of N2O5 to NO2 and O2 is a first order reaction with a rate constant of 4.8 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? Substituting the given information we obtain:

[N2O5 ]t ln  0.396 2 1.65 * 10 mol / L 49

A PROBLEM TO CONSIDER (4)  The

decomposition of N2O5 to NO2 and O2 is a first order reaction with a rate constant of 4.8* 10-4s-1. If the initial concentration of N2O5 is 1.65*10-2 mol/L, what is the concentration of N2O5 after 825 seconds?  Applying inverse natural logarithm of both equation sides we obtain:

[N2O5 ]t -0.396 e  0.673 2 1.65 * 10 mol / L

50

A PROBLEM TO CONSIDER (5) The

decomposition of N2O5 to NO2 and O2 is a first order reaction with a rate constant of 4.8*10-4 s-1. If the initial concentration of N2O5 is 1.65*10-2 mol/L, what is the concentration of N2O5 after 825 seconds? Solving for [N2O5] at 825 s it is obtained:

[N2O5 ]  (1.65 * 10 mol / L) * (0.673)  0.011 mol / L -2

51

CONCENTRATION-TIME EQUATIONS (1) SECOND ORDER RATE LAW  The rate law can be written in the following form:

D[ A ] 2 Rate    k * [A] Dt 52

CONCENTRATION-TIME EQUATIONS (2) 

SECOND ORDER RATE LAW Using calculus, it can be getting the following equation.

1 1 k*t  [ A ]t [A]o 

Where [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration. 53

HALF-LIFE (1) The

half-life of a reaction is the time required for the reactant concentration to decrease to onehalf of its initial value. • For a first-order reaction, the half-life is independent of the initial concentration of reactant. • In one half-life the amount of reactant decreases by one-half. Substituting into the first-order concentration-time equation, we get: 1 ln( 2 )

  k * t 21

54

HALF-LIFE (2) 



The HALF-LIFE of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. Solving for t1/2 it is obtained:

0.693 t 21  k 55

HALF-LIFE (3) 

Sulfuryl chloride (SO2Cl2) decomposes in a first-order reaction into SO2 and Cl2. SO2Cl2 (g)  SO2 (g)  Cl2 (g)



At 320oC, the rate constant is 2.2*10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature? Substituting the value of k into the relationship between k and t1/2. 0.693 t 21  56 k



HALF-LIFE (4) 





Sulfuryl chloride (SO2Cl2) decomposes in a first-order reaction into SO2 and Cl2. SO2Cl2 (g)  SO2 (g)  Cl2 (g) At 320oC, the rate constant is 2.2*10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature? Substituting the value of k into the relationship between k and t1/2.

0.693 t 21   5 -1 2.20 * 10 s

57

HALF-LIFE (5) 





Sulfuryl chloride (SO2Cl2) decomposes in a first-order reaction into SO2 and Cl2. SO2Cl2 (g)  SO2 (g)  Cl2 (g) At 320oC, the rate constant is 2.2*10-5 s-1. What is the half-life of SO2Cl2 vapor at this temperature? Substituting the value of k into the relationship between k and t1/2.

t 21  3.15 * 10 s 4

58

HALF-LIFE (6) For

a second order reaction, half-life depends on the initial concentration and becomes larger as time goes on.  Again, assuming that [A]t = ½[A]o after one half-life, it can be shown that: 1 t 21  k * [ A ]o 

Each succeeding half-life is twice the length of its predecessor. 59

GRAPHING KINETIC DATA (1)  

In addition to the initial rates’ method, rate laws can be deduced by graphical methods. If we rewrite the first order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line.

ln[ A ]t   kt  ln[ A ]o y 

= mx + b

It means that if we plot ln[A] versus time, It will be getting a straight line for a first order reaction (see next Figure). 60

Plot of ln[N2O5] versus reaction time

GRAPHING KINETIC DATA (3)  

In addition to the initial rates’ method, rate laws can be deduced by graphical methods. If we rewrite the second order concentration time equation in a slightly different form, it can be identified as the equation of a straight line.

