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Chapter 14 Chemical Kinetics John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation © 20 12 Pearson Education, Inc.

Kinetics:

Kinetics In kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates:

Factors That Affect Reaction Rates Physical state of the reactants. In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates:

Factors That Affect Reaction Rates Concentration of reactants . As the concentration of reactants increases , so does the likelihood that reactant molecules will collide . © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates:

Factors That Affect Reaction Rates Temperature At higher temperatures , reactant molecules have more kinetic energy , move faster, and collide more often and with greater energy . © 2012 Pearson Education, Inc.

Factors That Affect Reaction Rates:

Factors That Affect Reaction Rates Presence of a catalyst. Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. © 2012 Pearson Education, Inc.

Reaction Rates:

Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. © 2012 Pearson Education, Inc. A → B

A → B:

A → B

The Rate of a Chemical Reaction:

The Rate of a Chemical Reaction Rate of change of concentration with time . 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s [Fe 2+ ] = 0.0010 M Δ t = 38.5 s Δ [Fe 2+ ] = (0.0010 – 0) M Rate of formation of Fe 2+ = = = 2.6  10 -5 M s -1 Δ [Fe 2+ ] Δ t 0.0010 M 38.5 s

Reaction Rates :

Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) © 2012 Pearson Education, Inc.

Reaction Rates :

Reaction Rates The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) Average rate =  [C 4 H 9 Cl]  t © 2012 Pearson Education, Inc.

Reaction Rates :

Reaction Rates Note that the average rate decreases as the reaction proceeds . This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) © 2012 Pearson Education, Inc.

Reaction Rates :

Reaction Rates A plot of [C 4 H 9 Cl] versus time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) © 2012 Pearson Education, Inc.

Reaction Rates :

Reaction Rates All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. C 4 H 9 Cl( aq ) + H 2 O( l )  C 4 H 9 OH( aq ) + HCl( aq ) © 2012 Pearson Education, Inc.

Reaction Rates :

Reaction Rates In this reaction , the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH . C 4 H 9 Cl ( aq ) + H 2 O( l )  C 4 H 9 OH ( aq ) + HCl( aq ) Rate =   [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t © 2012 Pearson Education, Inc.

Reaction Rates and Stoichiometry:

Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI( g )  H 2 ( g ) + I 2 ( g ) In such a case , Rate =  1 2  [HI]  t =  [I 2 ]  t © 2012 Pearson Education, Inc.

Reaction Rates and Stoichiometry:

Reaction Rates and Stoichiometry To generalize, then, for the reaction a A + b B c C + d D Rate =  1 a  [A]  t =  1 b  [B]  t = 1 c  [C]  t 1 d  [D]  t = © 2012 Pearson Education, Inc.

Concentration and Rate:

Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration . © 2012 Pearson Education, Inc.

Concentration and Rate:

Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles , the initial rate doubles. NH 4 + ( aq ) + NO 2  ( aq ) N 2 ( g ) + 2 H 2 O( l ) © 2012 Pearson Education, Inc.

Concentration and Rate:

Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2  ] doubles , the initial rate doubles. NH 4 + ( aq ) + NO 2  ( aq ) N 2 ( g ) + 2 H 2 O( l ) © 2012 Pearson Education, Inc.

Concentration and Rate:

Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2  ] Rate  [NH 4 + ] [NO 2  ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2  ] This equation is called the rate law , and k is the rate constant . Therefore, © 2012 Pearson Education, Inc.

Rate Laws:

Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [NH 4 + ] [NO 2  ] the reaction is First-order in [NH 4 + ] and First-order in [NO 2  ] © 2012 Pearson Education, Inc.

Rate Laws:

Rate Laws Rate = k [NH 4 + ] [NO 2  ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall . © 2012 Pearson Education, Inc.

Determining Rate Law :

Determining Rate Law 8.4 x 10  4 0.030 0.10 3 4.2 x 10  4 0.015 0.20 2 2.1 x 10  4 0.015 0.10 1 Initial Rate ( M / s ) [B] ( M ) ‏ [A] ( M ) ‏ Exp.

Slide25:

In experiment 1 and 2; [B] is constant; [A] doubles and rate doubles - the reaction is 1st order with respect to [A ] In experiment 1 and 3; [A] is constant; [B] doubles but the rate quadruples! This means that the reaction is 2nd order with respect to [B]

Calculate the Rate Constant:

Calculate the Rate Constant Rate = k [A] [B] 2 The rxn is 1st order w/ respect to [A] The rxn is 2nd order w/ respect to [B] The rxn is 3rd order overall (1 + 2)

General effect of doubling the initial concentration of a particular reactant (with other reactant concentrations held constant).:

General effect of doubling the initial concentration of a particular reactant (with other reactant concentrations held constant) . • Zero order in the reactant —there is no effect on the initial rate of reaction. • First order in the reactant —the initial rate of reaction doubles. Second order in the reactant —the initial rate of reaction quadruples. • Third order in the reactant —the initial rate of reaction increases eightfold.

