Mechanism and Reactivity 1 Chemical Kinetics: Mechanism and Reactivity • Consider the problem below. • When a rock proceeds from A to B, its net potential energy decreases due to a reduction in height. The net change corresponds to the difference between states A and B. • Although the process (A Æ B) releases energy, the quantity of rocks the person could push over the hill per unit time ultimately depends on the height of the hill. • Thus, there are two separate concepts: o Thermodynamics: A Æ B net energy difference o Kinetics: speed of A Æ B conversion (hill-dependent)
Transcript
Mechanism and Reactivity 1
Chemical Kinetics: Mechanism and Reactivity
• Consider the problem below.
• When a rock proceeds from A to B, its net potential energy
decreases due to a reduction in height. The net change corresponds to the difference between states A and B.
• Although the process (A B) releases energy, the quantity of rocks the person could push over the hill per unit time ultimately depends on the height of the hill.
• Thus, there are two separate concepts:
o Thermodynamics: A B net energy difference
o Kinetics: speed of A B conversion (hill-dependent)
Mechanism and Reactivity 2
A. Collision Theory • Consider the combustion of CH4
CH4 + O2 CO2 + 2 H2O ΔH = −802 kJ/mol
• Thermodynamically, the reaction is highly exothermic, yet kinetically, it has a rate of zero at room temperature.
• Thermodynamics and kinetics are distinct. The former is a state function measuring the difference in energy between the reactants and the products. Whereas, the latter refers to the rate at which the reaction occurs.
• A reaction coordinate diagram illustrates the energy changes that occur on the route from reactants to products. It explains the exothermic, yet slow, reaction of CH4 at room temp.
reaction coordinate (reaction progress)
ener
gy
reactants
products
Mechanism and Reactivity 3
• In order for the chemical reaction to occur, the reactants must first overcome an activation barrier. The energy required to overcome this barrier is called the activation energy (Ea).
• The combustion of CH4 releases a net of 802 kJ/mol. Yet, the reaction does not occur, and no energy is released, unless the reactants have sufficient energy to pass over the barrier.
• The energy needed to overcome the activation barrier comes from heat, which is measured by temperature.
1. Heat is related to the kinetic energy of molecules.
2. However, recall that temperature is a measure of the average kinetic energy of a collection of molecules.
3. At any given temperature, there is a distribution of kinetic energies, as determined by the Boltzmann Distribution (which you may have seen in physics).
• For the combustion of methane, none of the reactants have enough energy to overcome the activation barrier at room temperature. When the temperature is increased, some of the reactant molecules now have enough energy to overcome Ea.
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kinetic energy of molecules
num
bero
fmol
ecul
esroomtemp
hightemp Ea needed
for reaction
• For the combustion of methane, the heat energy can be supplied by heating the entire mixture, by lighting the mixture with a match, or by introducing a spark.
• Once the reaction has started, the heat released from the exothermic reaction is enough to keep the reaction mixture hot and allow it to continue until the reactants are consumed.
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• Activation energy is just one aspect of Collision Theory.
o For a chemical reaction to occur, reactants must collide with sufficient energy to overcome the activation barrier;
o they must collide in a proper orientation; and,
o even then, the collision may not be successful, because at the transition state, the activated complex can proceed to products, or return to the reactants.
• Example: conversion of bromocyclohexane to cyclohexanol
reaction coordinate (reaction progress)
ener
gy
Br
OH
Br
OH
Br
OH
δ–
δ–
concerted bondbreakage andformation at TS
hydroxide must collidewith proper carbon andat a suitable angle
Mechanism and Reactivity 6
• Thus, the rate of a reaction is determined by these factors:
o Reactant concentration (higher = more collisions)
o Ea and temperature (higher temperature = more reactants have sufficient energy to overcome Ea)
o Some probability factor based on the probabilities of colliding in a proper collision and continuing to the products at the transition state.
Rate = number ofcollisions
fraction of collisionswith enough energy
to overcome Eaprobability
factor
concentration rate constant k
• The Arrhenius equation describes the effect of temperature, Ea, and probability factor on the rate constant k
RTEaAek /−= A = Arrhenius probability factor for a specific reaction Ea = activation energy for a specific reaction R = gas constant 8.314 J mol−1 K−1 T = temperature (Kelvin)
Mechanism and Reactivity 7
• Question: Four different reactions are shown on the reaction coordinate diagram below. Rank them in the order of increasing rate constant and increasing thermodynamic favourability. Assume constant temperature for all four.
reaction coordinate
ener
gy
A
B
C
D
Mechanism and Reactivity 8
B. Effect of Temperature or Activation Energy • For a given reaction, Ea is a constant.
• Ea can be determined, without knowing the probability factor, by performing two experiments at different temperatures while maintaining the same reactant concentrations.
