ELEMENTARY CHEMICAL KINETICS

Transcription

1 ELEMENTARY CHEMICAL KINETICS EDR Chapter a knowledge of the rate, or time dependence, of chemical change is of critical importance for the successful synthesis of new materials and for the utilization of the energy generated by a reaction. During the past century it has become clear that all macroscopic chemical processes consist of many elementary chemical reactions that are themselves simply a series of encounters between atomic or molecular species... to understand the time-dependence... chemical kineticists have... focussed on sorting out all of the elementary chemical reactions involved in a macroscopic chemical process and determining their respective rates. Yuan T. Lee, Nobel Prize in Chemistry, 986 AT OMS, MOLECULES, AND IONS MUST COLLIDE IN ORDER TO REACT Reaction Rates ( ) For a general chemical reaction with a, b, e, and f balanced coefficients aa + bb +... ee + f F +... introduce a variable x which measures the difference in concentration from the initial value: Note that for products x is defined by subtracting the initial concentrations so that x is always positive. x = a ([A] 0 [A]) = b ([B] 0 [B]) =... = e ([E] [E] 0) = f ([F] [F] 0) =... where the zero subscript denotes the initial concentration. Let the volume be constant. As the rate is the change in concentration per change in time take the change in the variable x per change in time, remembering that the initial concentrations are constant. One obtains for the rate In section 25.3 EDR introduces the advancement of a reaction, ξ, for the change in the number of moles. We usually consider molarity. So our variable x = ξ / V where V is the volume giving us concentrations in molarity as EDR shows in Eq. (25.8). R = dx = a = e d[a] d[e] = f = b d[f] d[b] =... =... In section 25. EDR uses ξ exactly like x [see Eqs. (25.98) and (25.99)]. This formulation is useful for several approaches for finding a rate law (25.3.2): initial rate method, isolation method and for perturbation-relaxation methods (25.). rate expressed in terms of reaction stoichiometry - there isonly one rate for a reaction at a specific T For the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) R = d[n 2] = d[h 2 ] 3 = d[nh 3 ] 2 EX. Consider the combustion of hydrogen gas: 2H 2 (g) + O 2 (g) 2H 2 O(g) If hydrogen is burning at the rate of 0.85 mol s,what is the rate of consumption of oxygen? [0.43 mol s ] What is the rate of formation of water vapor? [0.85 mol s ]

2 -2- Rate Laws (25.4) - Effect of Concentration on Rate (a way to categorize reactions) Many reactions are found experimentally to have rates that can be expressed in terms of a simple rate law with exponents that are usually integers or half-integers and can be negative R = k[a] α [B] β...[l] γ (25.9) Replace concentrations in Eq. (25.9) in terms of initial concentrations and the variable x from p. then examples R = k([a] 0 ax) α ([B] 0 bx) β...([l] 0 lx) λ (25.9 ) rate law (from experiment!) H 2 (g) + I 2 (g) 2HI(g) r = k[h 2 ][I 2 ] 2NO(g) + O 2 (g) 2NO 2 (g) r = k[no 2 ] 2 [O 2 ](note product NO 2 ) 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) r = k[n 2 O 5 ] H 2 PO 2(aq) + OH (aq) HPO 3(aq) + H 2 (g) r = k[h 2 PO 2][OH ] 2 H 2 (g) + Br 2 (g) 2HBr (g) r = rate constant k Determining Reaction Order ( ) k[h 2 ][Br 2 ] 3/2 (note product HBr) [Br 2 ] + k [HBr] reaction order: for A, B (α, β ); overall (α + β ); pseudo (isolation method). initial rate method - see example problem 25.2 At the beginning of a reaction (or for a sufficiently slow reaction) the extent of reaction, and thus x, isvery small so terms involving x can be ignored. Then Eq. (25.9 ) becomes R 0 = k[a] α 0 [B] β 0... [L] λ 0 As the figure on the right shows there are many ways to define a rate in terms of different instantaneous rates from tangents and different averages. We want one, unique number for the rate - initial rate method Appearance of Product NO 2 (g) + CO(g) NO(g) + CO 2 (g) Initial rate d [NO] = mol L s d t [NO t EX 2. Consider the gas-phase reaction between nitric oxide and bromine at 273 o C. 2NO(g) + Br 2 (g) 2NOBr (g) The following data for the initial rate of appearance of NOBr were obtained. Experiment [NO] [Br 2 ] Initial Rate 0.0 M 0.20 M 24 M s a) Determine the rate law k[no 2 ] 2 [Br 2 ] dd [NO ] b) Calculate the average value of the rate constant from the four data sets. c) How isthe rate of appearance of NOBr related to the rate of disappearance of Br 2? [R NOBr = 2R Br2 ] d) What is the rate of disappearance of Br 2 when [NO] = M and [Br 2 ]=0.25 M? [8.4 M s ]

