Chapter 13. Chemical Kinetics. Fu-Yin Hsu

Chapter 13 Chemical Kinetics Fu-Yin Hsu

Ectotherms ( 冷血動物 ) ectotherms animals whose body temperature matches their environment s temperature. Ex: Lizards ( 蜥蝪 ) The drop in body temperature immobilizes the lizard because its movement depends on chemical reactions that occur within its muscles, and the rates of those reactions are highly sensitive to temperature.

Chemical kinetics Chemical kinetics ( 化學動力學 ) is the study of the factors that affect the rates of chemical reactions, such as temperature. The speed of a chemical reaction is called its reaction rate. Rate is how much a quantity changes in a given period of time EX:

化學動力學 (Chemical Kinetics): 所關心的是由初狀態 ( 反應物 ) 變終狀態 ( 產物 ) 的過程 ( 路徑 ) 化學熱力學 ( thermodynamic): 所關心的是反應的初狀態與終狀態

Defining Reaction Rate The rate of a chemical reaction is measured as a change in the amounts of reactants or products (usually in concentration units) divided by the change in time. Example: rate of formation of product Rate [ P] t Appearance of product Or Rate A P [ A] t Disappearance of reactant

General rate of reaction General rate of reaction : calculated by dividing rate expressions by stoichiometric coefficients General rate of reaction = Ex: rate of disappearance of reactant or of formation of product stoichiometric coefficient of that reactant or product in the balanced equation We can define the rate of this reaction in the time interval t 1 to t 2 as follows: or

The Average Rate of the Reaction

Reactant and Product Concentrations as a Function of Time The instantaneous rate is the change in concentration at any one particular time [HI]

Summary

EXAMPLE 13.1 Expressing Reaction Rates Consider this balanced chemical equation

Sol: (a) (b)

Measuring Reaction Rates Continuous Monitoring Polarimetry this measures the change in the degree of rotation of plane-polarized light caused by one of the components over time. Spectrophotometry this measures the amount of light of a particular wavelength absorbed by one component over time. The component absorbs its complementary color. Total pressure the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction.

Spectrophotometry

13.3 The Rate Law: The Effect of Concentration on Reaction Rate Let s consider a reaction in which a single reactant, A, decomposes into products: As long as the rate of the reverse reaction is negligibly slow, we can express the relationship between the rate of the reaction and the concentration of the reactant called the rate law ( 反應速率定律 ) as follows: k is a constant of proportionality called the rate constant n is the reaction order.

Rate law If n = 0, the reaction is zero order and the rate is independent of the concentration of A. If n = 1, the reaction is first order and the rate is directly proportional to the concentration of A. If n = 2, the reaction is second order and the rate is proportional to the square of the concentration of A.

Rate law

units for k Overall reaction order units for k Zero MS -1 First S -1 Second M -1 S -1 Third M -2 S -1

Sol: 0.01= k [0.1] 2 R = k [0.5] 2 R =0.25 M/s

Rate law for multiple reactant aa + bb + cc products rate = k[a] m [B] n [C] p The overall order of a reaction is the sum of exponents in a rate law : m+n+. Note: The exponents in a rate law must be determined by experiment. They are not derived from the stoichiometric coefficients in an overall chemical equation, though in some instances the exponents and the coefficients may be the same.

Method of Initial Rates Rate = k[no] m [Cl 2 ] n 2.27x 10-5 = k (0.0125) m (0.0255) n (½)=(½) n n=1 4.55x 10-5 = k (0.0125) m (0.0510) n 9.08x 10-5 = k (0.0250) m (0.0255) (1/4)=(½) m m=2 n Rate = k[no] 2 [Cl 2 ]

Method of Initial Rates Rate = k [A] n When we double the concentration of a reactant A, there is no effect on the rate, the reaction is zeroorder in A the rate doubles, the reaction is first-order in A. the rate quadruples, the reaction is second-order in A. the rate increases eight times, the reaction is thirdorder in A.

EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction

Sol: Rate = k[no 2 ] m [CO] n x2 x4 m=2 x1 n=0 Rate = k[no 2 ] 2 [CO] 0 = k[no 2 ] 2

13.4 The Integrated Rate Law: The Dependence of Concentration on Time The integrated rate law ( 積分速率定律 ) is an equation that describes the concentration of a reactant as a function of time ( 在反應中物種的濃度是如何受時間的影響. ) Concentration = f (time) 濃度 =f( 時間 )

The integrated rate law: First- Order Reactions For the reaction A P The rate of reaction R = - ( [A]/ t) = k [A] - d[a] / dt = k [A] d[a] / [A] = - k dt [A] t d[a] / [A] = - k dt [A] 0 t 0 ln [A] t [A] 0 = kt

The integrated rate law: First- Order Reactions The integrated rate law: ln [A] t [A] 0 = kt ln [A] t ln[a] 0 = kt ln [A] t = kt + ln [A] 0 (y = mx + b)

EXAMPLE 13.3 The First-Order Integrated Rate Law: Using Graphical Analysis of Reaction Data

SOLUTION In order to show that the reaction is first order, prepare a graph of ln[so 2 Cl 2 ] versus time as shown. ln [A] t = kt + ln [A] 0 the rate constant is 2.9x10-4 s -1

EXAMPLE 13.4 The First-Order Integrated Rate Law: Determining the Concentration of a Reactant at a Given Time Sol: ln [A] t = kt + ln [A] 0

Second-Order Integrated Rate Law: For the reaction A P The rate of reaction R = - ( [A]/ t) = k [A] 2 - d[a] / dt = k [A] 2 d[a] / [A] 2 = - k dt [A] t d[a] / [A]2 = - k dt [A] 0 t 0-1/[A] t + 1/[A] o = -kt 1/[A] t = kt + 1/[A] o (y = mx + b)

EXAMPLE 13.5 The Second-Order Integrated Rate Law: Using Graphical Analysis of Reaction Data

Sol: ln [A] t = kt + ln [A] 0 1/[A] t = kt + 1/[A] o the rate constant is 0.255 M -1 s -1

Zero-Order Integrated Rate Law

Half-life of a Reaction The half-life (t 1/2 ) of a reaction is the time required for the concentration of a reactant to fall to one-half of its initial value.

First-Order Reaction Half-Life

EXAMPLE 13.6 Half-Life Molecular iodine dissociates at 625 K with a first-order rate constant of 0.271 s -1. What is the half-life of this reaction? SOLUTION:

Second-Order Reaction Half-Life For a second-order reaction, the integrated rate law is: At a time equal to the half-life :

Zero-Order Reaction Half-Life For a zero-order reaction, the integrated rate law is: At a time equal to the half-life :

Summary

13.5 The Effect of Temperature on Reaction Rate An increase in temperature generally results in an increase in k, which results in a faster rate. Swedish chemist Svante Arrhenius wrote a paper quantifying the temperature dependence of the rate constant. The modern form of the Arrhenius equation shows the relationship between the rate constant ( k ) and the temperature in kelvin ( T ):

where T is the temperature in kelvins. R is the gas constant in energy units, 8.314 J/(mol K). Ea is the activation energy, the extra energy needed to start the molecules reacting A is called the frequency factor, the rate the reactant energy approaches the activation energy.

Activation energy The activation energy E a is an energy barrier or hump that must be surmounted for the reactants to be transformed into products frequency factor ( A ) as the number of times that the reactants approach the activation barrier per unit time.

Methyl isonitrile rearranges to acetonitrile. For the reaction to occur, the H 3 C N bond must break, and a new H 3 C C bond must form.

Activation Energy and the Activated Complex There is an energy barrier to almost all reactions. The activation energy is the amount of energy needed to convert reactants into the activated complex. Also know as, transition state The activated complex is a chemical species with partially broken and partially formed bonds. Always very high in energy because of its partial bonds The frequency is the number of molecules that begin to form the activated complex in a given period of time.

