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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)
A. Collision Theory
Consider the combustion of CH (^4)
CH 4 + O 2 CO 2 + 2 H 2 O H = 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 CH 4 at room temp.
reaction coordinate (reaction progress)
reactants
products
kinetic energy of molecules
room temp
high temp (^) E a 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.
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)
Br
OH
Br
OH
Br
OH
concerted bond breakage and formation at TS
hydroxide must collide with proper carbon and at a suitable angle
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
A
B
C
D
B. Effect of Temperature or Activation Energy
For a given reaction, E a is a constant.
E a can be determined, without knowing the probability factor, by performing two experiments at different temperatures while maintaining the same reactant concentrations.
RateT1 = k T1 [A] [B] at Temp 1
Rate (^) T2 = k T2 [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.
1 2
a 1
2 T
T
R
ln
E
k
k OR 1 2
a 1
2 T
T
rate R
rate ln
E
Example: A reaction has an E a of 41.6 kJ mol 1 at 298 K. At what temperature will the reaction be thirty times faster?
C. Catalyst Effects
A catalyst, which is not consumed, provides an alternate pathway with lower E a, which in turn increases k.
reaction coordinate (reaction progress)
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. CH 2 OH
CH 2 OPO 3 2-
O
CHO
CH 2 OPO 3 2-
H OH
dihydroxyacetone phosphate
glyceraldehyde-3- phosphate
k uncat = 4.3 x 10-6^ s -1^ k cat = 4.4 x 10^3 s - Rate enhancement = factor of 1.0 x 10^9
Example: If the triose phosphate isomerise reaction is taking place at 298 K, by how much does the enzyme reduce E a?
Example: A catalyst lowers E a 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?
D. Reaction Mechanisms
Consider the reaction of NO with Cl (^2)
2 NO + Cl 2 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 Cl 2 actually react in a two-step mechanism:
Step 1: Cl 2 + NO NOCl 2 (an intermediate) Step 2: NOCl 2 + NO 2 NOCl Overall: 2 NO + Cl 2 2 NOCl
Each of the two elementary steps has an E a 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.
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
R
Br
OH
R Br
R
OH
Br
HO
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 = k 1F [R Br] = k 1R [R+^ ] [Br ]
Step 2 is a fast reaction between the carbocation and the nucleophile. Rate = k 2 [R+^ ] [OH ] R +^ + OH R OH
What would the rate law for the overall reaction be? R Br + OH R OH + Br
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 = k 1F [R Br] = k 1R [R+^ ] [Br ]
Step 2 is a slow reaction between the carbocation and the nucleophile. Rate = k 2 [R+^ ] [OH ] R +^ + OH R OH
The rate law for the overall reaction is based on the slow step, so is it Rate = k 2 [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.
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?
