Chemical Kinetics 化學動力學 Chapter 16 Chemical Kinetics 化學動力學

Outline The Rate of a Reaction Nature of the Reactants Concentrations of the Reactants: The Rate-Law Expression Concentration Versus Time: The Integrated Rate Equation Collision Theory of Reaction Rates Transition State Theory Reaction Mechanisms and the Rate-Law Expression Temperature: The Arrhenius Equation Catalysts

The Rate of a Reaction Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time. A reaction mechanism is the series of molecular steps by which a reaction occurs.

The Rate of a Reaction Thermodynamics (Chapter 15) determines if a reaction can occur. Kinetics (Chapter 16) determines how quickly a reaction occurs. Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.

aA(g) + bB(g)  cC(g) + dD(g) The Rate of Reaction Consider the hypothetical reaction, aA(g) + bB(g)  cC(g) + dD(g) equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:

The Rate of Reaction [A] is the symbol for the concentration of A in M ( mol/L). Note that the reaction does not go entirely to completion. The [A] and [B] > 0 plus the [C] and [D] < 1.

The Rate of Reaction Reaction rates are the rates at which reactants disappear or products appear. This movie is an illustration of a reaction rate.

The Rate of Reaction Mathematically, the rate of a reaction can be written as:

The Rate of Reaction The rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance. [A] is the concentration of A in molarity or moles/L. k is the specific rate constant. k is an important quantity in this chapter.

The Rate of Reaction For a simple expression like Rate = k[A] If the initial concentration of A is doubled, the initial rate of reaction is doubled. If the reaction is proceeding twice as fast, the amount of time required for the reaction to reach equilibrium would be: The same as the initial time. Twice the initial time. Half the initial time. If the initial concentration of A is halved the initial rate of reaction is cut in half.

The Rate of Reaction If more than one reactant molecule appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

The Rate of Reaction Rate Law Expressions must be determined experimentally. The rate law cannot be determined from the balanced chemical equation. This is a trap for new students of kinetics. The balanced reactions will not work because most chemical reactions are not one-step reactions. Other names for rate law expressions are: rate laws rate equations or rate expressions

The Rate of Reaction Important terminology for kinetics. The order of a reaction can be expressed in terms of either: each reactant in the reaction or the overall reaction. Order for the overall reaction is the sum of the orders for each reactant in the reaction. For example:

The Rate of Reaction A second example is:

The Rate of Reaction A final example of the order of a reaction is:

The Rate of Reaction Given the following one step reaction and its rate-law expression. Remember, the rate expression would have to be experimentally determined. Because it is a second order rate-law expression: If the [A] is doubled the rate of the reaction will increase by a factor of 4. 22 = 4 If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4

Factors That Affect Reaction Rates There are several factors that can influence the rate of a reaction: The nature of the reactants. The concentration of the reactants. The temperature of the reaction. The presence of a catalyst. We will look at each factor individually.

Nature of Reactants This is a very broad category that encompasses the different reacting properties of substances. For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

Nature of Reactants Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

Nature of Reactants The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

Nature of Reactants However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide. The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.

Concentrations of Reactants: The Rate-Law Expression This movie illustrates how changing the concentration of reactants affects the rate.

Concentrations of Reactants: The Rate-Law Expression This is a simplified representation of the effect of different numbers of molecules in the same volume. The increase in the molecule numbers is indicative of an increase in concentration. A(g) + B (g)  Products A B B A B A B A B 4 different possible A-B collisions 6 different possible A-B collisions 9 different possible A-B collisions

Concentrations of Reactants: The Rate-Law Expression Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A(g) + B(g) ® 3 C(g) Experiment Number Initial [A] (M) Initial [B] Initial rate of formation of C (M/s) 1 0.10 2.0 x 10-4 2 0.20 0.30 4.0 x 10-4 3

Concentrations of Reactants: The Rate-Law Expression Experiment Number Initial [A] (M) Initial [B] Initial rate of formation of C (M/s) 1 0.10 2.0 x 10-4 2 0.20 0.30 4.0 x 10-4 3

Concentrations of Reactants: The Rate-Law Expression Example 16-2: The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction? 2 A(g) + B(g) + 2 C(g) ® 3 D(g) + 2 E(g)

Concentrations of Reactants: The Rate-Law Expression Experiment Initial [A] (M) Initial [B] Initial [C] Initial rate of formation of D (M/s) 1 0.20 0.10 2.0 x 10-4 2 0.30 6.0 x 10-4 3 4 0.60 0.40 1.8 x 10-3

Concentrations of Reactants: The Rate-Law Expression Example 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks. You do it!

