Chemical Kinetics Chapter 14 Chemistry The Central Science

Chemical Kinetics Chapter 14 Chemistry, The Central Science, 7 th& 8 theditions Theodore L. Brown; H. Eugene Le. May, Jr. ; and Bruce E. Bursten 1.

What do we mean by kinetics? Kinetics refers to ¡ the rate at which chemical reactions occur. ¡ The reaction mechanism or pathway through which a reaction proceeds. 2.

Topics for study in kinetics Reaction Rates How we measure rates. Rate Laws How the reaction rate depends on amounts of reactants. Molecularity and order. Integrated Rate Laws How to calculate amount left or time to reach a given amount of reactant or product. Half-life How long it takes for 50% of reactants to react Arrhenius Equation How rate constant changes with temperature. Mechanisms How the reaction rate depends on the sequence of molecular scale processes. 3.

Factors That Affect Reaction Rates ¡ The Nature of the Reactants l ¡ Concentration of Reactants l ¡ As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature l ¡ Chemical compounds vary considerably in their chemical reactivities. At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Catalysts l Change the rate of a reaction by changing the mechanism. 4.

Reaction Rates The rate of a chemical reaction can be determined by monitoring the change in concentration of either reactants or the products as a function of time. [A] vs t 5.

Example 1: Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) [C 4 H 9 Cl] M In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times, t. Rate = [C 4 H 9 Cl] t 6.

Reaction Rates Calculation C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl (aq) Average Rate, M/s The average rate of the reaction over each interval is the change in concentration divided by the change in time: 7.

Reaction Rate Determination C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(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 the reacting molecules. 8.

Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) ¡ ¡ A plot of concentration 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. 9.

Reaction Rates C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) ¡ The reaction slows down with time because the concentration of the reactants decreases. 10.

Reaction Rates and Stoichiometry C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) ¡ ¡ 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. Rate = - [C 4 H 9 Cl] = t [C 4 H 9 OH] t 11.

Reaction Rates & Stoichiometry Suppose that the mole ratio is not 1: 1? Example H 2(g) + I 2(g) 2 HI(g) 2 moles of HI are produced for each mole of H 2 used. The rate at which H 2 disappears is only half of the rate at which HI is generated 12.

Concentration and Rate Each reaction has its own equation that gives its rate as a function of reactant concentrations. This is called its Rate Law The general form of the rate law is Rate = k[A]x[B]y Where k is the rate constant, [A] and [B] are the concentrations of the reactants. X and y are exponents known as rate orders that must be determined experimentally To determine the rate law we measure the rate at different starting concentrations. 13.

Concentration and Rate NH 4+ (aq) + NO 2 - (aq) N 2 (g) + 2 H 2 O (l) Compare Experiments 1 and 2: when [NH 4+] doubles, the initial rate doubles. 14.

Concentration and Rate NH 4+ (aq) + NO 2 - (aq) N 2 (g) + 2 H 2 O (l) Likewise, compare Experiments 5 and 6: when [NO 2 -] doubles, the initial rate doubles. 15.

Concentration and Rate NH 4+ (aq) + NO 2 - (aq) N 2 (g) + 2 H 2 O (l) This equation is called the rate law, and k is the rate constant. 16.

The Rate Law ¡ A rate law shows the relationship between the reaction rate and the concentrations of reactants. l For gas-phase reactants use PA instead of [A]. ¡ k is a constant that has a specific value for each reaction. ¡ The value of k is determined experimentally. Rate = K [NH 4+ ][NO 2 - ] “Constant” is relative herethe rate constant k is unique for each reaction and the value of k changes with temperature 17.

The Rate Law ¡ ¡ Exponents tell the order of the reaction with respect to each reactant. This reaction is First-order in [NH 4+] First-order in [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. Rate = K [NH 4+ ]1[NO 2 - ]1 18.

Determining the Rate constant and Order The following data was collected for the reaction of substances A and B to produce products C and D. Deduce the order of this reaction with respect to A and to B. Write an expression for the rate law in this reaction and calculate the value of the rate constant. [NO] mol dm-3 0. 40 0. 20 [O 2] mol dm-3 0. 50 0. 25 Rate mol dm-3 s-1 1. 6 x 10 -3 8. 0 x 10 -4 2. 0 x 10 -4 19.

The First Order Rate Equation Consider a simple 1 st order reaction: A B Rate = k[A] How much reactant A is left after time t? The rate equation as a function of time can be written as Where [A]t =[A]o e-kt Ln [A]t - Ln[A]o = - kt [A]t = the concentration of reactant A at time t [A]o = the concentration of reactant A at time t=0 K = the rate constant 20.

First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NC CH 3 CN How do we know this is a first order reaction? 21.

First-Order Processes CH 3 NC CH 3 CN This data was collected for this reaction at 198. 9°C. Does rate=k[CH 3 NC] for all time intervals? 22.

First-Order Processes ¡ When Ln P is plotted as a function of time, a straight line results. l The process is first-order. 23.

Half-Life of a Reaction ¡ ¡ Half-life is defined as the time required for onehalf of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A]t = 0. 5 [A]0. 24.

