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) - 226 - A. Collision Theory • Consider the combustion of CH4 CH4 + O2 Æ 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. energy reactants products reaction coordinate (reaction progress) - 227 - • 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. - 228 - room temp high temp Ea needed for reaction number of molecules kinetic energy of molecules • 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. - 229 - • 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 δ– Br concerted bond breakage and δ– OH formation at TS Br OH energy OH Br hydroxide must collide with proper carbon and at a suitable angle reaction coordinate (reaction progress) - 230 - • 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. fraction of collisions probability Rate = number of collisions with enough energy factor to overcome Ea concentration rate constant k • The Arrhenius equation describes the effect of temperature, Ea, and probability factor on the rate constant k −Ea / RT k = Ae 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) - 231 - • 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. A B energy C D reaction coordinate - 232 - 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. ⎛ k2 ⎞ Ea ⎛ 1 1 ⎞ ⎛rate2 ⎞ Ea ⎛ 1 1 ⎞ ln⎜ ⎟ = ⎜ − ⎟ OR ln⎜ ⎟ = ⎜ − ⎟ ⎝ k1 ⎠ R ⎝ T1 T2 ⎠ ⎝ rate1 ⎠ R ⎝ T1 T2 ⎠ - 233 - • So, using two rates or rate constants, obtained at different temperatures we can determine the activation energy. EXAMPLE 1: 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. - 234 - −1 EXAMPLE 2: A reaction has an Ea of 41.6 kJ mol at 298 K. At what temperature will the reaction be thirty times faster? - 235 - C. Effect of a Catalyst • A catalyst, which is not consumed in the reaction, provides an alternate pathway with lower Ea, which in turn increases k. energy reactants products reaction coordinate (reaction progress) • Enzymes are biological catalysts. One is triose phosphate isomerase, which interconverts two glycolysis intermediates, and a deficiency of which has serious clinical manifestations. CH2OH CHO O H OH 2- 2- CH2OPO3 CH2OPO3 dihydroxyacetone glyceraldehyde-3- phosphate phosphate -6 -1 3 -1 kuncat =4.3x10 s kcat =4.4x10 s 9 Rate enhancement = factor of 1.0 x 10 - 236 - • To determine the magnitude of the Ea reduction, keep all 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. - 237 - EXAMPLE 3: If the triose phosphate isomerise reaction is taking place at 298 K, by how much does the enzyme reduce Ea? EXAMPLE 4: A catalyst lowers Ea for a reaction from 100 to 70 kJmol―1 at 300 K. By what factor does the rate of the catalyzed reaction increase? - 238 - EXAMPLE 5: (Homework) With a catalyst present, a reaction is 65 times faster. If the catalyst is known to decrease the activation energy by 15.0 kJmol―1 , at what temp did the reaction occur? - 239 - D. Reaction Mechanisms • What is actually happening at the molecular level when reaction occurs ? Which atoms are colliding to produce the reaction? • Consider the reaction of NO with Cℓ2: 2 NO + Cℓ2 Æ 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 in a reaction mechanism 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 Step 2: NOCℓ2 + NO Æ 2 NOCℓ Overall: 2 NO + Cℓ2 Æ 2 NOCℓ • Notice that NOCℓ2 was formed in Step 1 and consumed in Step 2. This type of species is called a reaction intermediate. • 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. - 240 - • To understand the use of kinetic data for determining mechanism, the concept of molecularity is needed. • 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. A Æ products Rate = k [A] • A bimolecular process involves two species, either identical or different, and is second-order. For example: A + A Æ products Rate = k [A]2 A + B Æ products Rate = k [A] [B] • When reactions occur in two or more elementary steps, the steps can be of different molecularity, for example, the decomposition of O3. Step 1: O3 Æ O2 + O Rate = k1 [O3] Step 2: O3 + O Æ 2 O2 Rate = k2 [O3] [O] Overall: 2 O3 Æ 2 O2 • 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). - 241 - 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− Æ 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.