Concentrations of Reactants: The Rate-Law Expression Experiment Initial Rate (M/s) Initial [A] (M) Initial [B] 1 4.0 x 10-3 0.20 0.050 2 1.6 x 10-2 ? 3 3.2 x 10-2 0.40

Concentration vs. Time: The Integrated Rate Equation The integrated rate equation relates time and concentration for chemical and nuclear reactions. From the integrated rate equation we can predict the amount of product that is produced in a given amount of time. Initially we will look at the integrated rate equation for first order reactions. These reactions are 1st order in the reactant and 1st order overall.

Concentration vs. Time: The Integrated Rate Equation An example of a reaction that is 1st order in the reactant and 1st order overall is: a A ® products This is a common reaction type for many chemical reactions and all simple radioactive decays. Two examples of this type are: 2 N2O5(g)  2 N2O4(g) + O2(g) 238U  234Th + 4He

Concentration vs. Time: The Integrated Rate Equation The integrated rate equation for first order reactions is: where: [A]0= mol/L of A at time t=0. [A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation.

Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t. Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0.

Concentration vs. Time: The Integrated Rate Equation At time t = t1/2, the expression becomes:

Concentration vs. Time: The Integrated Rate Equation Example 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation. The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C.

Concentration vs. Time: The Integrated Rate Equation Example 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds? The integrated rate laws can be used for any unit that represents moles or concentration. In this example we will use grams rather than mol/L.

Concentration vs. Time: The Integrated Rate Equation Example 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours. CS2(g) ® CS(g) + S(g) You do it!

Concentration vs. Time: The Integrated Rate Equation

Concentration vs. Time: The Integrated Rate Equation For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is: Where: [A]0= mol/L of A at time t=0. [A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation.

Concentration vs. Time: The Integrated Rate Equation Second order reactions also have a half-life. Using the second order integrated rate-law as a starting point. At the half-life, t1/2 [A] = 1/2[A]0.

Concentration vs. Time: The Integrated Rate Equation If we solve for t1/2: Note that the half-life of a second order reaction depends on the initial concentration of A.

Concentration vs. Time: The Integrated Rate Equation Example 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?

Concentration vs. Time: The Integrated Rate Equation Example 16-7: The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?

Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH3CHO remain after 200 hours?

Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH3CHO remains after 200 hours?

Concentration vs. Time: The Integrated Rate Equation Example 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC? Note that the vessel size is increased by a factor of 10 which decreases the concentration by a factor of 10! You do it!

Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH3CHO remain after 200 hours? You do it!

Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH3CHO remains after 200 hours? You do it!

Concentration vs. Time: The Integrated Rate Equation Let us now summarize the results from the last two examples. Initial Moles CH3CHO [CH3CHO]0 (M) [CH3CHO] Moles of CH3CHO at 200 hr. % CH3CHO remaining Ex. 16-7 0.10 0.071 71% Ex. 16-8 0.010 0.0096 0.096 96%

Enrichment - Derivation of Integrated Rate Equations For the first order reaction a A  products the rate can be written as:

Enrichment - Derivation of Integrated Rate Equations For a first-order reaction, the rate is proportional to the first power of [A].

Enrichment - Derivation of Integrated Rate Equations In calculus, the rate is defined as the infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time.

Enrichment - Derivation of Integrated Rate Equations Rearrange the equation so that all of the [A] terms are on the left and all of the t terms are on the right.

Enrichment - Derivation of Integrated Rate Equations Express the equation in integral form.

Enrichment - Derivation of Integrated Rate Equations This equation can be evaluated as:

Enrichment - Derivation of Integrated Rate Equations Which rearranges to the integrated first order rate equation.

Enrichment - Derivation of Integrated Rate Equations Derive the rate equation for a reaction that is second order in reactant A and second order overall. The rate equation is:

Enrichment - Derivation of Integrated Rate Equations Separate the variables so that the A terms are on the left and the t terms on the right.

Enrichment - Derivation of Integrated Rate Equations Then integrate the equation over the limits as for the first order reaction.

Enrichment - Derivation of Integrated Rate Equations Which integrates the second order integrated rate equation.

