Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Chapter 13 Chemical Kinetics Kinetics • Kinetics is the study of the rates of chemical reactions. • Rate is the change of concentration (c) per unit time (t): c rate t Rate of Reaction • • Square brackets are used to denote molar concentration. Rate is expressed either as rate of appearance of product or rate of disappearance of reactant. c products reactants Rate t t t • Rate has units of M/s or molar/s or mol/(L·s). Reaction Rate Average Rates • • An average rate is a change in concentration measured over a non-zero time interval. Average rates are not very useful because they depend on the starting and ending times. Instantaneous Reaction Rate • The instantaneous rate of the reaction is equal to the slope of the line drawn tangent to the curve at time t. These graphs show how to determine the instantaneous rate at 10.0 s. Rate and Reaction Stoichiometry • • The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry. For the reaction 2HBr (g) H2 (g) + Br2 (g) two moles of HBr are consumed for every one mole of H2 that is formed so rate of change of [HBr] is double that of [H2]. 1 HBr H2 2 t t Rate and Reaction Stoichiometry • For any reaction aA+bBcC+dD the reaction rate is given by: 1 A 1 B rate a t b t 1 C 1 D c t d t • Note the signs as well as the coefficients. Example Problem For the reaction 5H2O2 + 2MnO4- + 6H+ → 2Mn2+ + 5O2 + 8H2O the experimentally determined rate of disappearance of MnO4- is 2.2 x 10-3 M/s. (a) Calculate the reaction rate. (b) What is the rate of appearance of O2? Relating Rate and Concentration • Experimental rate law: analysis of many experiments shows that the rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power. • For a reaction aA + bB products the rate law is the equation: x y rate = k[A] [B] Rate Law rate = k[A]x[B]y • x and y are the orders of the reaction in [A] and [B] respectively. • The overall order of the reaction is x + y. • x and y are usually small integers, but may be zero, negative, or fractions. • k is the specific rate constant. Relating Rate and Concentration • The reaction orders are determined by noting the effect that changing the concentration of each reactant have on the rate. • The rate constant, k, is evaluated once the orders in the rate law are known. Relating Rate and Concentration • Most often you will be given initial concentrations and rates and asked to determine the order, which is the exponent to which concentration is raised. • One way to learn to predict the order is to predict the rate given the order and concentration and see the relationship. Dependence of Rate on Order First order rate law rate = k[conc]1 Second order rate law rate = k[conc]2 Zero order rate law rate = k[conc]0 [Concentration] [Rate] 1 2 3 1 2 3 1 2 3 1 2 3 1 4 9 1 1 1 Initial Rate Method • The initial rate method measures the time during which a known small fraction of the reactants are consumed. • Experiments are performed in which initial concentrations of [A] and [B] are individually varied. • The time period of measurement is small enough that the measured rate is approximately equal to the instantaneous rate. Initial Rates • Initial rates are given below for the reaction F2 + 2ClO2 2FClO2 Determine the rate law and rate constant Trial Init. conc. Init. conc. [F2], M [ClO2], M Init. Rate Ms 1 0.10 0.010 1.2x10-3 2 0.10 0.040 4.8x10-3 3 0.20 0.010 2.4x10-3 Initial Rates • It is helpful to express concentrations and rates on a relative scale, by dividing the entries in each column by the smallest value. Trial [F2] Rel. conc. [ClO2] Rel. conc. Initial rate Rel. rate 1 0.10 1 0.010 1 1.2x10-3 1 2 0.10 1 0.040 4 1.8x10-3 4 3 0.20 2 0.010 1 2.4x10-3 2 Initial Rates • In trials 1 and 2, the concentration of F2 does not change, so the 4-fold change in [ClO2] causes the 4-fold change in rate; the order of ClO2 must be 1. • In trials 1 and 3, [ClO2] is the same, so the doubling of rate was caused by doubling [F2]; therefore the order of F2 is also 1. Initial Rates • The rate law is first order in both [F2] and [ClO2]. rate = k[F2][ClO2] • Solve the equation for k, and substitute the experimental concentrations and rates. 