1 1 k*t [ A ]t [ A ]o 



y = mx + b

It means that if we plot 1/[A] versus time, it will be getting a straight line for a second order reaction. 62

COLLISION THEORY (1) 

Rate constants use to vary with temperature.  Consequently, the actual rate of a reaction is significant temperature dependent.  The rate dependence on temperature can by explained by the “Collision Theory”. 63

COLLISION THEORY (2) 

The Collision Theory assumes that for a reaction to occur, reactant molecules must collide with enough energy and the proper orientation.  The minimum energy of collision required for two molecules to react is called Activation Energy (Ea). 64

TRANSITION-STATE THEORY (1) 

The Transition State Theory explains the reaction resulting from the collision of two molecules in terms of an Activated Complex. • An Activated Complex (Transition State) is an unstable grouping of atoms that can break up to form products. • A simple analogy would be the collision of three billiard balls on a billiard table. 65


The Transition State Theory explains the reaction resulting from the collision of two molecules in terms of an Activated Complex. • Suppose two balls are coated with a slightly stick adhesive. • We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair. 66


The Transition State Theory explains the reaction resulting from the collision of two molecules in terms of an Activated Complex. • The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing.” • At the instant of impact, when all three spheres are joined, we have an unstable “Transition State Complex”. 67


The Transition State Theory explains the reaction resulting from the collision of two molecules in terms of an Activated Complex. • If this scenario is repeated several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful. • It could be compared the energy we provided to the billiard balls to the 68 Activation Energy (Ea).

POTENTIAL-ENERGY DIAGRAMS FOR REACTIONS (1) 

To illustrate graphically the formation of a transition state, we can plot the EnergyPotential of a reaction versus Time. • Next figure illustrates the endothermic reaction of nitric oxide with chlorine gas. • Note that the forward activation energy is the energy necessary to form the “Activated Complex”. • The DH of the reaction is the net change in energy between reactants and products. 69

Energy-Potential Curve for the Endothermic Reaction of Nitric Oxide with Chlorine.

POTENTIAL-ENERGY DIAGRAMS FOR REACTIONS (2) 

The Energy-Potential Diagram for an exothermic reaction shows that the products are more stable than the reactants. • Next figure illustrates the Energy-Potential Diagram for an exothermic reaction. • It can be observed again that the forward activation energy is required to form the “Transition State Complex”. • In both of these graphs, the reverse reaction must still supply enough activation energy to form the “Activated Complex”. 71

Example of Energy – Potential Curve for an Exothermic Reaction.

COLLISION THEORY AND THE ARRHENIUS EQUATION (1) 

Collision theory states that the rate constant for a reaction is the product of three factors. 1. Z: The collision frequency; 2. F: The fraction of collisions with sufficient energy to react; 3. P: The fraction of collisions with the proper orientation to react.

k  Z *p * f 73


Z is a slightly temperature dependent. • This is illustrated using the kinetic theory of gases, which shows the relationship between the velocity of gas molecules and their absolute temperature. velocity 

3RTabs Mm

or

velocity  Tabs 74


Z is only slightly temperature dependent. • This alone does not account for the observed increases in rates with only small increases in temperature. • From the kinetic theory, it can be shown that a 10oC rise in temperature will produce only a 2% rise in collision frequency. 75


On the other hand, the fraction of molecules with sufficient activation energy (f), tends to be very temperature dependent. • It can be shown that f is related to Ea by the following expression:

fe



Ea RT

• Here e = 2.718… , and R is the ideal gas constant {R = 8.31 J/(mol.K)}. 76


On the other hand, the fraction of molecules with sufficient activation energy (f), tends to be very temperature dependent. This is the primary factor relating temperature increases to observed rate increases. Ea  RT

fe

• From this relationship, as temperature increases, f increases. • Also, a decrease in the activation energy (Ea), increases the value of f. 77


The reaction rate also depends on the fraction of collisions with the proper orientation (p). • This factor is independent of temperature changes. • So, with changes in temperature, Z and p remain fairly constant. • It can be used that fact to derive a mathematical relationship between the rate constant (k) and the absolute temperature. 78

THE ARRHENIUS EQUATION (1) 

If we were to combine the relatively constant terms, Z and p, into one constant, let’s call it A. We obtain the Arrhenius equation:

kAe



Ea RT

• The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy. 79


It is useful to recast the Arrhenius Equation into logarithmic form.  Taking the natural logarithm of both sides of the equation, we get:

ln k  ln A -

Ea RT 80

THE ARRHENIUS EQUATION (3) • It is useful to recast the Arrhenius equation into logarithmic form. • If we rearrange the general formula, a straight line equation can be obtained: 1 ln k  ln A - Ea ( )  R T

y = b+mx

A plot of ln k versus (1/T) should yield a straight line with a slope of (-Ea/R) and an intercept of ln A (see next Figure). 81

Plot of ln k versus 1/T.