Integrated Rate Laws:

Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 =  k t where [A] 0 is the initial concentration of A , and [A] t is the concentration of A at some time, t , during the course of the reaction . © 2012 Pearson Education, Inc. A → Products The rate law is : R = k[A]   But R = - , so = k[A]  [A]  t - [A]  t = - k d t [A] d[A ]  [A] 0 [A] t  0 t

Integrated Rate Laws:

Integrated Rate Laws Manipulating this equation produces ln [A] t [A] 0 =  kt ln [A] t  ln [A] 0 =  kt ln [A] t =  kt + ln [A] 0 which is in the form y = mx + b © 2012 Pearson Education, Inc.

H2O2(aq) → H2O(l) + ½ O2(g) :

H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g)

First-Order Processes:

First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line , and the slope of the line will be  k . ln [A] t =  kt + ln [A] 0 © 2012 Pearson Education, Inc.

First-Order Processes:

First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NC CH 3 CN © 2012 Pearson Education, Inc.

First-Order Processes:

First-Order Processes This data were collected for this reaction at 198.9  C. CH 3 NC CH 3 CN © 2012 Pearson Education, Inc.

First-Order Processes:

First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative of the slope: 5.1  10  5 s  1 . © 2012 Pearson Education, Inc.

Example :

Example The rate constant for the reaction 2A  B is 7.5 x 10  3 s  1 at 110  C . The reaction is 1st order in A . How long (in seconds) will it take for [A] to decrease from 1.25 M to 0.71 M ? ln [A] t [A] 0 =  k t ln 0.71 1.25 =  7.5 x 10  3 t t = 75 s

Example :

Example H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) The decomposition of hydrogen peroxide is first order reaction with k = 7.3 ×10 -4 S -1 . Calculate the percent of decomposition after 500 seconds? ln [H 2 O 2 ] t [H 2 O 2 ] 0 =  k t = - 7.3 ×10 -4 ×500 = -0.365 [H 2 O 2 ] t [H 2 O 2 ] 0 ln = -0.365 → [H 2 O 2 ] t [H 2 O 2 ] 0 = 0.694

Slide37:

= 0.694 [H 2 O 2 ] t [H 2 O 2 ] 0 [H 2 O 2 ] t = 0.694 [H 2 O 2 ] 0 The remaining of H 2 O 2 is 0.694 or 69.4 % of the [H 2 O 2 ] 0 This means that the percent of decomposition is 100 – 69.4% = 30.6 %

Second-Order Processes:

Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b © 2012 Pearson Education, Inc. A → Products d t = - k d[A] [A] 2  [A] 0 [A] t  0 t

Second-Order Processes:

Second-Order Processes So if a process is second-order in A, a plot of vs . t yields a straight line , and the slope of that line is k . 1 [A] t = kt + 1 [A] 0 1 [A] © 2012 Pearson Education, Inc.

Second-Order Processes:

Second-Order Processes The decomposition of NO 2 at 300 ° C is described by the equation NO 2 ( g ) NO( g ) + O 2 ( g ) and yields data comparable to this table: Time ( s ) [NO 2 ], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 1 2 © 2012 Pearson Education, Inc.

Second-Order Processes:

Second-Order Processes Plotting ln [NO 2 ] vs . t yields the graph at the right. Time ( s ) [NO 2 ], M ln [NO 2 ] 0.0 0.01000  4.610 50.0 0.00787  4.845 100.0 0.00649  5.038 200.0 0.00481  5.337 300.0 0.00380  5.573 The plot is not a straight line, so the process is not first-order in [A]. © 2012 Pearson Education, Inc.

Second-Order Processes:

Second-Order Processes Graphing ln vs. t , however, gives this plot Fig. 14.9(b). Time ( s ) [NO 2 ], M 1/[NO 2 ] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263 Because this is a straight line, the process is second-order in [A]. 1 [NO 2 ] © 2012 Pearson Education, Inc.

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Zero-Order Reactions

Half-Life:

Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0 . © 2012 Pearson Education, Inc.

Half-Life:

Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln =  kt 1/2 ln 0.5 =  kt 1/2  0.693 =  kt 1/2 = t 1/2 0.693 k Note: For a first-order process, then, the half-life does not depend on [A] 0 . © 2012 Pearson Education, Inc.

Half-Life:

Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2  1 [A] 0 = kt 1/2 1 [A] 0 = = t 1/2 1 k [A] 0 © 2012 Pearson Education, Inc. In this case the half-life depends on the initial concentration of reactant—the lower the initial concentration, the longer the half-life.

Slide48:

Zero-Order Reactions Half-Life

Example :

Example 2I ( g )  I 2( g ) The reaction is second order and has a rate constant of 7.0 x 10 9 M  1 s  1 at 23  C. If the initial [I] is 0.086 M , calculate the concentration after 2.0 min. b) Calculate the half-life of the reaction when the initial [I] is 0.60 M and when the [I] is 0.42 M .

