Theoretical Models for Chemical Kinetics Thus far we have calculated rate laws, rate constants, reaction orders, etc. based on observations of macroscopic properties, but what is happening at the molecular level? Collision Theory kinetic-molecular theory can calculate number of molecular collisions per unit time (the collision frequency) which is typically 10 30 collisions per second if each collision yielded a product molecule, the rate of reaction would be about 10 6 M s -1 gas-phase reactions generally have a rate of reaction around only 10-4 M s -1 ; Why so much slower? CHEM 1001 3.0 Chemical Kinetics 1 Collision Theory - continued only a fraction of collisions lead to a reaction (about 1 in 10 10!) energy from collisions must be redistributed to break appropriate bonds two slow-moving molecules that collide would not be expected to have enough kinetic energy to permit bond breakage the minimum energy above the average kinetic energy molecules must possess for a reaction to occur is known as the activation energy CHEM 1001 3.0 Chemical Kinetics
Collision Theory - continued -if the minimum energy required for a reaction to occur is indicated by the arrow in the accompanying figure, then the fraction of molecules possessing this energy will be greater at T than at T 1 - regardless of T, the fraction of molecules possessing enough energy will be small and thus the rate of reaction will be much less than the collision frequency CHEM 1001 3.0 Chemical Kinetics 3 Collision Theory - continued rate of reaction is also be limited by the orientation of molecules at the time of their collision Example: N O(g) + NO(g) N (g) + NO (g) CHEM 1001 3.0 Chemical Kinetics 4
Transition State Theory an unobserved, activated complex is proposed to exist as a transition state between reactants and products the activated complex is short-lived because it is highly reactive (may break apart to form products or revert to reactants) Example: N O(g) + NO(g) N (g) + NO (g) CHEM 1001 3.0 Chemical Kinetics 5 Transition State Theory - continued Reaction Profile CHEM 1001 3.0 Chemical Kinetics 6
Transition State Theory - continued Reaction profile features: the enthalpy change for the reaction is the difference in the activation energies of the forward and reverse reactions for an endothermic reaction, the activation energy must be greater than or equal to the enthalpy of the reaction (usually it is greater) CHEM 1001 3.0 Chemical Kinetics 7 Effect of Temperature on Reaction Rates Rate of a reaction generally increase with temperature (roughly doubles every 10ºC) Arrehenius expression illustrates this for rate constants k = or, taking logarithms, Ae E a / RT Ea ln k = + lna RT E a is the activation energy (J/mol) and A is the frequency or pre-exponential factor (same units as k) CHEM 1001 3.0 Chemical Kinetics 8
Temperature Effects - continued A - (frequency or pre-exponential factor) factor related to the collision frequency of a molecule; represents the limit to how fast two molecules can react (molecules cannot react unless they have enough energy) E a - (activation energy) even when molecules collide they cannot react unless they possess enough energy; E a is the minimum energy the reactants must possess in order to react CHEM 1001 3.0 Chemical Kinetics 9 Temperature Effects - continued A plot of ln k versus 1/T will yield a straight line with a slope of -E a /R (studied in first lab for this course) CHEM 1001 3.0 Chemical Kinetics 10
Temperature Effects - continued the Arrehenius equation can relate rate constants at two temperatures k Ea 1 1 ln = - k1 R T1 T (temperature in Kelvin and R in standard SI units) CHEM 1001 3.0 Chemical Kinetics 11 Reaction Mechanisms most reactions will not proceed in a single step the step-by-step pathway by which a reaction occurs is called the reaction mechanism a plausible reaction mechanism must be consistent with the 1) stoichiometry of the overall reaction ) experimentally determined rate law each step in the mechanism is called an elementary reaction (or process) CHEM 1001 3.0 Chemical Kinetics 1
