Theoretical Models for Chemical Kinetics Collision Theory ...

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Collision Theory. • kinetic-molecular theory can calculate number of molecular collisions per unit time (the collision frequency) which is typically 1030 collisions  ...

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 1030 collisions per second • if each collision yielded a product molecule, the rate of reaction would be about 106 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

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Collision Theory - continued • only a fraction of collisions lead to a reaction (about 1 in 1010!) • 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

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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 T2 than at T1 - 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

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Collision Theory - continued • rate of reaction is also be limited by the orientation of molecules at the time of their collision Example: N2O(g) + NO(g) → N2(g) + NO2(g)

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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: N2O(g) + NO(g) → N2(g) + NO2(g)

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Transition State Theory - continued Reaction Profile

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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)

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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 = Ae − Ea / RT or, taking logarithms,

lnk =

− Ea + lnA RT

Ea is the activation energy (J/mol) and A is the frequency or pre-exponential factor (same units as k) CHEM 1001 3.0

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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) Ea - (activation energy) even when molecules collide they cannot react unless they possess enough energy; Ea is the minimum energy the reactants must possess in order to react CHEM 1001 3.0

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Temperature Effects - continued A plot of ln k versus 1/T will yield a straight line with a slope of -Ea/R (studied in first lab for this course)

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Temperature Effects - continued the Arrehenius equation can relate rate constants at two temperatures

ln

k 2 Ea  1 1  =  -  k1 R  T1 T2 

(temperature in Kelvin and R in standard SI units)

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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 2) experimentally determined rate law • each step in the mechanism is called an elementary reaction (or process) CHEM 1001 3.0

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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

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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

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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. 2. 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 k2[B] and we can neglect with the term k2[B]. Then for the rate we will have: Rate = k2 k1[A]2[B]/k-1 If we assume that k = k2 k1/k-1 then the theoretical rate is the same as the experimental one: Rate = k[A]2[B] CHEM 1001 3.0

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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 2 (2) Fast I —k-1→ A + A (3)

Slow

I + B —k2→ 2C

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

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Rapid Equilibrium (1) Slow A + A —k1→ I (2) Fast I —k-1→ A + A (3) Slow I + B —k2→ 2C 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 (2) Slow

A + A —k1k → I -1 I + B —k2→ 2C CHEM 1001 3.0

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Steady-State Approximation continued • illustrate steady-state approximation with the reaction 2 NO + O2 → 2 NO2 making no assumptions about the relative rates of the steps in the mechanism NO + NO N2O2 N2O2 + O2

k1

k2 k3

N2O2 NO + NO 2 NO2

Note: the first reversible reaction has been written as two forward steps. CHEM 1001 3.0

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Steady-State Approximation continued The reaction 2 NO(g) + O2(g) → 2 NO2(g) has the experimentally determined rate law Rate = k[NO]2[O2] 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.

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General Approach to the Analysis of Simple Reaction Mechanisms Using Steady-State Approximation 1. Check the proposed mechanism for agreement with balanced chemical equation 2. 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 2 with [I] determined in step 3. Compare the theoretical and experimental rates of reaction. CHEM 1001 3.0

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1. Examine whether the proposed mechanism satisfies the balanced chemical equation • Experiment: 2 NO(g) + O2(g) → 2 NO2(g) Mechanism: NO + NO → N2O2 N2O2 → NO + NO N2O2 + O2 → 2 NO2

• Overall:

NO + NO + N2O2 + O2 → N2O2 + 2 NO2

2 NO + O2 → 2 NO2 • Note that reverse process is excluded from the sum • Thus, the proposed mechanism satisfies the balanced chemical equation CHEM 1001 3.0

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2. Write the expression for the reaction rate using the last elementary process in the reaction mechanism. The rate of the third reaction is: Rate = k3[N2O2][O2] The concentration of intermediate, N2O2, 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 k1[NO]2 = k2[N2O2] + k3[N2O2][O2] [N 2O 2 ] =

k1[NO]2 (k 2 + k3[O 2 ]) CHEM 1001 3.0

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4. Substitute [I] in the reaction rate with [I] determined using steady-state approximation. Compare the theoretical and experimental rates of reaction. Rate = k3[N2O2][O2]

[N 2O 2 ] = Rate =

k1[NO]2 (k 2 + k3 [O 2 ])

k1k3 [NO]2 [O 2 ] (k 2 + k3 [O 2 ])

This rate law appears to be different from the experimental one (i.e. Rate = k[NO]2[O2]), but we have not made any assumptions about the relative rates of the elementary steps in the reaction mechanism. CHEM 1001 3.0

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• If the second step is fast, and the third one is slow, then the rate of disappearance of N2O2 in the second step is greater than the rate of disappearance of N2O2 in the third step; in other words k2[N2O2] >> k3[N2O2][O2] or k2 >> k3[O2] • under these conditions, k2 + k3[O2] ≅ k2, and

k1k3 [NO]2 [O 2 ] k1k3 ≅ [ NO]2 [O 2 ] (k 2 + k3 [O 2 ]) k2 which is consistent with the experimental rate law if k = k1k3/ k2 Rate = k[NO]2[O2] Rate =

Aside: What happens if k3[O2] >> k2? Show that Rate = k1[NO]2 (i.e. the reaction is not dependent on the O2 concentration … recall exercise on pseudo-first-order reactions) CHEM 1001 3.0

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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 k1 A + B → I fast 1) k I →2 C slow k1

A + B → I slow k2 I→C fast

2)

Rate of Process = k1[A][B]

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Generalization of the Reaction Mechanisms - Continued 3) The 1st step must be slower than the second for the 3rd step to have any possible rate k

Ak+ B →1 I slow 2 I→ A + B faster than 1 k k k [A][B] Rate of Process = 1 3 I →3 C anything k2 + k3

4) Fast equilibrium in the first step followed by a slow k1 step: Ak + B → I slower than 2 k1 2 I k→ A + B fast A+B I fast equil. k 3 k3 2 I→C slower than 2 I→ C slow

Rate of Process = CHEM 1001 3.0

k1k 3 [A][B] k2

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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).

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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

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Homogeneous Catalysis example Acid-Catalyzed Decomposition of Formic Acid HCOOH(aq) → H2O(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

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Homogeneous Catalysis example - continued

• protonation of formic acid produces (HCOOH2)+ • 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

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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

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Enzymes as Catalysts continued • the following mechanism is common to virtually all enzyme-catalyzed reactions: k E+S ES k 1

-1

k2

ES E+P E - enzyme, S - substrate, P - product • the steady-state approximations yields the following rate law: k [E ][S] rate of reaction = 2 0 K + [S] [E0] is total enzyme concentration and K = (k-1+ k2)/k1 CHEM 1001 3.0

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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 = (k2/k1)(k-1+k2)[E0]) • at high substrate concentrations, [S] >> K and the rate of reaction is zero order (i.e. rate of reaction = k’, k’ = k2[E0])

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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 2) diffusion of reactants along the surface 3) reaction at an active site to form adsorbed product 4) desorption of product CHEM 1001 3.0

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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. 275-6) • automobiles are now equipped with catalytic converters containing a mixture of catalysts to reduce these emissions though the reaction 2 CO(g) + 2 NO(g) → 2 CO2(g) + N2(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

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Heterogeneous Catalysis example - continued a) adsorption of CO and NO b) diffusion and dissociation of NO c) combination of CO and O to form CO2, N atoms to form N2, along with desorption of products (dissociation & combination processes are equivalent to reaction step)

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