The Arrhenius Equation: The Exponential Factor The exponential factor in the Arrhenius equation is a number between 0 and 1. It represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier. Increasing the temperature increases the average kinetic energy of the molecules. increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier. increasing the temperature will increase the reaction rate.

Thermal Energy Distribution

Arrhenius Plots The Arrhenius equation can be algebraically solved to give the following form:

EXAMPLE 13.7 Using an Arrhenius Plot to Determine Kinetic Parameters

Solution To determine the frequency factor and activation energy, prepare a graph of the natural log of the rate constant (ln k ) versus the inverse of the temperature (1/T ).

Arrhenius Equation: Two-Point Form If you only have two (T,k) data points, the following forms of the Arrhenius equation can be used: From Arrhenius equation k = Ae Ea/RT -) ln k 1 = E a /RT 1 + ln A ln k 2 = E a /RT 2 + ln A ln k 1 -ln k 2 = E a /RT 1 + E a /RT 2 ln k = E a /RT + ln A

EXAMPLE 13.8 Using the Two-Point Form of the Arrhenius Equation

Solution:

The Collision Model: A Closer Look at the Frequency Factor For most reactions, for a reaction to take place, the reacting molecules must collide with each other. On average, about 10 9 collisions per second Once molecules collide they may react together or they may not, depending on two factors: Whether the collision has enough energy to "break the bonds holding reactant molecules together" Whether the reacting molecules collide in the proper orientation for new bonds to form

The collision frequency (z) is the number of collisions that occur per unit time, Under typical conditions, a single molecule undergoes on the order of 109 collisions every second. The orientation factor (p) is a number, usually between 0 and 1, which represents the fraction of collisions with an orientation that allows the reaction to occur.

Effective Collisions: Kinetic Energy Factor For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form.

Importance of Orientation the orientation factor is p = 0.16, This means that only 16 out of 100 sufficiently energetic collisions are actually successful in forming the products.

A few reactions have orientation factors greater than one. Ex: This reaction has an orientation factor of p = 4.8. there are more reactions than collisions the reactants do not even have to collide to react! at a certain close distance, an electron of K (the harpoon) is captured by Br 2 to give Br 2-. Once the ionic intermediates K + and Br 2 - are formed, they are strongly attracted towards each other and move towards one another. a potassium atom can actually transfer an electron to a bromine molecule without a collision.

Importance of Orientation One hydrogen atom can approach another from any direction Effective collision; the I atom can bond to the C atom to form CH 3 I and reaction will still occur; the spherical symmetry of the atoms means that orientation does not matter. Ineffective collision; orientation is important in this reaction.

Transition State Theory The configuration of the atoms at the time of the collision is called the transition state ( 暫態 ) The transitory species having this configuration is called the activated complex ( 活化複體 or 活化錯合物 ) I - δ- δ- + CH 3 -Br I-----CH 3 ------Br I-CH 3 +Br - Reactant Activated complex Product

Reaction profile A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction. CO(g) + NO 2 (g) CO 2 (g) + NO(g) activated complex Activation Energy ( 活化能 ) Heat of Reaction ( H)=Ea(forward)- Ea(reverse)

Reaction Mechanisms The reaction mechanism is the series of individual chemical steps by which an overall chemical reaction occurs. Ex: the reaction between hydrogen and iodine monochloride Each step in a reaction mechanism is an elementary step.

Reaction Mechanisms the overall reaction HI molecule appears in the reaction mechanism but not in the overall reaction equation. HI : reaction intermediates.

Elementary Reactions The molecularity ( 分子數 ) of an elementary reaction refers to the number of free atoms, ions, or molecules that enter into the reaction A+B+C products Termolecular

Rate Laws for Elementary Steps An elementary step ( 基本反應 ) represents a single stage in the progress of the overall reaction The exponents in the rate law for an elementary reaction are the same as the stoichiometric coefficients in the chemical equation for the elementary reaction

Rate-Determining Steps and Overall Reaction Rate Laws In most chemical reactions, one of the elementary steps called the rate-determining step is much slower than the others. The rate law of the rate determining step determines the rate law of the overall reaction.