3 -3- Rate Law from Linear Regression on Initial Rate vs Concentration Data NH 4 + (aq) + NO 2(aq) N 2 (g) + 2H 2 O(l) Can you verify that A = B =? For a given temperature the rate constant k is a constant. If a concentration does not change then it too is constant. Taking ln (or log) of the rate law gives separate terms for ln k and ln of each of the concen- 2. isolation method Consider a reaction where all reagents but one (say A) are present in very large excess. During the course of the reaction the concentrations of all reagents except A change very little from their initial values. For these reagents, which are in large excess, their concentrations can be approximated by their initial values. Then from Eq. (25.9 ), the rate law becomes r = k([a] 0 ax) α [B] β 0... [L] λ 0 = k [A] α trations where the latter is multiplied by the order of the species. Recognizing a straight line, y = mx + b, the order with respect to a specific species is the slope m and the intercept is the sum of ln k and the ln of the species whose concentration is constant. where k = k[b] β 0... [L]λ 0 is a constant. The reaction has been forced to be pseudo α order in A. EX 3. The reaction of A + B products has the rate law R = k[a][b] 2.Att = 0, [A] =.00 M and [B] = M are mixed. a) Give the rate law for [B] as a function of [A] 0,[B], and k and then simplify. b) In your rate law ofpart a) what is the order with respect to [A] and with respect to [B] R = k[a][b] 2 = k[a] 0 [B] 2 = k'[b] 2 [zero, pseudo 2nd order] 3. physical methods a) stopped flow b) flash photolysis CHEM 343 lab c) perturbation-relaxation (25.)

4 -4- Reaction Mechanisms (25.4) proposed mechanism: elementary step 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) N 2 O 5 <=> NO 2 + NO 3 NO 2 + NO 3 NO + NO 2 + O 2 NO + NO 3 2NO 2 Always check that sum of elementary steps gives the correct stoichiometric equation. mechanism A sequence of elementary steps which sum to the overall reaction (stoichiometric equation) giving the process by which the overall reaction occurs. It must be consistent with the experimental rate law. But remember that there may be one or more stoichiometric numbers to the elementary steps of the mechanism. reaction intermediate (26.) A species which does not appear in the overall mechanism as it is formed in one step and consumed in a subsequent step. In the above, NOand NO 3 are intermediates. (Later we will encounter a catalyst which also does not appear in the overall mechanism though it does appear in the rate law. stoichiometric number, s(26.) Number of times a step in the mechanism occurs for each occurrence of the overall reaction. The stoichiometric number of step is two. molecularity, (25.4) Power of species in elementary step "rate" law: uni-, bi-, termolecular, so it is the "order" of the elementary step. One should be cautious as it is only with an elementary step that the equivalence of molecularity and "order" is found. In general the molecularity of an individual step is different than the order of the over-all reaction. Step produces one NO 2 and one NO 3. However both steps 2 and 3 consume one NO 3 each. Therefore step must occur twice before step three can happen. Therefore the stoichiometric number, s, of step is 2. Now when you multiply both sides of step by 2 and add steps 2 and 3, you recover the correct stoichiometry. The stoichiometric number does not effect the elementary step rate law! While we can never guess the rate law for a chemical reaction by simply looking at the stoichiometric equation, the elementary steps of a reaction do give all the information one needs to determine the rate law for that elementary reaction. It is the combination of the rate laws of individual steps in an overall mechanism that ultimately gives the overall rate law, which can be extremely complicated. Here the above elementary step rate laws are: R = k [N 2 O 5 ] and R = k [NO 2 ][NO 3 ] R 2 = k 2 [NO 2 ][NO 3 ] R 3 = k 3 [NO][NO 3 ] It is easy to make a mistake writing the rate law for the reverse of a step or a reaction which is at equilibrium. Until you get used to thinking "backward" you might write the reverse reaction out and then write its rate law. So for step R - => NO 2 + NO 3 N 2 O 5 with rate constant k -

