Chapter 14 Chemical Kinetics

Chapter 14 Chemical Kinetics

14.1 Factors that Affect Reaction Rates

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).

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.

Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

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. P.575 GIST: How does increasing the partial pressures of the reactive components of a gaseous mixture affect the rate at which the components react with one another?

14.2 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.

A disappears over the time interval from 20 s to 40 s.
Sample Exercise 14.1 Calculating an Average Rate of Reaction From the data given in the caption of Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s. For the reaction pictured in Figure 14.3, calculate the average rate of appearance of B over the time interval from 0 to 40 s. Practice Exercise

Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.

Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate = [C4H9Cl] t p.577 GIST: Why do the rates of reactions decrease as concentrations decrease?

Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
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.

A plot of [C4H9Cl] vs. 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. p.578 GIST: This figure shows two triangles used to determine the slope of the curve at two different times. How do you determine how large to draw the triangle when determining the slope of a curve at a particular point?

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.

Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction
Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the initial rate).

Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s.
Practice Exercise

Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. Rate = -[C4H9Cl] t = [C4H9OH]

What if the ratio is not 1:1? 2 HI(g)  H2(g) + I2(g) In such a case, Rate = − 1 2 [HI] t = [I2]

To generalize, then, for the reaction aA + bB cC + dD Rate = − 1 a [A] t = − b [B] = c [C] d [D]

Sample Exercise 14.3 How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2O3(g) → 3O2(g)? If the rate at which O2 appears is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at the same time?

Practice The decomposition of N2O5 proceeds according to the following equation: 2N2O5(g) → 4NO2(g) + O2(g) If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of NO2? Of O2?

14.3 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.

NH4+(aq) + NO2−(aq) N2(g) + 2 H2O(l) If we compare Experiments 1 and 2, we see that when [NH4+] doubles, the initial rate doubles.

NH4+(aq) + NO2−(aq) N2(g) + 2 H2O(l) Likewise, when we compare Experiments 5 and 6, we see that when [NO2−] doubles, the initial rate doubles.

This means Rate  [NH4+] Rate  [NO2−] Rate  [NH4+] [NO2−] which, when written as an equation, becomes Rate = k [NH4+] [NO2−] This equation is called the rate law, and k is the rate constant. Therefore,

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 [NH4+] [NO2−] the reaction is First-order in [NH4+] and First-order in [NO2−].

Rate Laws Rate = k [NH4+] [NO2−]
The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

14.3 GIST What is a rate law? What is the name of the quantity k in any rate law? The experimentally determined rate law for the reaction 2NO + 2H2 → N2 + 2H2O is rate = k[NO]2[H2]. What are the reaction orders in this rate law? Does doubling the concentration of NO have the same effect on rate as doubling the concentration of H2?

Sample Exercise 14.4 Consider a reaction A + B → C for which rate = k[A][B]2. Each of the boxes represent a reaction mixture (A is red, B is purple). Rank these mixtures in order of increasing rate of reaction.

Practice Assuming the rate = k[A][B], rank the mixtures in order of increasing rate.

Units of Rate Constant Units of rate = (units of rate constant)(units of concentration)x So, Units of rate constant = units of rate / (units of concentration)x Units of k MEMORIZE Zero order: M/s First order: s-1 Second order: M-1s-1 Third order: M-2s-1

Sample Exercise 14.5 Determining Reaction Order and Units of Rate Constants
(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.10? (b) What are the units of the rate constant for the rate law in Equation 14.9? N2O5 → 4 NO2 + O2 Rate = k[N2O5] CHCl3 + Cl2 → CCl4 + HCl Rate = k[CHCl3][Cl2]1/2

(a) What is the reaction order of the reactant H2 in Equation 14. 11
(a) What is the reaction order of the reactant H2 in Equation 14.11? (b) What are the units of the rate constant for Equation 14.11? Practice Exercise H2 + I2 → 2HI Rate = k[H2][I2]

Method of Initial Rates
Used to find the form of the rate law Write the general form of the rate law, then determine order by doing the following. Choose one reactant to start with Find two experiments where the concentration of that reactant changes but all other reactants stay the same Write the rate laws for both experiments Divide the two rate laws Use log rules to solve for the order Follow the same technique for other reactants

Example Choose one reactant to start with
Find two experiments where the concentration of that reactant changes but all other reactants stay the same

Example

Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data
The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B, and the results are as follows: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = M and [B] = M.

Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data
The following data were measured for the reaction of nitric oxide with hydrogen: (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = M and [H2] = M Practice Exercise

14.4 The Change of Concentration With Time

Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A]t [A]0 = −kt 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.

