Chemical Kinetics

Chemical Kinetics David P. White University of North Carolina, Wilmington Chapter 14 Chapter 14

Chemical Kinetics • Kinetics is the study of how fast chemical reactions occur. • There are 4 important factors which affect rates of reactions: • reactant concentration, • temperature, • action of catalysts, and • surface area. • Goal: to understand chemical reactions at the molecular level. Chapter 14

Chemical Kinetics Reaction Rates • Speed of a reaction is measured by the change in concentration with time. • For a reaction A  B • Suppose A reacts to form B. Let us begin with 1.00 mol A. Chapter 14

Chemical Kinetics Reaction Rates Chapter 14

Chemical Kinetics Reaction Rates • At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present. • At t = 20 min, there is 0.54 mol A and 0.46 mol B. • At t = 40 min, there is 0.30 mol A and 0.70 mol B. • Calculating, Chapter 14

Chemical Kinetics Reaction Rates • For the reaction A  B there are two ways of measuring rate: • the speed at which the products appear (i.e. change in moles of B per unit time), or • the speed at which the reactants disappear (i.e. the change in moles of A per unit time). • A plot of number of moles versus time shows that as the reactants (red A) disappear, the products (blue B) appear. Chapter 14

Chemical Kinetics Reaction Rates Chapter 14

Chemical Kinetics Rates in Terms of Concentrations • For the reaction A  B there are two ways of • Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional. • Consider: C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Chapter 14

Chemical Kinetics Rates in Terms of Concentrations C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Chapter 14

Chemical Kinetics Rates in Terms of Concentrations C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) • We can calculate the average rate in terms of the disappearance of C4H9Cl. • The units for average rate are mol/L•s or M/s. • The average rate decreases with time. • We plot [C4H9Cl] versus time. • The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. • Instantaneous rate is different from average rate. • We usually call the instantaneous rate the rate. Chapter 14

Chemical Kinetics Rates in Terms of Concentrations Chapter 14

Chemical Kinetics Reaction Rates and Stoichiometry • For the reaction C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) we know • In general for aA + bB cC + dD Chapter 14

The Dependence of Rate on Concentration • In general rates increase as concentrations increase. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) Chapter 14

The Dependence of Rate on Concentration • For the reaction NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l) we note • as [NH4+] doubles with [NO2-] constant the rate doubles, • as [NO2-] doubles with [NH4+] constant, the rate doubles, • We conclude rate  [NH4+][NO2-]. • Rate law: Rate = k[NH4+][ NO2-]. • The constant k is the rate constant. Chapter 14

The Dependence of Rate on Concentration • For a general reaction with rate law Rate = k[reactant 1]m[reactant 2]n, we say the reaction is mth order in reactant 1 and nth order in reactant 2. • The overall order of reaction is m + n + …. • A reaction can be zeroth order if m, n, … are zero. • Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. Chapter 14

The Dependence of Rate on Concentration Using Initial Rates to Determine Rate Laws • A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. • A reaction is first order if doubling the concentration causes the rate to double. • A reaction is second order if doubling the concentration results in a 22 increase in rate. • A reacting is nth order if doubling the concentration causes an 2n increase in rate. • Note that the rate constant does not depend on concentration. Chapter 14

The Change of Concentration with Time First-Order Reactions • Goal: convert rate law into a convenient equation to give concentrations as a function of time. • For a first order reaction, the rate doubles as the concentration of a reactant doubles. • We can show that • A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0. • In the above we use the natural logarithm, ln, which is log to the base e. Chapter 14

The Change of Concentration with Time First-Order Reactions Chapter 14

The Change of Concentration with Time Half-Life • Half-life is the time taken for the concentration of a reactant to drop to half its original value. • That is, half life, t1/2 is the time taken for [A]0 to reach ½[A]0. • Mathematically, Chapter 14

The Change of Concentration with Time Half-Life Chapter 14

The Change of Concentration with Time Second-Order Reactions • For a second order reaction with just one reactant • A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0 • For a second order reaction, a plot of ln[A]t vs. t is not linear. Chapter 14

The Change of Concentration with Time Second-Order Reactions Chapter 14

The Change of Concentration with Time Second-Order Reactions • We can show that the half life • A reaction can have rate constant expression of the form rate = k[A][B], i.e., is second order overall, but has first order dependence on A and B. Chapter 14

Temperature and Rate • Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.) • When two light sticks are placed in water: one at room temperature and one in ice, the one at room temperature is brighter than the one in ice. • The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light. • As temperature increases, the rate increases. Chapter 14

Temperature and Rate Chapter 14

Temperature and Rate • As temperature increases, the rate increases. • Since the rate law has no temperature term in it, the rate constant must depend on temperature. • Consider the first order reaction CH3NC  CH3CN. • As temperature increases from 190 C to 250 C the rate constant increases from 2.52  10-5 s-1 to 3.16  10-3 s-1. • The temperature effect is quite dramatic. Why? Chapter 14

Temperature and Rate The Collision Model • Observations: rates of reactions are affected by concentration and temperature. • Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases. • The collision model: in order for molecules to react they must collide. • The greater the number of collisions the faster the rate. Chapter 14

Temperature and Rate The Collision Model Chapter 14

Temperature and Rate The Collision Model • The more molecules present, the greater the probability of collision and the faster the rate. • The higher the temperature, the more energy available to the molecules and the faster the rate. • Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. • In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products. Chapter 14