Half-Life of a First Order Reaction For a first-order process, set [A]t=0. 5 [A]0 in integrated rate equation: NOTE: For a first-order process, the half-life does not depend on the initial concentration, [A]0. 25.

First Order Rate Calculation Example 1: The decomposition of compound A is first order. If the initial [A]0 = 0. 80 mol dm-3. and the rate constant is 0. 010 s-1, what is the concentration of [A] after 90 seconds? 26.

First Order Rate Calculation Example 1: The decomposition of compound A is first order. If the initial [A]0 = 0. 80 mol dm-3. and the rate constant is 0. 010 s-1, what is the concentration of [A] after 90 seconds? Ln[A]t – Ln[A]o = -kt Ln[A]t – Ln[0. 80] = - (0. 010 s-1 )(90 s) Ln[A]t = - (0. 010 s-1 )(90 s) + Ln[0. 80] Ln[A]t = -0. 90 - 0. 2231 Ln[A]t = -1. 1231 [A]t = 0. 325 mol dm-3 27.

First Order Rate Calculations Example 2: A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2. 00 M to 1. 00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds. 28.

First Order Rate Calculations Example 2: A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2. 00 M to 1. 00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds. Solution k =0. 693/t 1/2 =0. 693/120 s =0. 005775 s-1 Ln[A] – Ln(2. 00) = -0. 005775 s-1 (80 s)= -0. 462 Ln A = - 0. 462 + 0. 693 = 0. 231 A = 1. 26 mol dm-3 29.

First Order Rate Calculations Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28. 8 years. If some strontium 90 were accidentally released, how long would it take for its concentration to fall to 1% of its original concentration? 30.

First Order Rate Calculations Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28. 8 years. If some strontium 90 were accidentally released, how long would it take for its concentration to fall to 1% of its original concentration? Solution k =0. 693/t 1/2 =0. 693/28. 8 yr =0. 02406 yr-1 Ln[1] – Ln(100) = - (0. 02406 yr-1)t = - 4. 065 t = - 4. 062. - 0. 02406 yr-1 t = 168. 8 years 31.

Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A: Rate = k[A]2 1 [A]t = kt + 1 [A]o Where [A]t = the concentration of reactant A at time t [A]o = the concentration of reactant A at time t=0 K = the rate constant 32.

Second-Order Rate Equation So if a process is second-order in A, a graph of 1/[A] vs. t will yield a straight line with a slope of k. 33.

Determining Reaction Order Distinguishing Between 1 st and 2 nd Order The decomposition of NO 2 at 300°C is described by the equation: NO 2 (g) NO (g) + 1/2 O 2 (g) A experiment with this reaction yields this data: Time (s) [NO 2], M 0. 01000 50. 00787 100. 00649 200. 00481 300. 00380 34.

Determining Reaction Order Distinguishing Between 1 st and 2 nd Order Graphing ln [NO 2] vs. t yields: • The graph is not a straight line, so this process cannot be first-order in [A]. Time (s) [NO 2], M ln [NO 2] 0. 01000 -4. 610 50. 00787 -4. 845 100. 00649 -5. 038 200. 00481 -5. 337 300. 00380 -5. 573 Does not fit the first order equation: 35.

Second-Order Reaction Kinetics A graph of 1/[NO 2] vs. t gives this plot. Time (s) [NO 2], M 1/[NO 2] 0. 01000 100 50. 00787 127 100. 00649 154 200. 00481 208 300. 00380 263 • This is a straight line. Therefore, the process is second-order in [NO 2]. • The slope of the line is the rate constant, k. 36.

Half-Life for 2 nd Order Reactions For a second-order process, set [A]t=0. 5 [A]0 in 2 nd order equation. In this case the half-life depends on the initial concentration of the reactant A. 37.

Sample Problem 1: Second Order Acetaldehyde, CH 3 CHO, decomposes by second-order kinetics with a rate constant of 0. 334 mol-1 dm 3 s-1 at 500 o. C. Calculate the amount of time it would take for 80 % of the acetaldehyde to decompose in a sample that has an initial concentration of 0. 00750 M. The final concentration will be 20% of the original 0. 00750 M or = 0. 00150 1. . = 0. 334 mol-1 dm 3 s-1 t + 1. 00150. 00750 666. 7 = 0. 334 t + 133. 33 0. 334 t = 533. 4 t = 1600 seconds 38.

Sample Problem 2: Second Order Acetaldehyde, CH 3 CHO, decomposes by second-order kinetics with a rate constant of 0. 334 mol-1 dm 3 s-1 at 500 o. C. If the initial concentration of acetaldehyde is 0. 00200 M. Find the concentration after 20 minutes (1200 seconds) Solution 1 [A]t . = 0. 334 mol-1 dm 3 s-1 (1200 s) + 1. 0. 00200 mol dm-3 . = 0. 334 mol-1 dm 3 s-1 (1200 s) + 500 mol-1 dm 3 = 900. 8 mol-1 dm 3 = 1 _____. = 0. 00111 mol dm-3 900. 8 mol-1 dm 3 39.