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)). • The two equations describing the relationship at each coordinate would be: 1 ln k1  ln A - Ea ( ) R T 1

Ea 1 ln k  ln A ( ) and 2 R T2 83


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)). • We can eliminate ln A by subtracting the two equations to obtain:

ln

k2 k1



Ea R

*



1 T1



1 T2



84


A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k1, (1/T1)) and (k2, (1/T2)). • With this form of the equation, given the Activation Energy and the rate constant (k1) at a given temperature (T1), it can be find the rate constant (k2) at other temperat. (T2).

ln

k2 k1



Ea R

*



1 T1



1 T2



85


The rate constant for the formation of Hydrogen Iodide from its elements is 2.7*10-4 L/(mol.s) at 600K and 3.5*10-3 L/(mol.s) at 650K. Find the activation energy, Ea.

H2 (g)  I2 (g)  2HI(g)



Substituting the given data into the Arrhenius equation. -3 3.5 * 10 Ea 1   1 ln  *   4 8.31 J/(mol  K)  600K 650K  2.7 * 10 86

A PROBLEM TO CONSIDER (2)  Simplifying, we get: Ea 4 ln (1.30 * 10 )  1.11  * (1.28 * 10 ) 8.31 J/(mol) 1

 Solving for Ea:

1.11 * 8.31J / mol 5 Ea   1 . 66 * 10 J  166 kJ 4 1.28 * 10 87

REACTION MECHANISMS 

Even though a balanced chemical equation may give the ultimate result of a reaction, what actually happens in the reaction may take place in several steps. • This “pathway” the reaction takes is referred to as the “Reaction Mechanism”. • The individual steps in the larger overall reaction are referred to as “Elementary Reactions”. 88

ELEMENTARY REACTIONS (1) Consider

the reaction of nitrogen dioxide with carbon monoxide.

NO2 (g)  CO(g)  NO(g)  CO2 (g) 

This reaction is believed to take place in two steps. NO2 (g)  NO2 (g)  NO3 (g)  NO(g)

(elementary reaction)

NO3 (g)  CO(g)  NO2 (g)  CO2 (g)

(elementary reaction) 89

ELEMENTARY REACTIONS (2) 

Each step is a singular molecular event resulting in the formation of products. • Note that NO3 does not appear in the overall equation, but is formed as a temporary “Reaction intermediate”. 90

ELEMENTARY REACTIONS (3) 

Each step is a singular molecular event resulting in the formation of products. • The overall chemical equation is obtained by adding the two steps together and canceling any species common to both sides. NO2 (g)  NO2 (g)  NO3 (g)  NO(g) NO3 (g)  CO(g)  NO2 (g)  CO2 (g) NO2 (g)  NO2 (g)  NO3 (g)  CO(g)  NO3 (g)  NO(g)  NO2 (g)  CO2 (g) 91

MOLECULARITY (1) 

Chemical reactions can be classified according to their Molecularity, that is, the number of molecules that must collide for the elementary reaction to occur into: • A uni-molecular reaction which involves only one reactant molecule. • A bi-molecular reaction that involves the collision of two reactant molecules. • A ter-molecular reaction that requires the collision of three reactant molecules. 92

MOLECULARITY (2)  We can classify reactions according to their Molecularity, that is, the number of molecules that must collide for the elementary reaction to occur.  Higher Molecularity are rare because of the small statistical probability that four or more molecules would all collide at the same instant. 93

RATE EQUATIONS FOR ELEMENTARY REACTIONS (1)  Since a chemical reaction may occur in several steps, there is no easily stated relationship between its overall reaction and its rate law.  For elementary reactions, the rate is proportional to the concentrations of all reactant molecules involved. 94