RateT1 = kT1 [A] [B] at Temp 1
RateT2 = kT2 [A] [B] at Temp 2
• Divide one by the other. [A] and [B] are the same in both experiments. Under these conditions, the rate of the reaction is proportional to the rate constant.
• Take the natural log to obtain two equations that will be
provided to you on the exam.
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
21
a
1
2
T1
T1
Rln E
kk OR ⎟⎟
⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
21
a
1
2
T1
T1
Rraterateln E
Mechanism and Reactivity 9
• So, using two rates or rate constants, obtained at different temperatures we can determine the activation energy.
• Example: When the reaction temperature is increased from 20 to 30 °C, the rate increases from 1.5 to 2.4 mol L−1 s−1. Calculate the activation energy for this reaction and the rate expected at 100 °C.
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• Example: A reaction has an Ea of 41.6 kJ mol−1 at 298 K. At what temperature will the reaction be thirty times faster?
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C. Catalyst Effects • A catalyst, which is not consumed, provides an alternate
pathway with lower Ea, which in turn increases k.
reaction coordinate (reaction progress)
ener
gy
reactants
products
• Enzymes are biological catalysts. One is triose phosphate isomerase, which interconverts two glycolysis intermediates, and a deficiency of which has serious clinical manifestations.
CH2OH
CH2OPO32-
O
CHO
CH2OPO32-
OHH
dihydroxyacetonephosphate
glyceraldehyde-3-phosphate
kuncat = 4.3 x 10-6 s-1 kcat = 4.4 x 103 s-1
Rate enhancement = factor of 1.0 x 109
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• To determine amount of Ea reduction, keep variables the same except for the absence or presence of the catalyst. i.e. both uncatalyzed and catalyzed reactions are taking place at the same temperature and same reactant concentration.
• A catalyst can never slow down a reaction, and a catalyst has no effect on the thermodynamics (ΔH). It also does not affect the equilibrium constant, but it allows a system to attain equilibrium faster.
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• Example: If the triose phosphate isomerise reaction is taking place at 298 K, by how much does the enzyme reduce Ea?
• Example: A catalyst lowers Ea for a reaction from 100 to 70 kJ/mol at 300 K. By what factor does the rate of the catalyzed reaction increase?
• Homework: With a catalyst present, a reaction is 65 times faster. If the catalyst is known to decrease the activation energy by 15.0 kJ/mol, at what temp did the reaction occur?
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D. Reaction Mechanisms • Consider the reaction of NO with Cl2
2 NO + Cl2 2 NOCl
• Although the balanced reaction shows three molecules on the reactant side, it is unlikely that all three molecules collide in the proper orientation and react together.
• Rather, chemical reactions often occur in multiple steps. A reaction mechanism describes the sequence of steps that occur. Each step is called an elementary step.
• Experimentally, it was determined that NO with Cl2 actually react in a two-step mechanism: Step 1: Cl2 + NO NOCl2 (an intermediate) Step 2: NOCl2 + NO 2 NOCl Overall: 2 NO + Cl2 2 NOCl
• Each of the two elementary steps has an Ea and a rate
constant. Elementary steps cannot be broken down further, as they are the simplest molecular events that are occurring.
• How was this mechanism determined? By examining the kinetic data and the rate law for the overall reaction.
• To understand the use of kinetic data for determining mechanism, the concept of molecularity is needed.
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1. Molecularity
• Molecularity refers to how many species react together in an elementary step.
• If a process involves only one reactant species, it is termed unimolecular, and must exhibit first-order kinetics. e.g. the decay of N2O5 was found to be unimolecular and first-order:
A products Rate = k [A]
• A bimolecular process involves two species, either identical or different, and is second-order. For example:
A + A products Rate = k [A]2
A + B products Rate = k [A] [B]
• When reactions occur in two or more elementary steps, the steps can be of different molecularity. e.g. O3 decomposition Step 1: O3 O2 + O Rate = k1 [O3] Step 2: O3 + O 2 O2 Rate = k2 [O3] [O] Overall: 2 O3 2 O2 Rate law exp. determined
• Two important consequences: o In an elementary step, and only in an elementary step,
the coefficients of the reactants become the exponents in the rate law for that step.
o The overall rate of a reaction is determined by the rate of the slowest, or the rate-determining, step (RDS).