3 rate 1.2 10 M / s 1 1 k 1.2M s F2 ClO 2 0.10M 0.010M Test Your Skill • Write the rate law for the reaction given the following data: 2NO + 2H2 N2 + 2H2O Init conc. Init conc. Init. Rate, Trial [NO], M [H2], M MIs 1 0.00570 0.140 7.01x10-5 2 0.00570 0.280 1.40x10-4 3 0.0114 0.140 2.81x10-4 Concentration-Time Dependence • A zero order rate law rate = k means that the reaction rate is independent of reactant concentration. Zero-Order Rate Laws • A plot of concentration vs. time yields a straight line • A plot of rate vs. time yields a straight line with a slope of zero Concentration-Time Dependence • First order rate law • Differential rate law rate = k[R] • Integrated rate law R Ro e kt or In R In Ro kt Concentration-Time Dependence • First order rate law: [R] = [R]o e -kt • A plot of concentration vs. time yields a curve. Concentration-Time Dependence • First order rate law: ln[R] = -kt + ln[R]o • A plot of ln(concentration) vs. time yields a straight line. First Order Rate Law • C12H22O11 + H2O C6H12O6 + C6H12O6 sucrose + water glucose + fructose The reaction is 1st order, k = 6.2 x 10-5 s-1. If [R]o = 0.40 M, what is [R] after 2 hr? Test Your Skill • How long did it take for the concentration in the same experiment to drop to 0.30 M? Half-Life • Half-life, t½, is the time required for the initial concentration to decrease by ½. Half-Life • Half-life, t½, is the time required for the initial concentration to decrease by ½. • For a first order rate law, the halflife is independent of the concentration. t1 2 [R] o 1 1 0.693 ln ln2 k 1/2[R] o k k Calculating Half-Life • k = 6.2 x 10-5 s-1 for the reaction C12H22O11 + H2O C6H12O6 + C6H12O6 Calculate the half-life. Radiocarbon Dating • The age of objects that were once living can be found by 14C dating, because: • The concentration 14C is a constant in the biosphere (the atmosphere and all living organisms). • When an organism dies, the 14C content decreases with first order kinetics (t½ = 5730 years). • Scientists calculate the age of an object from the concentration of 14C in a sample. Example: 14C Dating • A sample of wood has 58% of the 14C originally present. What is the age of the wood sample? Second Order Rate Law • For a second order rate law rate = k[R]2 1 1 kt [R] [R] o • A plot of 1/[R] vs. t is a straight line for a system described by second order kinetics. Example: Second Order Rate Law • The reaction 2NOCl 2NO + Cl2 obeys the rate law rate = 0.020 M-1s-1 [NOCl]2 Calculate the concentration of NOCl after 30 minutes, when the initial concentration was 0.050 M. Example: Order of Reaction Given the experimental data for the decomposition of 1,3-pentadiene shown below, determine the order of the reaction. Example: Order of Reaction (cont.) Review table 13.4 on page 534. Influence of Temperature on k • Reactions proceed at faster rates at higher temperatures. Collision Theory • The reaction rate is proportional to the collision frequency, Z, the number of molecular collisions per second. • Z depends on the temperature and the concentration of the colliding molecules. • Not all molecular collisions result in the formation of products. Collision Theory • Activation energy (Ea): the minimum collision energy required for reaction to occur. • Activated complex: the highest energy arrangement of atoms that occurs in the course of the reaction. The Activated Complex NO + O3 [activated complex]* NO2 + O2 Influence of Temperature on Kinetic Energy • The fraction of collisions with energy in excess of Ea is given by: fr e Ea / RT • The collision frequency is proportional to the concentrations of colliding species. • The reaction rate is proportional to the rate of collisions time the fraction of collisions with energy in excess of Ea. rate = Z × fr Orientation of Reactants The Steric Factor • The steric factor, p, is a number between 0 and 1 that is needed to account for factors other than energy before a reaction can occur. • The reaction rate is proportional to the steric factor times the collision frequency times the fraction of collisions with energy in excess of Ea: rate = p x Z x fr The Arrhenius Equation rate = p x Z x fr rate pZo colliding speciese Ea / RT Combine p and Zo into a term A: rate Acolliding speciese Ea / RT Experiments show that rate = k[colliding species], so k Ae Ea / RT The Arrhenius Equation k Ae Ea / RT • Take the natural log of both sides of the equation: Ea 1 In k In A R T • A plot of ln k vs. 1/T gives a straight line with a slope of -Ea/R and an intercept of ln A. Measuring Activation Energy • Determine Ea for the reaction, 2NO2 2NO + O2 given the data: k (M-1·s-1) T (°C) In k 1/T (K-1) 0.003 500 -5.8 2.00x10-3 0.037 550 -3.30 1.82x10-3 0.291 500 -1.234 1.67x10-3 1.66 650 0.507 1.54x10-3 7.39 700 2.000 1.43x10-3 Solution • Prepare a plot of ln k vs. 1/T slope 1.37 10 4 slope Ea /R Ea 1.37 10 4 8.314 Ea 1.14 105 J 1.14 10 2 kJ Example: Arrhenius Equation • The rate of a reaction exactly doubles, when the temperature is changed from 25.0o C to 36.2o C. Calculate the activation energy for this reaction. Catalysis • A catalyst is a substance that increases the reaction rate but is not consumed in the reaction. • A catalyst provides an alternate reaction path with a lower activation energy. Homogeneous Catalysis • A homogeneous catalyst is one that is present in the same phase as the reactants. • Bromide ion is a homogeneous catalyst for the decomposition of hydrogen peroxide. 