Characteristics of Elementary Reactions Elementary processes are either unimolecular (a single molecule dissociates) or bimolecular (two molecules collide); termolecular reactions (simultaneous collision of three molecules) are rare Orders of reaction in the rate law for the elementary reaction are the same as the stoichiometric coefficients in the balanced equation for the process - not true of the overall rate law and the overall balanced equation Elementary processes can be reversible reactions; sometimes rates of forward and reverse reactions are equal continued CHEM 1001 3.0 Chemical Kinetics 13 Characteristics of Elementary Reactions Intermediate species can be produced in an elementary reaction which do not appear in either the overall chemical reaction or the overall rate law; such species produced by one elementary reaction must be consumed by another One elementary reaction may occur much more slowly than the others and may determine the rate of the overall reaction; such a reaction is called the rate-determining step CHEM 1001 3.0 Chemical Kinetics 14
Intermediate Species in Elementary Processes 1. Intermediate species (I) can be produced in an elementary reaction. They do not appear in either the overall chemical reaction or the overall rate law; such species produced by one elementary reaction must be consumed by another.. An intermediate is a highly reactive species (often free radicals); its production is usually slower than its consumption 3. Due to its high reactivity an intermediate very rapidly reaches its steady-state concentration which is much lower than the concentrations of reactants and products during the steady-state period: [I] ss << [A] ss, [B] ss, [C] ss, 4. The last allows us to use so-called steady-state approximation CHEM 1001 3.0 Chemical Kinetics 15 Overall Reaction: A + B + C D k1 A + B I d[a] = k [A][B] 1 dt k d[d] I + C D = k [I][C] dt d[a] [I] ss << [A] ss, [B] ss, [C] ss, [D] ss = dt d[i] d[i] = dt dt prod d[i] + dt cons d[b] d[c] d[d] = dt dt dt d[a] d[d] = k1[a][b] k [I][C] = 0 dt dt Negative d[i] dt prod d[i] = dt cons CHEM 1001 3.0 Chemical Kinetics 16
Steady-State Approximation the assumption that (d[i]/dt) prod = - (d[i]/dt) cons is called steady-state approximation the steady-state approximation is the most common approach to the analysis of reaction mechanisms CHEM 1001 3.0 Chemical Kinetics 17 General Approach to the Analysis of Simple Reaction Mechanisms Using Steady-State Approximation 1. Check the proposed mechanism for agreement with balanced chemical equation.. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. It usually includes the concentration of intermediate, [I]. 3. Determine the steady state concentration of the intermediate using (d[i]/dt) prod = - (d[i]/dt) cons as a function of the concentrations of reagents. 4. Substitute [I] in the reaction rate found in step with [I] determined in step 3. Compare the theoretical and experimental rates of reaction. CHEM 1001 3.0 Chemical Kinetics 18
Examples of Steady-State Approximation Example 1 From the experiment: A + B C and Rate of reaction = k[a][b] Let s examine whether the following proposed mechanism is plausible: (1) A + B k1 I () I + B k C CHEM 1001 3.0 Chemical Kinetics 19 1. Examine whether the proposed mechanism satisfies the balanced chemical equation Experiment: A + B k C Mechanism: A + B k1 I I + B k C Overall: A + B + I + B k I + C A + B k C Thus, the proposed mechanism satisfies the balanced chemical equation CHEM 1001 3.0 Chemical Kinetics 0
. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. The rate of the second reaction is: Rate = k [I][B] The concentration of intermediate has to be eliminated 3. Determine the steady-state concentration of the intermediate using the steady-state approximation: (d[i]/dt) prod = - (d[i]/dt) cons k 1 [A][B] = - (-k [I][B]) k 1 [A][B] = k [I][B] [I] = k 1 [A][B]/k [B] = k 1 [A]/k CHEM 1001 3.0 Chemical Kinetics 1 4. Substitute [I] in the reaction rate with [I] determined using steady-state approximation. Compare the theoretical and experimental rates of reaction. Rate = k [I][B] [I] = k 1 [A]/k Rate = k 1 [A][B] It is the same as the experimental rate so that the proposed mechanism is plausible NOTE: The rate of the overall reaction does not depend on k and depends only on k l. In such a case the first reaction is called the rate-determining step. CHEM 1001 3.0 Chemical Kinetics