Reaction Mechanism NO 2(g) + CO (g) NO (g) + CO 2(g) Rate obs = k[no 2 ] 2 (experiment) 1. NO 2(g) + NO 2(g) NO 3(g) + NO (g) Rate = k 1 [NO 2 ] 2 Slow 2. NO 3(g) + CO (g) NO 2(g) + CO 2(g) Rate = k 2 [NO 3 ][CO] Fast The first step is slower than the second step because its activation energy is larger. The first step in this mechanism is the rate determining step. The rate law of the first step is the same as the rate law of the overall reaction.

Validating a Mechanism To validate (not prove) a mechanism, two conditions must be met: 1. The elementary steps must sum to the overall reaction. 2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law.

A mechanism with a slow step followed by a fast step Ex: I - 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) Slow step: H 2 O 2 + I - H 2 O + O I - Fast step: H 2 O 2 + OI - H 2 O + O 2 + I - Net equation: 2H 2 O 2 2H 2 O + O 2 I - ion is a catalyst ( 催化劑 ) and OI - is an intermediate ( 中間物 ) Slow step is the rate-determining step In the rate law, the reaction rate is equal to the rate of the slow step Rate = rate of slow step = k [H 2 O 2 ] [I - ] = k [H 2 O 2 ] ( [I - ] remains constant throughout the reaction)

A mechanism with a fast step followed by a slow step Mechanism for 2 NO (g) + O 2(g) 2 NO 2(g) Rate = k [NO] 2 [O 2 ] Experiment, not stoichiometry k First step : 2NO 1 N 2 O 2 k -1 Slow step : N 2 O 2 + O 2 2NO 2 Net equation : 2NO + O 2 2NO 2 Rate = k 2 [N 2 O 2 ][O 2 ]

A mechanism with a fast step followed by a slow step k 2NO 1 Rate N 2 O forward =k 1 [NO] 2 2 k -1 Rate reverse =k -1 [N 2 O 2 ] We assume that equilibrium is rapidly established in this elementary reaction Rate forward = Rate reverse k 1 [NO] 2 =k- 1 [N 2 O 2 ] [N 2 O 2 ] =(k 1 /k -1 ) [NO] 2 Rate = k 2 [N 2 O 2 ][O 2 ] = k 2 (k 1 /k -1 ) [NO] 2 [O 2 ] = k [NO] 2 [O 2 ]

EXAMPLE 13.9 Reaction Mechanisms

Solution

13.7 Catalysis Mechanism without catalyst: O 3(g) + O (g) 2 O 2(g) V. Slow reaction rates can be increased by using a catalyst, a substance that increases the rate of a chemical reaction but is not consumed by the reaction Catalysts work by providing an alternative mechanism for the reaction with a lower activation energy. Catalysts are consumed in an early mechanism step, and then made in a later step. Mechanism with catalyst: Cl (g) + O 3(g) O 2(g) + ClO (g) Fast ClO (g) + O (g) O 2(g) + Cl (g) Slow

Catalysts Homogeneous catalysts are in the same phase as the reactant particles. Cl(g) in the destruction of O 3 (g) Heterogeneous catalysts are in a different phase than the reactant particles. Solid catalytic converter in a car s exhaust system

Types of Catalysts

Catalytic Hydrogenation of Ethene Catalytic Hydrogenation: H 2 C=CH 2 + H 2 CH 3 CH 3

Enzymes: Biological Catalysts most biological reactions require a catalyst to proceed at a reasonable rate. Protein molecules that catalyze biological reactions are called enzymes. The reactant substance, called the substrate (S), attaches itself to an area on the enzyme (E) called the active site, to form an enzyme-substrate complex (ES). The enzyme substrate complex decomposes to form products (P), and the enzyme is regenerated.

Enzyme Substrate Binding