5 - 5 - Integrated Rate Law Expressions (25.5) Integrated Rate Law (Explain Dependence of Concentration on Time for: aa products). first-order reaction ( ) FIG - First-Order Reaction Plot N 2 O 5 (g) 2 NO 2 (g) + /2 O 2 (g) dd[a] aa dddd = kk[a] integrated rate ln[a] = ln[a] 0 - k a t or [A] = [A] 0 e -k a t where k a = ak a) sequential (consecutive) first-order reaction (25.7) A B C has two steps A B and B C where B is an intermediate. b) competing (parallel) first-order reaction (25.8) This is a branching reaction which has two steps A B and A C where A is consumed via two routes. c) reversible first-order reaction (25.0) half-life (25.5.2) general definition - the time t /2 that it takes [A(t)] = /2 [A] 0 EX 4. The first-order rate constant for the decomposition of N 2 O 5 (see example problem 25.3) N 2 O 5 (g) 2 NO 2 (g) + /2 O 2 (g) at 70 C is s -. Suppose we start with mol of N 2 O 5 (g) in 2.0 L. a) How many moles of N 2 O 5 will remain after 2.5 min? b) How many minutes will it take for the quantity of N 2 O 5 to drop to 0.00 mol? c) What is the half-life of N 2 O 5 at 70 C? [9.0 mol] [30 s] [02 s]

6 second-order reaction ( ) Second-Order Reaction Plot 2 NO 2 (g) 2 NO(g) + O 2 (g) dd[a] aa dddd = kk[a]2 integrated rate (Type I) half-life (25.5.4) slope = k a = 2k EX 5. At 300 C the rate constant for the second order reaction NO 2 (g) NO(g) + /2 O 2 (g) is M - s -. If the initial concentration of NO 2 in a closed vessel is M, what is the concentration remaining after hr? zero order [ M] 3. zero-order reaction dd[a] aa dddd = kk[a]0 = kk integrated rate Very common with catalysts. We will encounter these reactions in Ch 26, enzymes. half-life first order second order

7 -7- Rate Laws: A Summary It is not always easy to determine orders graphically A Reaction Plotted as Various Orders 2 NO 2 (g) 2 NO(g) + O 2 (g) 0th st ln[a] = ln[a] 0 k a t or [NO 2 ] ln[no 2 ] [A] = [A] 0 ee kk aatt [A] = [A] 0 k a t 2nd 3rd [NO 2 ] [A] = + kk [A] aa tt 0 [NO 2 ] 2 [A] 2 = 2 [A] + 2kk aatt 0 [NO 2 ]/M t/s A Comparison of Different Rate Laws For a general chemical reaction: aa products where a denotes the stoichiometric coefficient. The rate law, integrated rate law, half life, and form of concentration data giving a linear plot are summarized for various orders the reaction could follow. In the table k a = ak where k is the rate constant and a the stoichiometric coefficient rate law integrated rate law half-life, t /2 linear plot 0 Δ[A] aa Δtt = kk [A] = [A] o kk aa tt Δ[A] aa Δtt = kk[a] ln[a] = ln[a] o kk aa tt or [A] = [A] o ee kk aatt [A] o 2kk aa ln 2 kk aa [A] vs t lna vs t 2 Δ[A] aa Δtt = kk[a]2 [A] = [A] oo + kk aa tt vs tt kk aa [A] o [A] 3 Δ[A] aa Δtt = kk[a]3 [A] 2 = [A] o 2 + 2kk aatt 3 2kk aa [A] o 2 vs tt [A] 2 n > Δ[A] aa Δtt = kk[a]nn [A] nn = [A] o nn + (nn )kk aatt 2 nn (nn )kk aa [A] o nn vs tt [A] nn

8 -8- Sequential Reactions (25.7) The experimentally observed rate law provides information on the mechanism of a reaction since any proposed mechanism must yield this rate law. Often an exact determination of the rate law from the differential equations resulting from the rate equations of the elementary steps in a multistep mechanism is not feasible. Therefore one must rely upon some approximation, generally either the rate-determining step or the steady-state approximation are used. rate-determining step (rate-limiting step) approximation (25.7.2) If a particular step is known to be much slower than the other steps, it is the bottle-neck to the reaction and the rate law is the rate of this slow elementary step. EX 6. For the reaction NO 2 (g) + CO(g) NO(g) + CO 2 (g) a) Show that the following mechanism is consistent with the rate law and identify k exp Figure 25.9b illustrates the time evolution of such a system (below 500 K) r = k exp [NO 2 ] 2. NO 2 + NO 2 NO 3 + NO slow k NO 3 + CO NO 2 + CO 2 fast k 2 rate-limiting step b) What is the function of NO 3? steady-state approximation (25.7.3) - see example problem 25.6 When a reaction mechanism has several steps (possibly of comparable rates), the rate-determining step is often not obvious and can change under different sets of conditions. However, there will be an intermediate in some of the steps, a species that is neither one of the reactants nor one of the products. It is produced in some step and then consumed in a subsequent step. In such cases the steady-state approximation may be used. The method assumes that the concentration of an intermediate (possibly several) in the reaction mechanism remains the same at some stage of the reaction. So the system has reached a steady-state. As an example, the mechanism below, which includes the intermediate I, can be analytically solved for all concentrations. A I k I B k 2 steady state on d[i] c The steady-state approximation is applicable to the last case. It can be stated in different ways. The derivatives with respect to time of the concentrations of reactive intermediates = The steady-state concentrations of the reactive intermediates are constant. 3. The rates of formation of reactive intermediates equals the rates of their destruction. 4. The steady-state concentrations of the reactive intermediates are small. 5. The rates of all steps involving reactants, products, and intermediates are equal.

