Chemical Kinetics

Chemical kinetics - speed or rate at which a reaction occurs How are rates of reactions affected by Reactant concentration? Temperature? Reactant states? Catalysts?

The Instantaneous Reaction Rate Consider the following reaction A + B  C Define the instantaneous rate of consumption of reactant A, A

Reaction Rates and Reaction Stoichiometry Look at the reaction O 3 (g) + NO(g)  NO 2 (g) + O 2 (g)

Another Example 2 NOCl (g)  2 NO + 1 Cl 2 (g) WHY? 2 moles of NOCl disappear for every 1 mole Cl 2 formed.

The General Case a A + b B  c C + d D rate = -1 d[A] = -1 d[B] = +1 d[C] = +1 d[D] a dt b dt c dt d dt Why do we define our rate in this way? Removes ambiguity in the measurement of reaction rates Obtain a single rate for the entire equation,

Alternative Definition of the Rate Rate of conversion related to the advancement of the reaction, . V = solution volume

An Example Examine the following reaction 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) N2O5N2O5 NO 2 O2O2 Initial n ø 00 Change -2  +4  ++ Final n ø - 2  44

The N 2 O 5 Decomposition Note – constant volume system

The Rate Law Relates rate of the reaction to the reactant concentrations and rate constant For a general reaction a A + b B + c C  d D + e E

Rate Laws (Cont’d) The only way that we can determine the superscripts (x, y, and z) for a non- elementary chemical reaction is by experimentation. Use the isolation method (see first year textbooks).

For a general reaction  x + y + z = reaction order e.g. X = 1; Y = 1; Z = 0 2nd order reaction (x + y + z = 2) X = 0; Y = 0; Z = 1 (1st order reaction) X = 2; Y = 0; Z = 0 (2nd order)

Integrated Rate Laws The rate law gives us information about how the concentration of the reactant varies with time How much reactant remains after specified period of time? Use the integrated rate laws.

First Order Reaction A  product Rate = v = - d[A]/d  t = k[A] How does the concentration of the reactant depend on time? k has units of s -1

The Half-life of a First Order Reaction For a first order reaction, the half-life t 1/2 is calculated as follows.

Radioactive Decay Radioactive Samples decay according to first order kinetics. This is the half-life of samples containing e.g. 14 C, 239 Pu, 99 Tc. Example

Second Order Reaction A  productsv = k[A] 2 A + B  products v = k[A][B] Case 1 is 2 nd order in A Case 2 is 1 st order in A and B and 2 nd order overall

The Dependence of Concentration on Time For a second order process where v = k[A] 2

Half-life for This Second Order Reaction. [A] at t = t½ = ½ [A] 0

Other Second Order Reactions Examine the Case 2 from above A + B  productsv = k[A][B]

Reactions Approaching Equilibrium Examine the concentration profiles for the following simple process. A  B

Approaching Equilibrium Calculate the amounts of A and B at equilibrium.

The Equilibrium Condition At equilibrium, v A,eq = v B,eq.

Temperature Dependence of Reaction Rates Reaction rates generally increase with increasing temperature. Arrhenius Equation A = pre-exponential factor E a = the activation energy

Rate Laws for Multistep Processes Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism The slowest elementary step  the rate determining step (rds).

Elementary steps and the Molecularity Kinetics of the elementary step depends only on the number of reactant molecules in that step! Molecularity  the number of reactant molecules that participate in elementary steps

The Kinetics of Elementary Steps For the elementary step unimolecular step For elementary steps involving more than one reactant a bimolecular step

For the step a termolecular (three molecule) step. Termolecular (and higher) steps are not that common in reaction mechanisms.

The Steady-State Approximation Examine the following simple reaction mechanism Rate of product formation, v p, is proportional to the concentration of an intermediate.

What Is an Intermediate? B is an intermediate in the above reqction sequence. A species formed in one elementary step of a reaction mechanism and consumed in one or more later steps. Intermediates – generally small, indeterminate concentrations.

Applying the Steady State Approximation (SSA) Look for the intermediate in the mechanism. Step 1 – B is produced. Reverse of Step 1 – B is consumed. Step 2 – B is consumed.

The SSA (Cont’d) The SSA applied to the intermediate B.

SSA – The Final Step Substitute the expression for the concentration of B into the rate law v p.

Thermodynamic Formulation of Transition State Theory Activated complex theory Reactant molecules proceed through transition state TS falls apart unimolecularly to form products

The Equilibrium Constant Assume an equilibrium between the activated complex and the reactants

The Gibbs Energy of Activation

Relating E a and  H  For a bimolecular, gas-phase reaction For unimolecular, or solution-phase reactions

Comparing The Two k 2 Expressions

Know your ABC’s For all solution-phase reactions For gas-phase reactions