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

First Order ln[A] = -kt + ln[A]0
can calculate the concentration at any time can be determined by a graph where y = ln[A] x = t m = -k b = ln[A]0 ln[A] vs. t is a straight line

First Order half-life time is takes for a reactant concentration to reach half of its original concentration does not depend on the concentration

First-Order Processes
ln [A]t = -kt + ln [A]0 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.

Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN

CH3NC CH3CN This data was collected for this reaction at °C.

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. P.587 GIST: what do the y-intercepts in the above graphs represent?

Sample Exercise 14.7 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12 °C. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 × 10–7 g/cm3. Assume that the average temperature of the lake is 12 °C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to decrease to 3.0 × 10–7 g/cm3?

Sample Exercise 14.7 Using the Integrated First-Order Rate Law
The decomposition of dimethyl ether, (CH3)2O, at 510 °C is a first-order process with a rate constant of 6.8 × 10–4 s–1: If the initial pressure of (CH3)2O is 135 torr, what is its pressure after 1420 s? Practice Exercise

Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A]t = kt + [A]0 also in the form y = mx + b

1 [A]t = kt + [A]0 So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. 1 [A]

The decomposition of NO2 at 300°C is described by the equation 1 2 NO2 (g) NO (g) O2 (g) and yields data comparable to this: Time (s) [NO2], M 0.0 50.0 100.0 200.0 300.0

Plotting ln [NO2] vs. t yields the graph at the right. The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 −4.610 50.0 −4.845 100.0 −5.038 200.0 −5.337 300.0 −5.573

1 [NO2] Graphing ln vs. t, however, gives this plot. Because this is a straight line, the process is second-order in [A]. Time (s) [NO2], M 1/[NO2] 0.0 100 50.0 127 100.0 154 200.0 208 300.0 263

Second Order 1/[A] = kt + 1/[A]0
1/[A] vs. t is a straight line with a slope of k

Is the reaction first or second order in NO2?
Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300 °C, Is the reaction first or second order in NO2?

Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law

Consider again the decomposition of NO2 discussed in the Sample Exercise. The reaction is second order in NO2 with k = M–1 s–1. If the initial concentration of NO2 in a closed vessel is M, what is the remaining concentration after h? Practice Exercise

Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0.

Half-Life For a first-order process, this becomes 0.5 [A]0 [A]0 ln
= −kt1/2 ln 0.5 = −kt1/2 −0.693 = −kt1/2 = t1/2 0.693 k NOTE: For a first-order process, then, the half-life does not depend on [A]0.

Half-Life For a second-order process, 1 0.5 [A]0 = kt1/2 + [A]0 2 [A]0
2 − 1 [A]0 = kt1/2 1 = = t1/2 1 k[A]0

Second Order half-life does depend on the concentration for second order

14.5 GIST If a solution containing 10.0 g of a substance reacts by first-order kinetics, how many grams remain after 3 half-lives? How does the half-life of a second-order reaction change as the reaction proceeds?

Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction
The reaction of C4H9Cl with water is a first-order reaction. Figure 14.4 shows how the concentration of C4H9Cl changes with time at a particular temperature. (a) From that graph, estimate the half-life for this reaction. (b) Use the half-life from (a) to calculate the rate constant.

Using Equation 14.15, calculate t1/2 for the decomposition of the insecticide described in Sample Exercise The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr–1 at 12 °C How long does it take for the concentration of the insecticide to reach one-quarter of the initial value? Practice Exercise

Zero Order half-life does depend on the concentration

Example A certain first order reaction has t1/2 of 20.0 minutes Find k
How much time is required for this reaction to be 75 % complete?

14.5 Temperature and Rate

Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.

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.

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

Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea. 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.

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.

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.

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.

As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

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.

This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. -Ea RT f = e

Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: k = A e where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. -Ea RT

Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k = ( ) + ln A Ea R 1 T y = m x b Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. . 1 T

14.5 GIST What is the central idea of the collision model?
Why isn’t collision frequency the only factor affecting a reaction rate?

Sample Exercise 14.10 Relating Energy Profiles to Activation Energies and
Speed of Reaction Consider a series of reactions having the following energy profiles: Rank the reactions from slowest to fastest assuming that they have nearly the same frequency factors. Imagine that these reactions are reversed. Rank these reverse reactions from slowest to fastest. Practice Exercise

Sample Exercise 14.11 Determining the Energy of Activation
The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at K?

Sample Exercise 14.11 Determining the Energy of Activation
Using the data in Sample Exercise 14.11, calculate the rate constant for the rearrangement of methyl isonitrile at 280 °C. Practice Exercise

14.6 Reaction Mechanisms

Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process. Rate laws can be determined directly from elementary reactions. The rate law for the overall reaction is determined from the slowest elementary step in the mechanism

Reaction Mechanisms Example: Overall: NO2(g) + CO(g)  NO(g) + CO2(g)
2 Steps: NO2(g) + NO2(g)  NO3(g) + NO(g) NO3(g) + CO(g)  NO2(g) +CO2(g)

Reaction Mechanisms intermediate catalyst
species that is formed and consumed in the steps so does not appear in overall reaction appears as a product in one reaction and then reactant in following reaction catalyst species used to speed up the reaction is not in overall reaction appears as reactant and then as a product Intermediates and catalysts do not appear in the rate law.