9 -9- Temperature Dependence of Rate Constants (25.9) Arrhenius Equation (Explain Effect of Temperature on Rate) collision model - kinetic theory interpretation: to react molecules need. to collide 2. to be properly oriented Collision of Cl with NOCl 3. to have a minimum translational kinetic energy, E o (minimum collision energy reactants need in order to form products) Cl + ClNO effective Pressure vs Time for Methyl Isonitrile Rearrangement CH3N C(g) CH3C N(g) Cl + ONCl ineffective energy profiles of section 25.9 Reaction Coordinate (see p. 682) for Methyl Isonitrile Rearrangement CH 3 N C(g) CH 3 C N(g) these graphs indicate that the reaction is first order E a,f E a,r Maxwell-Boltzmann Distribution of Where energy comes from: Molecular Kinetic Energies, F 3 (E) Maxwell-Boltzmann Distribution of Molecular Speeds, F 3 (v) minimum energy, E o reactive fraction E o

10 -0- Kinetic Theory - Interpretation of Collisions, Rate Constant, E a (Levine Ch 4 and EDR) From the Maxwell-Boltzmann distribution of molecular speeds in an ideal gas at equilibrium (average speed) v = 8kkkk ππππ = 8RRkk ππππ Levine (4.47) Intuitively, for an elementary reaction: B + C products, the molecules must first collide (not only a useful concept for gas phase reactions; in a diffusion controlled reaction occurring in solution the reaction is only as fast as the molecules can come together => R D ~ kt/ηr as covered in section 25.3) (molecular) collision rate = (density) (cross sectional area) (relative speed) Z BC, the total number of collisions between B and C molecules per unit volume per unit time is Z BC = NN B VV NN C VV π dd2 BC v BC Levine (4.63) 2 where πdd BC is the collision cross sectional area and for v BC the molar mass M in v is replaced by the reduced mass In collision theory these collisions must be effective. So molecules need. to collide the collision rate Z BC 2. to be properly oriented steric factor p (will ignore but orientation covered in section 25.2) 3. to have a minimum kinetic energy, E o so that the rate R for molecular collisions becomes R = (collision rate) (fraction with E at least E o ) (molar) = (Z BC e -Eo/RT )/N o (Note: N B = n B N o and N C = n C N o ) = N o [B] [C] π dd 2 BC v BC e -Eo/RT = k [B] [C] compare with rate law for the elementary reaction where the above comparison with the rate law gives an expression for the molar rate constant k = N o π dd 2 BC v BC e -Eo/RT volume swept per unit time per mole = N o (cross sectional area) (relative speed) e -Eo/RT k = A e -Ea/RT the Arrhenius equation (25.82) A is the frequency factor or pre-exponential factor or frequency factor and E a is the Arrhenius activation energy. Both are constants and usually assumed to be independent of temperature. Arrhenius equation gives the effect of temperature on rate. What is the relationship between the empirical Arrhenius activation energy E a and the kinetic theory minimum energy required for a reactive collision E o? The activation energy is defined as E a = RT 2 ddddnnkk ddkk (25.27)

11 -- So that taking the derivative of the kinetic theory rate constant gives temperature dependence of k (compare with K) EX 7. What is the rate constant for the first-order isomerization of cyclopropane, C 3 H 6, to propene at 300 C if A = s - and E a = 272 kj/mol? [2.6 x0-0 s - ] EX 8. Solve for k(300) for the preceding example only using k(500) = s - and E a = 272 kj/mol (do not use the value for the pre-exponential factor, A) energy profiles for endothermic and exothermic reactions [2.57 x0-0 s - ] Reaction coordinate Reaction coordinate