Enrichment - Derivation of Integrated Rate Equations For a zero order reaction the rate expression is:

Enrichment - Derivation of Integrated Rate Equations Which rearranges to:

Enrichment - Derivation of Integrated Rate Equations Then we integrate as for the other two cases:

Enrichment - Derivation of Integrated Rate Equations Which gives the zeroeth order integrated rate equation.

Enrichment - Rate Equations to Determine Reaction Order Plots of the integrated rate equations can help us determine the order of a reaction. If the first-order integrated rate equation is rearranged. This law of logarithms, ln (x/y) = ln x - ln y, was applied to the first-order integrated rate-equation.

Enrichment - Rate Equations to Determine Reaction Order The equation for a straight line is: Compare this equation to the rearranged first order rate-law.

Enrichment - Rate Equations to Determine Reaction Order Now we can interpret the parts of the equation as follows: y can be identified with ln[A] and plotted on the y-axis. m can be identified with –ak and is the slope of the line. x can be identified with t and plotted on the x-axis. b can be identified with ln[A]0 and is the y-intercept.

Enrichment - Rate Equations to Determine Reaction Order Example 16-9: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant.

Enrichment - Rate Equations to Determine Reaction Order Time (min) 1 2 3 4 5 [C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37 ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99 1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7

Enrichment - Rate Equations to Determine Reaction Order We will make three different graphs of the data. Plot the [C2H5Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is zero order with respect to [C2H5Br]. Plot the ln [C2H5Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is first order with respect to [C2H5Br]. Plot 1/ [C2H5Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is second order with respect to [C2H5Br].

Enrichment - Rate Equations to Determine Reaction Order Plot of [C2H5Br] versus time. Is it linear or not?

Enrichment - Rate Equations to Determine Reaction Order Plot of ln [C2H5Br] versus time. Is it linear or not?

Enrichment - Rate Equations to Determine Reaction Order Plot of 1/[C2H5Br] versus time. Is it linear or not?

Enrichment - Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of ln[C2H5Br] vs. time. Thus this is a first order reaction with respect to [C2H5Br]. Next, we will determine the value of the rate constant from the slope of the line on the graph of ln[C2H5Br] vs. time. Remember slope = (y2-y1)/(x2-x1).

Enrichment - Rate Equations to Determine Reaction Order From the equation for a first order reaction we know that the slope = -a k. In this reaction a = 1.

Enrichment - Rate Equations to Determine Reaction Order The integrated rate equation for a reaction that is second order in reactant A and second order overall. This equation can be rearranged to:

Enrichment - Rate Equations to Determine Reaction Order Compare the equation for a straight line and the second order rate-law expression. Now we can interpret the parts of the equation as follows: y can be identified with 1/[A] and plotted on the y-axis. m can be identified with a k and is the slope of the line. x can be identified with t and plotted on the x-axis b can be identified with 1/[A]0 and is the y-intercept.

Enrichment - Rate Equations to Determine Reaction Order Example 16-10: Concentration-versus-time data for the decomposition of nitrogen dioxide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

Enrichment - Rate Equations to Determine Reaction Order Time (min) 1 2 3 4 5 [NO2] 1.0 0.53 0.36 0.27 0.22 0.18 ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7 1/[NO2] 1.9 2.8 3.7 4.6 5.5

Enrichment - Rate Equations to Determine Reaction Order Once again, we will make three different graphs of the data. Plot [NO2] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is zero order with respect to NO2. Plot ln [NO2] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is first order with respect to NO2. Plot 1/ [NO2] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is second order with respect to NO2.

Enrichment - Rate Equations to Determine Reaction Order Plot of [NO2] versus time. Is it linear or not?

Enrichment -Rate Equations to Determine Reaction Order Plot of ln [NO2] versus time. Is it linear or not?

Enrichment - Rate Equations to Determine Reaction Order Plot of 1/[NO2] versus time. Is it linear or not?

Enrichment - Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of 1/[NO2] vs. time. Thus this is a second order reaction with respect to [NO2]. Next, we will determine the value of the rate constant from the slope of the line on the graph of 1/[NO2] vs. time.

Enrichment - Rate Equations to Determine Reaction Order From the equation for a second order reaction we know that the slope = a k In this reaction a = 2.

Collision Theory of Reaction Rates Three basic events must occur for a reaction to occur the atoms, molecules or ions must: Collide. Collide with enough energy to break and form bonds. Collide with the proper orientation for a reaction to occur.