The Rate-Determining Step is a Slow Reaction The rate-determining elementary process is a bottle-neck in the overall process. It is often called a rate-determining step. In general, determining whether or not there is a rate-determining step requires the analysis that we have just done. Note:In the previous example the steady-state approximation requires that the first reaction is a slow rate-determining step and that the second process must be fast: slow A + B k1 I fast I + B k C CHEM 1001 3.0 Chemical Kinetics 3 Distinguish Reaction Intermediates And Transition States (Activated Complexes) A + B k1 I I + B k C A + B I + B C CHEM 1001 3.0 Chemical Kinetics 4
Illustration of Slow Step Followed by Fast Step H (g) + ICl(g) I (g) + HCl(g) rate of reaction = k[h ][ICl] (from experiment) Let s propose the following mechanism: (1) Slow: H + ICl HI + HCl () Fast: HI + ICl I + HCl Overall: H + ICl I + HCl Note: sum of steps in mechanism yields the correct overall reaction CHEM 1001 3.0 Chemical Kinetics 5 Slow Step - Fast Step Mechanism example continued each step in the mechanism is an elementary step (also bimolecular in this example) and rate law can be written directly rate of (1) = k 1 [H ][ICl] rate of () = k [HI][ICl] step (1) was assumed to be slow relative to step () (i.e. step (1) is the rate-determining step) so the overall reaction rate is linked to the rate of step (1) recall, from experiment, rate of reaction = k[h ][ICl] proposed mechanism is plausible CHEM 1001 3.0 Chemical Kinetics 6
Slow Step - Fast Step Mechanism Example continued H + ICl HI + HCl HI + ICl I + HCl CHEM 1001 3.0 Chemical Kinetics 7 Reaction Mechanism with a Reversible Reaction Example Experiment: A + B k C rate of reaction = k[a] [B] Proposed Mechanism: A + A k1 I I k-1 A + A (k -1 since it is reverse to the first) I + B k C What elementary processes should be fast and what should be slow to make the proposed mechanism plausible? CHEM 1001 3.0 Chemical Kinetics 8
General Approach to the Analysis of Simple Reaction Mechanisms Using Steady-State Approximation 1. Check the proposed mechanism for agreement with balanced chemical equation. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. It usually includes the concentration of intermediate, [I] 3. Determine the steady state concentration of the intermediate using (d[i]/dt) prod = - (d[i]/dt) cons as a function of the concentrations of reagents 4. Substitute [I] in the reaction rate found in step with [I] determined in step 3. Compare the theoretical and experimental rates of reaction. CHEM 1001 3.0 Chemical Kinetics 9 1. Examine whether the proposed mechanism satisfies the balanced chemical equation. Experiment: A + B k C Mechanism: A + A k1 I I k-1 A + A I + B k C Overall: A + A + I + B k I + C A + B k C Note that reverse process is excluded from the sum The proposed mechanism satisfies the balanced chemical equation CHEM 1001 3.0 Chemical Kinetics 30
. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. The rate of the second reaction is: Rate = k [I][B] The concentration of intermediate has to be eliminated 3. Determine the steady state concentration of the intermediate using the steady-state approximation: (d[i]/dt) prod = - (d[i]/dt) cons k 1 [A][A] = - (-k -1 [I] - k [I][B]) k 1 [A] = k -1 [I] + k [I][B] = [I](k -1 + k [B]) [I] = k 1 [A] /(k -1 + k [B]) CHEM 1001 3.0 Chemical Kinetics 31 4. Substitute [I] in the reaction rate with [I] determined using steady-state approximation. Compare the theoretical and experimental rates of reaction. Rate = k [I][B] [I] = k 1 [A] /(k -1 + k [B]) Rate = k k 1 [A] [B]/(k -1 + k [B]) This rate appears different from the experimental one (k[a] [B]). Does this imply that the proposed mechanism is not plausible? Do not hurry with conclusions! CHEM 1001 3.0 Chemical Kinetics 3
Assumptions About the Relative Rates Note that this rate: Rate = k k 1 [A] [B]/(k -1 + k [B]) was derived with no assumption on which elementary processes are slow and which are fast. However, if we assume that the second elementary process is faster than the third then k -1 >> k [B] and we can neglect with the term k [B]. Then for the rate we will have: Rate = k k 1 [A] [B]/k -1 If we assume that k = k k 1 /k -1 then the theoretical rate is the same as the experimental one: Rate = k[a] [B] CHEM 1001 3.0 Chemical Kinetics 33 One More Requirement for the Relative Rates Thus the proposed mechanism implies that the second reaction is fast and the third is slow (1)? A + A k1 I Always slower than () Fast I k-1 A + A (3) Slow I + B k C What about the first one? Note: The first one is always slower than the second one because the second involves the intermediate in the reverse process. CHEM 1001 3.0 Chemical Kinetics 34
