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## Chemical Kinetics Chapter 13

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**Chemical Kinetics Chapter 13**Topics 6 and 16 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.**Factors That Affect Reaction Rates**• The Nature of the Reactants • Chemical compounds vary considerably in their chemical reactivities. • 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. • Catalysts • 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**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) [C4H9Cl] M In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t. [C4H9Cl] t Rate = 6.**Reaction Rates Calculation**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, M/s 7.**Reaction Rate Determination**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 the reacting molecules. 8.**Reaction Rates**C4H9Cl(aq) + H2O(l) C4H9OH(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**C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • The reaction slows down with time because the concentration of the reactants decreases. 10.**-[C4H9Cl]**t Rate = = [C4H9OH] t 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. 11.**Reaction Rates & Stoichiometry**Suppose that the mole ratio is not 1:1? Example H2(g) + I2(g) 2 HI(g) 2 moles of HI are produced for each mole of H2 used. The rate at which H2 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**NH4+ (aq) + NO2- (aq) N2 (g) + 2H2O (l) Compare Experiments 1 and 2:when [NH4+] doubles, the initial rate doubles. 14.**Concentration and Rate**NH4+ (aq) + NO2- (aq) N2 (g) + 2H2O (l) Likewise, compare Experiments 5 and 6: when [NO2-] doubles, the initial rate doubles. 15.**Concentration and Rate**NH4+ (aq) + NO2- (aq) N2 (g) + 2H2O (l) This equation is called the rate law, andk is the rate constant. 16.**The Rate Law**• A rate law shows the relationship between the reaction rate and the concentrations of reactants. • 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 [NH4+][NO2- ] “Constant” is relative here- the 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 [NH4+] First-order in [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. Rate = K [NH4+]1[NO2- ]1 18.**[NO] mol dm-3**[O2] mol dm-3 Rate mol dm-3 s-1 0.40 0.50 1.6 x 10-3 0.40 0.25 8.0 x 10-4 0.20 0.25 2.0 x 10-4 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. 19.**The First Order Rate Equation**Consider a simple 1st 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 = the concentration of reactant A at time t [A]o = the concentration of reactant A at time t = 0 K = the rate constant [A]t =[A]o e-kt Ln [A]t - Ln[A]o = - kt 20.**CH3NC**CH3CN First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. How do we know this is a first order reaction? 21.**CH3NC**CH3CN First-Order Processes This data was collected for this reaction at 198.9°C. Does rate=k[CH3NC] for all time intervals? 22.**First-Order Processes**• When ln P is plotted as a function of time, a straight line results. • The process is first-order. • k is the negative slope: 5.1 10-5 s-1. 23.**Half-Life of a Reaction**• 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. 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/t1/2 =0.693/120s =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/t1/2 =0.693/28.8 yr =0.02406 yr-1 Ln[1] – ln(100) = - (0.02406 yr-1)t = - 4.60 t = - 4.60 . - 0.0241 yr-1 t = 191 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]o 1 [A]t = kt + 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. twill yield a straight line with a slope of k. 33.**NO2(g)**NO (g) + 1/2 O2(g) Determining Reaction OrderDistinguishing Between 1st and 2nd Order The decomposition of NO2 at 300°C is described by the equation: A experiment with this reaction yields this data: 34.**Determining Reaction OrderDistinguishing Between 1st and 2nd**Order Graphing ln [NO2] vs.t yields: • The graph is not a straight line, so this process cannot be first-order in [A]. Does not fit the first order equation: 35.**Second-Order Reaction Kinetics**A graph of 1/[NO2] vs. t gives this plot. • This is a straight line. Therefore, the process is second-order in [NO2]. • The slope of the line is the rate constant, k. 36.**Half-Life for 2nd Order Reactions**For a second-order process, set [A]t=0.5 [A]0 in 2nd order equation. In this case the half-life depends on the initial concentration of the reactant A. 37.**Sample Problem 1: Second Order**Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. 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 . .00150 1 . .00750 = 0.334 mol-1dm3s-1t + 666.7 = 0.334 t + 133.33 0.334 t = 533.4 t = 1600 seconds 38.**Sample Problem 2: Second Order**Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. If the initial concentration of acetaldehyde is 0.00200 M. Find the concentration after 20 minutes (1200 seconds) Solution 1 . [A]t 1 . 0.00200 mol dm-3 = 0.334 mol-1dm3s-1(1200s) + 1 . [A]t = 0.334 mol-1dm3 s-1(1200s) + 500 mol-1dm3 = 900.8 mol-1dm3 1 _____. 900.8 mol-1dm3 [A]t = = 0.00111 moldm-3 39.**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 10oC 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.