Collision Theory of Reaction Rates One method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules. As an example, look at the reaction of methane and oxygen: We must start the reaction with a match. This provides the initial energy necessary to break the first few bonds. Afterwards the reaction is self-sustaining.

Collision Theory of Reaction Rates Illustrate the proper orientation of molecules that is necessary for this reaction. X2(g) + Y2(g) 2 XY(g) Some possible ineffective collisions are : X Y Y Y X X X X Y Y

Collision Theory of Reaction Rates An example of an effective collision is: X Y X Y X Y +

Collision Theory of Reaction Rates This picture illustrates effective and ineffective molecular collisions.

Transition State Theory Transition state theory postulates that reactants form a high energy intermediate, the transition state, which then falls apart into the products. For a reaction to occur, the reactants must acquire sufficient energy to form the transition state. This energy is called the activation energy or Ea. Look at a mechanical analog for activation energy

Transition State Theory Cross section of mountain Boulder Eactivation Dh h2 h1 Epot=mgh2 Epot=mgh1 Height DEpot = mgDh

Transition State Theory Representation of a chemical reaction. Potential Energy Reaction Coordinate X2 + Y2 2 XY Eactivation - a kinetic quantity DE »DH a thermodynamic quantity

Transition State Theory

Transition State Theory The relationship between the activation energy for forward and reverse reactions is Forward reaction = Ea Reverse reaction = Ea + DE difference = DE

Transition State Theory The distribution of molecules possessing different energies at a given temperature is represented in this figure.

Reaction Mechanisms and the Rate-Law Expression Use the experimental rate-law to postulate a molecular mechanism. The slowest step in a reaction mechanism is the rate determining step.

Reaction Mechanisms and the Rate-Law Expression Use the experimental rate-law to postulate a mechanism. The slowest step in a reaction mechanism is the rate determining step. Consider the iodide ion catalyzed decomposition of hydrogen peroxide to water and oxygen.

Reaction Mechanisms and the Rate-Law Expression This reaction is known to be first order in H2O2 , first order in I- , and second order overall. The mechanism for this reaction is thought to be:

Reaction Mechanisms and the Rate-Law Expression Important notes about this reaction: One hydrogen peroxide molecule and one iodide ion are involved in the rate determining step. The iodide ion catalyst is consumed in step 1 and produced in step 2 in equal amounts. Hypoiodite ion has been detected in the reaction mixture as a short-lived reaction intermediate.

Reaction Mechanisms and the Rate-Law Expression Ozone, O3, reacts very rapidly with nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

Reaction Mechanisms and the Rate-Law Expression One possible mechanism is:

Reaction Mechanisms and the Rate-Law Expression A mechanism that is inconsistent with the rate-law expression is:

Reaction Mechanisms and the Rate-Law Expression Experimentally determined reaction orders indicate the number of molecules involved in: the slow step only, or the slow step and the equilibrium steps preceding the slow step.

Temperature: The Arrhenius Equation Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (Ea), and (3) the specific rate constant (k).

Temperature: The Arrhenius Equation This movie illustrates the effect of temperature on a reaction.

Temperature: The Arrhenius Equation If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 >T1.

Temperature: The Arrhenius Equation Subtract one equation from the other.

Temperature: The Arrhenius Equation Rearrange and solve for ln k2/k1.

Temperature: The Arrhenius Equation Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K). How much do the two rates differ?

Temperature: The Arrhenius Equation For reactions that have an Ea»50 kJ/mol, the rate approximately doubles for a 100C rise in temperature, near room temperature. Consider: 2 ICl(g) + H2(g) ® I2(g) + 2 HCl(g) The rate-law expression is known to be R=k[ICl][H2].

Catalysts Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

Catalysts Homogeneous catalysts exist in same phase as the reactants. Heterogeneous catalysts exist in different phases than the reactants. Catalysts are often solids.

Catalysts Examples of commercial catalyst systems include:

Catalysts This movie shows catalytic converter chemistry on the Molecular Scale

Catalysts A second example of a catalytic system is:

Catalysts A third examples of a catalytic system is:

Catalysts Look at the catalytic oxidation of CO to CO2 Overall reaction 2 CO(g)+ O2(g) ® 2CO2(g) Absorption CO(g) ® CO(surface) + O2(g) O2(g) ® O2(surface) Activation O2(surface) ® 2 O(surface) Reaction CO(surface) +O(surface) ® CO2(surface) Desorption CO2(surface) ® CO2(g)

Catalysts