Rapid Equilibrium (1) Slow A + A k1 I () Fast I k-1 A + A (3) Slow I + B k C If the second reverse reaction is faster than the first forward reaction then, these two reactions are said to establish rapid equilibrium. That can be written as: (1) Fast equilibrium A + A k1 I () Slow I + B k k -1 C CHEM 1001 3.0 Chemical Kinetics 35 Steady-State Approximation continued illustrate steady-state approximation with the reaction NO + O NO making no assumptions about the relative rates of the steps in the mechanism k NO + NO 1 N O N O k NO + NO k 3 N O + O NO Note: the first reversible reaction has been written as two forward steps. CHEM 1001 3.0 Chemical Kinetics 36
Steady-State Approximation continued The reaction NO(g) + O (g) NO (g) has the experimentally determined rate law Rate = k[no] [O ] Note: Although the rate law is consistent with the reaction proceeding as a one-step, termolecular process, this is highly unlikely - we will consider another mechanism. CHEM 1001 3.0 Chemical Kinetics 37 General Approach to the Analysis of Simple Reaction Mechanisms Using Steady-State Approximation 1. Check the proposed mechanism for agreement with balanced chemical equation. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. It usually includes the concentration of intermediate, [I] 3. Determine the steady state concentration of the intermediate using (d[i]/dt) prod = - (d[i]/dt) cons as a function of the concentrations of reagents 4. Substitute [I] in the reaction rate found in step with [I] determined in step 3. Compare the theoretical and experimental rates of reaction. CHEM 1001 3.0 Chemical Kinetics 38
1. Examine whether the proposed mechanism satisfies the balanced chemical equation Experiment: NO(g) + O (g) NO (g) Mechanism: NO + NO N O N O NO + NO N O + O NO Overall: NO + NO + N O + O N O + NO NO + O NO Note that reverse process is excluded from the sum Thus, the proposed mechanism satisfies the balanced chemical equation CHEM 1001 3.0 Chemical Kinetics 39. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. The rate of the third reaction is: Rate = k 3 [N O ][O ] The concentration of intermediate, N O, has to be eliminated. 3. Determine the steady state concentration of the intermediate using the steady-state approximation: (d[i]/dt) prod = - (d[i]/dt) cons k 1 [NO] = k [N O ] + k 3 [N O ][O ] k1[ NO] [NO] = ( k + k [O ]) 3 CHEM 1001 3.0 Chemical Kinetics 40
4. Substitute [I] in the reaction rate with [I] determined using steady-state approximation. Compare the theoretical and experimental rates of reaction. Rate = k 3 [N O ][O ] k1[ NO] [NO] = ( k + k [O ]) k1k3[ NO] [ O] Rate = ( k + k [ O ]) 3 3 This rate law appears to be different from the experimental one (i.e. Rate = k[no] [O ]), but we have not made any assumptions about the relative rates of the elementary steps in the reaction mechanism. CHEM 1001 3.0 Chemical Kinetics 41 If the second step is fast, and the third one is slow, then the rate of disappearance of N O in the second step is greater than the rate of disappearance of N O in the third step; in other words k [N O ] >> k 3 [N O ][O ] or k >> k 3 [O ] under these conditions, k + k 3 [O ] k, and k1k3[ NO] [ O] k1k3 Rate = [ NO] [ O] ( k + k3[ O]) k which is consistent with the experimental rate law if k = k 1 k 3 /k Rate = k[no] [O ] Aside: What happens if k 3 [O ] >> k? Show that Rate = k 1 [NO] (i.e. the reaction is not dependent on the O concentration recall exercise on pseudo-first-order reactions) CHEM 1001 3.0 Chemical Kinetics 4
Generalization of the Reaction Mechanisms Balanced reaction: A + B C may be described by many reaction mechanisms involving intermediates. Intermediate (I) is always a highly reactive species (i.e. its production is slower than at least one of the ways of its decay). In the steady-state approximation the following mechanisms have to be distinguished k 1 A + B I fast 1) k I C slow ) A + B I k I C k 1 slow fast Rate of Process = k 1 [A][B] CHEM 1001 3.0 Chemical Kinetics 43 Generalization of the Reaction Mechanisms - Continued 3) The 1 st step must be slower than the second for the 3 rd step to have any possible rate k 1 A + B I slow k I A + B faster than 1 k I 3 C anything Rate Process 4) Fast equilibrium in the first step followed by a slow k step: A + B 1 I slower than k A + B 1 k I fast equil. I A + B fast k k k I 3 C slow I 3 C slower than of k 1k = k k1k3[a][b] Rate of Process = k CHEM 1001 3.0 Chemical Kinetics 44 3 [A][B] + k 3