2H2O2(aq) 2H2O(l) + O2(g) • step 1: H2O2(aq) + 2Br-(aq) + 2H+(aq) Br2(aq) + 2H2O(l) • step 2: H2O2(aq) + Br2(aq) 2Br-(aq) + 2H+(aq) + O2(g) Heterogeneous Catalysis • A heterogeneous catalyst is one that is present in a different phase from the reactants. • The gas phase reaction of hydrogen with many organic compounds is catalyzed by solid platinum. Enzyme Catalysis • Enzymes are large molecules (macromolecules) which catalyze specific biochemical reactions. • Enzymes can increase the rates of 14 reactions by factors as large as 10 . • Enzymes are very specific in the reactions they catalyze. • Enzymes are active under mild reaction conditions. Reaction Mechanisms • A mechanism is a sequence of molecular-level steps that lead from reactants to products. • An elementary step is an equation that describes an actual molecular level event. • The concentration dependence in the rate law for an elementary step is given by the coefficients in the equation. Molecularity • The molecularity of an elementary step is the number of reactant species involved in that step. • Most elementary steps are either unimolecular (involving a single molecule) or bimolecular (collision of two species). Elementary Steps • The reaction 2NO + O2 2NO2 is believed to occur by the following sequence of elementary steps: 2NO N2O2 bimolecular reaction N2O2 + O2 2NO2 bimolecular reaction Rate Laws for Elementary Reactions • The rate of an elementary step is proportional to the concentration of each reactant species raised to the power of its coefficient in the equation: • step 1: 2NO N2O2 rate1 = k1[NO]2 • step 2: N2O2 + O2 2NO2 rate2 = k2[N2O2][O2] Test Your Skill • Write the rate law for the elementary step H2 + Cl H2Cl Rate-Limiting Steps • The overall rate of a multistep reaction is determined by its slowest step, called the rate-limiting step. • The rates of fast steps which follow the ratelimiting step have no effect on the overall rate law. • The rates of fast steps that precede the rate-limiting step usually affect the concentrations of the reactant species in the rate-determining step. Complex Reaction Mechanisms • R P (2 steps) • R intermediates • Intermediates P rate1 = k1[R] rate2 = k2[intermediates] • If step 1 is slow, then it determines rate • If step 2 is slow, how can we measure [intermediates]? • Many fast steps prior to slow step are fast and reversible Complex Reaction Mechanisms • Fast reversible steps help with [intermediate] • Consider the following reaction 2NO + 2H2 N2 + 2H2O • rate = k[NO]2[H2] • step 1: 2NO N2O2 fast and reversible • step 2: N2O2 + H2 N2O + H2O slow • step 3: N2O + H2 N2 + H2O fast Work on the board Complex Reaction Mechanisms • Consider the two-step reaction 2NO + O2 2NO2 • step 1: 2NO N2O2 rate1 = k1[NO]2 • step 2: N2O2 + O2 2NO2 rate2 = k2[N2O2][O2] Complex Reaction Mechanisms • If the first step is the rate-limiting step, the rate law is: rate = k1[NO]2 • If the first step is rapid and the second step is the rate limiting step, the rate law is: rate = k2[N2O2][O2] Complex Reaction Mechanisms If the first step reaches equilibrium rate1 (forward) = rate-1 (reverse): rate1 k1[NO] 2 rate1 k 1[N2 O 2 ] Because the rates at equilibriu m are equal : rate1 rate-1 k1[NO] k 1[N2 O 2 ] 2 k1 2 [N2 O 2 ] [NO] k 1 Complex Reaction Mechanisms • Substituting into the expression for step 2: rate = k2[N2O2][O2] k1 rate = k2 [NO]2[O2] k -1 Combine all the rate constants: • rate = k [NO]2[O2] • The reaction is second order in NO and first order in O2. Test Your Skill • For the reaction 2NO2 + O3 N2O5 + O2 the experimentally determined rate law is rate= k[NO2][O3]. • Identify the rate limiting step in the proposed two-step mechanism: NO2 + O3 NO3 + O2 NO3 + NO2 N2O5 step 1 step 2 The Hydrogen-Iodine Reaction • H2 + I2 2HI rate = k[H2][I2] • For many years this reaction was believed to occur as a single bimolecular step. • From more recent data, a very different mechanism is likely. I2 2I Fast, reversible I + H2 H2I Fast, reversible H2I + I 2HI Slow • Both mechanisms give the same rate law. Enzyme Catalysis • Most enzymes follow the Michaelis-Menten mechanism. • You will see this again in biochemistry!!