Temperature and Rate Activation Energy • Arrhenius: molecules must posses a minimum amount of energy to react. Why? • In order to form products, bonds must be broken in the reactants. • Bond breakage requires energy. • Activation energy, Ea, is the minimum energy required to initiate a chemical reaction. Chapter 14

Temperature and Rate Activation Energy Chapter 14

Temperature and Rate Activation Energy • Consider the rearrangement of acetonitrile: • In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state. • The energy required for the above twist and break is the activation energy, Ea. • Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond. Chapter 14

Temperature and Rate Activation Energy • The change in energy for the reaction is the difference in energy between CH3NC and CH3CN. • The activation energy is the difference in energy between reactants, CH3NC and transition state. • The rate depends on Ea. • Notice that if a forward reaction is exothermic (CH3NC  CH3CN), then the reverse reaction is endothermic (CH3CN  CH3NC). Chapter 14

Temperature and Rate Activation Energy • Consider the reaction between Cl and NOCl: • If the Cl collides with the Cl of NOCl then the products are Cl2 and NO. • If the Cl collided with the O of NOCl then no products are formed. • We need to quantify this effect. Chapter 14

Temperature and Rate The Arrhenius Equation • Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: • k is the rate constant, Eais the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. • A is called the frequency factor. • A is a measure of the probability of a favorable collision. • Both A and Ea are specific to a given reaction. • If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation: • If we do not have a lot of data, then we can use Chapter 14

Temperature and Rate The Arrhenius Equation • If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation: • If we do not have a lot of data, then we can use Chapter 14

Reaction Mechanisms • The balanced chemical equation provides information about the beginning and end of reaction. • The reaction mechanism gives the path of the reaction. • Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. Chapter 14

Reaction Mechanisms Elementary Steps • Elementary step: any process that occurs in a single step. • Molecularity: the number of molecules present in an elementary step. • Unimolecular: one molecule in the elementary step, • Bimolecular: two molecules in the elementary step, and • Termolecular: three molecules in the elementary step. • It is not common to see termolecular processes (statistically improbable). Chapter 14

Reaction Mechanisms Elementary Steps • Elementary steps must add to give the balanced chemical equation. • Intermediate: a species which appears in an elementary step which is not a reactant or product. Rate Laws of Elementary Steps • The rate law of an elementary step is determined by its molecularity: • Unimolecular processes are first order, • Bimolecular processes are second order, and • Termolecular processes are third order. Chapter 14

Reaction Mechanisms Rate Laws of Multistep Mechanisms • Rate-determining step: is the slowest of the elementary steps. • Therefore, the rate-determining step governs the overall rate law for the reaction. Mechanisms with an Initial Fast Step • It is possible for an intermediate to be a reactant. • Consider 2NO(g) + Br2(g)  2NOBr(g) Chapter 14

Reaction Mechanisms Mechanisms with an Initial Fast Step 2NO(g) + Br2(g)  2NOBr(g) • The experimentally determined rate law is Rate = k[NO]2[Br2] • Consider the following mechanism for which the rate law is (based on Step 2): Chapter 14

Reaction Mechanisms Mechanisms with an Initial Fast Step • The rate law is (based on Step 2): Rate = k2[NOBr2][NO] • The rate law should not depend on the concentration of an intermediate because intermediates are usually unstable. • Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBr and Br2 assuming there is an equilibrium in step 1 we have Chapter 14

Reaction Mechanisms Mechanisms with an Initial Fast Step • By definition of equilibrium: k1[NO][Br2] = k-1[NOBr2] • Therefore, the overall rate law becomes • Note the final rate law is consistent with the experimentally observed rate law. Chapter 14

Catalysis • A catalyst changes the rate of a chemical reaction. • There are two types of catalyst: • homogeneous, and • heterogeneous. • Chlorine atoms are catalysts for the destruction of ozone. Homogeneous Catalysis • The catalyst and reaction is in one phase. • Hydrogen peroxide decomposes very slowly: 2H2O2(aq)  2H2O(l) + O2(g). Chapter 14

Catalysis Homogeneous Catalysis 2H2O2(aq)  2H2O(l) + O2(g). • In the presence of the bromide ion, the decomposition occurs rapidly: • 2Br-(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l). • Br2(aq) is brown. • Br2(aq) + H2O2(aq)  2Br-(aq) + 2H+(aq) + O2(g). • Br- is a catalyst because it can be recovered at the end of the reaction. • Generally, catalysts operate by lowering the activation energy for a reaction. Chapter 14

Catalysis Homogeneous Catalysis Chapter 14

Catalysis Homogeneous Catalysis • Catalysts can operate by increasing the number of effective collisions. • That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing Ea. • A catalyst may add intermediates to the reaction. • Example: In the presence of Br-, Br2(aq) is generated as an intermediate in the decomposition of H2O2. • When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. Chapter 14

Catalysis Heterogeneous Catalysis • The catalyst is in a different phase than the reactants and products. • Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). • Most industrial catalysts are heterogeneous. • First step is adsorption (the binding of reactant molecules to the catalyst surface). • Adsorbed species (atoms or ions) are very reactive. • Molecules are adsorbed onto active sites (red spheres) on the catalyst surface. Chapter 14

Chemical Kinetics