Catalysis We saw previously that increasing the temperature generally increases the rate of of reaction (some molecules gain enough energy to exceed the activation energy). A catalyst can also speed up a reaction by providing an alternate reaction pathway with a lower activation energy (although molecules do not gain energy, more will exceed the activation energy). CHEM 1001 3.0 Chemical Kinetics 45 Catalysis continued A catalyst participates in a chemical reaction, but does not undergo a permanent change - overall, the catalyst is neither generated nor consumed two basic types of catalysis Homogeneous Catalysis - catalysts exists in same phase, or homogeneous mixture, as reacting molecules (usually in gas or liquid phase) Heterogeneous Catalysis - catalyst and reacting molecules in different phases (often gaseous reactants are adsorbed on to the surface of a solid catalyst) CHEM 1001 3.0 Chemical Kinetics 46
Homogeneous Catalysis example Acid-Catalyzed Decomposition of Formic Acid HCOOH(aq) H O(l) + CO(g) in uncatalyzed reaction, H atom must move from one part of the HCOOH molecule to another before the C-O bond can break - high activation energy for this atom transfer in the catalyzed reaction, H + from solution can add directly to this position - lower activation energy CHEM 1001 3.0 Chemical Kinetics 47 Homogeneous Catalysis example - continued protonation of formic acid produces (HCOOH ) + the protonation reaction could simply be reversed, or the C-O could break to release water if C-O bond breaks, intermediate species (HCO) + produced will release H + back into solution not the same H + that was absorbed, but no net change in [H + ] CHEM 1001 3.0 Chemical Kinetics 48
Enzymes as Catalysts the most impressive examples of homogeneous catalysis occur in nature where complex reactions are made possible by high molar mass molecules known as enzymes catalytic action of enzymes is extremely specific enzyme activity often described by a lock-and key model; only a reacting substance, the substrate (the key), that fits into an active site on the enzyme (the lock), will undergo a reaction CHEM 1001 3.0 Chemical Kinetics 49 Enzymes as Catalysts continued the following mechanism is common to virtually all enzyme-catalyzed reactions: k 1 E + S ES k -1 k ES E + P E - enzyme, S - substrate, P - product the steady-state approximations yields the following rate law: k [E 0][S] rate of reaction = K +[S] [E 0 ] is total enzyme concentration and K = (k -1 + k )/k 1 CHEM 1001 3.0 Chemical Kinetics 50
Enzymes as Catalysts continued at low substrate concentrations, K >> [S] and the rate of reaction is first order (i.e. rate of reaction = k[s], k = (k /k 1 )(k -1 +k )[E 0 ]) at high substrate concentrations, [S] >> K and the rate of reaction is zero order (i.e. rate of reaction = k, k = k [E 0 ]) CHEM 1001 3.0 Chemical Kinetics 51 Heterogeneous Catalysis many gaseous or solution phase reactions can be catalysed on an appropriate solid surface - many transition metals, or their compounds, are effective not all surface atoms are effective for catalysis; those that are effective are called active sites Proceeds by four basic steps 1) adsorption of reactants ) diffusion of reactants along the surface 3) reaction at an active site to form adsorbed product 4) desorption of product CHEM 1001 3.0 Chemical Kinetics 5
Heterogeneous Catalysis example carbon monoxide, CO, and nitric oxide, NO, found in automobile emissions, are partially responsible for the formation of photochemical smog (see Petrucci, pp. 75-6) automobiles are now equipped with catalytic converters containing a mixture of catalysts to reduce these emissions though the reaction CO(g) + NO(g) CO (g) + N (g) the method by which this process is believed to occur on a rhodium surface is illustrated on the next slide CHEM 1001 3.0 Chemical Kinetics 53 Heterogeneous Catalysis example - continued a) adsorption of CO and NO b) diffusion and dissociation of NO c) combination of CO and O to form CO, N atoms to form N, along with desorption of products (dissociation & combination processes are equivalent to reaction step) CHEM 1001 3.0 Chemical Kinetics 54