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2. Determining reaction mechanisms
• Consider a reaction of alkyl halides known as a nucleophilic substitution. The electron-bearing nucleophile (OH−) replaces the Br on the electrophilic (electronic-deficient) carbon atom.
R−Br + OH− R−OH + Br−
• To determine the mechanism, a chemist normally proposes the possible mechanisms. The experimental data are then compared to overall rate laws of the possible mechanisms.
• Possibility A o The reaction occurs in one step, so the overall reaction
equation is the only elementary step. Breakage of the R−Br bond is concomitant with R−OH bond formation.
reaction coordinate
ener
gy
RBr
OH
R Br
ROH
Br
HOδ–δ–
Mechanism and Reactivity 17
o Since the elementary step is the overall reaction: Reaction is bimolecular Rate law for the reaction is overall second-order, as two reactants are involved in the RDS
Rate = k [R−Br] [OH−]
• Possibility B o The reaction occurs in two steps.
Step 1 involves a slow cleavage of the alkyl halide to generate a carbocation intermediate in an equilibrium reaction.
R−Br R+ + Br−
In an equilibrium, the forward and reverse rates are equal. Hence, Rate = k1F [R−Br] = k1R [R+] [Br−]
Step 2 is a fast reaction between the carbocation and the nucleophile. Rate = k2 [R+] [OH−]
R+ + OH− R−OH
What would the rate law for the overall reaction be?
R−Br + OH− R−OH + Br−
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o Since there are two steps, the reaction coordinate diagram shows the intermediate and two TS. Note that the slow step is the one with the highest Ea.
reaction coordinate
ener
gy
R Br
R OH
δ–δ+
R Br
R
δ–δ+
R OH
OH Br
o What is the difference between an intermediate and a
transition state? An intermediate is a real species that can be found in the reaction mixture as the reaction progresses.
A transition state is a high-energy transitory structure comprised of reactants and products. It cannot be found in the mixture or observed.
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• Possibility C (like B, but slow/fast swapped) o The reaction occurs in two steps.
Step 1 involves a fast cleavage of the alkyl halide to generate a carbocation intermediate in an equilibrium reaction.
R−Br R+ + Br−
In an equilibrium, the forward and reverse rates are equal. Hence, Rate = k1F [R−Br] = k1R [R+] [Br−]
Step 2 is a slow reaction between the carbocation and the nucleophile. Rate = k2 [R+] [OH−]
R+ + OH− R−OH
The rate law for the overall reaction is based on the slow step, so is it Rate = k2 [R+] [OH−] ?
R−Br + OH− R−OH + Br−
Intermediates must not be in the overall rate law! We must rearrange the rate expression from the first step and substitute for R+ in the slow rate law.
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To derive the overall rate law:
This is an example of where Br− (a product) slows the reaction. High [Br−] reduces [R+], which in turn slows down step 2, the rate-determining step.
How would the reaction coordinate diagram look?
Mechanism and Reactivity 21
• So which possibility occurs? o Using tert-butyl bromide, which is a tertiary alkyl halide
(carbon bearing the Br is connected to three carbons), the experimental data below was obtained.
Br + OH OH + Br
Expt # [t-BuBr] [OH−] [Br−] Rate i 0.1 0.1 0.1 5 ii 0.1 0.2 0.1 5 iii 0.2 0.2 0.1 10 iv 0.2 0.4 0.2 10 What is the rate law? Is it mechanism A, B, or C? Name of reaction?
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o Using ethyl bromide, a primary alkyl halide, the experimental data below was obtained.
CH3CH2 Br + OH CH3CH2 OH + Br Expt # [EtBr] [OH−] [Br−] Rate i 0.1 0.1 0.1 5 ii 0.1 0.2 0.1 10 iii 0.2 0.2 0.1 20 iv 0.4 0.4 0.2 80 What is the rate law? Is it mechanism A, B, or C? Name of reaction?
o In second-year organic chemistry, you’ll learn why tertiary alkyl halides react via one mechanism, primary by another mechanism, and secondary could undergo both mechanisms.
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• Homework: the following reaction occurs in three steps