Slide52:

1 [A] t = kt + 1 [A] 0 1 [A] t = 7.0 x 10 9 × 120 s + 1 0.086 [A] = 1.2 x 10 -12 M

Temperature and Rate:

Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature-dependent. © 2012 Pearson Education, Inc.

The Collision Model:

The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. © 2012 Pearson Education, Inc.

The Collision Model:

The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. © 2012 Pearson Education, Inc.

Activation Energy:

Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy , E a . Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation-energy barrier. © 2012 Pearson Education, Inc.

Transition State Theory:

Transition State Theory The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

Reaction Coordinate Diagrams:

Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. © 2012 Pearson Education, Inc.

Reaction Coordinate Diagrams:

Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore,  E ). The high point on the diagram is the transition state . The species present at the transition state is called the activated complex . The energy gap between the reactants and the activated complex is the activation-energy barrier . © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions:

Maxwell – Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions:

Maxwell – Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus, at higher temperatures, a larger population of molecules has higher energy. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions:

Maxwell – Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation-energy barrier. As a result, the reaction rate increases. © 2012 Pearson Education, Inc.

Maxwell–Boltzmann Distributions:

Maxwell – Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e −E a / RT © 2012 Pearson Education, Inc.

Arrhenius Equation:

Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = Ae where A is the frequency factor , a number that represents the likelihood that collisions would occur with the proper orientation for reaction. −E a / RT © 2012 Pearson Education, Inc.

Arrhenius Equation:

Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k =  ( ) + ln A 1 T y = mx + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. . E a R 1 T © 2012 Pearson Education, Inc.

Arrhenius Plot:

Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2 × 10 4 K R - E a E a = 1.0 × 10 2 kJ mol -1

Example :

Example Use the data to calculate activation energy of the reaction mathematically 7.0 x 10  1 500 6.1 x 10  2 450 2.9 x 10  3 400 k (s  1 ) T (Kelvin)

Slide70:

ln = - R E a T 2 1 k 2 k 1 T 1 1

Reaction Mechanisms:

Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism . Step-by-step description of a reaction. Each step is called an elementary process . Reaction mechanism must be consistent with: Stoichiometry for the overall reaction. The experimentally determined rate law.

Elementary Processes:

Elementary Processes Unimolecular or bimolecular . Exponents for concentration terms are the same as the stoichiometric factors for the elementary process. Intermediates are produced in one elementary process and consumed in another. One elementary step is usually slower than all the others and is known as the rate determining step .

Reaction Mechanisms:

Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

Reaction Mechanisms:

Reaction Mechanisms The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. NO 2 ( g ) + CO( g )  NO( g ) + CO 2 ( g )

Reaction Mechanisms:

Reaction Mechanisms A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. NO 2 ( g ) + CO( g )  NO( g ) + CO 2 ( g )

Reaction Mechanisms:

Reaction Mechanisms The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] 2 NO( g ) + Br 2 ( g )  2 NOBr( g )

Reaction Mechanisms:

Reaction Mechanisms A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast) 2 NO( g ) + Br 2 ( g )  2 NOBr( g )

Fast Initial Step:

Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

Fast Initial Step:

Fast Initial Step NOBr 2 can react two ways: With NO to form NOBr . By decomposition to reform NO and Br 2 . The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

Fast Initial Step:

Fast Initial Step Because Rate f = Rate r , k 1 [NO] [Br 2 ] = k  1 [NOBr 2 ] Solving for [NOBr 2 ], gives us k 1 k  1 [NO] [Br 2 ] = [NOBr 2 ] Step 1: NO + Br 2 NOBr 2 (fast)

Fast Initial Step:

Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step, gives k 2 k 1 k  1 Rate = [NO] [Br 2 ] [NO] R = k [NO] 2 [Br 2 ] Rate = k 2 [NOBr 2 ] [NO] k 1 k  1 [NO] [Br 2 ] = [NOBr 2 ] The Steady State Approximation

Reaction Mechanism :

Reaction Mechanism H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) R= k [H 2 ][ICl] Proposed mechanism H 2 (g) + ICl(g) HI(g) + HCl(g) HI(g) + ICl(g) I 2 (g) + HCl(g) R= k 1 [H 2 ][ICl] R= k 2 [HI][ICl] H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) slow fast

Reaction Mechanism :

Reaction Mechanism

Example :

Example Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: O 3 ( g ) + 2 NO 2 ( g )  N 2 O 5 ( g ) + O 2 ( g ) The reaction is believed to occur in two steps: O 3 ( g ) + NO 2 ( g )  NO 3 ( g ) + O 2 ( g ) NO 3 ( g ) + NO 2 ( g )  N 2 O 5 ( g ) The experimental rate law is R = k [O 3 ][NO 2 ] Which of the two steps of the mechanism is the rate determining step?

Catalysts:

Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs.

Catalysts:

Catalysts

Catalysts:

Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

Enzymes:

Enzymes Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.

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