Summary of Kinetics Equations First order Second order Rate Laws Integrat ed Rate Laws complicated Half-life complicated 40.

Temperature and Rate ¡ ¡ ¡ Generally speaking, the reaction rate increases as the temperature increases. This is because k is temperature dependent. As a rule of thumb a reaction rate increases about 10 fold for each 10 o. C rise in temperature 41.

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. ¡ These collisions must occur with sufficient energy and at the appropriate orientation. ¡ 42.

The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bonds to break and new bonds to form 43.

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. 44.

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. 45.

Reaction Coordinate Diagrams ¡ ¡ It 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. 46.

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. 47.

Maxwell–Boltzmann Distributions ¡ ¡ As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy. 48.

Maxwell–Boltzmann Distributions ¡ If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases. 49.

Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression: where R is the gas constant and T is the temperature in Kelvin. 50.

Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. Ea is the activation energy. T is the Kelvin temperature and R is the universal thermodynamics (gas) constant. R = 8. 314 J mol-1 K-1 or 8. 314 x 10 -3 J mol-1 K-1 51.

Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes 1 RT y = mx + b When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of ln k vs. 1/T. 52.

Arrhenius Equation for 2 Temperatures When measurements are taken for two different temperatures the Arrhenius equation can be symplified as follows: Write the above equation twice, once for each of the two Temperatures and then subtract the lower temperature conditions from the higher temperature. The equation then becomes: 53.

Arrhenius Equation Sample Problem 1 The rate constant for the decomposition of hydrogen iodide was determined at two different temperatures 2 HI H 2 + I 2. At 650 K, k 1 = 2. 15 x 10 -8 dm 3 mol-1 s-1 At 700 K, k 2 = 2. 39 x 10 -7 dm 3 mol-1 s-1 Find the activation energy for this reaction. 2. 39 x 10 -7 Ea Ln -------- = - ------------ x 2. 15 x 10 -8 (8. 314 J mol-1 K-1) [ 1 1 ------ -- -----700 K 650 K ] Ea = 180, 000 J mol-1 = 180 k. J mol-1 54.

Overview of Kinetics Equations First order Second order Rate Laws Integrated Rate Laws Half-life complicated Rate and Temp (T) 55.

Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. 56.

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. 57.

Reaction Mechanisms • The molecularity of a process tells how many molecules are involved in the process. • The rate law for an elementary step is written directly from that step. 58.

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. 59.

Slow Initial Step NO 2 (g) + CO (g) NO (g) + CO 2 (g) ¡ 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. This suggests the reaction occurs in two steps. 60.

Slow Initial Step ¡ ¡ ¡ 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. 61.

Fast Initial Step ¡ The rate law for this reaction is found (experimentally) to be ¡ Because termolecular (= trimolecular) processes are rare, this rate law suggests a two-step mechanism. 62.

Fast Initial Step ¡ A proposed mechanism is Step 1 is an equilibriumit includes the forward and reverse reactions. 63.

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 ¡ But how can we find [NOBr 2]? ¡ 64.

Fast Initial Step ¡ ¡ ¡ NOBr 2 can react two ways: l With NO to form NOBr l By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Ratef = Rater 65.

Fast Initial Step ¡ Because Ratef = Rater , k 1 [NO] [Br 2] = k− 1 [NOBr 2] Solving for [NOBr 2] gives us k 1 [NO] [Br ] = [NOBr ] 2 2 k− 1 66.

Fast Initial Step Substituting this expression for [NOBr 2] in the rate law for the rate-determining step gives 67.

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. Some catalysts also make atoms line up in the correct orientation so as to enhance the reaction rate 68.

Catalysts may be either homogeneous or heterogeneous A homogeneous catalyst is in the same phase as the substances reacting. A heterogeneous catalyst is in a different phase 69.

Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. Heterogeneous catalysts often act in this way 70.

Catalysts Some catalysts help to lower the energy formation for the activated complex or provide a new activated complex with a lower activation energy Al. Cl 3 + Cl 2 Cl+ + Al. Cl 4 Cl+ + C 6 H 6 C 6 H 5 Cl + H+ H+ + Al. Cl 4 - HCl + Al. Cl 3 Overall reaction C 6 H 6 + Cl 2 C 6 H 5 Cl + HCl 71.

Catalysts & Stratospheric Ozone In the stratosphere, oxygen molecules absorb ultraviolet light and break into individual oxygen atoms known as free radicals The oxygen radicals can then combine with ordinary oxygen molecules to make ozone. Ozone can also be split up again into ordinary oxygen and an oxygen radical by absorbing ultraviolet light. 72.

Catalysts & Stratospheric Ozone The presence of chlorofluorcarbons in the stratosphere can catalyze the destruction of ozone. UV light causes a Chlorine free radical to be released The chlorine free radical attacks ozone and converts it Back to oxygen. It is then regenerated to repeat the Process. The result is that each chlorine free radical can Repeat this process many times. The result is that Ozone is destroyed faster than it is formed, causing its level to drop 73.

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. 74.

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