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## Principles of Reactivity: Chemical Kinetics

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**Principles of Reactivity: Chemical Kinetics**Chapter 15**Kinetics**• Chemical kinetics is the study of the rates of chemical reactions. • Macroscopic level: rates of reactions, reaction rate, factors affecting rates • Particulate level: reaction mechanism (the detailed pathway taken by atoms and molecules as a reaction proceeds)**15.1 Rates of Chemical Reactions**• reaction rate is a measure of the speed of a reaction • rate of a chemical reaction refers to the change in concentration of substance per unit of time • amounts of reactants decrease and amounts of products increase • often in units of M/s**Rates of Chemical Reactions**• 2N2O5 4NO2 + O2 • rate of rxn = - Δ[N2O5]/Δt • rate is expressed as a positive value; the sign on the rate indicates a decrease (reactant concentration) or increase (product concentration)**Rates of Chemical Reactions**• 2N2O5 4NO2 + O2 • The graph of concentration vs. time is not straight because the rate of reaction changes during the course of the reaction. (largest at time = 0) • Over a time period, the average rate can be calculated. For a single point in time, an instantaneous rate can also be calculated from the slope of the tangent line.**Practice Problem**• Compare the rates of appearance or disappearance of each product and reactant, in the decomposition of nitrosyl chloride, NOCl. 2NOCl(g) 2NO(g) + Cl2(g)**Practice Problem**• Sucrose decomposes to fructose and glucose in acid solution. A plot of the concentration of sucrose as a function of time is given on page 675. What is the rate of change of the sucrose concentration over the first 2 h? What is the rate of change over the last 2 h? Estimate the instantaneous rate at 4 h.**15.2 Reaction Conditions and Rate**• Reaction rate increases with more collisions and more energy. Factors affecting the speed of reaction: • concentration of reactants – increasing [reactant] causes more collisions therefore increasing reaction rate • surface area – more surface area results in more collisions (powdered medicine works more quickly than pill forms) • physical state of reactants – homogeneous mixtures of liquids and gases tend to react faster (tied to surface area) • temperature – higher temp increases KE therefore increases collisions • catalysts – accelerate reaction but are not transformed themselves; changes reaction mechanism • bio catalysts are called enzymes (BEANO!)**Homework**• After reading sections 15.1-15.2, you should be able to do the following… • p. 712 (1-5)**15.3 Effect of Concentration on Rate**• the rate of reaction is proportional to the reactant concentration • may be proportional to the concentrations of more than one reactant • catalyst can affect the rate**Rate Equations**• A rate equation or rate law expresses the relationship between reactant concentration and rate rate of reaction = k[N2O5] where k is the rate constant.**Rate Equations**• For a reaction such as aA + bB xX the rate equation is rate = k[A]m[B]n where m and n are experimentally determined exponents NOT necessarily the stoichiometric coefficients (usually positive whole numbers)**Order of a Reaction**• The order with respect to a reactant is the exponent of its concentration term in the rate expression. • Total reaction order is the sum of the exponents in the concentration terms.**Order of a Reaction**• A reactant is first order if doubling it (while holding the other reactants constant) doubles the rate. • A reactant is second order if doubling it quadruples the rate. • A reactant is third order if doubling it increases the rate by 8. • A reactant is zero order if doubling it has no effect on the rate.**Order of a Reaction**2NO(g) + Cl2(g) 2NOCl(g) Rate = k[NO]2[Cl2] This reaction is second order in NO, first order in Cl2, and third order overall. Compare experimental evidence to figure out the order respective to reactants.**Rate Constant, k**• The rate constant, k, is a proportionality constant relating rate and concentrations at a given temp. • Units of the rate constant have to match; depends on order of reaction. • 0 order = mol/L∙time • 1st order = time-1 • 2nd order = L/mol∙time • 3rd order = L2/mol2∙time**Determining a Rate Equation**• A rate equation must be determined experimentally. • “method of initial rates” • initial rate is the instantaneous reaction rate at the start of the reaction (at t = 0)**Practice Problem**• The initial rate of the reaction of nitrogen monoxide and oxygen 2NO(g) + O2(g) 2NO2(g) was measured at 25oC for various initial concentrations of NO and O2. Data are collected in the table on p. 683. Determine the rate equation from these data. What is the value of the rate constant and what are the appropriate units of k?**Practice Problem**• The rate constant k is 0.090 h-1 for the reaction Pt(NH3)2Cl2(aq) + H2O(l) [Pt(NH3)2(H2O)Cl]+(aq) + Cl-(aq) and the rate equation is rate = k[Pt(NH3)2Cl2] Calculate the rate of reaction when the concentration of Pt(NH3)2Cl2 is 0.020M. What is the rate of change in the concentration of Cl- under those conditions?**Homework**• After reading section 15.3, you should be able to do the following… • P. 713 (9-13)**15.4 Concentration-Time Relationships**• Integrated rate equation: ln([R]t/[R]o) = -kt [R]o and [R]t are concentrations of reactant at time t = 0 and at a later time, t. The ratio of concentrations is the fraction of reactant that remains after a given time has elapsed.**First Order Reactions**• ln[R]t/[R]o = -kt • For a first order reaction, k has units of time -1, which means that you can choose any unit for [R] such as M, moles, grams, atoms, molecules, or pressure.**Practice Problem**• Sucrose, a sugar, decomposes in acid solution to glucose and fructose. The reaction is first order in sucrose, and the rate constant at 25oC is k=0.21h-1. If the initial concentration of sucrose is 0.010 M, what is its concentration after 5.0 h.