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

How do you know if the proposed mechanism is correct?
1. The series of elementary steps must add up to give overall balanced equation 2. must agree with the experimentally determined rate law

p.598 GIST What is the molecularity of this elementary reaction?
NO(g) + Cl2(g) → NOCl(g) + Cl(g)

Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates
It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism: (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s).

Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates
For the reaction the proposed mechanism is (a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s). Practice Exercise

Sample Exercise 14.13 Predicting Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary reaction, predict its rate law: Consider the following reaction: 2 NO(g) + Br2(g) → 2 NOBr(g). (a) Write the rate law for the reaction, assuming it involves a single elementary reaction. (b) Is a single step mechanism likely for this reaction? Practice Exercise

Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step.

p.601 GIST Why can’t the rate law for a reaction generally be deduced from the balanced equation for the reaction?

Slow Initial Step NO2 (g) + CO (g)  NO (g) + CO2 (g)
The rate law for this reaction is found experimentally to be Rate = k [NO2]2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps.

Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) The NO3 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.

Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism: (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction.

Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:
Practice Exercise The reaction is believed to occur in two steps: The experimental rate law is rate = k[O3][NO2]. What can you say about the relative rates of the two steps of the mechanism?

Fast Initial Step The rate law for this reaction is found to be
2 NO (g) + Br2 (g)  2 NOBr (g) The rate law for this reaction is found to be Rate = k [NO]2 [Br2] Because termolecular processes are rare, this rate law suggests a two-step mechanism.

Fast Initial Step A proposed mechanism is Step 1: NO + Br2
NOBr2 (fast) Step 2: NOBr2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions.

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 = k2 [NOBr2] [NO] But how can we find [NOBr2]?

Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr By decomposition to reform NO and Br2 The reactants and products of the first step are in equilibrium with each other. Therefore, Ratef = Rater

Fast Initial Step Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us k1 k−1 [NO] [Br2] = [NOBr2]

Fast Initial Step Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives k2k1 k−1 Rate = [NO] [Br2] [NO] = k [NO]2 [Br2]

Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a
Fast Initial Step Show that the following mechanism for Equation (2NO + Br2 → 2NOBr) also produces a rate law consistent with the experimentally observed one:

The first step of a mechanism involving the reaction of bromine is
What is the expression relating the concentration of Br(g) to that of Br2(g)? Practice Exercise

14.7 Catalysis

Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction, more collisions have the minimum Ea. Catalysts change the mechanism by which the process occurs. is never consumed in the reaction so it is not considered a reactant

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

Catalysis heterogeneous enzyme homogeneous in different phases
same phase enzyme catalysts in body

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.

14.7 GIST How does a catalyst increase the rate of a reaction?
How does a homogeneous catalyst compare with a heterogeneous one regarding the ease of recovery of the catalyst from the reaction mixture? What names are given to the following aspects of enzyme catalysis: The place on the enzyme where catalysis occurs The substances that undergo catalysis

Sample Integrative Exercise Putting Concepts Together
Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures as follows: HCOOH(g) → CO2(g) + H2(g) The uncatalyzed decomposition reaction is determined to be first order. A graph of the partial pressure of HCOOH versus time for decomposition at 838 K is shown as the red curve in Figure When a small amount of solid ZnO is added to the reaction chamber, the partial pressure of acid versus time varies as shown by the blue curve in Figure (a) Estimate the half-life and first-order rate constant for formic acid decomposition. (b) What can you conclude from the effect of added ZnO on the decomposition of formic acid? (c) The progress of the reaction was followed by measuring the partial pressure of formic acid vapor at selected times. Suppose that, instead, we had plotted the concentration of formic acid in units of mol/L. What effect would this have had on the calculated value of k? (d) The pressure of formic acid vapor at the start of the reaction is 3.00 × 102 torr. Assuming constant temperature and ideal-gas behavior, what is the pressure in the system at the end of the reaction? If the volume of the reaction chamber is 436 cm3, how many moles of gas occupy the reaction chamber at the end of the reaction? (e) The standard heat of formation of formic acid vapor is ΔH°f = –378.6 kJ/mol. Calculate ΔH° for the overall reaction. If the activation energy (Ea) for the reaction is 184 kJ/mol, sketch an approximate energy profile for the reaction, and label Ea, ΔH°, and the transition state.

Figure 14.28

Chapter 14 Chemical Kinetics

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