RATE EQUATIONS FOR ELEMENTARY REACTIONS (2) For example, consider the generic equation below:

A  Products 

The rate is dependent only on the concentration of A, that is:

Rate  k * [A] 95

RATE EQUATIONS FOR ELEMENTARY REACTIONS (3) However, for the reaction:

A  B  Products 

The rate is dependent on the concentrations of both A and B:

Rate  k * [A] * [B] 96

RATE EQUATIONS FOR ELEMENTARY REACTIONS (4) 

For a Termolecular reaction:

A  B  C  Products 

The rate is dependent on the populations of all three participants.

ln k 2  ln A -

Ea 1 ( ) R T2 97


Note that if two molecules of a given reactant are required, it appears twice in the rate law. For example, the reaction:

2A  B  Products would have the rate law:

Rate  k * [A] * [A] * [B] or Rate  k * [A] * [B] 2

98




In essence, for an elementary reaction, the coefficient of each reactant becomes the power to which it is raised in the rate law for that reaction. Note that many reactions occur in multiple steps and it is, therefore, impossible to predict the rate law based solely on the overall reaction. 99

RATE LAWS AND MECHANISMS (1) 

Consider the reaction below: 2 NO2 (g)  F2 (g)  2 NO2F(g)



Experiments performed with reaction show that the rate law is

this

Rate  k * [NO2 ] * [F2 ] 

The reaction is first order with respect to each reactant, even though the coefficient for NO2 in the overall reaction is 2.

100

RATE LAWS AND MECHANISMS (2) 

Consider the reaction below: 2 NO2 (g)  F2 (g)  2 NO2F(g) •

Experiments performed with this reaction show that the rate law is:

Rate  k * [NO2 ] * [F2 ]

•

This implies that the reaction above is not an elementary reaction but rather the result of multiple steps.

101

RATE-DETERMINING STEP (1) 

In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest. • The overall reaction cannot proceed any faster than this slowest rate determining step. 102

RATE-DETERMINING STEP (2) In

multiple step reactions, one of the elementary reactions in the sequence is often slower than the rest. • Our previous example occurs in two elementary steps where the first step is much slower. NO2 (g)  F2 (g)  NO2F(g)  F(g) (slow) k2 NO2 (g)  F(g)  NO2F(g) (fast) k1

2 NO2 (g)  F2 (g)  2 NO2F(g) 103


In multiple-step reactions, one of the elementary reactions in the sequence is often slower than the rest. • Since the overall rate of this reaction is determined by the slow step, it seems logical that the observed rate law is:

Rate = k1*[NO2]*[F2]

NO2 (g)  F2 (g)  NO2F(g)  F(g) k1

(slow) 104


In a mechanism where the first elementary step is the ratedetermining step, the overall rate law is simply expressed as the elementary rate law for that slow step. • A more complicated scenario occurs when the rate-determining step contains a reaction intermediate, as you’ll see in the next section.

105

RATE-DETERMINING STEP (5)  Mechanisms with an Initial Fast Step. • There are cases where the ratedetermining step of a mechanism contains a reaction intermediate that does not appear in the overall reaction. • The experimental rate law, however, can be expressed only in terms of substances that appear in the overall reaction. 106


Consider the reduction of nitric oxide: 2NO(g)  2H2 (g)  N2 (g)  2H2O(g) • A proposed mechanism is: k (fast, equilibrium) 2NO N2O2 1

k-1

N2O2  H2     N2O  H2O k3 N2O  H2   N2  H2O k2

•

(slow) (fast)

It has been experimentally determined that the rate law is Rate = k *[NO]2*[H2] 107

RATE-DETERMINING STEP (7) The

rate determining step (step 2 in this case) generally outlines the rate law for the overall reaction:

Rate  k 2 * [N2O2 ] * [H2 ] (Rate law for the rate-determining step)



As mentioned earlier, the overall rate law can be expressed only in terms of substances represented in the overall reaction and cannot contain intermediates reactants. 108


The rate determining step (step 2 in this case) generally outlines the rate law for the overall reaction:

Rate  k 2 * [N2O2 ] * [H2 ] (Rate law for the rate determining step)

•

It is necessary to re-express this proposed rate law after eliminating intermediate reactant [N2O2].