**Practice Problem**• Gaseous NO2 decomposes when heated 2NO2(g) 2NO(g) + O2(g) • The disappearance of NO2 is a first order reaction with k = 3.6x10-3 s-1 at 300oC. • A sample of gaseous NO2 is placed in a flask and heated at 300 oC for 150 s. What fraction of the initial sample remains after this time? • How long must a sample be heated so that 99% of the sample has decomposed?**Second-Order Reaction**• For second-order reactions, the rate equation becomes 1/[R]t – 1/[R]o = kt same symbolism applies and k has units of L/mol∙time.**Practice Problem**• Using the rate constant for HI as k = 30. L/mol∙min, calculate the concentration of HI after 12 min if [HI]0 = 0.010 mol/L.**Zero-Order Reactions**• For a zero-order reaction, the integrated rate equation is [R]0 – [R]t = kt where the units of k are mol/(L∙time).**Graphical Methods**• Each of the integrated rate laws can be rearranged to take the form of a straight line y = mx + b where m is the slope and b is the y-intercept.**Graphical Methods**• Zero Order [R]t = -kt + [R]o • First Order ln[R]t = -kt + ln[R]o • Second Order 1/[R]t = kt + 1/[R]o**Half-Life and First-Order Reactions**• The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to one-half its initial value. • Longer half-life equals slower reaction. For first order reactions: t1/2 = 0.693/k half life is independent of concentration!**Half-Life and Other Reactions**• Zero Order t1/2 = [R]o/2k • Second Order t1/2 = 1/(k[R]o)**Practice Problem**• Americium is used in smoke detectors and in medicine for the treatment of certain malignancies. One isotope of americium, 241Am, has a rate constant, k, for radioactive decay of 0.0016 year -1. In contrast, radioactive iodine-125, which is used for studies of thyroid functioning, has a rate constant for decay of 0.011 day-1. a. What are the half-lives of these isotopes? b. Which element decays faster? c. If you begin a treatment with iodine-125, and have 1.6e15 atoms, how many remain after 2.0 days?**Homework**• After reading section 15.4, you should be able to do the following… • P. 714 (19-23)**15.5 Particulate View of Reaction Rates**• For a reaction to occur: • Collision Theory • Reacting molecules must collide with each other. • Reacting molecules must collide with sufficient energy. • Molecules must collide in an orientation that can lead to rearrangement of the atoms.**Collision Theory**• Increasing the concentration of the reactant particles will increase the number of collisions. • The number of collisions between the reactant molecules is directly proportional to the concentrations of each reactant.**Activation Energy**• An “energy barrier” must be overcome for a reaction to occur. • The energy required to overcome this barrier is called activation energy, Ea. • Fast reactions usually have low activation energies, and slow reactions have high activation energies.**Activation Energy**• As a reaction passes over the activation energy barrier, the transition state (also called an activated complex) is the arrangement of reactant molecules and atoms at the maximum point in the reaction coordinate diagram. (see page 695) • Must be inferred!**Effect of Temperature Increase**• Raising temperature increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier.**Effect of Molecular Orientation**• “steric factor” • The lower the probability of achieving the proper alignment, the lower the value of k and the slower the reaction.**Arrhenius Equation**• The observation that reaction rates depend on the energy and frequency of collisions between reacting molecules, on the temperature, and on whether the collisions have the correct geometry is summarized by the Arrhenius equation: k = Ae-Ea/RT**Arrhenius Equation**• k = Ae-Ea/RT • where R is the gas constant (8.314510e-3 kJ/Kmol) and T is the temperature in kelvins. The parameter A is called the frequency factor and is related to the number of collisions that have correct geometry. The other factor refers to the fraction of molecules having at least the minimum energy required for reaction.**Arrhenius Equation**• Can be used to calculate the value of activation energy from the temperature dependence of the rate constant AND calculate the rate constant for a given temperature if the activation energy and A are known. • Conceptual aspects of the Arrhenius equation and interpreting graphs are tested on the National Exam, but calculations will not be.**Arrenius Equation**• Can be rearranged to form an equation for a straight line relating ln k to 1/T. • k = Ae-Ea/RT • ln k = ln A – (Ea/R)(1/T) • y = mx + b**Arrhenius Equation**• We can also solve for Ea by comparing k at two different temperatures… ln k2/k1 = - Ea/R[1/T2 – 1/T1]**Practice Problem**• The colorless gas N2O4 decomposes to the brown gas NO2 in a first-order reaction: N2O4(g) 2NO2(g) The rate constant k = 4.5 x 103 s-1 at 274K and 1.00 x 104 s-1 at 283K. What is the energy of activation?**Effect of Catalysts**• Catalysts speed up the rate of a chemical reaction but are not consumed in the reaction. • Reaction intermediates are species formed in one step of a reaction that are then consumed in a later step.**Catalysis**• Acid-base catalysis: a reactant either gains or loses a proton; changing the rate of the reaction • Surface catalysis: either a new reaction intermediate is formed, or the probability of successful collisions is modified • Some enzymes accelerate reactions by binding to the reactants in a way that lowers activation energy; others react with reactant species to form a new intermediate**Homework**• After reading section 15.5, you should be able to do the following… • P. 715 (32-36)**15.6 Reaction Mechanisms**• Reaction mechanisms refer to the sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products.**Reaction Mechanisms**• Step 1: Br2 + NO Br2NO • Step 2: Br2NO + NO 2BrNO • Overall: Br2 + 2NO 2BrNO • Each step is an elementary step. Each step has its own Ea and k. • Intermediates are substances formed in one step and then consumed in another (don’t appear in overall reaction).