109



Rate  k 2 * [N2O2 ] * [H2 ] (Rate law for the rate determining step)

•

We can do this by looking at the first step, which is fast and establishes equilibrium.

110


The rate determining step (step 2 in this case) generally outlines the rate law for the overall reaction: Rate  k 2 * [N2O2 ] * [H2 ] (Rate law for the rate-determining step)

•

At equilibrium, the forward rate and the reverse rate are equal.

k1 * [NO]  k 1 * [N2O2 ] 2

111



Rate  k 2 * [N2O2 ] * [H2 ] (Rate law for the rate-determining step) 

Therefore,

[N2O2 ]  (k1 / k 1) * [NO]

2 112



Rate  k 2 * [N2O2 ] * [H2 ]

(Rate law for the rate-determining step)  If we substitute this into our proposed rate law we obtain:

k 2 * k1 2 Rate  * [NO] * [H2 ] k 1



If we replace the constants (k2k1/k-1) with k, we obtain the observed rate law: Rate = k*[NO]2*[H2].113

CATALYSTS (1) 

A Catalyst is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed during the reaction. • To avoid being consumed, the catalyst must participate in at least one step of the reaction and then be regenerated in a later step.

114

CATALYSTS (2) 

A Catalyst is a substance that provides a good “environment” for a reaction to occur, thereby increasing the reaction rate without being consumed during the reaction. • Its presence increases the rate of reaction by either increasing the frequency factor, A (from the Arrhenius equation) or lowering the activation energy (Ea). 115

CATALYSTS (3) 

Homogeneous catalysis is the use of a catalyst in the same phase as the reacting species. • The oxidation of sulfur dioxide using nitric oxide as a catalyst is an example where all species are in the gas phase. NO( g )

2SO2 (g)  O2 (g)    2SO3 (g) 116

CATALYSTS (4) 

Heterogeneous Catalysis is the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a liquid or gaseous solution of reactants. • Such surface catalysis is thought to occur by chemical adsorption of the reactants onto the surface of the catalyst. • Adsorption is the attraction of molecules to a surface (see next Figure). 117

Representation of the Heterogeneous Catalysis

ENZYME CATALYSIS 

Enzymes have enormous catalytic activity. • The substance whose reaction the enzyme catalyzes is called the substrate (see next Figure). • Second next Figure illustrates the reduction in activation energy resulting from the formation of an enzyme-substrate complex.

119

Enzyme action: Lock and key model

ENZYMES CATALYZED REACTION The

proper fit of a substrate (S) in an active site forms an enzyme-substrate Complex (ES): E + S ↔ ES Within the ES complex, the reaction occurs to convert substrate to product (P): ES → E + P The products, which are no longer attracted to the active site, are released. Overall, substrate is convert to product:

E + S ↔ ES → E + P

121

Energy-Potential Curves for the reaction of substrate (S) to products (P). Graphic shows the reduction of activation energy. (A): Activation energy barrier without Enzymes (B): Illustration of enzyme catalyzes reaction.

CONCLUSIONS (1) The main aspects summarized in this presentation were: 1. The different ways of expressing chemical reaction rates. 2. The factors that affects the rate of chemical reactions. 3. The form and meaning of a rate law including the ideas of reaction order and rate constant. 4. How calculate the average reaction rate from a series of experiments given. 123

CONCLUSIONS (2) The main aspects summarized in this presentation were: 5. The determining of the reaction order from the rate law. 6. The method to determining the rate law from initial rates. 7. The use of the concentration-time equation of first order reactions for rate law determining. 8. The relation of the reaction half-life with the rate constant. 124

CONCLUSIONS (3) The main aspects summarized in this presentation were: 9. Use the integrated form of a rate law to determine the concentration of a reactant at a given time. 10.How the activation energy affects a rate and be able to use the Arrhenius Equation. 11.How predict a rate law for a reaction having multistep mechanism given the individual steps in the mechanism. 12.How a catalyst works. 125

CONCLUSIONS (4) The

Arrhenius equation is useful to: 1. Writing the overall chemical equation from a mechanism. 2. Determining the Molecularity of an elementary reaction. 3. Writing the rate equation for an elementary reaction. 4. Determining the rate law from a mechanism.

126